xrstools
XRStools Documentation Contents:
- Provided by: python3-xrstools (Version: 0.15.0+git20210910+c147919d-2build1)
- Source: xrstools
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XRStools Documentation Contents:
git clone https://gitlab.esrf.fr/ixstools/xrstools
python setup.py install --prefix=~/packages
then to run the code you must do beforehand
export PYTHONPATH=/home/yourname/packages/lib/python2.7/site-packages export PATH=/home/yourname/bin:$PATH
export MYPREFIX=/REPLACE/WITH/YOUR/TARGET
cd ${MYPREFIX}
python3 -m venv myenv
source ${MYPREFIX}/myenv/bin/activate
pip install pip --upgrade
pip install setuptools --upgrade
git clone https://gitlab.esrf.fr/ixstools/xrstools
cd ${MYPREFIX}/xrstools/
pip install -r requirements.txt
python setup.py install
XRS_wizard --wroot ~/software/XRStoolsSuperResolution/XRStools/WIZARD/methods/
the wroot argument tells where extra workflow can be found. In the above instruction we give workflows in the home source directory. This is practical because the wizard allows to edit them online and the modification will remain in the sources. or to access extra workflows that are not coming with the main disribution.
The following documentation has been generated automatically from the comments found in the code.
embedded doc :
embedded doc :
embedded doc :
e_rois
Implementation after Holm and Ribberfors (citation ...).
This function is based on a Matlab implementation by S. Huotari of M. Krisch's Fortran programs.
Calculates absorption correction for given mu1 and mu2. Multiply the measured spectrum with this correction factor. This is a translation of Keijo Hamalainen's Matlab function (KH 30.05.96).
This is a translation of Keijo Hamalainen's Matlab function (KH 30.05.96).
Dynamical theory of diffraction.
function [U, S, V] = constrained_svd(M,U_ini,S_ini,V_ini,U_up,max_iter=10000,verbose=False)
Funktion adapted from Keijo Hamalainen.
Function adapted from Matlab function by S. Huotari.
class to hold all things dynamic theory of diffraction.
This function is translated from Keijo Hamalainen's Matlab implementation (KH 29.05.96).
here is how i opened the HH data: data = np.fromfile(f,np.int32) image = np.reshape(data,(dim,dim))
Returns atomic number of given element, if z is a string of the element symbol or string of element symbol of given atomic number z.
Based on Matlab function from A. Soininen.
% $Id: householder.m,v 1.1 2008-01-16 15:33:30 mike Exp $ % M. M. Sussman
ONLY WORKS FOR EQUALLY/EVENLY SPACED XC, XI!
from scipy.interpolate import Rbf x = arange(20) d = zeros(len(x)) d[10] = 1 xc = arange(0.5,19.5) rbfi = Rbf(x, d) di = rbfi(xc)
Based on Matlab function by S. Huotari.
Class to store some useful values from the detectors used. To be used for arranging the ROIs.
Calculates the photoelectric, elastic, and inelastic absorption of a chemical compound.
Returns the photoelectric absorption for a sum of different chemical compounds.
Calculates the photelectric , elastic, and inelastic absorption of an element Z. Z can be atomic number or element symbol.
Parses the constituing elements and stoichiometries from a given chemical sum formula.
Returns the incident energy for a given photon energy and scattering angle. This function is translated from Keijo Hamalainen's Matlab implementation (KH 29.05.96).
Examples
t = np.linspace(-4, 4, 500) y = np.exp( -t**2 ) + np.random.normal(0, 0.05, t.shape) ysg = savitzky_golay(y, window_size=31, order=4) import matplotlib.pyplot as plt plt.plot(t, y, label='Noisy signal') plt.plot(t, np.exp(-t**2), 'k', lw=1.5, label='Original signal') plt.plot(t, ysg, 'r', label='Filtered signal') plt.legend() plt.show()
Adapted from original Matlab function of Keijo Hamalainen.
Copied from S.I. of L. Zhang et al. "Anisotropic elasticity of silicon and its application to the modelling of X-ray optics." J. Synchrotron Rad. 21, no. 3 (2014): 507-517.
Adapted from original Matlab function of Keijo Hamalainen.
Function by S. Huotari, adopted for Python.
Function by S. Huotari (2007) adopted to Python.
The Beta distribution is a special case of the Dirichlet distribution, and is related to the Gamma distribution. It has the probability distribution function
f(x; a,b) = \frac{1}{B(\alpha, \beta)} x^{\alpha - 1}
(1 - x)^{\beta - 1},
where the normalization, B, is the beta function,
B(\alpha, \beta) = \int_0^1 t^{\alpha - 1}
(1 - t)^{\beta - 1} dt.
It is often seen in Bayesian inference and order statistics.
NOTE:
Generator.beta: which should be used for new code.
Samples are drawn from a binomial distribution with specified parameters, n trials and p probability of success where n an integer >= 0 and p is in the interval [0,1]. (n may be input as a float, but it is truncated to an integer in use)
NOTE:
Generator.binomial: which should be used for new code.
The probability density for the binomial distribution is
P(N) = \binom{n}{N}p^N(1-p)^{n-N},
where n is the number of trials, p is the probability of success, and N is the number of successes.
When estimating the standard error of a proportion in a population by using a random sample, the normal distribution works well unless the product p*n <=5, where p = population proportion estimate, and n = number of samples, in which case the binomial distribution is used instead. For example, a sample of 15 people shows 4 who are left handed, and 11 who are right handed. Then p = 4/15 = 27%. 0.27*15 = 4, so the binomial distribution should be used in this case.
Draw samples from the distribution:
>>> n, p = 10, .5 # number of trials, probability of each trial
>>> s = np.random.binomial(n, p, 1000)
# result of flipping a coin 10 times, tested 1000 times.
A real world example. A company drills 9 wild-cat oil exploration wells, each with an estimated probability of success of 0.1. All nine wells fail. What is the probability of that happening?
Let's do 20,000 trials of the model, and count the number that generate zero positive results.
>>> sum(np.random.binomial(9, 0.1, 20000) == 0)/20000.
# answer = 0.38885, or 38%.
% BROADEN_DIAGRAM2 Broaden a StoBe line diagram % % [ENE2,SQW2] = BROADEN_DIAGRAM2(ENE,SQW,PARAMS,NPOINTS) % % gives the broadened spectrum SQW2(ENE2) of the line-spectrum % SWQ(ENE). Each line is substituted with a Gaussian peak, % the FWHM of which is determined by PARAMS. ENE2 is a linear % scale of length NPOINTS (default 1000). % % PARAMS = [f_min f_max emin max] % % For ENE <= e_min, FWHM = f_min. % For ENE >= e_max, FWHM = f_min. % FWHM increases linearly from [f_min f_max] between [e_min e_max]. % % T Pylkkanen @ 2008-04-18 [17:37]
When df independent random variables, each with standard normal distributions (mean 0, variance 1), are squared and summed, the resulting distribution is chi-square (see Notes). This distribution is often used in hypothesis testing.
NOTE:
Generator.chisquare: which should be used for new code.
The variable obtained by summing the squares of df independent, standard normally distributed random variables:
Q = \sum_{i=0}^{\mathtt{df}} X^2_i
is chi-square distributed, denoted
Q \sim \chi^2_k.
The probability density function of the chi-squared distribution is
p(x) = \frac{(1/2)^{k/2}}{\Gamma(k/2)}
x^{k/2 - 1} e^{-x/2},
where \Gamma is the gamma function,
\Gamma(x) = \int_0^{-\infty} t^{x - 1} e^{-t} dt.
>>> np.random.chisquare(2,4)
array([ 1.89920014, 9.00867716, 3.13710533, 5.62318272]) # random
New in version 1.7.0.
NOTE:
randint, shuffle, permutation Generator.choice: which should be used in new code
Setting user-specified probabilities through p uses a more general but less efficient sampler than the default. The general sampler produces a different sample than the optimized sampler even if each element of p is 1 / len(a).
Sampling random rows from a 2-D array is not possible with this function, but is possible with Generator.choice through its axis keyword.
Generate a uniform random sample from np.arange(5) of size 3:
>>> np.random.choice(5, 3) array([0, 3, 4]) # random >>> #This is equivalent to np.random.randint(0,5,3)
Generate a non-uniform random sample from np.arange(5) of size 3:
>>> np.random.choice(5, 3, p=[0.1, 0, 0.3, 0.6, 0]) array([3, 3, 0]) # random
Generate a uniform random sample from np.arange(5) of size 3 without replacement:
>>> np.random.choice(5, 3, replace=False) array([3,1,0]) # random >>> #This is equivalent to np.random.permutation(np.arange(5))[:3]
Generate a non-uniform random sample from np.arange(5) of size 3 without replacement:
>>> np.random.choice(5, 3, replace=False, p=[0.1, 0, 0.3, 0.6, 0]) array([2, 3, 0]) # random
Any of the above can be repeated with an arbitrary array-like instead of just integers. For instance:
>>> aa_milne_arr = ['pooh', 'rabbit', 'piglet', 'Christopher']
>>> np.random.choice(aa_milne_arr, 5, p=[0.5, 0.1, 0.1, 0.3])
array(['pooh', 'pooh', 'pooh', 'Christopher', 'piglet'], # random
dtype='<U11')
Draw size samples of dimension k from a Dirichlet distribution. A Dirichlet-distributed random variable can be seen as a multivariate generalization of a Beta distribution. The Dirichlet distribution is a conjugate prior of a multinomial distribution in Bayesian inference.
NOTE:
Generator.dirichlet: which should be used for new code.
The Dirichlet distribution is a distribution over vectors x that fulfil the conditions x_i>0 and \sum_{i=1}^k x_i = 1.
The probability density function p of a Dirichlet-distributed random vector X is proportional to
p(x) \propto \prod_{i=1}^{k}{x^{\alpha_i-1}_i},
where \alpha is a vector containing the positive concentration parameters.
The method uses the following property for computation: let Y be a random vector which has components that follow a standard gamma distribution, then X = \frac{1}{\sum_{i=1}^k{Y_i}} Y is Dirichlet-distributed
Taking an example cited in Wikipedia, this distribution can be used if one wanted to cut strings (each of initial length 1.0) into K pieces with different lengths, where each piece had, on average, a designated average length, but allowing some variation in the relative sizes of the pieces.
>>> s = np.random.dirichlet((10, 5, 3), 20).transpose()
>>> import matplotlib.pyplot as plt
>>> plt.barh(range(20), s[0])
>>> plt.barh(range(20), s[1], left=s[0], color='g')
>>> plt.barh(range(20), s[2], left=s[0]+s[1], color='r')
>>> plt.title("Lengths of Strings")
class to analyze ERKALE XRS results.
Its probability density function is
f(x; \frac{1}{\beta}) = \frac{1}{\beta} \exp(-\frac{x}{\beta}),
for x > 0 and 0 elsewhere. \beta is the scale parameter, which is the inverse of the rate parameter \lambda = 1/\beta. The rate parameter is an alternative, widely used parameterization of the exponential distribution
[3]_.
The exponential distribution is a continuous analogue of the geometric distribution. It describes many common situations, such as the size of raindrops measured over many rainstorms
[1]_, or the time between page requests to Wikipedia
[2]_.
NOTE:
Generator.exponential: which should be used for new code.
Samples are drawn from an F distribution with specified parameters, dfnum (degrees of freedom in numerator) and dfden (degrees of freedom in denominator), where both parameters must be greater than zero.
The random variate of the F distribution (also known as the Fisher distribution) is a continuous probability distribution that arises in ANOVA tests, and is the ratio of two chi-square variates.
NOTE:
Generator.f: which should be used for new code.
The F statistic is used to compare in-group variances to between-group variances. Calculating the distribution depends on the sampling, and so it is a function of the respective degrees of freedom in the problem. The variable dfnum is the number of samples minus one, the between-groups degrees of freedom, while dfden is the within-groups degrees of freedom, the sum of the number of samples in each group minus the number of groups.
An example from Glantz[1], pp 47-40:
Two groups, children of diabetics (25 people) and children from people without diabetes (25 controls). Fasting blood glucose was measured, case group had a mean value of 86.1, controls had a mean value of 82.2. Standard deviations were 2.09 and 2.49 respectively. Are these data consistent with the null hypothesis that the parents diabetic status does not affect their children's blood glucose levels? Calculating the F statistic from the data gives a value of 36.01.
Draw samples from the distribution:
>>> dfnum = 1. # between group degrees of freedom
>>> dfden = 48. # within groups degrees of freedom
>>> s = np.random.f(dfnum, dfden, 1000)
The lower bound for the top 1% of the samples is :
>>> np.sort(s)[-10]
7.61988120985 # random
So there is about a 1% chance that the F statistic will exceed 7.62, the measured value is 36, so the null hypothesis is rejected at the 1% level.
Samples are drawn from a Gamma distribution with specified parameters, shape (sometimes designated "k") and scale (sometimes designated "theta"), where both parameters are > 0.
NOTE:
Generator.gamma: which should be used for new code.
The probability density for the Gamma distribution is
p(x) = x^{k-1}\frac{e^{-x/\theta}}{\theta^k\Gamma(k)},
where k is the shape and \theta the scale, and \Gamma is the Gamma function.
The Gamma distribution is often used to model the times to failure of electronic components, and arises naturally in processes for which the waiting times between Poisson distributed events are relevant.
Draw samples from the distribution:
>>> shape, scale = 2., 2. # mean=4, std=2*sqrt(2) >>> s = np.random.gamma(shape, scale, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> import scipy.special as sps
>>> count, bins, ignored = plt.hist(s, 50, density=True)
>>> y = bins**(shape-1)*(np.exp(-bins/scale) /
... (sps.gamma(shape)*scale**shape))
>>> plt.plot(bins, y, linewidth=2, color='r')
>>> plt.show()
Bernoulli trials are experiments with one of two outcomes: success or failure (an example of such an experiment is flipping a coin). The geometric distribution models the number of trials that must be run in order to achieve success. It is therefore supported on the positive integers, k = 1, 2, ....
The probability mass function of the geometric distribution is
f(k) = (1 - p)^{k - 1} p
where p is the probability of success of an individual trial.
NOTE:
Generator.geometric: which should be used for new code.
Draw ten thousand values from the geometric distribution, with the probability of an individual success equal to 0.35:
>>> z = np.random.geometric(p=0.35, size=10000)
How many trials succeeded after a single run?
>>> (z == 1).sum() / 10000. 0.34889999999999999 #random
Calculates the distance vector between two atoms from an arbitrary simulation box using the minimum image convention.
Calculates the distance of two atoms from an arbitrary simulation box using the minimum image convention.
For more details, see set_state.
If legacy is False, or the BitGenerator is not MT19937, then state is returned as a dictionary.
set_state
set_state and get_state are not needed to work with any of the random distributions in NumPy. If the internal state is manually altered, the user should know exactly what he/she is doing.
Draw samples from a Gumbel distribution with specified location and scale. For more information on the Gumbel distribution, see Notes and References below.
NOTE:
scipy.stats.gumbel_l scipy.stats.gumbel_r scipy.stats.genextreme weibull Generator.gumbel: which should be used for new code.
The Gumbel (or Smallest Extreme Value (SEV) or the Smallest Extreme Value Type I) distribution is one of a class of Generalized Extreme Value (GEV) distributions used in modeling extreme value problems. The Gumbel is a special case of the Extreme Value Type I distribution for maximums from distributions with "exponential-like" tails.
The probability density for the Gumbel distribution is
p(x) = \frac{e^{-(x - \mu)/ \beta}}{\beta} e^{ -e^{-(x - \mu)/
\beta}},
where \mu is the mode, a location parameter, and \beta is the scale parameter.
The Gumbel (named for German mathematician Emil Julius Gumbel) was used very early in the hydrology literature, for modeling the occurrence of flood events. It is also used for modeling maximum wind speed and rainfall rates. It is a "fat-tailed" distribution - the probability of an event in the tail of the distribution is larger than if one used a Gaussian, hence the surprisingly frequent occurrence of 100-year floods. Floods were initially modeled as a Gaussian process, which underestimated the frequency of extreme events.
It is one of a class of extreme value distributions, the Generalized Extreme Value (GEV) distributions, which also includes the Weibull and Frechet.
The function has a mean of \mu + 0.57721\beta and a variance of \frac{\pi^2}{6}\beta^2.
Draw samples from the distribution:
>>> mu, beta = 0, 0.1 # location and scale
>>> s = np.random.gumbel(mu, beta, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, 30, density=True)
>>> plt.plot(bins, (1/beta)*np.exp(-(bins - mu)/beta)
... * np.exp( -np.exp( -(bins - mu) /beta) ),
... linewidth=2, color='r')
>>> plt.show()
Show how an extreme value distribution can arise from a Gaussian process and compare to a Gaussian:
>>> means = [] >>> maxima = [] >>> for i in range(0,1000) : ... a = np.random.normal(mu, beta, 1000) ... means.append(a.mean()) ... maxima.append(a.max()) >>> count, bins, ignored = plt.hist(maxima, 30, density=True) >>> beta = np.std(maxima) * np.sqrt(6) / np.pi >>> mu = np.mean(maxima) - 0.57721*beta >>> plt.plot(bins, (1/beta)*np.exp(-(bins - mu)/beta) ... * np.exp(-np.exp(-(bins - mu)/beta)), ... linewidth=2, color='r') >>> plt.plot(bins, 1/(beta * np.sqrt(2 * np.pi)) ... * np.exp(-(bins - mu)**2 / (2 * beta**2)), ... linewidth=2, color='g') >>> plt.show()
Samples are drawn from a hypergeometric distribution with specified parameters, ngood (ways to make a good selection), nbad (ways to make a bad selection), and nsample (number of items sampled, which is less than or equal to the sum ngood + nbad).
NOTE:
Generator.hypergeometric: which should be used for new code.
The probability density for the Hypergeometric distribution is
P(x) = \frac{\binom{g}{x}\binom{b}{n-x}}{\binom{g+b}{n}},
where 0 \le x \le n and n-b \le x \le g
for P(x) the probability of x good results in the drawn sample, g = ngood, b = nbad, and n = nsample.
Consider an urn with black and white marbles in it, ngood of them are black and nbad are white. If you draw nsample balls without replacement, then the hypergeometric distribution describes the distribution of black balls in the drawn sample.
Note that this distribution is very similar to the binomial distribution, except that in this case, samples are drawn without replacement, whereas in the Binomial case samples are drawn with replacement (or the sample space is infinite). As the sample space becomes large, this distribution approaches the binomial.
Draw samples from the distribution:
>>> ngood, nbad, nsamp = 100, 2, 10
# number of good, number of bad, and number of samples
>>> s = np.random.hypergeometric(ngood, nbad, nsamp, 1000)
>>> from matplotlib.pyplot import hist
>>> hist(s)
# note that it is very unlikely to grab both bad items
Suppose you have an urn with 15 white and 15 black marbles. If you pull 15 marbles at random, how likely is it that 12 or more of them are one color?
>>> s = np.random.hypergeometric(15, 15, 15, 100000)
>>> sum(s>=12)/100000. + sum(s<=3)/100000.
# answer = 0.003 ... pretty unlikely!
Reads structure files from Keith's SiO2 simulations.
The Laplace distribution is similar to the Gaussian/normal distribution, but is sharper at the peak and has fatter tails. It represents the difference between two independent, identically distributed exponential random variables.
NOTE:
Generator.laplace: which should be used for new code.
It has the probability density function
f(x; \mu, \lambda) = \frac{1}{2\lambda}
\exp\left(-\frac{|x - \mu|}{\lambda}\right).
The first law of Laplace, from 1774, states that the frequency of an error can be expressed as an exponential function of the absolute magnitude of the error, which leads to the Laplace distribution. For many problems in economics and health sciences, this distribution seems to model the data better than the standard Gaussian distribution.
Draw samples from the distribution
>>> loc, scale = 0., 1.
>>> s = np.random.laplace(loc, scale, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, 30, density=True)
>>> x = np.arange(-8., 8., .01)
>>> pdf = np.exp(-abs(x-loc)/scale)/(2.*scale)
>>> plt.plot(x, pdf)
Plot Gaussian for comparison:
>>> g = (1/(scale * np.sqrt(2 * np.pi)) *
... np.exp(-(x - loc)**2 / (2 * scale**2)))
>>> plt.plot(x,g)
Samples are drawn from a logistic distribution with specified parameters, loc (location or mean, also median), and scale (>0).
NOTE:
Generator.logistic: which should be used for new code.
The probability density for the Logistic distribution is
P(x) = P(x) = \frac{e^{-(x-\mu)/s}}{s(1+e^{-(x-\mu)/s})^2},
where \mu = location and s = scale.
The Logistic distribution is used in Extreme Value problems where it can act as a mixture of Gumbel distributions, in Epidemiology, and by the World Chess Federation (FIDE) where it is used in the Elo ranking system, assuming the performance of each player is a logistically distributed random variable.
Draw samples from the distribution:
>>> loc, scale = 10, 1
>>> s = np.random.logistic(loc, scale, 10000)
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, bins=50)
# plot against distribution
>>> def logist(x, loc, scale):
... return np.exp((loc-x)/scale)/(scale*(1+np.exp((loc-x)/scale))**2)
>>> lgst_val = logist(bins, loc, scale)
>>> plt.plot(bins, lgst_val * count.max() / lgst_val.max())
>>> plt.show()
Draw samples from a log-normal distribution with specified mean, standard deviation, and array shape. Note that the mean and standard deviation are not the values for the distribution itself, but of the underlying normal distribution it is derived from.
NOTE:
Generator.lognormal: which should be used for new code.
A variable x has a log-normal distribution if log(x) is normally distributed. The probability density function for the log-normal distribution is:
p(x) = \frac{1}{\sigma x \sqrt{2\pi}}
e^{(-\frac{(ln(x)-\mu)^2}{2\sigma^2})}
where \mu is the mean and \sigma is the standard deviation of the normally distributed logarithm of the variable. A log-normal distribution results if a random variable is the product of a large number of independent, identically-distributed variables in the same way that a normal distribution results if the variable is the sum of a large number of independent, identically-distributed variables.
Draw samples from the distribution:
>>> mu, sigma = 3., 1. # mean and standard deviation
>>> s = np.random.lognormal(mu, sigma, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, 100, density=True, align='mid')
>>> x = np.linspace(min(bins), max(bins), 10000)
>>> pdf = (np.exp(-(np.log(x) - mu)**2 / (2 * sigma**2))
... / (x * sigma * np.sqrt(2 * np.pi)))
>>> plt.plot(x, pdf, linewidth=2, color='r')
>>> plt.axis('tight')
>>> plt.show()
Demonstrate that taking the products of random samples from a uniform distribution can be fit well by a log-normal probability density function.
>>> # Generate a thousand samples: each is the product of 100 random
>>> # values, drawn from a normal distribution.
>>> b = []
>>> for i in range(1000):
... a = 10. + np.random.standard_normal(100)
... b.append(np.product(a))
>>> b = np.array(b) / np.min(b) # scale values to be positive
>>> count, bins, ignored = plt.hist(b, 100, density=True, align='mid')
>>> sigma = np.std(np.log(b))
>>> mu = np.mean(np.log(b))
>>> x = np.linspace(min(bins), max(bins), 10000)
>>> pdf = (np.exp(-(np.log(x) - mu)**2 / (2 * sigma**2))
... / (x * sigma * np.sqrt(2 * np.pi)))
>>> plt.plot(x, pdf, color='r', linewidth=2)
>>> plt.show()
Samples are drawn from a log series distribution with specified shape parameter, 0 < p < 1.
NOTE:
Generator.logseries: which should be used for new code.
The probability density for the Log Series distribution is
P(k) = \frac{-p^k}{k \ln(1-p)},
where p = probability.
The log series distribution is frequently used to represent species richness and occurrence, first proposed by Fisher, Corbet, and Williams in 1943 [2]. It may also be used to model the numbers of occupants seen in cars [3].
Draw samples from the distribution:
>>> a = .6
>>> s = np.random.logseries(a, 10000)
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s)
# plot against distribution
>>> def logseries(k, p):
... return -p**k/(k*np.log(1-p))
>>> plt.plot(bins, logseries(bins, a)*count.max()/
... logseries(bins, a).max(), 'r')
>>> plt.show()
The multinomial distribution is a multivariate generalization of the binomial distribution. Take an experiment with one of p possible outcomes. An example of such an experiment is throwing a dice, where the outcome can be 1 through 6. Each sample drawn from the distribution represents n such experiments. Its values, X_i = [X_0, X_1, ..., X_p], represent the number of times the outcome was i.
NOTE:
In other words, each entry out[i,j,...,:] is an N-dimensional value drawn from the distribution.
Generator.multinomial: which should be used for new code.
Throw a dice 20 times:
>>> np.random.multinomial(20, [1/6.]*6, size=1) array([[4, 1, 7, 5, 2, 1]]) # random
It landed 4 times on 1, once on 2, etc.
Now, throw the dice 20 times, and 20 times again:
>>> np.random.multinomial(20, [1/6.]*6, size=2)
array([[3, 4, 3, 3, 4, 3], # random
[2, 4, 3, 4, 0, 7]])
For the first run, we threw 3 times 1, 4 times 2, etc. For the second, we threw 2 times 1, 4 times 2, etc.
A loaded die is more likely to land on number 6:
>>> np.random.multinomial(100, [1/7.]*5 + [2/7.]) array([11, 16, 14, 17, 16, 26]) # random
The probability inputs should be normalized. As an implementation detail, the value of the last entry is ignored and assumed to take up any leftover probability mass, but this should not be relied on. A biased coin which has twice as much weight on one side as on the other should be sampled like so:
>>> np.random.multinomial(100, [1.0 / 3, 2.0 / 3]) # RIGHT array([38, 62]) # random
not like:
>>> np.random.multinomial(100, [1.0, 2.0]) # WRONG Traceback (most recent call last): ValueError: pvals < 0, pvals > 1 or pvals contains NaNs
The multivariate normal, multinormal or Gaussian distribution is a generalization of the one-dimensional normal distribution to higher dimensions. Such a distribution is specified by its mean and covariance matrix. These parameters are analogous to the mean (average or "center") and variance (standard deviation, or "width," squared) of the one-dimensional normal distribution.
NOTE:
In other words, each entry out[i,j,...,:] is an N-dimensional value drawn from the distribution.
Generator.multivariate_normal: which should be used for new code.
The mean is a coordinate in N-dimensional space, which represents the location where samples are most likely to be generated. This is analogous to the peak of the bell curve for the one-dimensional or univariate normal distribution.
Covariance indicates the level to which two variables vary together. From the multivariate normal distribution, we draw N-dimensional samples, X = [x_1, x_2, ... x_N]. The covariance matrix element C_{ij} is the covariance of x_i and x_j. The element C_{ii} is the variance of x_i (i.e. its "spread").
Instead of specifying the full covariance matrix, popular approximations include:
This geometrical property can be seen in two dimensions by plotting generated data-points:
>>> mean = [0, 0] >>> cov = [[1, 0], [0, 100]] # diagonal covariance
Diagonal covariance means that points are oriented along x or y-axis:
>>> import matplotlib.pyplot as plt
>>> x, y = np.random.multivariate_normal(mean, cov, 5000).T
>>> plt.plot(x, y, 'x')
>>> plt.axis('equal')
>>> plt.show()
Note that the covariance matrix must be positive semidefinite (a.k.a. nonnegative-definite). Otherwise, the behavior of this method is undefined and backwards compatibility is not guaranteed.
>>> mean = (1, 2)
>>> cov = [[1, 0], [0, 1]]
>>> x = np.random.multivariate_normal(mean, cov, (3, 3))
>>> x.shape
(3, 3, 2)
The following is probably true, given that 0.6 is roughly twice the standard deviation:
>>> list((x[0,0,:] - mean) < 0.6)
[True, True] # random
Samples are drawn from a negative binomial distribution with specified parameters, n successes and p probability of success where n is > 0 and p is in the interval [0, 1].
NOTE:
Generator.negative_binomial: which should be used for new code.
The probability mass function of the negative binomial distribution is
P(N;n,p) = \frac{\Gamma(N+n)}{N!\Gamma(n)}p^{n}(1-p)^{N},
where n is the number of successes, p is the probability of success, N+n is the number of trials, and \Gamma is the gamma function. When n is an integer, \frac{\Gamma(N+n)}{N!\Gamma(n)} = \binom{N+n-1}{N}, which is the more common form of this term in the the pmf. The negative binomial distribution gives the probability of N failures given n successes, with a success on the last trial.
If one throws a die repeatedly until the third time a "1" appears, then the probability distribution of the number of non-"1"s that appear before the third "1" is a negative binomial distribution.
Draw samples from the distribution:
A real world example. A company drills wild-cat oil exploration wells, each with an estimated probability of success of 0.1. What is the probability of having one success for each successive well, that is what is the probability of a single success after drilling 5 wells, after 6 wells, etc.?
>>> s = np.random.negative_binomial(1, 0.1, 100000)
>>> for i in range(1, 11):
... probability = sum(s<i) / 100000.
... print(i, "wells drilled, probability of one success =", probability)
The noncentral \chi^2 distribution is a generalization of the \chi^2 distribution.
NOTE:
Changed in version 1.10.0: Earlier NumPy versions required dfnum > 1.
Generator.noncentral_chisquare: which should be used for new code.
The probability density function for the noncentral Chi-square distribution is
P(x;df,nonc) = \sum^{\infty}_{i=0}
\frac{e^{-nonc/2}(nonc/2)^{i}}{i!} P_{Y_{df+2i}}(x),
where Y_{q} is the Chi-square with q degrees of freedom.
Draw values from the distribution and plot the histogram
>>> import matplotlib.pyplot as plt
>>> values = plt.hist(np.random.noncentral_chisquare(3, 20, 100000),
... bins=200, density=True)
>>> plt.show()
Draw values from a noncentral chisquare with very small noncentrality, and compare to a chisquare.
>>> plt.figure()
>>> values = plt.hist(np.random.noncentral_chisquare(3, .0000001, 100000),
... bins=np.arange(0., 25, .1), density=True)
>>> values2 = plt.hist(np.random.chisquare(3, 100000),
... bins=np.arange(0., 25, .1), density=True)
>>> plt.plot(values[1][0:-1], values[0]-values2[0], 'ob')
>>> plt.show()
Demonstrate how large values of non-centrality lead to a more symmetric distribution.
>>> plt.figure()
>>> values = plt.hist(np.random.noncentral_chisquare(3, 20, 100000),
... bins=200, density=True)
>>> plt.show()
Samples are drawn from an F distribution with specified parameters, dfnum (degrees of freedom in numerator) and dfden (degrees of freedom in denominator), where both parameters > 1. nonc is the non-centrality parameter.
NOTE:
Changed in version 1.14.0: Earlier NumPy versions required dfnum > 1.
Generator.noncentral_f: which should be used for new code.
When calculating the power of an experiment (power = probability of rejecting the null hypothesis when a specific alternative is true) the non-central F statistic becomes important. When the null hypothesis is true, the F statistic follows a central F distribution. When the null hypothesis is not true, then it follows a non-central F statistic.
In a study, testing for a specific alternative to the null hypothesis requires use of the Noncentral F distribution. We need to calculate the area in the tail of the distribution that exceeds the value of the F distribution for the null hypothesis. We'll plot the two probability distributions for comparison.
>>> dfnum = 3 # between group deg of freedom
>>> dfden = 20 # within groups degrees of freedom
>>> nonc = 3.0
>>> nc_vals = np.random.noncentral_f(dfnum, dfden, nonc, 1000000)
>>> NF = np.histogram(nc_vals, bins=50, density=True)
>>> c_vals = np.random.f(dfnum, dfden, 1000000)
>>> F = np.histogram(c_vals, bins=50, density=True)
>>> import matplotlib.pyplot as plt
>>> plt.plot(F[1][1:], F[0])
>>> plt.plot(NF[1][1:], NF[0])
>>> plt.show()
The probability density function of the normal distribution, first derived by De Moivre and 200 years later by both Gauss and Laplace independently
[2]_, is often called the bell curve because of its characteristic shape (see the example below).
The normal distributions occurs often in nature. For example, it describes the commonly occurring distribution of samples influenced by a large number of tiny, random disturbances, each with its own unique distribution
[2]_.
NOTE:
Generator.normal: which should be used for new code.
The probability density for the Gaussian distribution is
p(x) = \frac{1}{\sqrt{ 2 \pi \sigma^2 }}
e^{ - \frac{ (x - \mu)^2 } {2 \sigma^2} },
where \mu is the mean and \sigma the standard deviation. The square of the standard deviation, \sigma^2, is called the variance.
The function has its peak at the mean, and its "spread" increases with the standard deviation (the function reaches 0.607 times its maximum at x + \sigma and x - \sigma
[2]_). This implies that normal is more likely to return samples lying close to the mean, rather than those far away.
Draw samples from the distribution:
>>> mu, sigma = 0, 0.1 # mean and standard deviation
>>> s = np.random.normal(mu, sigma, 1000)
Verify the mean and the variance:
>>> abs(mu - np.mean(s))
0.0 # may vary
>>> abs(sigma - np.std(s, ddof=1))
0.1 # may vary
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, 30, density=True)
>>> plt.plot(bins, 1/(sigma * np.sqrt(2 * np.pi)) *
... np.exp( - (bins - mu)**2 / (2 * sigma**2) ),
... linewidth=2, color='r')
>>> plt.show()
Two-by-four array of samples from N(3, 6.25):
>>> np.random.normal(3, 2.5, size=(2, 4))
array([[-4.49401501, 4.00950034, -1.81814867, 7.29718677], # random
[ 0.39924804, 4.68456316, 4.99394529, 4.84057254]]) # random
The Lomax or Pareto II distribution is a shifted Pareto distribution. The classical Pareto distribution can be obtained from the Lomax distribution by adding 1 and multiplying by the scale parameter m (see Notes). The smallest value of the Lomax distribution is zero while for the classical Pareto distribution it is mu, where the standard Pareto distribution has location mu = 1. Lomax can also be considered as a simplified version of the Generalized Pareto distribution (available in SciPy), with the scale set to one and the location set to zero.
The Pareto distribution must be greater than zero, and is unbounded above. It is also known as the "80-20 rule". In this distribution, 80 percent of the weights are in the lowest 20 percent of the range, while the other 20 percent fill the remaining 80 percent of the range.
NOTE:
Generator.pareto: which should be used for new code.
The probability density for the Pareto distribution is
p(x) = \frac{am^a}{x^{a+1}}
where a is the shape and m the scale.
The Pareto distribution, named after the Italian economist Vilfredo Pareto, is a power law probability distribution useful in many real world problems. Outside the field of economics it is generally referred to as the Bradford distribution. Pareto developed the distribution to describe the distribution of wealth in an economy. It has also found use in insurance, web page access statistics, oil field sizes, and many other problems, including the download frequency for projects in Sourceforge
[1]_. It is one of the so-called "fat-tailed" distributions.
Draw samples from the distribution:
>>> a, m = 3., 2. # shape and mode
>>> s = (np.random.pareto(a, 1000) + 1) * m
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, _ = plt.hist(s, 100, density=True)
>>> fit = a*m**a / bins**(a+1)
>>> plt.plot(bins, max(count)*fit/max(fit), linewidth=2, color='r')
>>> plt.show()
Parses an OCEAN input file and returns lattice vectors, atom names, and relative atom positions.
Parses a PWSCF file and returns a xyzBox object.
Parses a VASPS file and returns a xyzBox object.
If x is a multi-dimensional array, it is only shuffled along its first index.
NOTE:
Generator.permutation: which should be used for new code.
>>> np.random.permutation(10) array([1, 7, 4, 3, 0, 9, 2, 5, 8, 6]) # random
>>> np.random.permutation([1, 4, 9, 12, 15]) array([15, 1, 9, 4, 12]) # random
>>> arr = np.arange(9).reshape((3, 3)) >>> np.random.permutation(arr) array([[6, 7, 8], # random [0, 1, 2], [3, 4, 5]])
The Poisson distribution is the limit of the binomial distribution for large N.
NOTE:
Generator.poisson: which should be used for new code.
The Poisson distribution
f(k; \lambda)=\frac{\lambda^k e^{-\lambda}}{k!}
For events with an expected separation \lambda the Poisson distribution f(k; \lambda) describes the probability of k events occurring within the observed interval \lambda.
Because the output is limited to the range of the C int64 type, a ValueError is raised when lam is within 10 sigma of the maximum representable value.
Draw samples from the distribution:
>>> import numpy as np
>>> s = np.random.poisson(5, 10000)
Display histogram of the sample:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, 14, density=True)
>>> plt.show()
Draw each 100 values for lambda 100 and 500:
>>> s = np.random.poisson(lam=(100., 500.), size=(100, 2))
Also known as the power function distribution.
NOTE:
Generator.power: which should be used for new code.
The probability density function is
P(x; a) = ax^{a-1}, 0 \le x \le 1, a>0.
The power function distribution is just the inverse of the Pareto distribution. It may also be seen as a special case of the Beta distribution.
It is used, for example, in modeling the over-reporting of insurance claims.
Draw samples from the distribution:
>>> a = 5. # shape
>>> samples = 1000
>>> s = np.random.power(a, samples)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> count, bins, ignored = plt.hist(s, bins=30)
>>> x = np.linspace(0, 1, 100)
>>> y = a*x**(a-1.)
>>> normed_y = samples*np.diff(bins)[0]*y
>>> plt.plot(x, normed_y)
>>> plt.show()
Compare the power function distribution to the inverse of the Pareto.
>>> from scipy import stats
>>> rvs = np.random.power(5, 1000000)
>>> rvsp = np.random.pareto(5, 1000000)
>>> xx = np.linspace(0,1,100)
>>> powpdf = stats.powerlaw.pdf(xx,5)
>>> plt.figure()
>>> plt.hist(rvs, bins=50, density=True)
>>> plt.plot(xx,powpdf,'r-')
>>> plt.title('np.random.power(5)')
>>> plt.figure()
>>> plt.hist(1./(1.+rvsp), bins=50, density=True)
>>> plt.plot(xx,powpdf,'r-')
>>> plt.title('inverse of 1 + np.random.pareto(5)')
>>> plt.figure()
>>> plt.hist(1./(1.+rvsp), bins=50, density=True)
>>> plt.plot(xx,powpdf,'r-')
>>> plt.title('inverse of stats.pareto(5)')
NOTE:
Create an array of the given shape and populate it with random samples from a uniform distribution over [0, 1).
random
>>> np.random.rand(3,2)
array([[ 0.14022471, 0.96360618], #random
[ 0.37601032, 0.25528411], #random
[ 0.49313049, 0.94909878]]) #random
Return random integers from the "discrete uniform" distribution of the specified dtype in the "half-open" interval [low, high). If high is None (the default), then results are from [0, low).
NOTE:
New in version 1.11.0.
Generator.integers: which should be used for new code.
>>> np.random.randint(2, size=10) array([1, 0, 0, 0, 1, 1, 0, 0, 1, 0]) # random >>> np.random.randint(1, size=10) array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
Generate a 2 x 4 array of ints between 0 and 4, inclusive:
>>> np.random.randint(5, size=(2, 4))
array([[4, 0, 2, 1], # random
[3, 2, 2, 0]])
Generate a 1 x 3 array with 3 different upper bounds
>>> np.random.randint(1, [3, 5, 10]) array([2, 2, 9]) # random
Generate a 1 by 3 array with 3 different lower bounds
>>> np.random.randint([1, 5, 7], 10) array([9, 8, 7]) # random
Generate a 2 by 4 array using broadcasting with dtype of uint8
>>> np.random.randint([1, 3, 5, 7], [[10], [20]], dtype=np.uint8)
array([[ 8, 6, 9, 7], # random
[ 1, 16, 9, 12]], dtype=uint8)
NOTE:
NOTE:
If positive int_like arguments are provided, randn generates an array of shape (d0, d1, ..., dn), filled with random floats sampled from a univariate "normal" (Gaussian) distribution of mean 0 and variance 1. A single float randomly sampled from the distribution is returned if no argument is provided.
standard_normal : Similar, but takes a tuple as its argument. normal : Also accepts mu and sigma arguments. Generator.standard_normal: which should be used for new code.
For random samples from N(\mu, \sigma^2), use:
sigma * np.random.randn(...) + mu
>>> np.random.randn() 2.1923875335537315 # random
Two-by-four array of samples from N(3, 6.25):
>>> 3 + 2.5 * np.random.randn(2, 4)
array([[-4.49401501, 4.00950034, -1.81814867, 7.29718677], # random
[ 0.39924804, 4.68456316, 4.99394529, 4.84057254]]) # random
Return random integers of type np.int_ from the "discrete uniform" distribution in the closed interval [low, high]. If high is None (the default), then results are from [1, low]. The np.int_ type translates to the C long integer type and its precision is platform dependent.
This function has been deprecated. Use randint instead.
Deprecated since version 1.11.0.
To sample from N evenly spaced floating-point numbers between a and b, use:
a + (b - a) * (np.random.random_integers(N) - 1) / (N - 1.)
>>> np.random.random_integers(5) 4 # random >>> type(np.random.random_integers(5)) <class 'numpy.int64'> >>> np.random.random_integers(5, size=(3,2)) array([[5, 4], # random [3, 3], [4, 5]])
Choose five random numbers from the set of five evenly-spaced numbers between 0 and 2.5, inclusive (i.e., from the set {0, 5/8, 10/8, 15/8, 20/8}):
>>> 2.5 * (np.random.random_integers(5, size=(5,)) - 1) / 4. array([ 0.625, 1.25 , 0.625, 0.625, 2.5 ]) # random
Roll two six sided dice 1000 times and sum the results:
>>> d1 = np.random.random_integers(1, 6, 1000) >>> d2 = np.random.random_integers(1, 6, 1000) >>> dsums = d1 + d2
Display results as a histogram:
>>> import matplotlib.pyplot as plt >>> count, bins, ignored = plt.hist(dsums, 11, density=True) >>> plt.show()
Results are from the "continuous uniform" distribution over the stated interval. To sample Unif[a, b), b > a multiply the output of random_sample by (b-a) and add a:
(b - a) * random_sample() + a
NOTE:
Generator.random: which should be used for new code.
>>> np.random.random_sample() 0.47108547995356098 # random >>> type(np.random.random_sample()) <class 'float'> >>> np.random.random_sample((5,)) array([ 0.30220482, 0.86820401, 0.1654503 , 0.11659149, 0.54323428]) # random
Three-by-two array of random numbers from [-5, 0):
>>> 5 * np.random.random_sample((3, 2)) - 5
array([[-3.99149989, -0.52338984], # random
[-2.99091858, -0.79479508],
[-1.23204345, -1.75224494]])
The \chi and Weibull distributions are generalizations of the Rayleigh.
NOTE:
Generator.rayleigh: which should be used for new code.
The probability density function for the Rayleigh distribution is
P(x;scale) = \frac{x}{scale^2}e^{\frac{-x^2}{2 \cdotp scale^2}}
The Rayleigh distribution would arise, for example, if the East and North components of the wind velocity had identical zero-mean Gaussian distributions. Then the wind speed would have a Rayleigh distribution.
Draw values from the distribution and plot the histogram
>>> from matplotlib.pyplot import hist
>>> values = hist(np.random.rayleigh(3, 100000), bins=200, density=True)
Wave heights tend to follow a Rayleigh distribution. If the mean wave height is 1 meter, what fraction of waves are likely to be larger than 3 meters?
>>> meanvalue = 1
>>> modevalue = np.sqrt(2 / np.pi) * meanvalue
>>> s = np.random.rayleigh(modevalue, 1000000)
The percentage of waves larger than 3 meters is:
>>> 100.*sum(s>3)/1000000.
0.087300000000000003 # random
% READSTF Load StoBe fort.11 (XAS output) data % % [E,P,S,PX,PY,PZ] = READXAS(FILENAME) % % E energy transfer [eV] % P dipole transition intensity % S r^2 transition intensity % PX dipole transition intensity along x % PY dipole transition intensity along y % PZ dipole transition intensity along z % % as line diagrams. % % T Pylkkanen @ 2011-10-17
This is a convenience, legacy function.
The best practice is to not reseed a BitGenerator, rather to recreate a new one. This method is here for legacy reasons. This example demonstrates best practice.
>>> from numpy.random import MT19937
>>> from numpy.random import RandomState, SeedSequence
>>> rs = RandomState(MT19937(SeedSequence(123456789)))
# Later, you want to restart the stream
>>> rs = RandomState(MT19937(SeedSequence(987654321)))
For use if one has reason to manually (re-)set the internal state of the bit generator used by the RandomState instance. By default, RandomState uses the "Mersenne Twister"
[1]_
If state is a dictionary, it is directly set using the BitGenerators state property.
get_state
set_state and get_state are not needed to work with any of the random distributions in NumPy. If the internal state is manually altered, the user should know exactly what he/she is doing.
For backwards compatibility, the form (str, array of 624 uints, int) is also accepted although it is missing some information about the cached Gaussian value: state = ('MT19937', keys, pos).
This function only shuffles the array along the first axis of a multi-dimensional array. The order of sub-arrays is changed but their contents remains the same.
NOTE:
None
Generator.shuffle: which should be used for new code.
>>> arr = np.arange(10) >>> np.random.shuffle(arr) >>> arr [1 7 5 2 9 4 3 6 0 8] # random
Multi-dimensional arrays are only shuffled along the first axis:
>>> arr = np.arange(9).reshape((3, 3)) >>> np.random.shuffle(arr) >>> arr array([[3, 4, 5], # random [6, 7, 8], [0, 1, 2]])
Also known as the Lorentz distribution.
NOTE:
Generator.standard_cauchy: which should be used for new code.
The probability density function for the full Cauchy distribution is
P(x; x_0, \gamma) = \frac{1}{\pi \gamma \bigl[ 1+
(\frac{x-x_0}{\gamma})^2 \bigr] }
and the Standard Cauchy distribution just sets x_0=0 and \gamma=1
The Cauchy distribution arises in the solution to the driven harmonic oscillator problem, and also describes spectral line broadening. It also describes the distribution of values at which a line tilted at a random angle will cut the x axis.
When studying hypothesis tests that assume normality, seeing how the tests perform on data from a Cauchy distribution is a good indicator of their sensitivity to a heavy-tailed distribution, since the Cauchy looks very much like a Gaussian distribution, but with heavier tails.
Draw samples and plot the distribution:
>>> import matplotlib.pyplot as plt
>>> s = np.random.standard_cauchy(1000000)
>>> s = s[(s>-25) & (s<25)] # truncate distribution so it plots well
>>> plt.hist(s, bins=100)
>>> plt.show()
standard_exponential is identical to the exponential distribution with a scale parameter of 1.
NOTE:
Generator.standard_exponential: which should be used for new code.
Output a 3x8000 array:
>>> n = np.random.standard_exponential((3, 8000))
Samples are drawn from a Gamma distribution with specified parameters, shape (sometimes designated "k") and scale=1.
NOTE:
Generator.standard_gamma: which should be used for new code.
The probability density for the Gamma distribution is
p(x) = x^{k-1}\frac{e^{-x/\theta}}{\theta^k\Gamma(k)},
where k is the shape and \theta the scale, and \Gamma is the Gamma function.
The Gamma distribution is often used to model the times to failure of electronic components, and arises naturally in processes for which the waiting times between Poisson distributed events are relevant.
Draw samples from the distribution:
>>> shape, scale = 2., 1. # mean and width
>>> s = np.random.standard_gamma(shape, 1000000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> import scipy.special as sps
>>> count, bins, ignored = plt.hist(s, 50, density=True)
>>> y = bins**(shape-1) * ((np.exp(-bins/scale))/
... (sps.gamma(shape) * scale**shape))
>>> plt.plot(bins, y, linewidth=2, color='r')
>>> plt.show()
NOTE:
Generator.standard_normal: which should be used for new code.
For random samples from N(\mu, \sigma^2), use one of:
mu + sigma * np.random.standard_normal(size=...) np.random.normal(mu, sigma, size=...)
>>> np.random.standard_normal() 2.1923875335537315 #random
>>> s = np.random.standard_normal(8000)
>>> s
array([ 0.6888893 , 0.78096262, -0.89086505, ..., 0.49876311, # random
-0.38672696, -0.4685006 ]) # random
>>> s.shape
(8000,)
>>> s = np.random.standard_normal(size=(3, 4, 2))
>>> s.shape
(3, 4, 2)
Two-by-four array of samples from N(3, 6.25):
>>> 3 + 2.5 * np.random.standard_normal(size=(2, 4))
array([[-4.49401501, 4.00950034, -1.81814867, 7.29718677], # random
[ 0.39924804, 4.68456316, 4.99394529, 4.84057254]]) # random
A special case of the hyperbolic distribution. As df gets large, the result resembles that of the standard normal distribution (standard_normal).
NOTE:
Generator.standard_t: which should be used for new code.
The probability density function for the t distribution is
P(x, df) = \frac{\Gamma(\frac{df+1}{2})}{\sqrt{\pi df}
\Gamma(\frac{df}{2})}\Bigl( 1+\frac{x^2}{df} \Bigr)^{-(df+1)/2}
The t test is based on an assumption that the data come from a Normal distribution. The t test provides a way to test whether the sample mean (that is the mean calculated from the data) is a good estimate of the true mean.
The derivation of the t-distribution was first published in 1908 by William Gosset while working for the Guinness Brewery in Dublin. Due to proprietary issues, he had to publish under a pseudonym, and so he used the name Student.
From Dalgaard page 83
[1]_, suppose the daily energy intake for 11 women in kilojoules (kJ) is:
>>> intake = np.array([5260., 5470, 5640, 6180, 6390, 6515, 6805, 7515, \
... 7515, 8230, 8770])
Does their energy intake deviate systematically from the recommended value of 7725 kJ? Our null hypothesis will be the absence of deviation, and the alternate hypothesis will be the presence of an effect that could be either positive or negative, hence making our test 2-tailed.
Because we are estimating the mean and we have N=11 values in our sample, we have N-1=10 degrees of freedom. We set our significance level to 95% and compute the t statistic using the empirical mean and empirical standard deviation of our intake. We use a ddof of 1 to base the computation of our empirical standard deviation on an unbiased estimate of the variance (note: the final estimate is not unbiased due to the concave nature of the square root).
>>> np.mean(intake)
6753.636363636364
>>> intake.std(ddof=1)
1142.1232221373727
>>> t = (np.mean(intake)-7725)/(intake.std(ddof=1)/np.sqrt(len(intake)))
>>> t
-2.8207540608310198
We draw 1000000 samples from Student's t distribution with the adequate degrees of freedom.
>>> import matplotlib.pyplot as plt
>>> s = np.random.standard_t(10, size=1000000)
>>> h = plt.hist(s, bins=100, density=True)
Does our t statistic land in one of the two critical regions found at both tails of the distribution?
>>> np.sum(np.abs(t) < np.abs(s)) / float(len(s))
0.018318 #random < 0.05, statistic is in critical region
The probability value for this 2-tailed test is about 1.83%, which is lower than the 5% pre-determined significance threshold.
Therefore, the probability of observing values as extreme as our intake conditionally on the null hypothesis being true is too low, and we reject the null hypothesis of no deviation.
class to analyze StoBe results
The triangular distribution is a continuous probability distribution with lower limit left, peak at mode, and upper limit right. Unlike the other distributions, these parameters directly define the shape of the pdf.
NOTE:
Generator.triangular: which should be used for new code.
The probability density function for the triangular distribution is
P(x;l, m, r) = \begin{cases}
\frac{2(x-l)}{(r-l)(m-l)}& \text{for $l \leq x \leq m$},\\
\frac{2(r-x)}{(r-l)(r-m)}& \text{for $m \leq x \leq r$},\\ 0&
\text{otherwise}. \end{cases}
The triangular distribution is often used in ill-defined problems where the underlying distribution is not known, but some knowledge of the limits and mode exists. Often it is used in simulations.
Draw values from the distribution and plot the histogram:
>>> import matplotlib.pyplot as plt
>>> h = plt.hist(np.random.triangular(-3, 0, 8, 100000), bins=200,
... density=True)
>>> plt.show()
Samples are uniformly distributed over the half-open interval [low, high) (includes low, but excludes high). In other words, any value within the given interval is equally likely to be drawn by uniform.
NOTE:
randint : Discrete uniform distribution, yielding integers. random_integers : Discrete uniform distribution over the closed
random_sample : Floats uniformly distributed over [0, 1). random : Alias for random_sample. rand : Convenience function that accepts dimensions as input, e.g.,
Generator.uniform: which should be used for new code.
The probability density function of the uniform distribution is
p(x) = \frac{1}{b - a}
anywhere within the interval [a, b), and zero elsewhere.
When high == low, values of low will be returned. If high < low, the results are officially undefined and may eventually raise an error, i.e. do not rely on this function to behave when passed arguments satisfying that inequality condition. The high limit may be included in the returned array of floats due to floating-point rounding in the equation low + (high-low) * random_sample(). For example:
>>> x = np.float32(5*0.99999999) >>> x 5.0
Draw samples from the distribution:
>>> s = np.random.uniform(-1,0,1000)
All values are within the given interval:
>>> np.all(s >= -1) True >>> np.all(s < 0) True
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt >>> count, bins, ignored = plt.hist(s, 15, density=True) >>> plt.plot(bins, np.ones_like(bins), linewidth=2, color='r') >>> plt.show()
Samples are drawn from a von Mises distribution with specified mode (mu) and dispersion (kappa), on the interval [-pi, pi].
The von Mises distribution (also known as the circular normal distribution) is a continuous probability distribution on the unit circle. It may be thought of as the circular analogue of the normal distribution.
NOTE:
Generator.vonmises: which should be used for new code.
The probability density for the von Mises distribution is
p(x) = \frac{e^{\kappa cos(x-\mu)}}{2\pi I_0(\kappa)},
where \mu is the mode and \kappa the dispersion, and I_0(\kappa) is the modified Bessel function of order 0.
The von Mises is named for Richard Edler von Mises, who was born in Austria-Hungary, in what is now the Ukraine. He fled to the United States in 1939 and became a professor at Harvard. He worked in probability theory, aerodynamics, fluid mechanics, and philosophy of science.
Draw samples from the distribution:
>>> mu, kappa = 0.0, 4.0 # mean and dispersion
>>> s = np.random.vonmises(mu, kappa, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> from scipy.special import i0
>>> plt.hist(s, 50, density=True)
>>> x = np.linspace(-np.pi, np.pi, num=51)
>>> y = np.exp(kappa*np.cos(x-mu))/(2*np.pi*i0(kappa))
>>> plt.plot(x, y, linewidth=2, color='r')
>>> plt.show()
As the scale approaches infinity, the distribution becomes more like a Gaussian. Some references claim that the Wald is an inverse Gaussian with mean equal to 1, but this is by no means universal.
The inverse Gaussian distribution was first studied in relationship to Brownian motion. In 1956 M.C.K. Tweedie used the name inverse Gaussian because there is an inverse relationship between the time to cover a unit distance and distance covered in unit time.
NOTE:
Generator.wald: which should be used for new code.
The probability density function for the Wald distribution is
P(x;mean,scale) = \sqrt{\frac{scale}{2\pi x^3}}e^
\frac{-scale(x-mean)^2}{2\cdotp mean^2x}
As noted above the inverse Gaussian distribution first arise from attempts to model Brownian motion. It is also a competitor to the Weibull for use in reliability modeling and modeling stock returns and interest rate processes.
Draw values from the distribution and plot the histogram:
>>> import matplotlib.pyplot as plt
>>> h = plt.hist(np.random.wald(3, 2, 100000), bins=200, density=True)
>>> plt.show()
Draw samples from a 1-parameter Weibull distribution with the given shape parameter a.
X = (-ln(U))^{1/a}
Here, U is drawn from the uniform distribution over (0,1].
The more common 2-parameter Weibull, including a scale parameter \lambda is just X = \lambda(-ln(U))^{1/a}.
NOTE:
scipy.stats.weibull_max scipy.stats.weibull_min scipy.stats.genextreme gumbel Generator.weibull: which should be used for new code.
The Weibull (or Type III asymptotic extreme value distribution for smallest values, SEV Type III, or Rosin-Rammler distribution) is one of a class of Generalized Extreme Value (GEV) distributions used in modeling extreme value problems. This class includes the Gumbel and Frechet distributions.
The probability density for the Weibull distribution is
p(x) = \frac{a}
{\lambda}(\frac{x}{\lambda})^{a-1}e^{-(x/\lambda)^a},
where a is the shape and \lambda the scale.
The function has its peak (the mode) at \lambda(\frac{a-1}{a})^{1/a}.
When a = 1, the Weibull distribution reduces to the exponential distribution.
Draw samples from the distribution:
>>> a = 5. # shape
>>> s = np.random.weibull(a, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> x = np.arange(1,100.)/50.
>>> def weib(x,n,a):
... return (a / n) * (x / n)**(a - 1) * np.exp(-(x / n)**a)
>>> count, bins, ignored = plt.hist(np.random.weibull(5.,1000))
>>> x = np.arange(1,100.)/50.
>>> scale = count.max()/weib(x, 1., 5.).max()
>>> plt.plot(x, weib(x, 1., 5.)*scale)
>>> plt.show()
xyzAtom
Class to hold information about and manipulate a single atom in xyz-style format.
xyzBox
Class to hold information about and manipulate a xyz-periodic cubic box.
Counts number of neighbors (of name2) around atom of name1.
Calculates the distance vector between two atoms from an arbitrary simulation box using the minimum image convention.
UNTESTED!!!
Brings back all atoms into the original box using periodic boundary conditions and minimal image convention.
xyzMolecule
Class to hold information about and manipulate an xyz-style molecule.
Brings back all atoms into the original box using periodic boundary conditions and minimal image convention.
Samples are drawn from a Zipf distribution with specified parameter a > 1.
The Zipf distribution (also known as the zeta distribution) is a continuous probability distribution that satisfies Zipf's law: the frequency of an item is inversely proportional to its rank in a frequency table.
NOTE:
Generator.zipf: which should be used for new code.
The probability density for the Zipf distribution is
p(x) = \frac{x^{-a}}{\zeta(a)},
where \zeta is the Riemann Zeta function.
It is named for the American linguist George Kingsley Zipf, who noted that the frequency of any word in a sample of a language is inversely proportional to its rank in the frequency table.
Draw samples from the distribution:
>>> a = 2. # parameter
>>> s = np.random.zipf(a, 1000)
Display the histogram of the samples, along with the probability density function:
>>> import matplotlib.pyplot as plt
>>> from scipy import special
Truncate s values at 50 so plot is interesting:
>>> count, bins, ignored = plt.hist(s[s<50], 50, density=True)
>>> x = np.arange(1., 50.)
>>> y = x**(-a) / special.zetac(a)
>>> plt.plot(x, y/max(y), linewidth=2, color='r')
>>> plt.show()
dataset A class to hold all information from HF Compton profiles necessary to subtract background from the experiment.
edge_extraction Class to destill core edge spectra from x-ray Raman scattering experiments.
Args:
guess (list): [position, FWHM, shape, intensity, ax, b, scale ]
Args
valence_CP Class to organize information about extracted experimental valence Compton profiles.
AtomProfile
Class to construct and handle Hartree-Fock atomic Compton Profile of a single atoms.
Apply absorption correction to the Compton profiles on energy loss scale.
Args: E0 : float
Class for multiple HF Compton profiles.
This class should hold one or more instances of the ComptonProfile class and have methods to return profiles from single atoms, single shells, all atoms. It should be able to apply corrections etc. on those...
FormulaProfile
Class to construct and handle Hartree-Fock atomic Compton Profile of a single chemical compound.
HFProfile
Class to construct and handle Hartree-Fock atomic Compton Profile of sample composed of several chemical compounds.
Attributes
Implementation after Holm and Ribberfors (citation ...).
Reads in tabulated HF Compton profiles from the Biggs paper, interpolates them, and normalizes them to the # of electrons in the shell.
Class to build a S(q,w) prediction based on HF Compton Profiles.
Attributes:
Uses the PzProfile function to read read in Biggs HF profiles and converts them onto energy loss scale. The profiles are cut at the respective electron binding energies and are normalized to the f-sum rule (i.e. S(q,w) is in units of [1/eV]).
Translates to and from spectroscopic edge notation and the convention of the Biggs database.
occupation = occupation number of the according shells bindingen = binding energies of the accorting shells colnames = strings of column names as used in the file
Implementation after Holm and Ribberfors (citation ...).
This function is based on a Matlab implementation by S. Huotari of M. Krisch's Fortran programs.
Calculates absorption correction for given mu1 and mu2. Multiply the measured spectrum with this correction factor. This is a translation of Keijo Hamalainen's Matlab function (KH 30.05.96).
This is a translation of Keijo Hamalainen's Matlab function (KH 30.05.96).
Dynamical theory of diffraction.
function [U, S, V] = constrained_svd(M,U_ini,S_ini,V_ini,U_up,max_iter=10000,verbose=False)
Funktion adapted from Keijo Hamalainen.
Function adapted from Matlab function by S. Huotari.
class to hold all things dynamic theory of diffraction.
This function is translated from Keijo Hamalainen's Matlab implementation (KH 29.05.96).
here is how i opened the HH data: data = np.fromfile(f,np.int32) image = np.reshape(data,(dim,dim))
Returns atomic number of given element, if z is a string of the element symbol or string of element symbol of given atomic number z.
Based on Matlab function from A. Soininen.
% $Id: householder.m,v 1.1 2008-01-16 15:33:30 mike Exp $ % M. M. Sussman
ONLY WORKS FOR EQUALLY/EVENLY SPACED XC, XI!
from scipy.interpolate import Rbf x = arange(20) d = zeros(len(x)) d[10] = 1 xc = arange(0.5,19.5) rbfi = Rbf(x, d) di = rbfi(xc)
Based on Matlab function by S. Huotari.
Class to store some useful values from the detectors used. To be used for arranging the ROIs.
Calculates the photoelectric, elastic, and inelastic absorption of a chemical compound.
Returns the photoelectric absorption for a sum of different chemical compounds.
Calculates the photelectric , elastic, and inelastic absorption of an element Z. Z can be atomic number or element symbol.
Parses the constituing elements and stoichiometries from a given chemical sum formula.
Returns the incident energy for a given photon energy and scattering angle. This function is translated from Keijo Hamalainen's Matlab implementation (KH 29.05.96).
Examples
t = np.linspace(-4, 4, 500) y = np.exp( -t**2 ) + np.random.normal(0, 0.05, t.shape) ysg = savitzky_golay(y, window_size=31, order=4) import matplotlib.pyplot as plt plt.plot(t, y, label='Noisy signal') plt.plot(t, np.exp(-t**2), 'k', lw=1.5, label='Original signal') plt.plot(t, ysg, 'r', label='Filtered signal') plt.legend() plt.show()
Adapted from original Matlab function of Keijo Hamalainen.
Copied from S.I. of L. Zhang et al. "Anisotropic elasticity of silicon and its application to the modelling of X-ray optics." J. Synchrotron Rad. 21, no. 3 (2014): 507-517.
Adapted from original Matlab function of Keijo Hamalainen.
Function by S. Huotari, adopted for Python.
Function by S. Huotari (2007) adopted to Python.
Christoph Sahle, Alessandro Mirone
2022, Christoph Sahle, Alessandro Mirone