Fast arbitrary distribution random sampling (inverse transform sampling)


The random module ( has several fixed functions to randomly sample from. For example random.gauss will sample random point from a normal distribution with a given mean and sigma values.

I’m looking for a way to extract a number N of random samples between a given interval using my own distribution as fast as possible in python. This is what I mean:

def my_dist(x):
    # Some distribution, assume c1,c2,c3 and c4 are known.
    f = c1*exp(-((x-c2)**c3)/c4)
    return f

# Draw N random samples from my distribution between given limits a,b.
N = 1000
N_rand_samples = ran_func_sample(my_dist, a, b, N)

where ran_func_sample is what I’m after and a, b are the limits from which to draw the samples. Is there anything of that sort in python?

Asked By: Gabriel



You need to use Inverse transform sampling method to get random values distributed according to a law you want. Using this method you can just apply inverted function
to random numbers having standard uniform distribution in the interval [0,1].

After you find the inverted function, you get 1000 numbers distributed according to the needed distribution this obvious way:

[inverted_function(random.random()) for x in range(1000)]

More on Inverse Transform Sampling:

Also, there is a good question on StackOverflow related to the topic:

Answered By: Igor Chubin

This code implements the sampling of n-d discrete probability distributions. By setting a flag on the object, it can also be made to be used as a piecewise constant probability distribution, which can then be used to approximate arbitrary pdf’s. Well, arbitrary pdfs with compact support; if you efficiently want to sample extremely long tails, a non-uniform description of the pdf would be required. But this is still efficient even for things like airy-point-spread functions (which I created it for, initially). The internal sorting of values is absolutely critical there to get accuracy; the many small values in the tails should contribute substantially, but they will get drowned out in fp accuracy without sorting.

class Distribution(object):
    draws samples from a one dimensional probability distribution,
    by means of inversion of a discrete inverstion of a cumulative density function

    the pdf can be sorted first to prevent numerical error in the cumulative sum
    this is set as default; for big density functions with high contrast,
    it is absolutely necessary, and for small density functions,
    the overhead is minimal

    a call to this distibution object returns indices into density array
    def __init__(self, pdf, sort = True, interpolation = True, transform = lambda x: x):
        self.shape          = pdf.shape
        self.pdf            = pdf.ravel()
        self.sort           = sort
        self.interpolation  = interpolation
        self.transform      = transform

        #a pdf can not be negative

        #sort the pdf by magnitude
        if self.sort:
            self.sortindex = np.argsort(self.pdf, axis=None)
            self.pdf = self.pdf[self.sortindex]
        #construct the cumulative distribution function
        self.cdf = np.cumsum(self.pdf)
    def ndim(self):
        return len(self.shape)
    def sum(self):
        """cached sum of all pdf values; the pdf need not sum to one, and is imlpicitly normalized"""
        return self.cdf[-1]
    def __call__(self, N):
        """draw """
        #pick numbers which are uniformly random over the cumulative distribution function
        choice = np.random.uniform(high = self.sum, size = N)
        #find the indices corresponding to this point on the CDF
        index = np.searchsorted(self.cdf, choice)
        #if necessary, map the indices back to their original ordering
        if self.sort:
            index = self.sortindex[index]
        #map back to multi-dimensional indexing
        index = np.unravel_index(index, self.shape)
        index = np.vstack(index)
        #is this a discrete or piecewise continuous distribution?
        if self.interpolation:
            index = index + np.random.uniform(size=index.shape)
        return self.transform(index)

if __name__=='__main__':
    shape = 3,3
    pdf = np.ones(shape)
    dist = Distribution(pdf, transform=lambda i:i-1.5)
    print dist(10)
    import matplotlib.pyplot as pp

And as a more real-world relevant example:

x = np.linspace(-100, 100, 512)
p = np.exp(-x**2)
pdf = p[:,None]*p[None,:]     #2d gaussian
dist = Distribution(pdf, transform=lambda i:i-256)
print dist(1000000).mean(axis=1)    #should be in the 1/sqrt(1e6) range
import matplotlib.pyplot as pp
Answered By: Eelco Hoogendoorn
import numpy as np
import scipy.interpolate as interpolate

def inverse_transform_sampling(data, n_bins, n_samples):
    hist, bin_edges = np.histogram(data, bins=n_bins, density=True)
    cum_values = np.zeros(bin_edges.shape)
    cum_values[1:] = np.cumsum(hist*np.diff(bin_edges))
    inv_cdf = interpolate.interp1d(cum_values, bin_edges)
    r = np.random.rand(n_samples)
    return inv_cdf(r)

So if we give our data sample that has a specific distribution, the inverse_transform_sampling function will return a dataset with exactly the same distribution. Here the advantage is that we can get our own sample size by specifying it in the n_samples variable.

Answered By: Srivatsan

I was in a similar situation but I wanted to sample from a multivariate distribution, so, I implemented a rudimentary version of Metropolis-Hastings (which is an MCMC method).

def metropolis_hastings(target_density, size=500000):
    burnin_size = 10000
    size += burnin_size
    x0 = np.array([[0, 0]])
    xt = x0
    samples = []
    for i in range(size):
        xt_candidate = np.array([np.random.multivariate_normal(xt[0], np.eye(2))])
        accept_prob = (target_density(xt_candidate))/(target_density(xt))
        if np.random.uniform(0, 1) < accept_prob:
            xt = xt_candidate
    samples = np.array(samples[burnin_size:])
    samples = np.reshape(samples, [samples.shape[0], 2])
    return samples

This function requires a function target_density which takes in a data-point and computes its probability.

For details check-out this detailed answer of mine.

Answered By: Abdul Fatir

Here is a rather nice way of performing inverse transform sampling with a decorator.

import numpy as np
from scipy.interpolate import interp1d

def inverse_sample_decorator(dist):
    def wrapper(pnts, x_min=-100, x_max=100, n=1e5, **kwargs):
        x = np.linspace(x_min, x_max, int(n))
        cumulative = np.cumsum(dist(x, **kwargs))
        cumulative -= cumulative.min()
        f = interp1d(cumulative/cumulative.max(), x)
        return f(np.random.random(pnts))
    return wrapper

Using this decorator on a Gaussian distribution, for example:

def gauss(x, amp=1.0, mean=0.0, std=0.2):
    return amp*np.exp(-(x-mean)**2/std**2/2.0)

You can then generate sample points from the distribution by calling the function. The keyword arguments x_min and x_max are the limits of the original distribution and can be passed as arguments to gauss along with the other key word arguments that parameterise the distribution.

samples = gauss(5000, mean=20, std=0.8, x_min=19, x_max=21)

Alternatively, this can be done as a function that takes the distribution as an argument (as in your original question),

def inverse_sample_function(dist, pnts, x_min=-100, x_max=100, n=1e5, 
    x = np.linspace(x_min, x_max, int(n))
    cumulative = np.cumsum(dist(x, **kwargs))
    cumulative -= cumulative.min()
    f = interp1d(cumulative/cumulative.max(), x)
    return f(np.random.random(pnts))
Answered By: JS00N

I’m using numpy.random.choice to simulate coinflips based on any pdf, which is arguably not very fast but it is very simple.

import numpy as np
rng = np.random.default_rng(seed=43)

def gaussian(x, mu, sig):
    return 1 / (np.sqrt(2 * np.pi) * sig) * np.exp(-np.power((x - mu) / sig, 2) / 2)

x = np.linspace(-4, 4, num=250, endpoint=True)
y = gaussian(x, 0, 1.)

Then normalize to convert to pdf and drawing samples:

pdf = y / y.sum()
N = 1000
samples = np.array([rng.choice(x, p=pdf) for i in range(N)])

The timing for my use case is acceptable:

%timeit samples = np.array([rng.choice(x, p=pdf) for i in range(N)])
>>> 25.3 ms ± 915 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Inspired by @Eelco I tried the cdf approach:

cdf = np.cumsum(pdf)
x_sample = rng.uniform(0, 1, size=1000)
indices = np.searchsorted(cdf, x_sample)
cdf_samples = x[indices]

This is much faster:

%timeit x_sample = rng.uniform(0, 1, size=1000)
# 7.08 µs ± 179 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

%timeit indices = np.searchsorted(cdf, x_sample)
# 37.9 µs ± 933 ns per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

%timeit cdf_samples = x[indices]
1.45 µs ± 17 ns per loop (mean ± std. dev. of 7 runs, 1,000,000 loops each)

In total about 50 µs. This approach is probably a bit more restrictive compared to what the Distribution class offers such as the requirement of having an actual pdf which sums to one, but I think that is a good constraint to impose anyway.

If you want more samples a meaningful histogram with 'fd' binning, you also need to increase the sampling of the pdf (ie the np.linspace size).

Plotting the result:

import proplot as pplt

fig, ax = pplt.subplots()
ax.hist(samples, bins='fd', density=True, histtype='step', lw=2, label='rng.choice')
ax.hist(rng.normal(0, 1, N), bins='fd', density=True, histtype='step', lw=2, label='rng.normal')
ax.hist(cdf_samples,  bins='fd', density=True, histtype='step', lw=2, label='rng.uniform')
ax.plot(x, y, color='k', label='pdf')
ax.format(xlabel='x', ylabel='y')
ax.legend(loc='t', ncols=2)

Sampled Gaussian PDF

Answered By: xyzzyqed
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