How to get a normal distribution within a range in numpy? [duplicate]

Answer 1

The parametrization of truncnorm is complicated, so here is a function that translates the parametrization to something more intuitive:

from scipy.stats import truncnorm

def get_truncated_normal(mean=0, sd=1, low=0, upp=10):
    return truncnorm(
        (low - mean) / sd, (upp - mean) / sd, loc=mean, scale=sd)

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How to use it?

1. Instance the generator with the parameters: mean, standard deviation, and truncation range:

   >>> X = get_truncated_normal(mean=8, sd=2, low=1, upp=10)
   

2. Then, you can use X to generate a value:

   >>> X.rvs()
   6.0491227353928894
   

3. Or, a numpy array with N generated values:

   >>> X.rvs(10)
   array([ 7.70231607,  6.7005871 ,  7.15203887,  6.06768994,  7.25153472,
           5.41384242,  7.75200702,  5.5725888 ,  7.38512757,  7.47567455])
   

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A Visual Example

Here is the plot of three different truncated normal distributions:

X1 = get_truncated_normal(mean=2, sd=1, low=1, upp=10)
X2 = get_truncated_normal(mean=5.5, sd=1, low=1, upp=10)
X3 = get_truncated_normal(mean=8, sd=1, low=1, upp=10)

import matplotlib.pyplot as plt
fig, ax = plt.subplots(3, sharex=True)
ax[0].hist(X1.rvs(10000), normed=True)
ax[1].hist(X2.rvs(10000), normed=True)
ax[2].hist(X3.rvs(10000), normed=True)
plt.show()

Answer 2

Besides @bakkal suggestion (+1) you might also want to take a look into Vincent Mazet recipe for achieving this, rewritten as py-rtnorm module by Christoph Lassner.

Answer 3

If you're looking for the Truncated normal distribution, SciPy has a function for it called truncnorm

The standard form of this distribution is a standard normal truncated to the range [a, b] — notice that a and b are defined over the domain of the standard normal. To convert clip values for a specific mean and standard deviation, use:

a, b = (myclip_a - my_mean) / my_std, (myclip_b - my_mean) / my_std

truncnorm takes a and b as shape parameters.

>>> from scipy.stats import truncnorm
>>> truncnorm(a=-2/3., b=2/3., scale=3).rvs(size=10)
array([-1.83136675,  0.77599978, -0.01276925,  1.87043384,  1.25024188,
        0.59336279, -0.39343176,  1.9449987 , -1.97674358, -0.31944247])

The above example is bounded by -2 and 2 and returns 10 random variates (using the .rvs() method)

>>> min(truncnorm(a=-2/3., b=2/3., scale=3).rvs(size=10000))
-1.9996074381484044
>>> max(truncnorm(a=-2/3., b=2/3., scale=3).rvs(size=10000))
1.9998486576228549

Here's a histogram plot for -6, 6:

Answer 4

You can subdivide your targeted range (by convention) to equal partitions and then calculate the integration of each and all area, then call uniform method on each partition according to the surface. It's implemented in python:

quad_vec(eval('scipy.stats.norm.pdf'), 1, 4,points=[0.5,2.5,3,4],full_output=True)