如何获得numpy范围内的正态分布?

问题描述

在机器学习任务中.我们应该得到一组带约束的随机w.r.t正态分布.我们可以使用np.random.normal()来获得正态分布号，但它不提供任何绑定参数.我想知道该怎么做?

In machine learning task. We should get a group of random w.r.t normal distribution with bound. We can get a normal distribution number with np.random.normal() but it does't offer any bound parameter. I want to know how to do that?

推荐答案

truncnorm的参数化很复杂，因此这里提供了一个将参数化转换为更直观的功能的函数:

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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如何使用它?

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How to use it?

1. 使用以下参数实例化生成器:平均值，标准偏差和截断范围:

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

然后，您可以使用X生成一个值:

Then, you can use X to generate a value:

>>> X.rvs()
6.0491227353928894

或者，一个具有N个生成值的numpy数组:

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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视觉示例

这是三个不同的截断正态分布的图:

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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()

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