使用 numpy.random.normal 时如何指定上下限 [英] How to specify upper and lower limits when using numpy.random.normal

问题描述

我希望能够从仅介于 0 和 1 之间的正态分布中选取值.在某些情况下，我希望能够基本上返回完全随机的分布，而在其他情况下，我想返回值呈高斯形状.

I want to be able to pick values from a normal distribution that only ever fall between 0 and 1. In some cases I want to be able to basically just return a completely random distribution, and in other cases I want to return values that fall in the shape of a gaussian.

目前我正在使用以下功能:

At the moment I am using the following function:

def blockedgauss(mu,sigma):
    while True:
        numb = random.gauss(mu,sigma)
        if (numb > 0 and numb < 1):
            break
    return numb

它从正态分布中选择一个值，如果它不在 0 到 1 的范围内，则将其丢弃，但我觉得必须有更好的方法来做到这一点.

It picks a value from a normal distribution, then discards it if it falls outside of the range 0 to 1, but I feel like there must be a better way of doing this.

推荐答案

听起来你想要一个 截断正态分布.使用 scipy，您可以使用 scipy.stats.truncnorm 从这样的分布中生成随机变量:

It sounds like you want a truncated normal distribution. Using scipy, you could use scipy.stats.truncnorm to generate random variates from such a distribution:

import matplotlib.pyplot as plt
import scipy.stats as stats

lower, upper = 3.5, 6
mu, sigma = 5, 0.7
X = stats.truncnorm(
    (lower - mu) / sigma, (upper - mu) / sigma, loc=mu, scale=sigma)
N = stats.norm(loc=mu, scale=sigma)

fig, ax = plt.subplots(2, sharex=True)
ax[0].hist(X.rvs(10000), normed=True)
ax[1].hist(N.rvs(10000), normed=True)
plt.show()

上图显示截断的正态分布，下图显示具有相同均值mu和标准差sigma的正态分布.

The top figure shows the truncated normal distribution, the lower figure shows the normal distribution with the same mean mu and standard deviation sigma.

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