Python Matplotlib：如何绘制标准偏差

1. 代码中没有错误:平均值为5，标准值为2，所以阴影区域介于5 - 2 = 3和5 + 2 = 7之间.

2. 在kde图中有一个小的峰值，因为它是用X表示的数据分布，实际上，X不是正态分布.您可以使用真实的正态分布来判断这一点:

   mean = 5
   std = 2
   X = np.random.randn(10000)
   X = (X - X.mean())/X.std()*std + mean
   

   

3. 您可以使用大于i的for循环绘制其他标准折旧.x1是左侧，x2是中心部分(然后设置为np.nan)，最后x3是分布的右侧.然后必须将要排除的区域设置为np.nan(对应于x2):

   N = 10
   for i in [1, 2, 3]:
       x1 = np.linspace(mean - i*std, mean - (i - 1)*std, N)
       x2 = np.linspace(mean - (i - 1)*std, mean + (i - 1)*std, N)
       x3 = np.linspace(mean + (i - 1)*std, mean + i*std, N)
       x = np.concatenate((x1, x2, x3))
       x = np.where((mean - (i - 1)*std < x) & (x < mean + (i - 1)*std), np.nan, x)
       y = norm.pdf(x, mean, std)
       ax.fill_between(x, y, alpha=0.5)
   

完整代码

import numpy as np
from scipy.stats import norm

import matplotlib.pyplot as plt
import seaborn as sns

# Line width: Maximum 130 characters in the output, post which it will continue in next line.
np.set_printoptions(linewidth=130)

sns.set_context("paper", font_scale=1.5)

# Distribution
X = [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9]


mean = np.mean(X)
var = np.var(X)
std = np.std(X)

print("Mean:", mean)
print("Variance:", var)
print("Standard Deviation:", std)

"""
Mean: 5.0
Variance: 4.0
Standard Deviation: 2.0
"""

plt.figure(figsize=(10, 5))

ax = sns.kdeplot(X, shade=True)

N = 10
for i in [1, 2, 3]:
    x1 = np.linspace(mean - i*std, mean - (i - 1)*std, N)
    x2 = np.linspace(mean - (i - 1)*std, mean + (i - 1)*std, N)
    x3 = np.linspace(mean + (i - 1)*std, mean + i*std, N)
    x = np.concatenate((x1, x2, x3))
    x = np.where((mean - (i - 1)*std < x) & (x < mean + (i - 1)*std), np.nan, x)
    y = norm.pdf(x, mean, std)
    ax.fill_between(x, y, alpha=0.5)

plt.xlabel("Random variable X")
plt.ylabel("Probability Density Function")
plt.xticks(ticks=range(0, 10))
plt.grid()

plt.show()

情节