使用Matplotlib绘制正态分布图

Question

请帮我绘制以下数据的正态分布图：

数据：

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

h = [186, 176, 158, 180, 186, 168, 168, 164, 178, 170, 189, 195, 172,
     187, 180, 186, 185, 168, 179, 178, 183, 179, 170, 175, 186, 159,
     161, 178, 175, 185, 175, 162, 173, 172, 177, 175, 172, 177, 180]

std = np.std(h) 
mean = np.mean(h)    
plt.plot(norm.pdf(h,mean,std))

输出：

Standard Deriviation = 8.54065575872 
mean = 176.076923077

情节不正确，我的代码有什么问题？

- Adel

Answer 1

• 注意：此解决方案使用pylab而不是matplotlib.pyplot

您可以尝试使用hist将数据信息与拟合曲线放在一起，如下所示:

import numpy as np
import scipy.stats as stats
import pylab as pl

h = sorted([186, 176, 158, 180, 186, 168, 168, 164, 178, 170, 189, 195, 172,
     187, 180, 186, 185, 168, 179, 178, 183, 179, 170, 175, 186, 159,
     161, 178, 175, 185, 175, 162, 173, 172, 177, 175, 172, 177, 180])  #sorted

fit = stats.norm.pdf(h, np.mean(h), np.std(h))  #this is a fitting indeed

pl.plot(h,fit,'-o')

pl.hist(h,normed=True)      #use this to draw histogram of your data

pl.show()                   #use may also need add this 

- Developer

---

normed已被弃用，现应该用density代替，但它仍然有效： pl.hist(h, density=True) - KateYoak

Answer 2

假设你从 scipy.stats 中获取了 norm，你可能只需要对列表进行排序：

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

h = [186, 176, 158, 180, 186, 168, 168, 164, 178, 170, 189, 195, 172,
     187, 180, 186, 185, 168, 179, 178, 183, 179, 170, 175, 186, 159,
     161, 178, 175, 185, 175, 162, 173, 172, 177, 175, 172, 177, 180]
h.sort()
hmean = np.mean(h)
hstd = np.std(h)
pdf = stats.norm.pdf(h, hmean, hstd)
plt.plot(h, pdf) # including h here is crucial

于是我得到了：

- Paul H