python - 对直方图中绘制的给定数据标准化威 bool 分布

Question

我正在尝试复制此图 https://wind-data.ch/tools/weibull.php

我编写的代码是

import matplotlib.pyplot as plt
import numpy as np
import matplotlib.mlab as mlab
import math
import scipy.stats as stats
from scipy.stats import dweibull
import pandas as pd
import seaborn as sns

frequency =     [2.75,7.80,11.64,13.79,14.20,13.15,
11.14,8.72,6.34,4.30,2.73,1.62,0.91,0.48,
0.24,0.11,0.05,0.02,0.01,0.00]
k = 2.00
lambd =6.00
mu = 0
dist = dweibull(k,mu,lambd)
x = np.linspace(-20,20, 1000)

sns.set_style('darkgrid')
sns.distplot(frequency, fit=stats.dist.pdf(x), kde=False)
sns.show()

我想我可能有一些逻辑错误，需要帮助来纠正它(我对这些东西完全陌生)

最初，我尝试使用 matplotlib 作为:

plt.plot(x, 210 * dist.pdf(x),label=r'$k=%.1f,\ \ lambd=%i$' % (k,   lambd))
plt.xlim(0, 21)
plt.ylim(0, 15.0)
plt.hist(frequency,bins ='auto')
plt.show()

(而不是之前代码中的最后三行，没有成功)

Answer 1

不完全确定 iiuc，但这将是我的第一个方法:

import matplotlib.pyplot as plt
import numpy as np

def wd(x, k, A):
    return k/A*(x/A)**(k-1) * np.exp(-(x/A)**k)

frequency = [2.75, 7.80, 11.64, 13.79, 14.20, 13.15, 11.14, 8.72, 6.34, 4.30, 2.73, 1.62, 0.91, 0.48, 0.24, 0.11, 0.05, 0.02, 0.01, 0.00]
k = 2.00
lambd = 6.00

plt.figure(figsize=(8, 4))
plt.step(range(len(frequency)), frequency, where='post')
plt.plot(wd(np.arange(len(frequency)), k, lambd)*100)

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