对绘制在直方图中的给定数据进行归一化威布尔分布

Question

I am trying to replicate this graph https://wind-data.ch/tools/weibull.php

The code I worked on, is

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()

I think I might had some logical error, need help in correcting it (I am totally new to this stuff)

Initially, I tried matplotlib as:

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()

(instead of last three lines in prior code, didn't worked out)

Answer 1

Not completely sure iiuc, but this would be my first approach:

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)

creates