Clash Royale CLAN TAG

Weibull distribution calculation Runtimewarning overflow

I was trying to use the code below to calculate the Weibull distribution, but I am getting a runtime warning of overflow exp: given that my data is a list of 6 members , but containing big numbers, it seems that the calculation takes a lot and my plot is not like what I see in the link.
I used the fit and got the error :

too many values to unpack (expected 2)

Weibull example code 1

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

def weib(x,n,a):
return (a / n) * (x / n)**(a - 1) * np.exp(-(x / n)**a)

data = [ 79000 , 85900 , 95000, 101000, 155250 , 280000]

(loc, scale) = s.exponweib.fit_loc_scale(data, 1, 1)
print loc, scale

x = np.linspace(data[0], data[-1], 1000)
plt.plot(x, weib(x, loc, scale))
plt.hist(data, data[-1], normed=True)
plt.show()


Method 2 :
I tried this code : using this source ( code 2 )

data = [ 79000 , 85900 , 95000, 101000, 155250 , 280000]
xdata = np.asarray(data)

a = 1. # shape
s1 = np.random.weibull(a, 1000)
x = xdata
def weib(x,n,a):
return (a / n) * (x / n)**(a - 1) * np.exp(-(x / n)**a)

count, bins, ignored = plt.hist(np.random.weibull(5.,1000))
x = xdata
scale = count.max()/weib(x, 1., 5.).max()
plt.plot(x, weib(x, 1., 5.)*scale)
plt.show()


I got some results like these :



I don't know why I don't get the line curve together with the bands, and I am not sure whether s1 also should be a random parameter or should be chosen repectively

Update 1 :

Searching for another method, I used the following code, but the problem here is that I get the x and y values as NAN

# Expoential Weibull PDF
def expweibPDF(x, k, lam, alpha):
return (alpha * (k/lam) *
((x/lam)**(k-1)) *
((1 - exp(-(x/lam)**k))**(alpha-1)) *
exp(-(x/lam)**k))

# Expoential Weibull CDF
def exp_cdf(x, k, lam, alpha):
return (1 - exp(-(x / lam)**k))**alpha

# Expoential Weibull Inverse CDF
def exp_inv_cdf(p, k, lam, alpha):
return lam * ( - log( (1 - p)**(1/alpha) ))**(1/k)

# parameters for the fit - alpha = 1.0 reduces to normal Webull
# the shape parameters k = 5.0 and lam = 1.0 are demonstrated on Wikipedia:
# https://en.wikipedia.org/wiki/Weibull_distribution

alpha = 1.0
k0 = 5.0
lam0 = 1.0
x =
y =

# create a Weibull distribution
random.seed(123)
n = len(data)
for i in range(1,n) :

p=data[i]
x0 = exp_inv_cdf(p,k0,lam0,alpha)
x += [ x0 ]
y += [ expweibPDF(x0,k0,lam0,alpha) ]


# now fit the Weibull using python library
# setting f0=1 should set alpha = 1.0
# so, shape parameters should be the k0 = 5.0 and lam = 1.0

(exp1, k1, loc1, lam1) = stats.exponweib.fit(y,floc=0, f0=1)

print (exp1, k1, loc1, lam1)

fig, ax = plt.subplots(2, 1)
ax[0].plot(x,y, 'ro', lw=2)
ax[1].plot(x,stats.exponweib.pdf(x,exp1,k1,loc1,lam1), 'ro', lw=2)
plt.show()


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