Python 中的 Hosmer-Lemeshow 拟合优度检验

Question

我估计 python 有 glm。我怎样才能表现出 Hosmer-Lemeshow 的优点

在 python 中对该模型进行适合性测试？

Answer 1

我找到了一种方法，代码质量不是最好，但它有效：

import pandas as pd
import numpy as np
from scipy.stats import chi2
pihat=model.predict()
pihatcat=pd.cut(pihat, np.percentile(pihat,[0,25,50,75,100]),labels=False,include_lowest=True) #here I've chosen only 4 groups


meanprobs =[0]*4 
expevents =[0]*4
obsevents =[0]*4 
meanprobs2=[0]*4 
expevents2=[0]*4
obsevents2=[0]*4 

for i in range(4):
   meanprobs[i]=np.mean(pihat[pihatcat==i])
   expevents[i]=np.sum(pihatcat==i)*np.array(meanprobs[i])
   obsevents[i]=np.sum(data.r[pihatcat==i])
   meanprobs2[i]=np.mean(1-pihat[pihatcat==i])
   expevents2[i]=np.sum(pihatcat==i)*np.array(meanprobs2[i])
   obsevents2[i]=np.sum(1-data.r[pihatcat==i]) 


data1={'meanprobs':meanprobs,'meanprobs2':meanprobs2}
data2={'expevents':expevents,'expevents2':expevents2}
data3={'obsevents':obsevents,'obsevents2':obsevents2}
m=pd.DataFrame(data1)
e=pd.DataFrame(data2)
o=pd.DataFrame(data3)

tt=sum(sum((np.array(o)-np.array(e))**2/np.array(e))) #the statistic for the test, which follows,under the null hypothesis, the chi-squared distribution with degrees of freedom equal to amount of groups - 2 
pvalue=1-chi2.cdf(tt,2)
pvalue

Answer 2

import matplotlib.pyplot as plt

import pandas as pd 
import numpy as np
from scipy.stats import chi2

# This could be made into a neat function of Hosmer-Lemeshow. 
def HosmerLemeshow (model,Y):
    pihat=model.predict()
    pihatcat=pd.cut(pihat, np.percentile(pihat,[0,25,50,75,100]),labels=False,include_lowest=True) #here we've chosen only 4 groups


    meanprobs =[0]*4 
    expevents =[0]*4
    obsevents =[0]*4 
    meanprobs2=[0]*4 
    expevents2=[0]*4
    obsevents2=[0]*4 

    for i in range(4):
       meanprobs[i]=np.mean(pihat[pihatcat==i])
       expevents[i]=np.sum(pihatcat==i)*np.array(meanprobs[i])
       obsevents[i]=np.sum(Y[pihatcat==i])
       meanprobs2[i]=np.mean(1-pihat[pihatcat==i])
       expevents2[i]=np.sum(pihatcat==i)*np.array(meanprobs2[i])
       obsevents2[i]=np.sum(1-Y[pihatcat==i]) 


    data1={'meanprobs':meanprobs,'meanprobs2':meanprobs2}
    data2={'expevents':expevents,'expevents2':expevents2}
    data3={'obsevents':obsevents,'obsevents2':obsevents2}
    m=pd.DataFrame(data1)
    e=pd.DataFrame(data2)
    o=pd.DataFrame(data3)
    
    # The statistic for the test, which follows, under the null hypothesis,
    # The chi-squared distribution with degrees of freedom equal to amount of groups - 2. Thus 4 - 2 = 2
    tt=sum(sum((np.array(o)-np.array(e))**2/np.array(e))) 
    pvalue=1-chi2.cdf(tt,2)

    return pd.DataFrame([[chi2.cdf(tt,2).round(2), pvalue.round(2)]],columns = ["Chi2", "p - value"])
    
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