python 基于卡方值分箱算法的实现示例_Python

python 基于卡方值分箱算法的实现示例

2020-07-17 23:17wyzwyzwyzo Python

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原理很简单，初始分20箱或更多，先确保每箱中都含有0，1标签，对不包含0，1标签的箱向前合并，计算各箱卡方值，对卡方值最小的箱向后合并，代码如下

									import pandas as pd

									import numpy as np

									import scipy

									from scipy import stats

									def chi_bin(DF,var,target,binnum=5,maxcut=20):

									  '''

									  DF:data

									  var:variable

									  target:target / label

									  binnum: the number of bins output

									  maxcut: initial bins number 

									  '''

									  data=DF[[var,target]]

									  #equifrequent cut the var into maxcut bins

									  data["cut"],breaks=pd.qcut(data[var],q=maxcut,duplicates="drop",retbins=True)

									  #count 1,0 in each bin

									  count_1=data.loc[data[target]==1].groupby("cut")[target].count()

									  count_0=data.loc[data[target]==0].groupby("cut")[target].count()

									  #get bins value: min,max,count 0,count 1

									  bins_value=[*zip(breaks[:maxcut-1],breaks[1:],count_0,count_1)]

									  #define woe

									  def woe_value(bins_value):

									    df_woe=pd.DataFrame(bins_value)

									    df_woe.columns=["min","max","count_0","count_1"]

									    df_woe["total"]=df_woe.count_1+df_woe.count_0

									    df_woe["bad_rate"]=df_woe.count_1/df_woe.total

									    df_woe["woe"]=np.log((df_woe.count_0/df_woe.count_0.sum())/(df_woe.count_1/df_woe.count_1.sum()))

									    return df_woe

									  #define iv

									  def iv_value(df_woe):

									    rate=(df_woe.count_0/df_woe.count_0.sum())-(df_woe.count_1/df_woe.count_1.sum())

									    iv=np.sum(rate * df_woe.woe)

									    return iv

									  #make sure every bin contain 1 and 0

									  ##first bin merge backwards

									  for i in range(len(bins_value)):

									    if 0 in bins_value[0][2:]:

									      bins_value[0:2]=[(

									        bins_value[0][0],

									        bins_value[1][1],

									        bins_value[0][2]+bins_value[1][2],

									        bins_value[0][3]+bins_value[1][3])]

									      continue

									  ##bins merge forwards

									    if 0 in bins_value[i][2:]:

									      bins_value[i-1:i+1]=[(

									        bins_value[i-1][0],

									        bins_value[i][1],

									        bins_value[i-1][2]+bins_value[i][2],

									        bins_value[i-1][3]+bins_value[i][3])]

									      break

									    else:

									      break

									  #calculate chi-square merge the minimum chisquare    

									  while len(bins_value)>binnum:

									    chi_squares=[]

									    for i in range(len(bins_value)-1):

									      a=bins_value[i][2:]

									      b=bins_value[i+1][2:]

									      chi_square=scipy.stats.chi2_contingency([a,b])[0]

									      chi_squares.append(chi_square)

									  #merge the minimum chisquare backwards

									    i = chi_squares.index(min(chi_squares))

									    bins_value[i:i+2]=[(

									      bins_value[i][0],

									      bins_value[i+1][1],

									      bins_value[i][2]+bins_value[i+1][2],

									      bins_value[i][3]+bins_value[i+1][3])]

									    df_woe=woe_value(bins_value)

									  #print bin number and iv

									    print("箱数：{},iv:{:.6f}".format(len(bins_value),iv_value(df_woe)))

									  #return bins and woe information 

									  return woe_value(bins_value)

以下是效果：

初始分成10箱，目标为3箱

1	`chi_bin(data,"age","SeriousDlqin2yrs",binnum=3,maxcut=10)`

箱数：8,iv:0.184862
箱数：7,iv:0.184128
箱数：6,iv:0.179518
箱数：5,iv:0.176980
箱数：4,iv:0.172406
箱数：3,iv:0.160015
min max count_0 count_1 total bad_rate woe
0 0.0 52.0 70293 7077 77370 0.091470 -0.266233
1 52.0 61.0 29318 1774 31092 0.057056 0.242909
2 61.0 72.0 26332 865 27197 0.031805 0.853755

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原文链接：https://blog.csdn.net/wyzwyzwyzo/article/details/107354019