信用风控模型问题，用iv的时候为什么总提示“keyerror"?

Question

问题遇到的现象和发生背景

一个信用风控问题，给出了训练集和测试集，在通过分箱时，计算iv值发生了问题

import pandas as pd
import numpy as np
X_train_df0=pd.read_csv("train_X.csv")
y_train_df0=pd.read_csv("train_y.csv")
X_train_df0

train_df0=pd.merge(X_train_df0,y_train_df0,on='ID')
train_df0


![img](https://img-mid.csdnimg.cn/release/static/image/mid/ask/553695992346191.png "#left")

运行结果及报错内容

##定义分箱函数，计算iv值
def cal_iv(data,cut_num,feature,target):
    #1.数据分箱
    data_cut=pd.cut(data[feature],cut_num)
    #2.统计各个分箱的违约样本，未违约样本
    cut_group_all=data[target].groupby(data_cut).count()
    cut_group_y=data[target].groupby(data_cut).sum()
    cut_group_n=cut_group_all=cut_group_y
    #3.统计样本比率
    df=pd.DataFrame()
    df['bad']=list(cut_group_y)
    df['good']=list(cut_group_n)
    df['all']=list(cut_group_all)
    df['违约%']=df['bad'] /df['bad'].sum()
    df['不违约%']=df['good'] /df['good'].sum()
    #4.计算WOE
    df['WOE']=np.log(df['违约%']/df['不违约%'])
    df=df.replace({'WOE':{np.inf:0,-np.inf:0}})
    #5.计算各个分箱的iv值
    df['IV']=df['WOE']*(df['违约%']-df['不违约%'])
    iv=df['IV'].sum()
    return iv
print(cal_iv(train_df0,10,'x2','y_1'))
iv_list=[]
df_iv=pd.DataFrame()
for i in train_df0.columns[:-1]:
    a=cal_iv(train_df0,10,i,'y-1')
    iv_list.append(a)
df_iv['feature']=list(train_df0.columns[:-1])
df_iv['IV']=iv_list
b=df_iv.sort_values(by='IV',ascending=False)
b.head(10)

报错结果：

KeyError                                  Traceback (most recent call last)
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
   2894             try:
-> 2895                 return self._engine.get_loc(casted_key)
   2896             except KeyError as err:

pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()

pandas\_libs\hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

pandas\_libs\hashtable_class_helper.pxi in pandas._libs.hashtable.PyObjectHashTable.get_item()

KeyError: 'y_1'

The above exception was the direct cause of the following exception:

KeyError                                  Traceback (most recent call last)
<ipython-input-31-837000a62aae> in <module>
     21     iv=df['IV'].sum()
     22     return iv
---> 23 print(cal_iv(train_df0,10,'x2','y_1'))
     24 iv_list=[]
     25 df_iv=pd.DataFrame()

<ipython-input-31-837000a62aae> in cal_iv(data, cut_num, feature, target)
      4     data_cut=pd.cut(data[feature],cut_num)
      5     #2.统计各个分箱的违约样本，未违约样本
----> 6     cut_group_all=data[target].groupby(data_cut).count()
      7     cut_group_y=data[target].groupby(data_cut).sum()
      8     cut_group_n=cut_group_all=cut_group_y

~\anaconda3\lib\site-packages\pandas\core\frame.py in __getitem__(self, key)
   2900             if self.columns.nlevels > 1:
   2901                 return self._getitem_multilevel(key)
-> 2902             indexer = self.columns.get_loc(key)
   2903             if is_integer(indexer):
   2904                 indexer = [indexer]

~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
   2895                 return self._engine.get_loc(casted_key)
   2896             except KeyError as err:
-> 2897                 raise KeyError(key) from err
   2898 
   2899         if tolerance is not None:

KeyError: 'y_1'

我的解答思路和尝试过的方法

我是照着别人的代码扒的，原代码数据名称和我不一样，y-1我理解的是，train-y中有一列是y,好用户是0，坏用户是1.原代码此处为'Loan_status_Charged Off'