python - 当其他列满足某些条件时，如何使用 fillna() 估算列中的值

Question

我计算了credit_history 具有 NaN 值时的计数。

Credit_History 为 NaN 时的输出:

Self_Employed
Yes  532
No   32

Married
No   398
Yes  21

对于数值，我计算了所有列的平均值

当 Credit_History 为 NaN 时非数值的输出:

Mean Applicant Income: 54003.1232
LoanAmount: 35435.12
Loan_Amount_Term: 360
ApplicantIncome: 30000

在这些情况下我现在如何使用 fillna():

情况 1:当 Self_Employed = Y 且 Married = N 时； Credit_History 应为 0

情况 2:当 Self_Employed = N 且 ApplicantIncome > 20000 时； Credit_History 应为 1

案例3:当Self_Employed = Y，Married = N且ApplicantIncome > 2000时； Credit_History 应为 1

另外，当使用 fillna() 在某些条件下不太明显时，我们可以使用数据透视表来计算中值，然后使用 fillna() 进行插补吗？

提前致谢。

Answer 1

使用numpy.select如果所有条件均为False，则输出由参数default定义:

from  itertools import  product
c = ['Self_Employed','Married','ApplicantIncome']
df =  pd.DataFrame(list(product(list('NY'), list('NY'), [10000, 30000])), 
                   columns=c)


m1 = (df.Self_Employed == 'Y') & (df.Married == 'N')
m2 = (df.Self_Employed == 'N') & (df.ApplicantIncome > 20000)
m3 = m1 & (df.ApplicantIncome > 20000)

df['Credit_History'] = np.select([m1, m2, m3], [0,1,1], default=2)
print (df)
  Self_Employed Married  ApplicantIncome  Credit_History
0             N       N            10000               2
1             N       N            30000               1
2             N       Y            10000               2
3             N       Y            30000               1
4             Y       N            10000               0
5             Y       N            30000               0
6             Y       Y            10000               2
7             Y       Y            30000               2

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但如果想按条件替换，请添加 fillna :

c = ['Self_Employed','Married','ApplicantIncome']
df =  pd.DataFrame(list(product(list('NY'), list('NY'), [10000, 30000])), 
                   columns=c).assign(Credit_History=[np.nan,1,0, np.nan] *2)
print (df)
  Self_Employed Married  ApplicantIncome  Credit_History
0             N       N            10000             NaN
1             N       N            30000             1.0
2             N       Y            10000             0.0
3             N       Y            30000             NaN
4             Y       N            10000             NaN
5             Y       N            30000             1.0
6             Y       Y            10000             0.0
7             Y       Y            30000             NaN

m1 = (df.Self_Employed == 'Y') & (df.Married == 'N')
m2 = (df.Self_Employed == 'N') & (df.ApplicantIncome > 20000)
m3 = m1 & (df.ApplicantIncome > 20000)

s = pd.Series(np.select([m1, m2, m3], [0,1,1], default=2), index=df.index)
df['Credit_History'] = df['Credit_History'].fillna(s)
print (df)
  Self_Employed Married  ApplicantIncome  Credit_History
0             N       N            10000             2.0
1             N       N            30000             1.0
2             N       Y            10000             0.0
3             N       Y            30000             1.0
4             Y       N            10000             0.0
5             Y       N            30000             1.0
6             Y       Y            10000             0.0
7             Y       Y            30000             2.0