Python程序用True和False替换'qualify'列值'yes'和'no'

时间:2018-02-10 09:19:05

标签: python pandas replace merge concatenation

import numpy as np
import pandas as pd

exam_data =pd.DataFrame( {'name': ['Anastasia', 'Dima', 'Katherine', 
'James', 'Emily', 'Michael', 'Matthew', 'Laura', 'Kevin', 'Jonas'],
'score': [12.5, 9, 16.5, np.nan, 9, 20, 14.5, np.nan, 8, 19],
'attempts': [1, 3, 2, 3, 2, 3, 1, 1, 2, 1],
'qualify': ['yes', 'no', 'yes', 'no', 'no', 'yes', 'yes', 'no', 'no', 
'yes']})

 exam_data.set_index([['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j'], 
 'name'])



 r1 = exam_data.replace('yes', 'true')
 r2 = exam_data.replace('no', 'false')
 r1

我希望结果是

attempts    name        qualify score
       a    1           Anastasia   true    12.5
       b    3           Dima        false   9.0
       c    2           Katherine   true    16.5
       d    3           James       false   NaN
       e    2           Emily       false   9.0
       f    3           Michael     true    20.0
       g    1           Matthew     true    14.5
       h    1           Laura       false   NaN
       i    2           Kevin       false   8.0
       j    1           Jonas       true    19.0

1 个答案:

答案 0 :(得分:1)

最简单的是按yes比较值:

exam_data['qualify'] = exam_data['qualify'] == 'yes'
print (exam_data)
   attempts       name  qualify  score
0         1  Anastasia     True   12.5
1         3       Dima    False    9.0
2         2  Katherine     True   16.5
3         3      James    False    NaN
4         2      Emily    False    9.0
5         3    Michael     True   20.0
6         1    Matthew     True   14.5
7         1      Laura    False    NaN
8         2      Kevin    False    8.0
9         1      Jonas     True   19.0

如果要使用replace - 如果dict中定义的其他值未更改:

exam_data['qualify'] = exam_data['qualify'].replace({'yes':True, 'no':False})

map - 如果dict中定义的其他值已替换为NaN s:

exam_data['qualify'] = exam_data['qualify'].map({'yes':True, 'no':False})