Question

如果我想在Pandas中使用以下语法来使用比较运算符创建if条件，是否需要将or条件包装在其自己的括号内？

没有多余的括号

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       (df['Col3'] == 'No') |
       (df['Col3'] == 'Yes') |
       (df['Col3'] == 'Maybe'),
       ['Result']] = 'CORRECT'

带有额外的括号

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       ((df['Col3'] == 'No') |
       (df['Col3'] == 'Yes') |
       (df['Col3'] == 'Maybe')),
       ['Result']] = 'CORRECT'

还是都有效？很难从正确的文档中得出结论。

Answer 1

我认为最好使用isin：

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       (df['Col3'].isin(['No', 'Yes', 'Maybe'])), 'Result'] = 'CORRECT'

但是您的陈述是不同的：

@DeepSpace-评论：

它与熊猫没有直接关系，但与operator precedence in Python无关。按位和（＆）优先于按位或（|），因此在这种情况下需要括号。

np.random.seed(1245)

a = ['No', 'Yes', 'Maybe']
df = pd.DataFrame(np.random.choice(a, size=(10, 3)), columns=['Col1','Col2','Col3'])

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       (df['Col3'] == 'No') |
       (df['Col3'] == 'Yes') |
       (df['Col3'] == 'Maybe'),
       'Result1'] = 'CORRECT'

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       ((df['Col3'] == 'No') |
       (df['Col3'] == 'Yes') |
       (df['Col3'] == 'Maybe')),
       'Result2'] = 'CORRECT'

df.loc[(df['Col1'] == 'Yes') &
       (df['Col2'] == 'Yes') &
       (df['Col3'].isin(['No', 'Yes', 'Maybe'])), 'Result3'] = 'CORRECT'

print (df)
    Col1   Col2   Col3  Result1  Result2  Result3
0  Maybe    Yes    Yes  CORRECT      NaN      NaN
1     No  Maybe     No      NaN      NaN      NaN
2    Yes     No  Maybe  CORRECT      NaN      NaN
3  Maybe    Yes     No      NaN      NaN      NaN
4    Yes    Yes  Maybe  CORRECT  CORRECT  CORRECT
5    Yes  Maybe    Yes  CORRECT      NaN      NaN
6  Maybe  Maybe    Yes  CORRECT      NaN      NaN
7  Maybe     No    Yes  CORRECT      NaN      NaN
8  Maybe  Maybe  Maybe  CORRECT      NaN      NaN
9    Yes    Yes     No  CORRECT  CORRECT  CORRECT

使用| |的多个if语句的Pandas语法和＆运算子

1 个答案: