Question

我有一个包含3列的DF，我想执行布尔检查。代码如下所示。

import pandas as pd
df = pd.DataFrame({
    'Col1':[A,A,A,B,B,C,C,C],
    'Col2':[[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],[1, 2, 3, 4, 5],[1, 2, 3, 4, 5], [1, 2, 3, 4, 5,6,7] ,[1, 2, 3, 4, 5,6,7],[1, 2, 3, 4, 5,6,7] ]
    'Col3': [[1, 2, 3, 4],[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],[1, 2, 3, 4, 5, 6, 7, 8, 9],[1, 2, 3, 4, 5],[1, 2, 3], [1, 2, 3, 4, 5,6,7] ,[1, 2, 3],[1, 2, 3, 4] ]
})

我想要一个给我DF1唯一值的DF，如果Col2，Col3的一行相等，则打印True，否则打印False。我想看到的结果是：

'Col1' 'Col2'                         'Col3'          'Col4'                               
  A    [1,2,3,4,5,6,7,8,9,10,11,12]  [1,2,3,4]         False

print(map(type, df['Col2']) == map(type,df['Col3']))

这使我返回true值，但我希望它作为df中的新列Col4

对于我尝试的多个代码，我总是发现错误为unhashable type: 'list'

Answer 1

转换为字符串以进行比较：

    df['Col4'] = df.Col2.map(str) == df.Col3.map(str)

Answer 2

如果要对每组Col1应用此逻辑，请使用groupby.apply：

df['Col4'] = df.groupby('Col1').apply(lambda x: x['Col2'].eq(x['Col3'])).values

  Col1                                     Col2                                     Col3   Col4
0    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]                             [1, 2, 3, 4]  False
1    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]   True
2    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]              [1, 2, 3, 4, 5, 6, 7, 8, 9]  False
3    B                          [1, 2, 3, 4, 5]                          [1, 2, 3, 4, 5]   True
4    B                          [1, 2, 3, 4, 5]                                [1, 2, 3]  False
5    C                    [1, 2, 3, 4, 5, 6, 7]                    [1, 2, 3, 4, 5, 6, 7]   True
6    C                    [1, 2, 3, 4, 5, 6, 7]                                [1, 2, 3]  False
7    C                    [1, 2, 3, 4, 5, 6, 7]                             [1, 2, 3, 4]  False

不考虑组，在这种情况下，使用eq

给出相同的输出

df['Col4'] = df['Col2'].eq(df['Col3'])

  Col1                                     Col2                                     Col3   Col4
0    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]                             [1, 2, 3, 4]  False
1    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]   True
2    A  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]              [1, 2, 3, 4, 5, 6, 7, 8, 9]  False
3    B                          [1, 2, 3, 4, 5]                          [1, 2, 3, 4, 5]   True
4    B                          [1, 2, 3, 4, 5]                                [1, 2, 3]  False
5    C                    [1, 2, 3, 4, 5, 6, 7]                    [1, 2, 3, 4, 5, 6, 7]   True
6    C                    [1, 2, 3, 4, 5, 6, 7]                                [1, 2, 3]  False
7    C                    [1, 2, 3, 4, 5, 6, 7]                             [1, 2, 3, 4]  False

Answer 3

引入一种新方法tuple

df.Col2.map(tuple)==df.Col3.map(tuple)
Out[646]: 
0    False
1     True
2    False
3     True
4    False
5     True
6    False
7    False
dtype: bool

我有2列的DF。我想对其进行布尔检查

3 个答案: