检查多行 pandas 中标志列的有效性 [英] check validity for flag column's in multiple rows pandas

问题描述

我有一个数据框:

a      id   flag1    flag2
abc    1     1          0
123    1     0          1
xyz    2     1          0
111    2     0          1
qwe    3     1          0
qwe    3     1          0
mmm    4     1          0
222    4     0          1

我想找到flag1和flag2均为1的id数字.

i want to find the id number where both flag1 and flag2 are 1.

例如 对于id 1，在第一行中，标志1 = 1和标志2 = 0，在第二行中，标志1 = 0和标志2 = 1.

for eg. for id 1, in the first row, flag1 = 1 and flag2 = 0, and second row, flag1 = 0 and flag2 = 1.

我的最终输出应该是这样

my final output should look like this

a    id    flag1   flag2
abc    1     1          0
123    1     0          1
xyz    2     1          0
111    2     0          1
mmm    4     1          0
222    4     0          1

或仅id列也可以在列表中的[1,2,4]中使用

or only id column would also work [1,2,4] in a list

由于id = 3，所以id = 3的两行的flag1均为1，flag 2的均为0，所以我不得不忽略它.

since for id = 3, flag1 was 1 in both the rows with id = 3 and flag 2 was 0, so i have to neglect it.

我试图写一个func，但是失败了.

i was trying to write a func, but failed.

def checkValidTransactionRow(frame):
df['id'][(df['flag1']==1) & (df['flag2']==1) ].unique()

推荐答案

尝试以下方法:

In [23]: ids = df.groupby('id')['flag1','flag2'].apply(lambda x: x.eq(1).any()).all(1)

In [24]: ids
Out[24]:
id
1     True
2     True
3    False
4     True
dtype: bool

In [25]: ids.index[ids]
Out[25]: Int64Index([1, 2, 4], dtype='int64', name='id')

说明:

In [26]: df.groupby('id')['flag1','flag2'].apply(lambda x: x.eq(1).any())
Out[26]:
   flag1  flag2
id
1   True   True
2   True   True
3   True  False
4   True   True

x.eq(1).any()与(x == 1).any()相同-即，如果x系列中的至少一个值等于1，则返回True，否则返回False

x.eq(1).any() is the same as (x == 1).any() - i.e. return True if at least one value in x series equals to 1, otherwise return False

更新:

In [34]: ids.index[ids].values
Out[34]: array([1, 2, 4], dtype=int64)

In [35]: ids.index[ids].values.tolist()
Out[35]: [1, 2, 4]

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