在 Pandas 数据框中查找所有重复的行
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Find all duplicate rows in a pandas dataframe
提问by Nico
I would like to be able to get the indices of all the instances of a duplicated row in a dataset without knowing the name and number of columns beforehand. So assume I have this:
我希望能够获得数据集中重复行的所有实例的索引，而无需事先知道列的名称和数量。所以假设我有这个：
col
1 | 1
2 | 2
3 | 1
4 | 1
5 | 2
I'd like to be able to get [1, 3, 4]and [2, 5]. Is there any way to achieve this? It sounds really simple, but since I don't know the columns beforehand I can't do something like df[col == x...].
我希望能够获得[1, 3, 4]和[2, 5]。有没有办法实现这一目标？这听起来很简单，但因为我事先不知道列，所以我不能做类似df[col == x...].
回答by jezrael
First filter all duplicatedrows and then groupbywith applyor convert indexto_series:
首先过滤所有duplicated行，然后 groupby使用apply或转换indexto_series：
df = df[df.col.duplicated(keep=False)]
a = df.groupby('col').apply(lambda x: list(x.index))
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object
a = df.index.to_series().groupby(df.col).apply(list)
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object
And if need nested lists:
如果需要嵌套列表：
L = df.groupby('col').apply(lambda x: list(x.index)).tolist()
print (L)
[[1, 3, 4], [2, 5]]
If need use only first column is possible selected by position with iloc:
如果需要仅使用第一列，则可以按位置选择iloc：
a = df[df.iloc[:,0].duplicated(keep=False)]
.groupby(df.iloc[:,0]).apply(lambda x: list(x.index))
print (a)
col
1 [1, 3, 4]
2 [2, 5]
dtype: object
