pandas 标签重复项 [英] Pandas Label Duplicates

问题描述

给出以下数据框:

import pandas as pd
d=pd.DataFrame({'label':[1,2,2,2,3,4,4],
               'values':[3,5,7,2,5,8,3]})
d
    label   values
0     1       3
1     2       5
2     2       7
3     2       2
4     3       5
5     4       8
6     4       3

我知道如何计算像这样的唯一值:

I know how to count the unique values like this:

d['dup']=d.groupby('label')['label'].transform('count')

这将导致:

    label   values  dup
0     1     3       1
1     2     5       3
2     2     7       3
3     2     2       3
4     3     5       1
5     4     8       2
6     4     3       2

但是我想让一列具有以下值: 如果每个标签列有1 unique行，则为1；如果存在duplicates，并且该行为该行的first，则为2；如果该行是duplicate的行，则为0原件.像这样:

But what I would like is a column to have the following values: 1 if there is 1 unique row per the label column, 2 if there are duplicates and the row in question is the first of such, and 0 if the row is a duplicate of an original. Like this:

    label   values  dup  status
0     1     3       1     1
1     2     5       3     2
2     2     7       3     0
3     2     2       3     0
4     3     5       1     1
5     4     8       2     2
6     4     3       2     0

提前谢谢！

推荐答案

我认为您可以使用

I think you can use loc with condition created by function duplicated:

d['status'] = 2
d.loc[d.dup == 1, 'status'] = 1
d.loc[d.label.duplicated(), 'status'] = 0 
print (d)

   label  values  dup  status
0      1       3    1       1
1      2       5    3       2
2      2       7    3       0
3      2       2    3       0
4      3       5    1       1
5      4       8    2       2
6      4       3    2       0

或加倍 numpy.where :

d['status1'] = np.where(d.dup == 1, 1,
               np.where(d.label.duplicated(), 0, 2))

print (d)  
   label  values  dup  status  status1
0      1       3    1       1        1
1      2       5    3       2        2
2      2       7    3       0        0
3      2       2    3       0        0
4      3       5    1       1        1
5      4       8    2       2        2
6      4       3    2       0        0           

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