在 Pandas 数据框中随机排列行，将重复项放在一起

Question

I have a data like this:

A  B  C  D  E  F
35 1  2  35 25 65
40 5  7  47 57 67
20 1  8  74 58 63
35 1  2  37 28 69
40 5  7  49 58 69
20 1  8  74 58 63
35 1  2  47 29 79
40 5  7  55 77 87
20 1  8  74 58 63

Here we can see that Columns A,B and C have replicas that are repeated in various rows. I want to shuffle all the rows and have the replicas in consecutive rows, without deleting any of them. The output should look like this:

A  B  C  D  E  F
35 1  2  35 25 65
35 1  2  37 28 69
35 1  2  47 29 79
40 5  7  47 57 67
40 5  7  49 58 69
40 5  7  55 77 87
20 1  8  74 58 63
20 1  8  74 58 63
20 1  8  74 58 63

When I use pandas.DataFrame.duplicated , it can give me duplicated rows. How can I keep all the identical rows using groupby ?

Answer 1

Here is code that achieves the result you asked for (which doesn't require either explicit shuffling or sorting, but merely grouping your existing df by columns A,B,C):

df_shuf = pd.concat( group[1] for group in df.groupby(['A','B','C'], sort=False) )

print(df_shuf.to_string(index=False))

A  B  C   D   E   F
35  1  2  35  25  65
35  1  2  37  28  69
35  1  2  47  29  79
40  5  7  47  57  67
40  5  7  49  58  69
40  5  7  55  77  87
20  1  8  74  58  63
20  1  8  74  58  63
20  1  8  74  58  63

Notes:

• I couldn't figure out how to do df.reindex in-place on the grouped object. But we can get by without it.
• You don't need pandas.DataFrame.duplicated , since df.groupby(['A','B','C'] puts all duplicates in the same group already.
• df.groupby(... sort=False) is faster, use it whenever you don't need the groups sorted by default.