按列值复制 Pandas 数据框中的行
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Replicating rows in a pandas data frame by a column value
提问by Mersenne Prime
I want to replicate rows in a Pandas Dataframe. Each row should be repeated n times, where n is a field of each row.
我想复制 Pandas 数据框中的行。每行应重复 n 次，其中 n 是每行的一个字段。
import pandas as pd
what_i_have = pd.DataFrame(data={
'id': ['A', 'B', 'C'],
'n' : [ 1, 2, 3],
'v' : [ 10, 13, 8]
})
what_i_want = pd.DataFrame(data={
'id': ['A', 'B', 'B', 'C', 'C', 'C'],
'v' : [ 10, 13, 13, 8, 8, 8]
})
Is this possible?
这可能吗？
回答by DSM
You could use np.repeatto get the repeated indices and then use that to index into the frame:
您可以使用np.repeat来获取重复的索引，然后使用它来索引到框架中：
>>> df2 = df.loc[np.repeat(df.index.values,df.n)]
>>> df2
id n v
0 A 1 10
1 B 2 13
1 B 2 13
2 C 3 8
2 C 3 8
2 C 3 8
After which there's only a bit of cleaning up to do:
之后，只需进行一些清理工作：
>>> df2 = df2.drop("n",axis=1).reset_index(drop=True)
>>> df2
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
Note that if you might have duplicate indices to worry about, you could use .ilocinstead:
请注意，如果您可能需要担心重复的索引，则可以.iloc改用：
In [86]: df.iloc[np.repeat(np.arange(len(df)), df["n"])].drop("n", axis=1).reset_index(drop=True)
Out[86]:
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
which uses the positions, and not the index labels.
它使用位置，而不是索引标签。
回答by Zero
You could use set_indexand repeat
你可以使用set_index和repeat
In [1057]: df.set_index(['id'])['v'].repeat(df['n']).reset_index()
Out[1057]:
id v
0 A 10
1 B 13
2 B 13
3 C 8
4 C 8
5 C 8
Details
细节
In [1058]: df
Out[1058]:
id n v
0 A 1 10
1 B 2 13
2 C 3 8
