Imagine I have dataframe
A B C id
0 1 2 3 1
1 2 3 4 1
given list of values for id, for example [1,2,3] how can I obtain
A B C id
0 1 2 3 1
1 2 3 4 1
2 1 2 3 2
3 2 3 4 2
4 1 2 3 3
5 2 3 4 3
in a most convenient and efficient way?
Try repeat the index, then repeat the id list:
id_list = [1,2,3]
(df.loc[df.index.repeat(len(id_list))]
.assign(id=np.repeat(id_list, len(df)))
)
Output:
A B C id
0 1 2 3 1
0 1 2 3 1
0 1 2 3 2
1 2 3 4 2
1 2 3 4 3
1 2 3 4 3
Another technique (just for learning purpose, not efficient):
concat and then multiple the id with list element after grouping them together.
idx = [1,2,3]
y = pd.concat([df]*len(idx))
y.assign(id=y.groupby(y.index)['id'].transform(lambda x:x.mul(idx))).reset_index(drop=True)
| A | B | C | id | |
|---|---|---|---|---|
| 0 | 1 | 2 | 3 | 1 |
| 1 | 2 | 3 | 4 | 1 |
| 2 | 1 | 2 | 3 | 2 |
| 3 | 2 | 3 | 4 | 2 |
| 4 | 1 | 2 | 3 | 3 |
| 5 | 2 | 3 | 4 | 3 |
