Pandas Groupby return incremental numbers of groups as new column
发布于2025-01-05 08:57 阅读(115) 评论(0) 点赞(29) 收藏(3)
Original DF
| performance | driver_id | season |
|-------------|-----------|--------|
| 1 | 1 | 17 |
| 2 | 1 | 17 |
| 3 | 2 | 17 |
| 4 | 2 | 18 |
| 5 | 2 | 18 |
| 6 | 2 | 19 |
| 7 | 1 | 17 |
| 8 | 1 | 18 |
| 9 | 1 | 18 |
Desired DF
| performance | driver_id | season | season_order |
|-------------|-----------|--------|--------------|
| 1 | 1 | 17 | 1 |
| 2 | 1 | 17 | 1 |
| 3 | 2 | 17 | 1 |
| 4 | 2 | 18 | 2 |
| 5 | 2 | 18 | 2 |
| 6 | 2 | 19 | 3 |
| 7 | 1 | 17 | 1 |
| 8 | 1 | 18 | 2 |
| 9 | 1 | 18 | 2 |
I have a DF where I want to set drivers performances as either first, second, third seasons etc. but this must be differnet for a given driver.
I have tried using cumsum and rank but only help within groups I want between groups ordering/ranking. These methods seem to work better for single column groupbys from my experimenting.
df["season_number"] = (df
.groupby(["driver_id ", "season"])
.season
.transform("rank")
)
解决方案
I think you need GroupBy.rank:
df["season_number"] = df.groupby("driver_id").season.rank(method='dense').astype(int)
print (df)
performance driver_id season season_number
0 1 1 17 1
1 2 1 17 1
2 3 2 17 1
3 4 2 18 2
4 5 2 18 2
5 6 2 19 3
6 7 1 17 1
7 8 1 18 2
8 9 1 18 2
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作者：黑洞官方问答小能手
链接：https://www.pythonheidong.com/blog/article/2046771/febb0a03e50872b9c58c/
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