Python dataframe 根據其他id列創建索引列

Question

我有一個像這樣的 dataframe：

ID                  Price
000afb96ded6677c    1514.5
000afb96ded6677c    13.0
000afb96ded6677c    611.0
000afb96ded6677c    723.0
000afb96ded6677c    2065.0
ffea14e87a4e1269    2286.0
ffea14e87a4e1269    1150.0
ffea14e87a4e1269    80.0
fff455057ad492da    650.0
fff5fc66c1fd66c2    450.0

我需要一個 ID 列，它從 1 迭代到有多少行，但我需要它像下面的代碼一樣：

ID                  Price    ID 2
000afb96ded6677c    1514.5   1
000afb96ded6677c    13.0     1
000afb96ded6677c    611.0    1
000afb96ded6677c    723.0    1
000afb96ded6677c    2065.0   1
ffea14e87a4e1269    2286.0   2
ffea14e87a4e1269    1150.0   2
ffea14e87a4e1269    80.0     2
fff455057ad492da    650.0    3
fff5fc66c1fd66c2    450.0    4

Answer 1

嘗試groupby ngroup + 1：

df['ID_2'] = df.groupby('ID').ngroup() + 1

或與Rank ：

df['ID_2'] = df['ID'].rank(method='dense').astype(int)

或使用pd.factorize ：

df['ID_2'] = pd.factorize(df['ID'])[0] + 1

df ：

                 ID   Price  ID_2
0  000afb96ded6677c  1514.5     1
1  000afb96ded6677c    13.0     1
2  000afb96ded6677c   611.0     1
3  000afb96ded6677c   723.0     1
4  000afb96ded6677c  2065.0     1
5  ffea14e87a4e1269  2286.0     2
6  ffea14e87a4e1269  1150.0     2
7  ffea14e87a4e1269    80.0     2
8  fff455057ad492da   650.0     3
9  fff5fc66c1fd66c2   450.0     4