按顺序计算每个组的熊猫

Question

我有一个df,我按两列分组.我想按顺序计算每个组.下面的代码按顺序计算组中的每一行.这似乎比我想象的要容易,但无法弄清楚.

df = pd.DataFrame({
    'Key': ['10003', '10009', '10009', '10009',
            '10009', '10034', '10034', '10034'], 
    'Date1': [20120506, 20120506, 20120506, 20120506,
              20120620, 20120206, 20120206, 20120405],
    'Date2': [20120528, 20120507, 20120615, 20120629,
              20120621, 20120305, 20120506, 20120506]
})


df['Count'] = df.groupby(['Key','Date1']).cumcount() + 1

预期结果:

    Date1       Date2       Key    Count
0   20120506    20120528    10003  1
1   20120506    20120507    10009  2
2   20120506    20120615    10009  2
3   20120506    20120629    10009  2
4   20120620    20120621    10009  3
5   20120206    20120305    10034  4
6   20120206    20120506    10034  4
7   20120405    20120506    10034  5

Answer 1

你在寻找groupby+ ngroup:

df['Count'] = df.groupby(['Key','Date1']).ngroup() + 1
df

      Date1     Date2    Key  Count
0  20120506  20120528  10003      1
1  20120506  20120507  10009      2
2  20120506  20120615  10009      2
3  20120506  20120629  10009      2
4  20120620  20120621  10009      3
5  20120206  20120305  10034      4
6  20120206  20120506  10034      4
7  20120405  20120506  10034      5

ngroup 只需给每个组一个标签.