Python pandas 的排名/排序基于另一列，每列输入均不同 [英] Python pandas rank/sort based on another column that differs for each input

问题描述

我想根据前三个提出以下第四栏:

I would like to come up with the 4th column below based on the first three:

user    job  time  Rank
A   print   1559   2
A   print   1540   2
A   edit    1520   1
A   edit    1523   1
A   deliver 9717   3
B   edit    1717   2
B   edit    1716   2
B   edit    1715   2
B   deliver 1527   1
B   deliver 1524   1

第4列中的排名对于每个用户(第1列)都是独立的.对于每个用户，我想根据第三列的值对第二列进行排名.例如.对于用户A，他/她有3个职位待定.因为编辑"的时间值最小，然后编辑下一个，则交付最大，因此这三个时间的等级分别是编辑"，"1"，打印" -2和交付-3".

The ranking in the 4th columns is independent for each user (1st column). For each user, I would like to rank the second column based on the value of the 3rd column. Eg. for user A, s/he has three jobs to be ranks. Because the time value of 'edit' is the smallest and edit the next and deliver the largest, the ranking for the three is edit - 1, print - 2 and deliver -3.

我知道我应该从第一列开始groupby，但是以某种方式无法弄清楚如何根据第三行对第二列进行排名，而第三行的每一行都不相同.

I know I should start with groupby the first column, but somehow cannot figure how to rank the 2nd column based on the 3rd that's different for each row.

推荐答案

首先，分配一个新列，其中包含用户-作业对的最短时间:

First, assign a new column which contains the minimum time for user-job pairs:

df['min_time'] = df.groupby(['user', 'job'])['time'].transform('min')

然后按每个用户分组并对其进行排名:

Then group by each user and rank them:

df.groupby('user')['min_time'].rank(method='dense').astype(int)
Out: 
0    2
1    2
2    1
3    1
4    3
5    2
6    2
7    2
8    1
9    1
Name: min_time, dtype: int64

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