使用python-pandas在组内排名

Question

我正在solver使用一组实例比较一组八个算法(列),每个实例对每个算法执行一次,参数级别D从1到10执行.因此,结果数据框将如下所示:

         instance  D    z             solver
0   1000_ep0.0075  1  994         threatened
1   1000_ep0.0075  1  993               desc
2   1000_ep0.0075  1  994             degree
3   1000_ep0.0075  1  993    threatened_desc
4   1000_ep0.0075  1  993  threatened_degree
5   1000_ep0.0075  1  994         desc_later
6   1000_ep0.0075  1  994       degree_later
7   1000_ep0.0075  1  993         dyn_degree
8   1000_ep0.0075  2  986         threatened
9   1000_ep0.0075  2  987               desc
10  1000_ep0.0075  2  988             degree
11  1000_ep0.0075  2  987    threatened_desc
12  1000_ep0.0075  2  986  threatened_degree
13  1000_ep0.0075  2  987         desc_later
14  1000_ep0.0075  2  988       degree_later
15  1000_ep0.0075  2  987         dyn_degree
....

凡z列对应于由算法找到的值(越小越好).

我想在数据帧中添加一列,根据z每个组合的值对应每个算法的等级<instance, D>.对于上面的例子,将是这样的:

         instance  D    z             solver z_rank
0   1000_ep0.0075  1  994         threatened 2
1   1000_ep0.0075  1  993               desc 1
2   1000_ep0.0075  1  994             degree 2
3   1000_ep0.0075  1  993    threatened_desc 1
4   1000_ep0.0075  1  993  threatened_degree 1
5   1000_ep0.0075  1  994         desc_later 2
6   1000_ep0.0075  1  994       degree_later 2
7   1000_ep0.0075  1  993         dyn_degree 1
8   1000_ep0.0075  2  986         threatened 1
9   1000_ep0.0075  2  987               desc 2
10  1000_ep0.0075  2  988             degree 3
11  1000_ep0.0075  2  987    threatened_desc 2
12  1000_ep0.0075  2  986  threatened_degree 1
13  1000_ep0.0075  2  987         desc_later 2
14  1000_ep0.0075  2  988       degree_later 3
15  1000_ep0.0075  2  987         dyn_degree 2
...

使用python-pandas,这是我到目前为止所能得到的:

df.loc[:, 'z_rank'] = df_rg.groupby(['instance', 'D'])['z'].rank()
df.head(16)
         instance  D    z             solver  z_rank
0   1000_ep0.0075  1  994         threatened    47.5
1   1000_ep0.0075  1  993               desc    16.5
2   1000_ep0.0075  1  994             degree    47.5
3   1000_ep0.0075  1  993    threatened_desc    16.5
4   1000_ep0.0075  1  993  threatened_degree    16.5
5   1000_ep0.0075  1  994         desc_later    47.5
6   1000_ep0.0075  1  994       degree_later    47.5
7   1000_ep0.0075  1  993         dyn_degree    16.5
8   1000_ep0.0075  2  986         threatened     7.0
9   1000_ep0.0075  2  987               desc    18.5
10  1000_ep0.0075  2  988             degree    44.5
11  1000_ep0.0075  2  987    threatened_desc    18.5
12  1000_ep0.0075  2  986  threatened_degree     7.0
13  1000_ep0.0075  2  987         desc_later    18.5
14  1000_ep0.0075  2  988       degree_later    44.5
15  1000_ep0.0075  2  987         dyn_degree    18.5

这显然不是我想要的.

有人可以帮我吗？

Answer 1

您需要method=dense在SeriesGroupBy.rank()组之间增加1的位置:

df['z_rank'] = df.groupby(['instance', 'D'])['z'].rank(method='dense').astype(int)