Question

我的虚拟数据帧如下：

+--------+------+------+------+------+
| item   |   p1 |   p2 |   p3 |   p4 |
|--------+------+------+------+------|
| a      |    1 |    0 |    1 |    1 |
| b      |    0 |    1 |    1 |    0 |
| c      |    1 |    0 |    1 |    1 |
| d      |    0 |    0 |    0 |    1 |
| e      |    1 |    0 |    1 |    1 |
| f      |    1 |    1 |    1 |    1 |
| g      |    1 |    0 |    0 |    0 |
+--------+------+------+------+------+

我想找到组合使用或不组合使用参数p1,p2,p3,p4的方式。预期结果如下：

+--------+------+--------+--------+--------+
| Length | P-groups(s)   | Count  | Items  |
+--------+---------------+--------+--------+
| 1      | p1            |  1     | g      |
|        | p4            |  1     | d      |
|        |               |        |        |
| 2      | p2,p3         |  1     | b      |
|        |               |        |        |
| 3      | p1,p2,p3      |  3     | [a,c,e]|
|        |               |        |        |
| 4      | p1,p2,p3,p4   |  1     | f      |
+--------+---------------+--------+--------+

所以，我的原始代码如下：

import pandas as pd
from itertools import chain, combinations

df= pd.DataFrame({'item': ['a','b','c','d','e','f','g'], 
        'p1': [1,0,1,0,1,1,1],
        'p2': [0,1,0,0,0,1,0],
        'p3': [1,1,1,0,1,1,0],
        'p4': [1,0,1,1,1,1,0]})


def all_subsets(ss):
    return chain(*map(lambda x: combinations(ss, x), range(0, len(ss)+1)))


subsets = []

for subset in all_subsets(list(df)[1:]):
    subsets.append(list(subset))

for grp in subsets[1:]: #subset[1:] is to exclude empty set
    print df.groupby(grp).size().reset_index().rename(columns={0:'count'})

我想知道是否有任何熊猫方法可以达到预期效果？

Answer 1

将pd.groupby与pd.filter一起使用：

import pandas as pd

tmp = df.filter(like='p')
new = tmp.replace(1, pd.Series(tmp.columns, tmp.columns)).copy(deep=True)
df['length'] = tmp.sum(1)
df['groups'] = new.apply(lambda x:','.join(s for s in x if s), 1)

gdf = df.groupby(['length', 'groups'])['item'].agg(['count', list])
print(gdf)

输出：

                    count       list
length groups                       
1      p1               1        [g]
       p4               1        [d]
2      p2,p3            1        [b]
3      p1,p3,p4         3  [a, c, e]
4      p1,p2,p3,p4      1        [f]

如果要打开gdf['list']的包装，请添加以下行：

gdf['list'] = [l[0] if len(l)==1 else l for l in gdf['list']]

其效果与期望的输出相同：

                    count       list
length groups                       
1      p1               1          g
       p4               1          d
2      p2,p3            1          b
3      p1,p3,p4         3  [a, c, e]
4      p1,p2,p3,p4      1          f

熊猫：按列子集的所有组合分组

1 个答案: