Pandas,按计数分组并将计数添加到原始数据框中？

Question

尝试计算数据框中具有类似"种类"的行时:

import pandas as pd

items = [('aaa','aaa text 1'), ('aaa','aaa text 2'), ('aaa','aaa text 3'),
         ('bb', 'bb text 1'), ('bb', 'bb text 2'), ('bb', 'bb text 3'), 
         ('bb', 'bb text 4'),
         ('cccc','cccc text 1'), ('cccc','cccc text 2'),
         ('dd', 'dd text 1'),
         ('e', 'e text 1'),
         ('fff', 'fff text 1'),
        ]

df = pd.DataFrame(items, columns=['kind', 'msg'])
df

    kind    msg
0   aaa     aaa text 1
1   aaa     aaa text 2
2   aaa     aaa text 3
3   bb      bb text 1
4   bb      bb text 2
5   bb      bb text 3
6   bb      bb text 4
7   cccc    cccc text 1
8   cccc    cccc text 2
9   dd      dd text 1
10  e       e text 1
11  fff     fff text 1

此代码有效:

df = df[['kind']].groupby(['kind'])['kind'] \
                         .count() \
                         .reset_index(name='count') \
                         .sort_values(['count'], ascending=False) \
                         .head(5)

df

导致:

    kind      count
    0   aaa   1
    1   bb    1
    2   cccc  1
    3   dd    1
    4   e     1

然而,如何才能获得包含所有列的数据框,如原始列和"计数"列？那么结果应该按顺序排列'kind','msg','count'列？

另外,如何按计数的降序对结果数据帧进行排序？

Answer 1

IIUC

In [247]: df['count'] = df.groupby('kind').transform('count')

In [248]: df
Out[248]:
    kind          msg  count
0    aaa   aaa text 1      3
1    aaa   aaa text 2      3
2    aaa   aaa text 3      3
3     bb    bb text 1      4
4     bb    bb text 2      4
5     bb    bb text 3      4
6     bb    bb text 4      4
7   cccc  cccc text 1      2
8   cccc  cccc text 2      2
9     dd    dd text 1      1
10     e     e text 1      1
11   fff   fff text 1      1

排序:

In [249]: df.sort_values('count', ascending=False)
Out[249]:
    kind          msg  count
3     bb    bb text 1      4
4     bb    bb text 2      4
5     bb    bb text 3      4
6     bb    bb text 4      4
0    aaa   aaa text 1      3
1    aaa   aaa text 2      3
2    aaa   aaa text 3      3
7   cccc  cccc text 1      2
8   cccc  cccc text 2      2
9     dd    dd text 1      1
10     e     e text 1      1
11   fff   fff text 1      1

Answer 2

这是计算频率并在数据帧中添加一列的简单代码。

df['count'] = df.groupby('kind')['Kind'].transform('count')