这适用于重复的值

如果你在前n个值中有重复的值，并且只想要唯一的值，你可以这样做:

import pandas as pd

ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])

    id first_name last_name department  salary
24  12   Shandler      Bing      Audit  110000
25  14      Jason       Tom      Audit  100000
26  16     Celine    Anston      Audit  100000
27  15    Michale   Jackson      Audit   70000

If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:

(df.groupby('department')['salary']
 .apply(lambda ser: ser.drop_duplicates().nlargest(3))
 .droplevel(level=1)
 .sort_index()
 .reset_index()
)

This gives

department  salary
0   Audit   110000
1   Audit   100000
2   Audit   70000
3   Management  250000
4   Management  200000
5   Management  150000
6   Sales   220000
7   Sales   200000
8   Sales   150000

df.groupby('id').apply(lambda x : x.sort_values(by = 'value', ascending = False).head(2).reset_index(drop = True))

这里的排序值升序为false，类似于nbiggest, True类似于nminimal。 头部内的值与我们在nbiggest内给出的值相同，以获得每个组要显示的值的数量。 Reset_index是可选的，不是必需的。

有时，提前排序整个数据非常耗时。 我们可以先分组，然后对每一组做topk:

g = df.groupby(['id']).apply(lambda x: x.nlargest(topk,['value'])).reset_index(drop=True)

这适用于重复的值

如果你在前n个值中有重复的值，并且只想要唯一的值，你可以这样做:

import pandas as pd

ifile = "https://raw.githubusercontent.com/bhishanpdl/Shared/master/data/twitter_employee.tsv"
df = pd.read_csv(ifile,delimiter='\t')
print(df.query("department == 'Audit'")[['id','first_name','last_name','department','salary']])

    id first_name last_name department  salary
24  12   Shandler      Bing      Audit  110000
25  14      Jason       Tom      Audit  100000
26  16     Celine    Anston      Audit  100000
27  15    Michale   Jackson      Audit   70000

If we do not remove duplicates, for the audit department we get top 3 salaries as 110k,100k and 100k.
If we want to have not-duplicated salaries per each department, we can do this:

(df.groupby('department')['salary']
 .apply(lambda ser: ser.drop_duplicates().nlargest(3))
 .droplevel(level=1)
 .sort_index()
 .reset_index()
)

This gives

department  salary
0   Audit   110000
1   Audit   100000
2   Audit   70000
3   Management  250000
4   Management  200000
5   Management  150000
6   Sales   220000
7   Sales   200000
8   Sales   150000

要获得每个组的前N行，另一种方法是通过groupby().nth[:N]。此调用的结果与groupby().head(N)相同。例如，对于每个id的前2行，调用:

N = 2
df.groupby('id', as_index=False).nth[:N]

要获得每个组中最大的N个值，首先按“id”和“value”排序(确保通过适当地使用升序参数将id按升序排序，将“value”按降序排序)，然后调用groupby().nth[]。

N = 2
df.sort_values(by=['id', 'value'], ascending=[True, False]).groupby('id', as_index=False).nth[:N]

这比在这里(1,2)的其他答案中建议的groupby().apply()调用快得多。在一个包含100k行和8000个组的示例中，%timeit测试表明它比这些解决方案快24-70倍。

此外，除了切片，你还可以将list/tuple/range传递给.nth()调用:

df.groupby('id', as_index=False).nth([0,1])
df.groupby('id', as_index=False).nth([0,2])  # <-- doesn't even have to be consecutive; 
                                             #     this returns 1st and 3rd row of each id

你试过了吗

df.groupby('id').head(2)

输出生成:

       id  value
id             
1  0   1      1
   1   1      2 
2  3   2      1
   4   2      2
3  7   3      1
4  8   4      1

(请记住，你可能需要排序/排序之前，取决于你的数据)

编辑:正如提问者提到的，使用

df.groupby('id').head(2).reset_index(drop=True)

删除MultiIndex并平化结果:

    id  value
0   1      1
1   1      2
2   2      1
3   2      2
4   3      1
5   4      1