Python Pandas:在groupby中选择第二个最小值 [英] Python pandas: select 2nd smallest value in groupby

问题描述

我有一个示例DataFrame，如下所示:

I have an example DataFrame like the following:

import pandas as pd
import numpy as np
df = pd.DataFrame({'ID':[1,2,2,2,3,3,], 'date':array(['2000-01-01','2002-01-01','2010-01-01','2003-01-01','2004-01-01','2008-01-01'],dtype='datetime64[D]')})

我试图在每个ID组中获得第二天最早的成绩.所以我写了以下函数:

I am trying to get the 2nd earliest day in each ID group. So I wrote the following funciton:

def f(x):
    if len(x)==1:
        return x[0]
    else:
        x.sort()
        return x[1]

然后我写道:

df.groupby('ID').date.apply(lambda x:f(x))

结果是错误.

您能找到一种使这项工作成功的方法吗?

Could you find a way to make this work?

推荐答案

这需要0.14.1.并且会非常有效，尤其是在您有大型群组的情况下(因为这不需要对它们进行完全排序).

This requires 0.14.1. And will be quite efficient, especially if you have large groups (as this doesn't require fully sorting them).

In [32]: df.groupby('ID')['date'].nsmallest(2)
Out[32]: 
ID   
1   0   2000-01-01
2   1   2002-01-01
    3   2003-01-01
3   4   2004-01-01
    5   2008-01-01
dtype: datetime64[ns]

In [33]: df.groupby('ID')['date'].nsmallest(2).groupby(level='ID').last()
Out[33]: 
ID
1    2000-01-01
2    2003-01-01
3    2008-01-01
dtype: datetime64[ns]

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