通过A列获取B列分组的组大小和最小值 [英] Getting both group size and min of column B grouping by column A

问题描述

给定一个DataFrame df ，我可以通过A列获得列A的大小

Given a DataFrame df, I can obtain the size of groups by column A with

df.groupby(['columnA']).size()

df.groupby(['columnA']).agg({'columnB':min})

或（我猜是语法糖）

df.groupby(['columnA'])['columnB'].min()

但是我如何直接获得具有这两列的DataFrame？

but how can I obtain directly a DataFrame with such 2 columns?

在SQL中，如果您愿意，

In SQL, if you comfortable with it, this would be as simple as

SELECT count(columnA), min(columnB) FROM table GROUP BY columnA

预先感谢您提供任何线索。

Thanks in advance for any clue.

推荐答案

将它们都放在 agg 应该起作用，因为 agg 允许使用功能列表。

Placing both of them in agg should work, since agg permits a list of functions.

>>> df
  columnA  columnB  columnC
0    cat1        3      400
1    cat1        2       20
2    cat1        5     3029
3    cat2        1      492
4    cat2        4       30
5    cat3        2      203
6    cat3        6      402
7    cat3        4      391

>>> df.groupby(['columnA']).columnB.agg(['size', 'min'])
         size  min
columnA           
cat1        3    2
cat2        2    1
cat3        3    2

>>> df.groupby(['columnA']).agg(['size', 'min'])
        columnB     columnC     
           size min    size  min
columnA                         
cat1          3   2       3   20
cat2          2   1       2   30
cat3          3   2       3  203

agg 也接受列的dict->函数，因此，如果您想将一个函数分别映射到每个列，则可以使用某些方法（此处没有必要）

agg also accepts a dict of columns -> functions, and so were you to want to map a function to each column individually you could do so with something (not necessary here) like

df.groupby(['columnA']).agg({'columnA':'size','columnB':'min‌​'})

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