python pandas自定义agg函数 [英] python pandas custom agg function

问题描述

Dataframe:
  one two
a  1  x
b  1  y
c  2  y
d  2  z
e  3  z

grp = DataFrame.groupby('one')
grp.agg(lambda x: ???) #or equivalent function

grp.agg所需的输出:

Desired output from grp.agg:

one two
1   x|y
2   y|z
3   z

在集成数据帧之前，我的agg函数是"|".join(sorted(set(x))).理想情况下，我希望组中有任意数量的列，并且agg为每个列项目返回"|".join(sorted(set())，就像上面的两个一样.我也尝试过np.char.join().

My agg function before integrating dataframes was "|".join(sorted(set(x))). Ideally I want to have any number of columns in the group and agg returns the "|".join(sorted(set()) for each column item like two above. I also tried np.char.join().

爱熊猫，这使我从800线复杂的程序带到了缩放公园中的400线步行.谢谢:)

Love Pandas and it has taken me from a 800 line complicated program to a 400 line walk in the park that zooms. Thank you :)

推荐答案

您是如此亲密:

In [1]: df.groupby('one').agg(lambda x: "|".join(x.tolist()))
Out[1]:
     two
one
1    x|y
2    y|z
3      z

扩展答案以处理排序并仅接受集合:

Expanded answer to handle sorting and take only the set:

In [1]: df = DataFrame({'one':[1,1,2,2,3], 'two':list('xyyzz'), 'three':list('eecba')}, index=list('abcde'), columns=['one','two','three'])

In [2]: df
Out[2]:
   one two three
a    1   x     e
b    1   y     e
c    2   y     c
d    2   z     b
e    3   z     a

In [3]: df.groupby('one').agg(lambda x: "|".join(x.order().unique().tolist()))
Out[3]:
     two three
one
1    x|y     e
2    y|z   b|c
3      z     a

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