python pandas自定义agg函数
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 :)
答
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