我正在尝试从groupby
pandas
返回df
。我希望不对输出值求和merged
。但是以下merges
是适当的lists
。
import pandas as pd
d = ({
'Id' : [1,2,2,1],
'Val' : ['A','B','B','A'],
'Output' : [[1,2,3,4,5],[5,3,3,2,1],[6,7,8,9,1],[6,7,8,9,1]],
})
df = pd.DataFrame(data = d)
df = df.groupby(['Id','Val']).agg({'Output':'sum'}, axis = 1)
出局:
Output
Id Val
1 A [1, 2, 3, 4, 5, 6, 7, 8, 9, 1]
2 B [5, 3, 3, 2, 1, 6, 7, 8, 9, 1]
预期输出:
Output
Id Val
1 A [7,9,11,13,6]
2 B [11,10,11,11,2]
答案 0 :(得分:3)
或使用单线转换为np.array
:
df = df.groupby(['Id','Val']).apply(lambda x: x.Output.apply(np.array).sum())
print(df)
输出:
Id Val
1 A [7, 9, 11, 13, 6]
2 B [11, 10, 11, 11, 2]
dtype: object
答案 1 :(得分:2)
您可以将list
更改为numpy
array
,然后
df.Output=df.Output.apply(np.array)
df.groupby(['Id','Val']).Output.apply(lambda x : np.sum(x))
Out[389]:
Id Val
1 A [7, 9, 11, 13, 6]
2 B [11, 10, 11, 11, 2]
Name: Output, dtype: object
答案 2 :(得分:1)
另一种使用zip而不是应用apply的解决方案,
df.groupby(['Id','Val']).Output.apply(lambda x: [sum(i) for i in list(zip(*x))])
Id Val
1 A [7, 9, 11, 13, 6]
2 B [11, 10, 11, 11, 2]