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Pandas groupby列表中的值

[英]Pandas groupby values in a list

I am trying to return a groupby from a pandas df . 我试图从pandas df返回一个groupby I want the output values to be summed not merged . 我希望将输出值相加而不merged But the following merges the appropriate lists . 但以下内容merges了相应的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)

Out: 日期:

                                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]

Intended Output: 预期产出:

                                Output
Id Val                                
1  A    [7,9,11,13,6]
2  B    [11,10,11,11,2]

Or use a one-liner which converts to np.array : 或者使用转换为np.array

df = df.groupby(['Id','Val']).apply(lambda x: x.Output.apply(np.array).sum())
print(df)

Output: 输出:

Id  Val
1   A        [7, 9, 11, 13, 6]
2   B      [11, 10, 11, 11, 2]
dtype: object

You can change the list to numpy array then 您可以将list更改为numpy array

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

Another solution using using zip rather than using apply twice, 使用zip而不是使用apply两次的另一个解决方案,

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]

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