在 pandas 组的字典中仅保留没有None值的键 [英] keep only keys without None values in dictionary from pandas groups

问题描述

>>> df = pd.DataFrame({'a': [1,1,1,2,2,3,3,3,3,4,4,5,5], 
'b': [0,1,1,0,1,0,0,1,4,1,0,3,0], 
'v': [2,4,3,7,6,5,9,3,2,4,5,2,3]})
>>> df
    a  b  v
0   1  0  2
1   1  1  4
2   1  1  3
3   2  0  7
4   2  1  6
5   3  0  5
6   3  0  9
7   3  1  3
8   3  4  2
9   4  1  4
10  4  0  5
11  5  3  2
12  5  0  3

>>> df.groupby(by =['a', 'b']).v.apply(list).unstack().to_dict('index')
{1: {0: [2], 1: [4, 3], 3: None, 4: None}, 2: {0: [7], 1: [6], 3: None, 4: 
None}, 3: {0: [5, 9], 1: [3], 3: None, 4: [2]}, 4: {0: [5], 1: [4], 3: None, 4: 
None}, 5: {0: [3], 1: None, 3: [2], 4: None}}

如何在输出字典中避免使用None值的键?在当前条件下，我的字典的大小比使用所需的键大了20倍.

How can the keys with None values be avoided in the output dictionary? In the present condition, my dictionary ends up 20x bigger than it should just with the needed keys.

推荐答案

使用相同的想法，只需两次to_dict

Using same idea , just need to_dict twice

df.groupby(by =['a', 'b']).v.apply(list).groupby(level=0).agg(lambda x : x.reset_index(level=0,drop=True).to_dict()).to_dict()
Out[1092]: 
{1: {0: [2], 1: [4, 3]},
 2: {0: [7], 1: [6]},
 3: {0: [5, 9], 1: [3], 4: [2]},
 4: {0: [5], 1: [4]},
 5: {0: [3], 3: [2]}}

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