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python - 多索引 Pandas 数据框到字典

转载 作者:太空宇宙 更新时间:2023-11-03 12:55:18 26 4
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我有一个数据框如下:

raw_data = {'regiment': ['Nighthawks', 'Nighthawks', 'Nighthawks', 'Nighthawks', 'Dragoons', 'Dragoons', 'Dragoons', 'Dragoons', 'Scouts', 'Scouts', 'Scouts', 'Scouts'],
'company': ['1st', '1st', '2nd', '2nd', '1st', '1st', '2nd', '2nd','1st', '1st', '2nd', '2nd'],
'name': ['Miller', 'Jacobson', 'Ali', 'Milner', 'Cooze', 'Jacon', 'Ryaner', 'Sone', 'Sloan', 'Piger', 'Riani', 'Ali'],
'preTestScore': [4, 24, 31, 2, 3, 4, 24, 31, 2, 3, 2, 3],
'postTestScore': [25, 94, 57, 62, 70, 25, 94, 57, 62, 70, 62, 70]}

df = pd.DataFrame(raw_data, columns = ['regiment', 'company', 'name', 'preTestScore', 'postTestScore'])

如果我按两列分组并计算大小,

df.groupby(['regiment','company']).size()

我得到以下信息:

regiment    company
Dragoons 1st 2
2nd 2
Nighthawks 1st 2
2nd 2
Scouts 1st 2
2nd 2
dtype: int64

我想要的输出是一个字典,如下所示:

{'Dragoons':{'1st':2,'2nd':2},
'Nighthawks': {'1st':2,'2nd':2},
... }

我尝试了不同的方法,但都无济于事。有没有相对干净的方法来实现上述目标?

非常感谢您!!!!

最佳答案

您可以添加 Series.unstackDataFrame.to_dict :

d = df.groupby(['regiment','company']).size().unstack().to_dict(orient='index')
print (d)
{'Dragoons': {'2nd': 2, '1st': 2},
'Nighthawks': {'2nd': 2, '1st': 2},
'Scouts': {'2nd': 2, '1st': 2}}

另一个解决方案,与另一个答案非常相似:

from collections import Counter

df = {i: dict(Counter(x['company'])) for i, x in df.groupby('regiment')}
print (df)
{'Dragoons': {'2nd': 2, '1st': 2},
'Nighthawks': {'2nd': 2, '1st': 2},
'Scouts': {'2nd': 2, '1st': 2}}

但如果使用第一种解决方案,NaN 就会有问题(这取决于数据)

示例:

raw_data = {'regiment': ['Nighthawks', 'Nighthawks', 'Nighthawks', 'Nighthawks', 'Dragoons', 'Dragoons', 'Dragoons', 'Dragoons', 'Scouts', 'Scouts', 'Scouts', 'Scouts'],
'company': ['1st', '1st', '2nd', '2nd', '1st', '1st', '2nd', '2nd','1st', '1st', '2nd', '3rd'],
'name': ['Miller', 'Jacobson', 'Ali', 'Milner', 'Cooze', 'Jacon', 'Ryaner', 'Sone', 'Sloan', 'Piger', 'Riani', 'Ali'],
'preTestScore': [4, 24, 31, 2, 3, 4, 24, 31, 2, 3, 2, 3],
'postTestScore': [25, 94, 57, 62, 70, 25, 94, 57, 62, 70, 62, 70]}

df = pd.DataFrame(raw_data, columns = ['regiment', 'company', 'name', 'preTestScore', 'postTestScore'])
print (df)
regiment company name preTestScore postTestScore
0 Nighthawks 1st Miller 4 25
1 Nighthawks 1st Jacobson 24 94
2 Nighthawks 2nd Ali 31 57
3 Nighthawks 2nd Milner 2 62
4 Dragoons 1st Cooze 3 70
5 Dragoons 1st Jacon 4 25
6 Dragoons 2nd Ryaner 24 94
7 Dragoons 2nd Sone 31 57
8 Scouts 1st Sloan 2 62
9 Scouts 1st Piger 3 70
10 Scouts 2nd Riani 2 62
11 Scouts 3rd Ali 3 70

df1 = df.groupby(['regiment','company']).size().unstack()
print (df1)
company 1st 2nd 3rd
regiment
Dragoons 2.0 2.0 NaN
Nighthawks 2.0 2.0 NaN
Scouts 2.0 1.0 1.0

d = df1.to_dict(orient='index')
print (d)
{'Dragoons': {'3rd': nan, '2nd': 2.0, '1st': 2.0},
'Nighthawks': {'3rd': nan, '2nd': 2.0, '1st': 2.0},
'Scouts': {'3rd': 1.0, '2nd': 1.0, '1st': 2.0}}

然后是必要的使用:

d = {i: dict(Counter(x['company'])) for i, x in df.groupby('regiment')}
print (d)
{'Dragoons': {'2nd': 2, '1st': 2},
'Nighthawks': {'2nd': 2, '1st': 2},
'Scouts': {'3rd': 1, '2nd': 1, '1st': 2}}

或另一个John Galt回答。

关于python - 多索引 Pandas 数据框到字典,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44823418/

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