[英]multi-index pandas dataframe to a dictionary
我有一個數據幀如下:
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'])
如果我按兩列分組並計算大小,
df.groupby(['regiment','company']).size()
我得到以下內容:
regiment company
Dragoons 1st 2
2nd 2
Nighthawks 1st 2
2nd 2
Scouts 1st 2
2nd 2
dtype: int64
我想要的輸出是字典如下:
{'Dragoons':{'1st':2,'2nd':2},
'Nighthawks': {'1st':2,'2nd':2},
... }
我嘗試了不同的方法但無濟於事。 是否有相對干凈的方式來實現上述目標?
非常感謝你!!!!
您可以添加Series.unstack
與DataFrame.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}}
另一個解決方案,與另一個答案非常相似
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}}
但是如果使用第一個解決方案,那就是NaN
的問題(它取決於數據)
樣品:
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}}
然后是必要的用途:
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}}
或者另一個John Galt回答。
您可以在分組后重置索引,並根據需要透視數據。 下面的代碼給出了所需的輸出。
df = df.groupby(['regiment','company']).size().reset_index()
print(pd.pivot_table(df, values=0, index='regiment', columns='company').to_dict(orient='index'))
輸出:
{'Nighthawks': {'2nd': 2, '1st': 2}, 'Scouts': {'2nd': 2, '1st': 2}, 'Dragoons': {'2nd': 2, '1st': 2}}
如何用群體理解創建詞典。
In [409]: {g:v['company'].value_counts().to_dict() for g, v in df.groupby('regiment')}
Out[409]:
{'Dragoons': {'1st': 2, '2nd': 2},
'Nighthawks': {'1st': 2, '2nd': 2},
'Scouts': {'1st': 2, '2nd': 2}}
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