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python-熊猫:从分层数据创建字典

2019-11-09 01:57:31  阅读:10  来源: 互联网

标签:pandas python



说我有以下数据框df:

      A            B       
0     mother1      NaN
1     NaN          child1
2     NaN          child2
3     mother2      NaN
4     NaN          child1
5     mother3      NaN
6     NaN          child1
7     NaN          child2
8     NaN          child3

你怎么能把它变成字典,产生:

结果= {‘mother1’:[‘child1′,’child2’],’mother2’:[‘child1’],’mother3’:[‘child1′,’child2′,’child3’]}

我对此:

import pandas as pd
import numpy as np

results={}

for index1,row1 in df.iterrows():
    if row1['A'] is not np.nan:
        children=[]
        for index2,row2 in df.iterrows():
            if row2['B'] is not np.nan:
                children.append(row2['B'])
        results[row1['A']]=children

但是,结果是错误的:

In[1]: results
Out[1]: 
{'mother1': ['child1', 'child2', 'child1', 'child1', 'child2', 'child3'],
 'mother2': ['child1', 'child2', 'child1', 'child1', 'child2', 'child3'],
 'mother3': ['child1', 'child2', 'child1', 'child1', 'child2', 'child3']}

解决方法:

这是一种方法:

df['A'].fillna(method='ffill', inplace=True)

给予:

         A       B
0  mother1     NaN
1  mother1  child1
2  mother1  child2
3  mother2     NaN
4  mother2  child1
5  mother3     NaN
6  mother3  child1
7  mother3  child2
8  mother3  child3

然后删除子NA:

df.dropna(subset=['B'], inplace=True)

给予:

         A       B
1  mother1  child1
2  mother1  child2
4  mother2  child1
6  mother3  child1
7  mother3  child2
8  mother3  child3

然后,您可以使用groupby和字典理解功能来获得最终结果:

results = {k: v['B'].tolist() for k, v in df.groupby('A')}

结果:

{'mother1': ['child1', 'child2'],
 'mother2': ['child1'],
 'mother3': ['child1', 'child2', 'child3']}


标签:pandas,python

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