Question

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

你怎么能把它变成一个字典，产生：

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

我接受它：

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

然而，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']}

Answer 1

以这种方式：

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

，并提供：

         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

然后放弃孩子NAs：

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

，并提供：

         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']}

熊猫：从分层数据创建字典

1 个答案: