Pandas：从分层数据创建字典

Question

假设我有以下数据框 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

你怎样将此转化为一个返回以下结果的字典：
results={'mother1':['child1','child2'],'mother2':['child1'],'mother3':['child1','child2','child3']} 我的理解：

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']}

- FaCoffee

Answer 1

这里有一种方法：

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

给定：

         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

然后删除缺失值:

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

给定：

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

你可以使用groupby和字典推导式来得到最终结果：

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

结果：

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

- sjw

---

它有效，并且比我想出的嵌套for循环快得多。我相信这可以扩展到祖父母->父母->孩子的情况。 - FaCoffee