如何从回复 ID（Python）中获取线程/对话？

Question

我是 python 的相对新手，我正在尝试从具有 ID 列表的数据帧中重建对话/线程。

我目前有一个推文/reddit 帖子的 pandas 数据框，其格式大致如下：

id	text	parent_id	replies
id1	blah blah	_ post _	id2, id3, id4, id5, id6, id7
id2	blah blah	id1	id4, id5, id6, id7
id3	blah blah	id1
id4	blah blah	id2	id6, id7
id5	blah blah	id2
id6	blah blah	id4	id7
id7	blah blah	id6

我的目标是根据 id 将数据分成线程/对话。这意味着，从上面的例子中，得到以下序列作为输出：

[id1, id2, id4, id6],

[id1, id2, id4, id7],

[id1, id2, id5], &

[id1，id3]。

拥有这些列表之后，我就可以查看整个线程。目前我的代码非常复杂，看起来像这样：

out_list = []
for i, row in df.iterrows():
    id_ = row["id"]
    # create our output file 
    sequence = [id_]
    replies = list(row['replies'])
    # creates a new dataframe from the replies to the topline comment in question
    reply_df= df.loc[df['id'].isin(replies)]
    reply_df = reply_df[reply_df.Parent_id2 == id_]
    #check if ends at topline
    if reply_df.empty == False:
        
        def turn_recursion(df, reply_df):
            for j, row_ in reply_df.iterrows():
                replies_2 = reply_df.loc[j, 'replies']
                id_2 = row_["id"]

                reply_df2 =  df.loc[df['id'].isin(replies_2)]
                reply_df2 = reply_df2[reply_df2.Parent_id2 == id_2]

                nonlocal sequence
                nonlocal out_list
                            
                if reply_df2.empty == False:
                    sequence.append(id_2)
                    return(turn_recursion(df, reply_df2))
                
                else:
                    sequence.append(id_2)
                    out_list.append(sequence)
        
        turn_recursion(test2, reply_df)
    else:
        out_list.append(sequence)
    

目前这给了我半准确的结果，但不是得到：[[id1，id2，id4，id6]，[id1，id2，id4，id7]]，我得到：[id1，id2，id4，id6 , id7]。

我意识到我可能有点昏暗并且有一个简单的解决方案，但是对于我的生活，我似乎无法找到一种方法来做到这一点，以便它可以正常工作并适用于任何线程长度。

提前感谢您的任何建议。 :)

Answer 1

使用networkx实现你想要的：

import pandas as pd
import networkx as nx
from collections import defaultdict

data = defaultdict(list)

# Build graph from pandas
G = nx.from_pandas_edgelist(df, source='parent_id', target='id', 
                            create_using=nx.DiGraph)

# Find leaves (id3, id5, id7)
leaves = [node for node, degree in G.out_degree() if degree == 0]

# Enumerate all possible paths
for node in df['id']:
    for leaf in leaves:
        for path in nx.all_simple_paths(G, node, leaf):
            data[node].append(path)

输出：

>>> data
defaultdict(list,
            {'id1': [['id1', 'id3'],
              ['id1', 'id2', 'id5'],
              ['id1', 'id2', 'id4', 'id6', 'id7']],
             'id2': [['id2', 'id5'], ['id2', 'id4', 'id6', 'id7']],
             'id4': [['id4', 'id6', 'id7']],
             'id6': [['id6', 'id7']]})

如果要将字典合并到数据框：

df['replies'] = df['id'].map(data)
print(df)

# Output:
    id       text parent_id                                            replies
0  id1  blah blah  _ post _  [[id1, id3], [id1, id2, id5], [id1, id2, id4, ...
1  id2  blah blah       id1                 [[id2, id5], [id2, id4, id6, id7]]
2  id3  blah blah       id1                                                 []
3  id4  blah blah       id2                                  [[id4, id6, id7]]
4  id5  blah blah       id2                                                 []
5  id6  blah blah       id4                                       [[id6, id7]]
6  id7  blah blah       id6                                                 []

现在你可以分解你的数据框了：

df = df.explode('replies')
print(df)

# Output:
    id       text parent_id                    replies
0  id1  blah blah  _ post _                 [id1, id3]
0  id1  blah blah  _ post _            [id1, id2, id5]
0  id1  blah blah  _ post _  [id1, id2, id4, id6, id7]
1  id2  blah blah       id1                 [id2, id5]
1  id2  blah blah       id1       [id2, id4, id6, id7]
2  id3  blah blah       id1                        NaN
3  id4  blah blah       id2            [id4, id6, id7]
4  id5  blah blah       id2                        NaN
5  id6  blah blah       id4                 [id6, id7]
6  id7  blah blah       id6                        NaN