如何從回復id(Python)中獲取執行緒/對話？

我是 python 的相對新手，我正在嘗試從具有 ID 串列的資料幀重建對話/執行緒。

我目前有一個推文/reddit帖子的pandas資料框，其格式大致如下：

我的目標是根據 id 將資料分成執行緒/對話。這意味著，從上面的例子中，得到以下序列作為輸出：

[id1, id2, id4, id6],

[id1, id2, id4, id7],

[id1, id2, id5], &

[id1，id3]。

擁有這些串列將使我能夠完整地查看執行緒。目前我的代碼非常復雜，看起來像這樣：

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)
    

這目前給了我半準確的結果，但不是得到：[[id1, id2, id4, id6],[id1, id2, id4, id7]]，我得到：[id1, id2, id4, id6, id7] .

我意識到我可能有點昏昏欲睡，并且有一個簡單的解決方案，但對于我的生活，我似乎無法找到一種方法來做到這一點，以便它可以正常作業并且適用于任何執行緒長度。

預先感謝您的任何建議。:)

uj5u.com熱心網友回復：

使用networkx以達到你想要什么：

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)

輸出：

>>> 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']]})

如果要將字典合并到資料框：

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                                                 []

現在你可以分解你的資料框：

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

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