从 python 中列表的列表创建列表

Question

我正在创建一个树状结构，其中每个叶节点都有 5 个文档。要获取父节点的文档，子节点的所有文档都将分配给它。

例如A 是父节点，B、C 是子节点，它们各有 5 个文档。因此，A 的文档将是 5+5=10。同理，A的父节点会得到A的10个文档+A的兄弟节点的一个文档的no。重复此操作，直到到达根节点。

我想将 A 的文档存储为大小为 10 的列表，类似地，将 A 的父级存储为他们子级的文档总数。但它将其存储为大小为 2 的列表，并且在每个列表下，每个都有 5 个文档。 A 的父级也将 A 的文档存储为 3 个列表，而不是 3*5=15 我想要的。

如何将每个节点的文档存储为文档总数而不是列表列表？下面是我正在使用的代码。

from anytree import Node, RenderTree
import pandas as pd
import numpy as np

class Node(Node):
    Node.documents = None
    Node.vector = None

### Creating tree by giving documnets to leaf ###
### Tree Creation ###    
# L1    
Finance = Node("Finance")
# L2
Credit_and_Lending = Node("Credit and Lending", parent=Finance)
# L3
Credit_Cards = Node("Credit Cards", parent=Credit_and_Lending)

Loans = Node("Loans", parent=Credit_and_Lending)

# L4
Low_Interest_and_No_Interest_Credit_Cards = Node("Low Interest & No Interest Credit Cards", parent=Credit_Cards, documents=[(fvc.loc[(fvc['keyword']=='low interest & no interest credit cards') & (fvc['organic_rank']==1)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='low interest & no interest credit cards') & (fvc['organic_rank']==2)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='low interest & no interest credit cards') & (fvc['organic_rank']==3)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='low interest & no interest credit cards') & (fvc['organic_rank']==4)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='low interest & no interest credit cards') & (fvc['organic_rank']==5)])['vocab'].tolist()[0]])

Rewards_Cards = Node("Rewards Cards", parent=Credit_Cards, documents=[(fvc.loc[(fvc['keyword']=='rewards cards') & (fvc['organic_rank']==1)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='rewards cards') & (fvc['organic_rank']==2)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='rewards cards') & (fvc['organic_rank']==3)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='rewards cards') & (fvc['organic_rank']==4)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='rewards cards') & (fvc['organic_rank']==5)])['vocab'].tolist()[0]])

Student_Credit_Cards = Node("Student Credit Cards", parent=Credit_Cards, documents=[(fvc.loc[(fvc['keyword']=='student credit cards') & (fvc['organic_rank']==1)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='student credit cards') & (fvc['organic_rank']==2)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='student credit cards') & (fvc['organic_rank']==3)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='student credit cards') & (fvc['organic_rank']==4)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='student credit cards') & (fvc['organic_rank']==5)])['vocab'].tolist()[0]])

Auto_Financing = Node("Auto Financing", parent=Loans, documents=[(fvc.loc[(fvc['keyword']=='auto financing') & (fvc['organic_rank']==1)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='auto financing') & (fvc['organic_rank']==2)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='auto financing') & (fvc['organic_rank']==3)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='auto financing') & (fvc['organic_rank']==4)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='auto financing') & (fvc['organic_rank']==5)])['vocab'].tolist()[0]])
Commercial_Lending = Node("Commercial Lending", parent=Loans, documents=[(fvc.loc[(fvc['keyword']=='commercial lending') & (fvc['organic_rank']==1)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='commercial lending') & (fvc['organic_rank']==2)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='commercial lending') & (fvc['organic_rank']==3)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='commercial lending') & (fvc['organic_rank']==4)])['vocab'].tolist()[0]
                            , (fvc.loc[(fvc['keyword']=='commercial lending') & (fvc['organic_rank']==5)])['vocab'].tolist()[0]])

##### Visualizing the created tree #####
for pre, fill, node in RenderTree(Finance):
    print("%s%s" % (pre, node.name))

##### Getting documents for parent nodes #####
def get_documents(node):    
    if node.documents is not None:
        return node.documents
    else:
        child_nodes = node.children
        lis = []
        for child in child_nodes:
            child_docs = get_documents(child)
            lis.append(child_docs)
        node.documents = lis
        return lis


get_documents(Finance)

Answer 1

您可以使用以下语法：

lis = lis + child_docs

而不是

 lis.append(child_docs)

从 python 中列表的列表创建列表

Creating a list from list of a list in python

python

nlp

text-mining