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,每个节点都有5个文档。因此,A 的文档将为 5+5=10。同样,A 的父节点将获取 A 的 10 个文档 + A 的兄弟节点的文档编号。我们将重复此操作,直到到达根节点。
我想将 A 的文档存储为大小为 10 的列表,同样将 A 的父级存储为其子级的文档总数。但它将其存储为大小为 2 的列表,并且每个列表下各有 5 个文档。 A 的父级也将 A 的文档存储为 3 的列表,而不是我想要的 3*5=15
。
如何将每个节点上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)
最佳答案
您可以使用以下语法:
lis = lis + child_docs
而不是
lis.append(child_docs)
关于python - 在Python中从列表的列表创建列表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49377971/