通过将它们与字典进行比较来定义网络节点的属性

如何解决通过将它们与字典进行比较来定义网络节点的属性

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我需要在网络的每个节点上添加一个名为 class 的属性，它代表一个社会经济分类。我有一个字典，包含节点 id 和类，如下：

dict_users = {"5B5F2C071D12AF13219DF5EBE05132AF": 3,"9FB3B96B6D5E16C9DD564AA3E84F1954": 2,"B3299B0E587D7275E3E4D530E9EECF50": 3,"6432F1DF21BA38368D9A165C739EEBB3": 2,"85D5C50A6D882CA8E4BB00BCA3574417": 3,"0D8583F810B9720A8032BB939F12B3FF": 2,"6C10A9E9F325CAA3CCB7F9A0D6983D2A": 3,"B0C50ED1DEC9E06E4C64E7419DDC4B09": 2}

但是，当我通过字典遍历网络时，所有节点都会收到字典中最后一个的值。完整代码如下：

import pandas as pd
import networkx as nx
G = nx.read_gml('/home/gustavo/Desktop/MesTrado/mesTrado_dados/Redes/Teste.gml')

dict_users = {"5B5F2C071D12AF13219DF5EBE05132AF": 3,"B0C50ED1DEC9E06E4C64E7419DDC4B09": 3}

for i in G.nodes:
    for j in dict_users.keys():
        if i == j:
            nx.set_node_attributes(G,dict_users.get(j),'class')

谁能告诉我我应该怎么做才能让每个节点都像字典中那样接收与其对应的值？另外，有什么方法可以降低网络和字典之间的迭代计算成本吗？真正的 .gml 文件要大得多，字典也是。

解决方法

我找到了一个答案，但它的计算成本仍然非常高：

import pandas as pd
import networkx as nx
G = nx.read_gml('/home/gustavo/Desktop/Mestrado/mestrado_dados/Redes/Teste.gml')

dict_users = {"5B5F2C071D12AF13219DF5EBE05132AF": 3,"9FB3B96B6D5E16C9DD564AA3E84F1954": 2,"B3299B0E587D7275E3E4D530E9EECF50": 3,"6432F1DF21BA38368D9A165C739EEBB3": 2,"85D5C50A6D882CA8E4BB00BCA3574417": 3,"0D8583F810B9720A8032BB939F12B3FF": 2,"6C10A9E9F325CAA3CCB7F9A0D6983D2A": 3,"B0C50ED1DEC9E06E4C64E7419DDC4B09": 3}

for i in G.nodes:
    for j in dict_users.keys():
        if i == j:
            G.nodes[i]['class'] = dict_users.get(j)