使用 Networkx 从图演算处理字典生成器
Handling dictionary generator from graph calculus with Networkx
这是我的问题：
假设这张图：
import networkx as nx
import pandas as pd
user_network_G = user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893])
nan_edge = [(u, v) for (u, v, d) in user_network_G.edges(data=True) if math.isnan(d["Trust"])]
real_edge = [(u, v) for (u, v, d) in user_network_G.edges(data=True) if math.isnan(d["Trust"]) == False]
pos = nx.kamada_kawai_layout(user_network_G)
# nodes
nx.draw_networkx_nodes(user_network_G, pos, node_size=700)
# edges
nx.draw_networkx_edges(user_network_G, pos, edgelist = real_edge, width=2)
nx.draw_networkx_edges(user_network_G, pos, edgelist = nan_edge, width=2, style="dashed")
# labels
nx.draw_networkx_labels(user_network_G, pos, font_size=12, font_family="sans-serif")
nx.draw_networkx_edge_labels(user_network_G, pos, edge_labels = nx.get_edge_attributes(user_network_G,'Trust') , font_size=10, font_family='sans-serif', label_pos = 0.6)
nx.draw_networkx_edge_labels(user_network_G, pos, edge_labels = nx.get_edge_attributes(user_network_G,'Intersection') , font_size=10, font_family='sans-serif', label_pos = 0.4)
plt.axis("off")
plt.show()
给予：
其中：
- 边上的整数是两个用户（节点）之间共有的产品集合的基数
- 边上的小数点是两个节点（即两个用户）之间的真值分数
现在我想计算两个节点之间（例如节点 30 和节点 40 之间）不存在边的边值。
为了实现这一点，我首先创建了我的图表（如上图所示）：
user_trust_cardinality_network = nx.from_pandas_edgelist(user_local_trust_computation(normalized_user_rating_matrix)[1].reset_index(level = ['User_U','User_V']), 'User_U', 'User_V', ['Trust', 'Intersection', 'Pondered_Trust'])
#Creation of the graph from a table containing for each existing paire of nodes the Trust Score, the Cardinality and the Pondered_Trust_Score
user_trust_cardinality_network.remove_edges_from(nx.selfloop_edges(user_trust_cardinality_network))
#Remove selfloop edges
edges_to_predict = [(u, v) for (u, v, d) in user_trust_cardinality_network.edges(data=True) if math.isnan(d["Trust"]) == True]
#Storing the set of edges to predict (edges where trust is NaN)
edges_to_predict_sub = [(u, v) for (u, v, d) in user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]).edges(data=True) if math.isnan(d["Trust"]) == True]
#Edges to predict on my subgraph (for quick example computation)
user_trust_cardinality_network.remove_edges_from([(u, v) for (u, v, d) in user_trust_cardinality_network.edges(data=True) if math.isnan(d["Trust"]) == True])
#Remove edges to predict, to add them after with the newly computed score
然后我计算删除边的分数：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = {}
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_dict = {'Path%s' % [u,v] : sum(pondered_trust)/sum(intersection)}
print(short_path_trust_dict)
给我这个 :
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
{'Path[40, 30]': 0.08499999999999999}
{'Path[40, 30]': 0.16333333333333333}
{'Path[40, 638]': 0.07833333333333332}
{'Path[40, 638]': 0.10333333333333333}
{'Path[40, 638]': 0.1572}
{'Path[145, 30]': 0.075}
{'Path[145, 30]': 0.05444444444444444}
{'Path[30, 638]': 0.017499999999999998}
{'Path[30, 638]': 0.01}
{'Path[30, 893]': 0.16555555555555557}
{'Path[30, 893]': 0.19486486486486487}
{'Path[893, 638]': 0.1581578947368421}
{'Path[893, 638]': 0.062}
{'Path[893, 638]': 0.19}
我的问题是结果似乎是一个字典生成器，因此是一组独立的字典。这不是字典的字典（我已经在研究这种可能性），这里是证明：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = {}
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_dict = {'Path%s' % [u,v] : sum(pondered_trust)/sum(intersection)}
print(type(short_path_trust_dict))
Return :
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
这是我的主要问题，如果我要求创建数据库：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = []
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust = [((u,v), sum(pondered_trust)/sum(intersection))]
short_path_trust_df = pd.DataFrame(short_path_trust, columns= ['Nodes','Path_Trust'])
print(short_path_trust_df)
我得到：
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
Nodes Path_Trust
0 (40, 30) 0.085
Nodes Path_Trust
0 (40, 30) 0.163333
Nodes Path_Trust
0 (40, 638) 0.078333
Nodes Path_Trust
0 (40, 638) 0.103333
Nodes Path_Trust
0 (40, 638) 0.1572
Nodes Path_Trust
0 (145, 30) 0.075
Nodes Path_Trust
0 (145, 30) 0.054444
Nodes Path_Trust
0 (30, 638) 0.0175
Nodes Path_Trust
0 (30, 638) 0.01
Nodes Path_Trust
0 (30, 893) 0.165556
Nodes Path_Trust
0 (30, 893) 0.194865
Nodes Path_Trust
0 (893, 638) 0.158158
Nodes Path_Trust
0 (893, 638) 0.062
Nodes Path_Trust
0 (893, 638) 0.19
然而，我的目标是得到一个像这样的 DataFrame :
Nodes Path_Trust
0 (40, 30) 0.085
1 (40, 30) 0.163333
2 (40, 638) 0.078333
3 (40, 638) 0.103333
4 (40, 638) 0.1572
5 (145, 30) 0.075
6 (145, 30) 0.054444
7 (30, 638) 0.0175
8 (30, 638) 0.01
9 (30, 893) 0.165556
10 (30, 893) 0.194865
11 (893, 638) 0.158158
12 (893, 638) 0.062
13 (893, 638) 0.19
或者，更好：
Node1 Node 2 Trust
0 40 30 0.085
1 40 30 0.163333
2 40 638 0.078333
3 40 638 0.103333
4 40 638 0.1572
5 145 30 0.075
6 145 30 0.054444
7 30 638 0.0175
8 30 638 0.01
9 30 893 0.165556
10 30 893 0.194865
11 893 638 0.158158
12 893 638 0.062
13 893 638 0.19
我希望我很好地解释了我的问题。
如果您需要更多信息，我会尽力提供。
感谢您的帮助:)
在您的代码中，您为每个节点对创建了一个与旧字典同名的新字典，因此您正在覆盖您的数据。
即使您修复了该问题，您也会 运行 在比较相同节点时遇到具有重复键的问题。
相反，我建议使用列表，如下所示：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_list = []
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_list.append([u,v, sum(pondered_trust)/sum(intersection)])
return pd.DataFrame(short_path_trust_list, columns=['node1', 'node2', 'trust'])
一般来说，虽然你的 post 非常详细，但我无法重现你的问题，因为你没有定义从中提取子图的图。
这是我的问题：
假设这张图：
import networkx as nx
import pandas as pd
user_network_G = user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893])
nan_edge = [(u, v) for (u, v, d) in user_network_G.edges(data=True) if math.isnan(d["Trust"])]
real_edge = [(u, v) for (u, v, d) in user_network_G.edges(data=True) if math.isnan(d["Trust"]) == False]
pos = nx.kamada_kawai_layout(user_network_G)
# nodes
nx.draw_networkx_nodes(user_network_G, pos, node_size=700)
# edges
nx.draw_networkx_edges(user_network_G, pos, edgelist = real_edge, width=2)
nx.draw_networkx_edges(user_network_G, pos, edgelist = nan_edge, width=2, style="dashed")
# labels
nx.draw_networkx_labels(user_network_G, pos, font_size=12, font_family="sans-serif")
nx.draw_networkx_edge_labels(user_network_G, pos, edge_labels = nx.get_edge_attributes(user_network_G,'Trust') , font_size=10, font_family='sans-serif', label_pos = 0.6)
nx.draw_networkx_edge_labels(user_network_G, pos, edge_labels = nx.get_edge_attributes(user_network_G,'Intersection') , font_size=10, font_family='sans-serif', label_pos = 0.4)
plt.axis("off")
plt.show()
给予：
其中：
- 边上的整数是两个用户（节点）之间共有的产品集合的基数
- 边上的小数点是两个节点（即两个用户）之间的真值分数
现在我想计算两个节点之间（例如节点 30 和节点 40 之间）不存在边的边值。
为了实现这一点，我首先创建了我的图表（如上图所示）：
user_trust_cardinality_network = nx.from_pandas_edgelist(user_local_trust_computation(normalized_user_rating_matrix)[1].reset_index(level = ['User_U','User_V']), 'User_U', 'User_V', ['Trust', 'Intersection', 'Pondered_Trust'])
#Creation of the graph from a table containing for each existing paire of nodes the Trust Score, the Cardinality and the Pondered_Trust_Score
user_trust_cardinality_network.remove_edges_from(nx.selfloop_edges(user_trust_cardinality_network))
#Remove selfloop edges
edges_to_predict = [(u, v) for (u, v, d) in user_trust_cardinality_network.edges(data=True) if math.isnan(d["Trust"]) == True]
#Storing the set of edges to predict (edges where trust is NaN)
edges_to_predict_sub = [(u, v) for (u, v, d) in user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]).edges(data=True) if math.isnan(d["Trust"]) == True]
#Edges to predict on my subgraph (for quick example computation)
user_trust_cardinality_network.remove_edges_from([(u, v) for (u, v, d) in user_trust_cardinality_network.edges(data=True) if math.isnan(d["Trust"]) == True])
#Remove edges to predict, to add them after with the newly computed score
然后我计算删除边的分数：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = {}
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_dict = {'Path%s' % [u,v] : sum(pondered_trust)/sum(intersection)}
print(short_path_trust_dict)
给我这个 :
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
{'Path[40, 30]': 0.08499999999999999}
{'Path[40, 30]': 0.16333333333333333}
{'Path[40, 638]': 0.07833333333333332}
{'Path[40, 638]': 0.10333333333333333}
{'Path[40, 638]': 0.1572}
{'Path[145, 30]': 0.075}
{'Path[145, 30]': 0.05444444444444444}
{'Path[30, 638]': 0.017499999999999998}
{'Path[30, 638]': 0.01}
{'Path[30, 893]': 0.16555555555555557}
{'Path[30, 893]': 0.19486486486486487}
{'Path[893, 638]': 0.1581578947368421}
{'Path[893, 638]': 0.062}
{'Path[893, 638]': 0.19}
我的问题是结果似乎是一个字典生成器，因此是一组独立的字典。这不是字典的字典（我已经在研究这种可能性），这里是证明：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = {}
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_dict = {'Path%s' % [u,v] : sum(pondered_trust)/sum(intersection)}
print(type(short_path_trust_dict))
Return :
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
<class 'dict'>
这是我的主要问题，如果我要求创建数据库：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_dict = []
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust = [((u,v), sum(pondered_trust)/sum(intersection))]
short_path_trust_df = pd.DataFrame(short_path_trust, columns= ['Nodes','Path_Trust'])
print(short_path_trust_df)
我得到：
short_path_trust(user_trust_cardinality_network.subgraph([30,40,71,145,638,783,893]), edges_to_predict_sub, 2)
Nodes Path_Trust
0 (40, 30) 0.085
Nodes Path_Trust
0 (40, 30) 0.163333
Nodes Path_Trust
0 (40, 638) 0.078333
Nodes Path_Trust
0 (40, 638) 0.103333
Nodes Path_Trust
0 (40, 638) 0.1572
Nodes Path_Trust
0 (145, 30) 0.075
Nodes Path_Trust
0 (145, 30) 0.054444
Nodes Path_Trust
0 (30, 638) 0.0175
Nodes Path_Trust
0 (30, 638) 0.01
Nodes Path_Trust
0 (30, 893) 0.165556
Nodes Path_Trust
0 (30, 893) 0.194865
Nodes Path_Trust
0 (893, 638) 0.158158
Nodes Path_Trust
0 (893, 638) 0.062
Nodes Path_Trust
0 (893, 638) 0.19
然而，我的目标是得到一个像这样的 DataFrame :
Nodes Path_Trust
0 (40, 30) 0.085
1 (40, 30) 0.163333
2 (40, 638) 0.078333
3 (40, 638) 0.103333
4 (40, 638) 0.1572
5 (145, 30) 0.075
6 (145, 30) 0.054444
7 (30, 638) 0.0175
8 (30, 638) 0.01
9 (30, 893) 0.165556
10 (30, 893) 0.194865
11 (893, 638) 0.158158
12 (893, 638) 0.062
13 (893, 638) 0.19
或者，更好：
Node1 Node 2 Trust
0 40 30 0.085
1 40 30 0.163333
2 40 638 0.078333
3 40 638 0.103333
4 40 638 0.1572
5 145 30 0.075
6 145 30 0.054444
7 30 638 0.0175
8 30 638 0.01
9 30 893 0.165556
10 30 893 0.194865
11 893 638 0.158158
12 893 638 0.062
13 893 638 0.19
我希望我很好地解释了我的问题。 如果您需要更多信息，我会尽力提供。 感谢您的帮助:)
在您的代码中，您为每个节点对创建了一个与旧字典同名的新字典，因此您正在覆盖您的数据。 即使您修复了该问题，您也会 运行 在比较相同节点时遇到具有重复键的问题。
相反，我建议使用列表，如下所示：
def short_path_trust (graph, edges_to_predict,cutoff) :
short_path_trust_list = []
for (u,v) in edges_to_predict :
for edges_path in nx.all_simple_edge_paths(graph, u, v, cutoff):
pondered_trust = [d['Pondered_Trust'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
intersection = [d['Intersection'] for (x,y,d) in graph.edges.data() if (x,y) in edges_path or (y,x) in edges_path]
short_path_trust_list.append([u,v, sum(pondered_trust)/sum(intersection)])
return pd.DataFrame(short_path_trust_list, columns=['node1', 'node2', 'trust'])
一般来说，虽然你的 post 非常详细，但我无法重现你的问题，因为你没有定义从中提取子图的图。