从节点列表中提取边缘和社区

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我有超过5万个节点的数据集,我正在尝试从中提取可能的边缘和群落。我确实尝试使用一些图形工具(例如gephi,cytoscape,socnet,nodexl等)来可视化和识别边缘和社区,但对于这些工具而言,节点列表太大。因此,我正在尝试编写脚本以精确定位边缘和社区。其他列是带有GPS位置的连接开始日期时间和结束日期时间。

输入:

id,开始时间,结束时间,gps1,gps2

0022d9064bc,1073260801,1073260803,819251,440006
00022d9064bc,1073260803,1073260810,819213,439954
00904b4557d3,1073260803,1073261920,817526,439458
00022de73863,1073260804,1073265410,817558,439525
00904b14b494,1073260804,1073262625,817558,439525
00904b14b494,1073260804,1073265163,817558,439525
00904b14b494,1073260804,1073263786,817558,439525
00022d1406df,1073260807,1073260809,820428,438735
00022d1406df,1073260807,1073260878,820428,438735
00022d623dfe,1073260810,1073276346,819251,440006
00022d7317d7,1073260810,1073276155,819251,440006
00022d9064bc,1073260810,1073272525,819251,440006
00022d9064bc,1073260810,1073260999,819251,440006
00022d9064bc,1073260810,1073260857,819251,440006
0030650c9eda,1073260811,1073260813,820356,439224
00022d0e0cec,1073260813,1073262843,820187,439271
00022d176cf3,1073260813,1073260962,817721,439564
000c30d8d2e8,1073260813,1073260902,817721,439564
00904b243bc4,1073260813,1073260962,817721,439564
00904b2fc34d,1073260813,1073260962,817721,439564
00904b52b839,1073260813,1073260962,817721,439564
00904b9a5a51,1073260813,1073260962,817721,439564
00904ba8b682,1073260813,1073260962,817721,439564
00022d3be9cd,1073260815,1073261114,819269,439403
00022d80381f,1073260815,1073261114,819269,439403
00022dc1b09c,1073260815,1073261114,819269,439403
00022d36a6df,1073260817,1073260836,820761,438607
00022d36a6df,1073260817,1073260845,820761,438607
003065d2d8b6,1073260817,1073267560,817735,439757
00904b0c7856,1073260817,1073265149,817735,439757
00022de73863,1073260825,1073260879,817558,439525
00904b14b494,1073260825,1073260879,817558,439525
00904b312d9e,1073260825,1073260879,817558,439525
00022d15b1c7,1073260826,1073260966,820353,439280
00022dcbe817,1073260826,1073260966,820353,439280

我正在尝试实现无向加权/非加权图。

andrew_reece

使用Pandas将数据获取到成对节点列表中,其中每行代表一个基于边缘条件的边缘。然后迁移到networkx对象中进行图分析。

两个节点共享一条边的标准包括:

  1. 相同的位置假设这意味着相同的gps1AND gps2
  2. “几乎相同的开始和结束时间”这有点模棱两可。出于此答案的目的,我已将此标准简化为“以相同的5秒间隔开始时间”groupby如果要在边缘上应用其他时间条件,扩展我在这里采用的方法应该不太困难

由于我们希望基于时间戳来操作数据,转换startenddatetime dtype

df.start = pd.to_datetime(df.start, unit="s")
df.end = pd.to_datetime(df.end, unit="s")

df.start.describe()
count                      35
unique                     11
top       2004-01-05 00:00:13
freq                        8
first     2004-01-05 00:00:01
last      2004-01-05 00:00:26
Name: start, dtype: object

df.head()
             ID               start                 end    gps1    gps2
0   0022d9064bc 2004-01-05 00:00:01 2004-01-05 00:00:03  819251  440006
1  00022d9064bc 2004-01-05 00:00:03 2004-01-05 00:00:10  819213  439954
2  00904b4557d3 2004-01-05 00:00:03 2004-01-05 00:18:40  817526  439458
3  00022de73863 2004-01-05 00:00:04 2004-01-05 01:16:50  817558  439525
4  00904b14b494 2004-01-05 00:00:04 2004-01-05 00:30:25  817558  439525

样本观测值在几秒钟之内发生,因此我们将分组频率设置为仅几秒钟:

near = "5s" 

现在groupby确定查找连接节点的位置和开始时间:

edges = (df.groupby(["gps1",
                     "gps2",
                     pd.Grouper(key="start", 
                                freq=near, 
                                closed="right", 
                                label="right")], 
                   as_index=False)
           .agg({"ID":','.join,
                 "start":"min",
                 "end":"max"})
            .reset_index()
            .rename(columns={"index":"edge",
                             "start":"start_min", 
                             "end":"end_max"})
        )

edges.ID = edges.ID.str.split(",")

edges.head()

   edge    gps1    gps2                                                 ID  \
0     0  817526  439458                                     [00904b4557d3]   
1     1  817558  439525  [00022de73863, 00904b14b494, 00904b14b494, 009...   
2     2  817558  439525         [00022de73863, 00904b14b494, 00904b312d9e]   
3     3  817721  439564  [00022d176cf3, 000c30d8d2e8, 00904b243bc4, 009...   
4     4  817735  439757                       [003065d2d8b6, 00904b0c7856]   

            start_min             end_max  
0 2004-01-05 00:00:03 2004-01-05 00:18:40  
1 2004-01-05 00:00:04 2004-01-05 01:16:50  
2 2004-01-05 00:00:25 2004-01-05 00:01:19  
3 2004-01-05 00:00:13 2004-01-05 00:02:42  
4 2004-01-05 00:00:17 2004-01-05 01:52:40 

现在,每一行代表一个唯一的边缘类别。ID是所有共享该边缘的节点列表。将此列表放入节点对的新结构中有些棘手;我求助于一些老式的嵌套for循环。这里可能有一些Pandas-fu可以提高效率:

注意:对于单例节点,我已经None为其对分配了一个值。如果您不想跟踪单例,则只需忽略if not len(combos): ...逻辑。

pairs = []
idx = 0
for e in edges.edge.values:
    nodes = edges.loc[edges.edge==e, "ID"].values[0]
    attrs = edges.loc[edges.edge==e, ["gps1","gps2","start_min","end_max"]]
    combos = list(combinations(nodes, 2))
    if not len(combos):
        pair = [e, nodes[0], None]
        pair.extend(attrs.values[0])
        pairs.append(pair)
        idx += 1
    else:
        for combo in combos:
            pair = [e, combo[0], combo[1]]
            pair.extend(attrs.values[0])
            pairs.append(pair)
            idx += 1
cols = ["edge","nodeA","nodeB","gps1","gps2","start_min","end_max"]
pairs_df = pd.DataFrame(pairs, columns=cols)    

pairs_df.head()

   edge         nodeA         nodeB    gps1    gps2           start_min  \
0     0  00904b4557d3          None  817526  439458 2004-01-05 00:00:03   
1     1  00022de73863  00904b14b494  817558  439525 2004-01-05 00:00:04   
2     1  00022de73863  00904b14b494  817558  439525 2004-01-05 00:00:04   
3     1  00022de73863  00904b14b494  817558  439525 2004-01-05 00:00:04   
4     1  00904b14b494  00904b14b494  817558  439525 2004-01-05 00:00:04   

              end_max  
0 2004-01-05 00:18:40  
1 2004-01-05 01:16:50  
2 2004-01-05 01:16:50  
3 2004-01-05 01:16:50  
4 2004-01-05 01:16:50      

现在,数据可以适合一个networkx对象:

import networkx as nx

g = nx.from_pandas_dataframe(pairs_df, "nodeA", "nodeB", edge_attr=True)

# access edge attributes by node pairing:
test_A = "00022de73863"
test_B = "00904b14b494"
g[test_A][test_B]["start_min"]
# output:
Timestamp('2004-01-05 00:00:25')

对于社区检测,有几种选择。考虑networkx社区算法以及community基于本机networkx功能构建模块

我认为您的问题主要与将数据处理为适合网络分析的格式有关。由于此答案已经足够长,我将留给您以寻求社区检测策略-与我在此处链接的模块可以直接使用几种方法。

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