基因网络拓扑重叠Python [英] Gene Network Topological Overlap Python

问题描述

在复杂的基因网络中，我们如何找到拓扑重叠.

In a complex gene network how can we find a topological overlap.

输入数据如下

code  code weight

3423 3455   3453
2344 2353   45
3432 3453   456
3235 4566   34532
2345 8687   356
2466 6467   3567
3423 2344   564
3455 2353   4564
3432 3423   456

节点列为col [0]和col [1]，花费的连接时间为col [2]

The node columns are col[0] and col[1] and the time spent connected is col[2]

代码:

import networkx as nx
import numpy as np

data = np.loadtxt("USC_Test.txt")
col = []
edge_list = zip[col[0],col[1]]

G = nx.Graph()
G.add_edges_from(edge_list)
components = nx.connected_components(G)

print components

错误

edge_list = zip[col[0],col[1]]
IndexError: list index out of range

推荐答案

我必须承认我不熟悉术语拓扑重叠，所以我不得不查找它:

I must confess that I wasn't familiar with the term topological overlap so I had to look it up:

如果网络中的一对节点都牢固地连接到同一组节点，则它们具有较高的拓扑重叠. (源)

A pair of nodes in a network is said to have high topological overlap if they are both strongly connected to the same group of nodes. (Source)

NetworkX似乎没有内置的方法，该方法可以让您查找具有拓扑重叠的节点对，但可以轻松地找到牢固连接的组件.例如:

NetworkX doesn't seem to have a builtin method which lets you find pairs of nodes with topological overlap but it can easily find strongly connected components. For example:

In [1]: import networkx as nx
In [2]: edge_list = [(1, 2), (2, 1), (3, 1), (1, 3), (2, 4), (1, 4), (5, 6)]
In [3]: G = nx.DiGraph()
In [4]: G.add_edges_from(edge_list)
In [5]: components = nx.strongly_connected_components(G)
In [6]: components
Out[6]: [[1, 3, 2], [4], [6], [5]]

如果您有无向图，则可以改用nx.connected_components.

If you have an undirected graph you can use nx.connected_components instead.

现在有了组件，可以轻松找到拓扑重叠的所有对对列表.例如，从components:

Now you have components, it is straightforward to find all lists of pairs with toplogical overlap. For example, generate all pairs of nodes from lists in components:

In [7]: from itertools import combinations
In [8]: top_overlap = [list(combinations(c, 2)) for c in components if len(c) > 1]
In [9]: top_overlap = [item for sublist in top_overlap for item in sublist]
In [10]: top_overlap
Out[10]: [(1, 3), (1, 2), (3, 2)]

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