对数据集的所有连接节点进行分组 [英] Grouping all connected nodes of a dataset

问题描述

这不是以下内容的重复:

This is not a duplicate of:

对熊猫数据框执行复杂搜索的最快方法

注意:pandas 0.23.4 版

Note: pandas ver 0.23.4

假设:数据可以按任何顺序排列.

Assumptions: data can be laid out in any order.

我有一个清单:

L = ['A', 'B', 'C', 'D', 'L', 'M', 'N', 'O']

我也有一个数据框.Col1 和 Col2 有几个关联的列，其中包含我希望保留的相关信息.信息随意，所以我没有填写.

I also have a dataframe. Col1 and Col2 have several associated columns that have related info I wish to keep. The information is arbitrary so I have not filled it in.

Col1  Col2  Col1Info  Col2Info  Col1moreInfo  Col2moreInfo
 A     B       x         x            x             x
 B     C
 D     C
 L     M
 M     N
 N     O

我正在尝试为列表中的每个元素执行搜索和分组".例如，如果我们对列表中的元素D"执行搜索，则会返回以下组.

I am trying to perform a 'search and group' for each element of the list. For example, if we performed a search on an element of the list, 'D', the following group would be returned.

To    From  Col1Info  Col2Info  Col1moreInfo  Col2moreInfo
 A     B       x         x            x             x
 B     C
 D     C

我一直在玩 networkx，但它是一个非常复杂的包.

I have been playing around with networkx but it is a very complex package.

推荐答案

您可以使用两列中的值作为边来定义图形，并查找 connected_components.下面是一种使用 NetworkX 的方法:

You could define a graph using the values from both columns as edges, and look for the connected_components. Here's a way using NetworkX:

import networkx as nx

G=nx.Graph()
G.add_edges_from(df.values.tolist())
cc = list(nx.connected_components(G))
# [{'A', 'B', 'C', 'D'}, {'L', 'M', 'N', 'O'}]

现在假设你想通过 D 过滤，你可以这样做:

Now say for instance you want to filter by D, you could then do:

component = next(i for i in cc if 'B' in i)
# {'A', 'B', 'C', 'D'}

并索引来自两列的值都在component中的数据框:

And index the dataframe where the values from both columns are in component:

df[df.isin(component).all(1)]

   Col1 Col2
0    A    B
1    B    C
2    D    C

<小时>

通过生成数据框列表，上述内容可以扩展到列表中的所有项目.然后我们只需使用给定项目在 L 中的位置进行索引:

L = ['A', 'B', 'C', 'D', 'L', 'M', 'N', 'O']

dfs = [df[df.isin(i).all(1)] for j in L for i in cc if j in i]
print(dfs[L.index('D')])

   Col1 Col2
0    A    B
1    B    C
2    D    C

print(dfs[L.index('L')])

   Col1 Col2
3    L    M
4    M    N
5    N    O

这篇关于对数据集的所有连接节点进行分组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！