Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
713 views
in Technique[技术] by (71.8m points)

postgresql - How to aggregate matching pairs into "connected components" in Python

Real-world problem:

I have data on directors across many firms, but sometimes "John Smith, director of XYZ" and "John Smith, director of ABC" are the same person, sometimes they're not. Also "John J. Smith, director of XYZ" and "John Smith, director of ABC" might be the same person, or might not be. Often examination of additional information (e.g., comparison of biographical data on "John Smith, director of XYZ" and "John Smith, director of ABC") makes it possible to resolve whether two observations are the same person or not.

Conceptual version of the problem:

In that spirit, am collecting data that will identify matching pairs. For example, suppose I have the following matching pairs: {(a, b), (b, c), (c, d), (d, e), (f, g)}. I want to use the transitivity property of the relation "is the same person as" to generate "connected components" of {{a, b, c, d, e}, {f, g}}. That is {a, b, c, d, e} is one person and {f, g} is another. (An earlier version of the question referred to "cliques", which are apparently something else; this would explain why find_cliques in networkx was giving the "wrong" results (for my purposes).

The following Python code does the job. But I wonder: is there a better (less computationally costly) approach (e.g., using standard or available libraries)?

There are examples here and there that seem related (e.g., Cliques in python), but these are incomplete, so I am not sure what libraries they are referring to or how to set up my data to use them.

Sample Python 2 code:

def get_cliques(pairs):
    from sets import Set

    set_list = [Set(pairs[0])]

    for pair in pairs[1:]:
        matched=False
        for set in set_list:
            if pair[0] in set or pair[1] in set:
                set.update(pair)
                matched=True
                break
        if not matched:
            set_list.append(Set(pair))

    return set_list

pairs = [('a', 'b'), ('b', 'c'), ('c', 'd'), ('d', 'e'), ('f', 'g')]

print(get_cliques(pairs))

This produces the desired output: [Set(['a', 'c', 'b', 'e', 'd']), Set(['g', 'f'])].

Sample Python 3 code:

This produces [set(['a', 'c', 'b', 'e', 'd']), set(['g', 'f'])]):

def get_cliques(pairs):

    set_list = [set(pairs[0])]

    for pair in pairs[1:]:
        matched=False
        for a_set in set_list:
            if pair[0] in a_set or pair[1] in a_set:
                a_set.update(pair)
                matched=True
                break
        if not matched:
            set_list.append(set(pair))

    return set_list

pairs = [('a', 'b'), ('b', 'c'), ('c', 'd'), ('d', 'e'), ('f', 'g')]

print(get_cliques(pairs))
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

With networkX:

import networkx as nx
G1=nx.Graph()
G1.add_edges_from([("a","b"),("b","c"),("c","d"),("d","e"),("f","g")])
sorted(nx.connected_components(G1), key = len, reverse=True)

giving:

[['a', 'd', 'e', 'b', 'c'], ['f', 'g']]

You have to check the fastest algorithm now ...

OP:

This works great! I have this in my PostgreSQL database now. Just organize pairs into a two-column table, then use array_agg() to pass to PL/Python function get_connected(). Thanks.

CREATE OR REPLACE FUNCTION get_connected(
    lhs text[],
    rhs text[])
  RETURNS SETOF text[] AS
$BODY$
    pairs = zip(lhs, rhs)

    import networkx as nx
    G=nx.Graph()
    G.add_edges_from(pairs)
    return sorted(nx.connected_components(G), key = len, reverse=True)

$BODY$ LANGUAGE plpythonu;

(Note: I edited answer, as I thought showing this step might be helpful addendum, but too long for a comment.)


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...