获取Triad节点列表，这些节点属于单个Triadic人口普查类别 [英] Get the list of Triad nodes , who fall under the category of individual Triadic Census

问题描述

通过执行Networkx triadic_census算法，我可以获得有关每种三重普查类型的节点数的字典

By executing Networkx triadic_census Algorithm, I'm able to get the dictionary of the number of nodes falling on each type of triadic census

triad_census_social=nx.triadic_census(social_graph.to_directed())

现在，我想返回黑社会列表，这些黑社会都遵循人口普查代码"201"，"120U"或16种现有类型中的任何一种. 如何获得普查计数下的那些节点列表?

Now, I'd like to return the list of triads, who all follow the pattern of census code "201", "120U", or any one of the 16 existing types. How can I get those node lists under a census count?

推荐答案

networkx中没有允许您执行的功能，因此应手动实现.我修改了networkx.algorithms.triads代码以供您返回三合会，而不是它们的计数:

There is no function in networkx that allow you to do it, so you should implement it manually. I modified the networkx.algorithms.triads code for you to return triads, not their count:

import networkx as nx

G = nx.DiGraph()
G.add_nodes_from([1,2,3,4,5])
G.add_edges_from([(1,2),(2,3),(2,4),(4,5)])

triad_census_social=nx.triadic_census(G)
# '003': 2,
# '012': 4,
# '021C': 3,
# '021D': 1,
# another: 0



#: The integer codes representing each type of triad.
#:
#: Triads that are the same up to symmetry have the same code.
TRICODES = (1, 2, 2, 3, 2, 4, 6, 8, 2, 6, 5, 7, 3, 8, 7, 11, 2, 6, 4, 8, 5, 9,
            9, 13, 6, 10, 9, 14, 7, 14, 12, 15, 2, 5, 6, 7, 6, 9, 10, 14, 4, 9,
            9, 12, 8, 13, 14, 15, 3, 7, 8, 11, 7, 12, 14, 15, 8, 14, 13, 15,
            11, 15, 15, 16)

#: The names of each type of triad. The order of the elements is
#: important: it corresponds to the tricodes given in :data:`TRICODES`.
TRIAD_NAMES = ('003', '012', '102', '021D', '021U', '021C', '111D', '111U',
               '030T', '030C', '201', '120D', '120U', '120C', '210', '300')


#: A dictionary mapping triad code to triad name.
TRICODE_TO_NAME = {i: TRIAD_NAMES[code - 1] for i, code in enumerate(TRICODES)}


def _tricode(G, v, u, w):
    """Returns the integer code of the given triad.

    This is some fancy magic that comes from Batagelj and Mrvar's paper. It
    treats each edge joining a pair of `v`, `u`, and `w` as a bit in
    the binary representation of an integer.

    """
    combos = ((v, u, 1), (u, v, 2), (v, w, 4), (w, v, 8), (u, w, 16),
              (w, u, 32))
    return sum(x for u, v, x in combos if v in G[u])


census = {name: set([]) for name in TRIAD_NAMES}
n = len(G)
m = {v: i for i, v in enumerate(G)}
for v in G:
    vnbrs = set(G.pred[v]) | set(G.succ[v])
    for u in vnbrs:
        if m[u] <= m[v]:
            continue
        neighbors = (vnbrs | set(G.succ[u]) | set(G.pred[u])) - {u, v}
        # Calculate dyadic triads instead of counting them.
        for w in neighbors:
            if v in G[u] and u in G[v]:
                census['102'].add(tuple(sorted([u, v, w])))
            else:
                census['012'].add(tuple(sorted([u, v, w])))
        # Count connected triads.
        for w in neighbors:
            if m[u] < m[w] or (m[v] < m[w] < m[u] and
                               v not in G.pred[w] and
                               v not in G.succ[w]):
                code = _tricode(G, v, u, w)
                census[TRICODE_TO_NAME[code]].add(tuple(sorted([u, v, w])))

# null triads, I implemented them manually because the original algorithm computes
# them as _number_of_all_possible_triads_ - _number_of_all_found_triads_
for v in G:
    vnbrs = set(G.pred[v]) | set(G.succ[v])
    not_vnbrs = set(G.nodes()) - vnbrs
    for u in not_vnbrs:
        unbrs = set(G.pred[u]) | set(G.succ[u])
        not_unbrs = set(G.nodes()) - unbrs
        for w in not_unbrs:
            wnbrs = set(G.pred[w]) | set(G.succ[w])
            if v not in wnbrs and len(set([u, v, w])) == 3:
                census['003'].add(tuple(sorted([u, v, w])))

# '003': {(1, 3, 4), (1, 3, 5)},
# '012': {(1, 2, 3), (1, 2, 4), (2, 3, 4), (2, 4, 5)},
# '021C': {(1, 2, 3), (1, 2, 4), (2, 4, 5)},
# '021D': {(2, 3, 4)},
# another: empty

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