python-3.x - NetworkX：在 DAG 中找到最长的路径，返回最大的所有关系

Question

我无法弄清楚如何更新 networkx dag_find_longest_path() 算法以返回“N”表示关系，而不是返回找到的第一个最大边，或者返回与最大权重相关的所有边的列表。

我首先从 pandas 数据帧创建了一个 DAG，其中包含如下子集的边缘列表：

edge1        edge2          weight
115252161:T  115252162:A     1.0
115252162:A  115252163:G     1.0
115252163:G  115252164:C     3.0
115252164:C  115252165:A     5.5
115252165:A  115252166:C     5.5
115252162:T  115252163:G     1.0
115252166:C  115252167:A     7.5
115252167:A  115252168:A     7.5
115252168:A  115252169:A     6.5
115252165:A  115252166:G     0.5

然后我使用下面的代码对图进行拓扑排序，然后根据边的权重找到最长的路径：

 G = nx.from_pandas_edgelist(edge_df, source="edge1", 
                                target="edge2", 
                                edge_attr=['weight'], 
                                create_using=nx.OrderedDiGraph())

longest_path = pd.DataFrame(nx.dag_longest_path(G))

这很好用，除非当最大加权边缘存在联系时，它会返回找到的第一个最大边缘，而我希望它只返回一个表示“Null”的“N”。所以目前，输出是：

115252161   T
115252162   A
115252163   G
115252164   C
115252165   A
115252166   C

但我真正需要的是：

115252161   T
115252162   N (or [A,T] )
115252163   G
115252164   C
115252165   A
115252166   C

寻找最长路径的算法是：

def dag_longest_path(G):

    dist = {}  # stores [node, distance] pair
    for node in nx.topological_sort(G):
        # pairs of dist,node for all incoming edges
        pairs = [(dist[v][0] + 1, v) for v in G.pred[node]]
        if pairs:
            dist[node] = max(pairs)
        else:
            dist[node] = (0, node)
    node, (length, _) = max(dist.items(), key=lambda x: x[1])
    path = []
    while length > 0:
        path.append(node)
        length, node = dist[node]
    return list(reversed(path))

的可复制粘贴定义G。

import pandas as pd
import networkx as nx
import numpy as np

edge_df = pd.read_csv(
    pd.compat.StringIO(
        """edge1        edge2          weight
115252161:T  115252162:A     1.0
115252162:A  115252163:G     1.0
115252163:G  115252164:C     3.0
115252164:C  115252165:A     5.5
115252165:A  115252166:C     5.5
115252162:T  115252163:G     1.0
115252166:C  115252167:A     7.5
115252167:A  115252168:A     7.5
115252168:A  115252169:A     6.5
115252165:A  115252166:G     0.5"""
    ),
    sep=r" +",
)


G = nx.from_pandas_edgelist(
    edge_df,
    source="edge1",
    target="edge2",
    edge_attr=["weight"],
    create_using=nx.OrderedDiGraph(),
)

longest_path = pd.DataFrame(nx.dag_longest_path(G))

score 1 · Accepted Answer

函数内的这一行似乎丢弃了你想要的路径；因为max只返回一个结果：

node, (length, _) = max(dist.items(), key=lambda x: x[1])

我会保留最大值，然后根据它搜索所有项目。然后重用代码以找到所需的路径。一个例子是这样的：

def dag_longest_path(G):
    dist = {}  # stores [node, distance] pair
    for node in nx.topological_sort(G):
        # pairs of dist,node for all incoming edges
        pairs = [(dist[v][0] + 1, v) for v in G.pred[node]]
        if pairs:
            dist[node] = max(pairs)
        else:
            dist[node] = (0, node)
    # store max value inside val variable
    node, (length, val) = max(dist.items(), key=lambda x: x[1])
    # find all dictionary items that have the maximum value
    nodes = [(item[0], item[1][0]) for item in dist.items() if item[1][1] == val]
    paths = []
    # iterate over the different nodes and append the paths to a list
    for node, length in nodes:
        path = []
        while length > 0:
            path.append(node)
            length, node = dist[node]
        paths.append(list(reversed(path)))
    return paths

PS。我还没有测试过这段代码以知道它是否正确运行。

score 1 · Accepted Answer

从您的示例来看，每个节点都由位置 ID（之前的编号:）确定，并且连接不同碱基的两个节点是相同的，用于计算路径长度。如果这是正确的，则无需修改算法，您可以通过操作顶点标签来获得结果。

基本上，删除中分号后的所有内容edge_df，计算最长路径并附加原始数据中的基本标签。

edge_df_pos = pd.DataFrame(
    {
        "edge1": edge_df.edge1.str.partition(":")[0],
        "edge2": edge_df.edge2.str.partition(":")[0],
        "weight": edge_df.weight,
    }
)

vert_labels = dict()
for col in ("edge1", "edge2"):
    verts, lbls = edge_df[col].str.partition(":")[[0, 2]].values.T
    for vert, lbl in zip(verts, lbls):
        vert_labels.setdefault(vert, set()).add(lbl)


G_pos = nx.from_pandas_edgelist(
    edge_df_pos,
    source="edge1",
    target="edge2",
    edge_attr=["weight"],
    create_using=nx.OrderedDiGraph(),
)

longest_path_pos = nx.dag_longest_path(G_pos)

longest_path_df = pd.DataFrame([[node, vert_labels[node]] for node in longest_path_pos])

结果

# 0       1
# 0  115252161     {T}
# 1  115252162  {A, T}
# 2  115252163     {G}
# 3  115252164     {C}
# 4  115252165     {A}
# 5  115252166  {G, C}
# 6  115252167     {A}
# 7  115252168     {A}
# 8  115252169     {A}

如果我的解释不正确，我怀疑是否存在基于拓扑排序的算法的简单扩展。问题是一个图可以接受多种拓扑排序。如果您按照示例dist中的定义进行打印dag_longest_path，您将得到如下信息：

{'115252161:T': (0, '115252161:T'),
 '115252162:A': (1, '115252161:T'),
 '115252162:T': (0, '115252162:T'),
 '115252163:G': (2, '115252162:A'),
 '115252164:C': (3, '115252163:G'),
 '115252165:A': (4, '115252164:C'),
 '115252166:C': (5, '115252165:A'),
 '115252166:G': (5, '115252165:A'),
 '115252167:A': (6, '115252166:C'),
 '115252168:A': (7, '115252167:A'),
 '115252169:A': (8, '115252168:A')}

请注意，这'115252162:T'发生在第三行而不是其他任何地方。因此，dist无法区分您的示例和'115252162:T'作为不相交组件出现的另一个示例。因此，'115252162:T'仅使用dist.

score 1 · Accepted Answer

我最终只是在 defaultdict 计数器对象中对行为进行建模。

from collections import defaultdict, Counter

我将边缘列表修改为（位置、核苷酸、重量）的元组：

test = [(112,"A",23.0), (113, "T", 27), (112, "T", 12.0), (113, "A", 27), (112,"A", 1.0)]

然后使用 defaultdict(counter) 快速总结每个核苷酸在每个位置的出现次数：

nucs = defaultdict(Counter)

for key, nuc, weight in test:
    nucs[key][nuc] += weight

然后遍历字典以提取所有等于最大值的核苷酸：

for key, nuc in nucs.items():
    seq_list = []
    max_nuc = []
    max_val = max(nuc.values())
    for x, y in nuc.items():
        if y == max_val:
            max_nuc.append(x)

    if len(max_nuc) != 1:
        max_nuc = "N"
    else:
        max_nuc = ''.join(max_nuc)

    seq_list.append(max_nuc)
    sequence = ''.join(seq_list)

这将返回找到最大值的核苷酸的最终序列，并在平局的位置返回 N：

TNGCACAAATGCTGAAAGCTGTACCATANCTGTCTGGTCTTGGCTGAGGTTTCAATGAATGGAATCCCGTAACTCTTGGCCAGTTCGTGGGCTTGTTTTGTATCAACTGTCCTTGTTGGCAAATCACACTTGTTTCCCACTAGCACCAT

然而，这个问题困扰着我，所以我最终在 networkx 中使用节点属性作为将每个节点标记为平局或不平局的方法。现在，当以最长路径返回节点时，我可以检查“tie”属性，如果已标记节点名称，则将其替换为“N”：

def get_path(self, edge_df):

    G = nx.from_pandas_edgelist(
        edge_df,
        source="edge1",
        target="edge2",
        edge_attr=["weight"],
        create_using=nx.OrderedDiGraph()
    )

    # flag all nodes as not having a tie
    nx.set_node_attributes(G, "tie", False)

    # check nodes for ties
    for node in G.nodes:

        # create list of all edges for each node
        edges = G.in_edges(node, data=True)

        # if there are multiple edges
        if len(edges) > 1:

            # find max weight
            max_weight = max([edge[2]['weight'] for edge in edges])

            tie_check = []
            for edge in edges:
                # pull out all edges that match max weight
                if edge[2]["weight"] == max_weight:
                    tie_check.append(edge)

            # check if there are ties       
            if len(tie_check) > 1:
                for x in tie_check:

                    # flag node as being a tie
                    G.node[x[0]]["tie"] = True

    # return longest path
    longest_path = nx.dag_longest_path(G)     

    return longest_path

python-3.x - NetworkX：在 DAG 中找到最长的路径，返回最大的所有关系

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