如何从 NetworkX 图转换为 ete3 树对象?

How to convert from NetworkX graph to ete3 Tree object?

我想知道如何从 networkx 有向图构建 ete3.Tree 对象?我按照我认为会产生预期结果的方式添加了每个 child,但我遇到了麻烦。

edges = [('lvl-1', 'lvl-2.1'), ('lvl-1', 'lvl-2.2'), ('lvl-2.1', 'lvl-3.1'), ('lvl-2.1', 2), ('lvl-2.2', 4), ('lvl-2.2', 6), ('lvl-3.1', 'lvl-4.1'), ('lvl-3.1', 5), ('lvl-4.1', 1), ('lvl-4.1', 3), ('input', 'lvl-1')]
graph = nx.OrderedDiGraph()
graph.add_edges_from(edges)
nx.draw(graph, pos=nx.nx_agraph.graphviz_layout(graph, prog="dot"), with_labels=True, node_size=1000, node_color="lightgray")

tree = ete3.Tree()
for parent, children in itertools.groupby(graph.edges(), lambda edge:edge[0]):
    subtree = ete3.Tree(name=parent)
    for child in children:
        subtree.add_child(name=child[1])
    tree.add_child(child=subtree, name=parent)
print(tree) 
#       /-lvl-2.1
#    /-|
#   |   \-lvl-2.2
#   |
#   |   /-lvl-3.1
#   |--|
#   |   \-2
#   |
#   |   /-4
#   |--|
# --|   \-6
#   |
#   |   /-lvl-4.1
#   |--|
#   |   \-5
#   |
#   |   /-1
#   |--|
#   |   \-3
#   |
#    \- /-lvl-1

我也试过以下但没有用:

tree = ete3.Tree()
for parent, child in graph.edges():
    if parent not in tree:
        tree.add_child(name=parent)
    subtree = tree.search_nodes(name=parent)[0]
    subtree.add_child(name=child)
print(tree)
#                /-1
#             /-|
#          /-|   \-3
#         |  |
#       /-|   \-5
#      |  |
#    /-|   \-2
#   |  |
#   |  |   /-4
# --|   \-|
#   |      \-6
#   |
#    \- /-lvl-1

子树和从 networkX object 中读取都没有问题,问题是您将所有子树直接添加到原始 tree 实例中。在ete3中,Treeclass是in fact just a Node(包括指向其后代的指针,如果有的话),所以tree.add_child直接添加新的childnodes/subtrees根节点。

你应该做的是 iterate over the leaves of ete tree,找到 node.name == parent 所在的那个,然后将所有 children 附加到它上面。此外,您应该将它们一一附加,而不是 pre-generate 一个子树。否则,您将获得带有单个 parent 和单个 child.

的额外内部节点

编辑:

您的代码的第二个版本几乎是正确的,但是您没有考虑到如果根不是,则节点永远不会附加到树(ie 根)他们的实际 parent。这可能就是为什么您将 lvl-1 作为一个单独的节点,而不是其他节点的 parent。另外,我不确定 networkX 图形遍历顺序,这可能很重要。更安全(如果更丑)的版本看起来像这样:

# Setting up a root node for lvl-1 to attach to
tree.add_child(name='input')
# A copy in a list, because you may not want to edit the original graph
edges = list(graph.edges)
while len(edges) > 0:
    for parent, child in edges:
        # check if this edge's parent is in the tree
        for leaf it tree.get_leaves(): 
            if leaf.name == parent:
                # if it is, add child and thus create an edge
                leaf.add_child(name=child)
            # Wouldn't want to add the same edge twice, would you?
            edges.remove((parent, child))
    # Now if there are edges still unplaced, try again.

里面可能有几个错别字,而且速度肯定超级慢。边数大约为 O(n**2) 或更糟,所有迭代和列表删除都是如此。可能有一种方法可以将图形从根部遍历到叶子部,这不需要边缘列表的副本(并且可以在单次迭代中工作)。但它最终会产生一个正确的树。

# Graph
edges = [('lvl-1', 'lvl-2.1'), ('lvl-1', 'lvl-2.2'), ('lvl-2.1', 'lvl-3.1'), ('lvl-2.1', 2), ('lvl-2.2', 4), ('lvl-2.2', 6), ('lvl-3.1', 'lvl-4.1'), ('lvl-3.1', 5), ('lvl-4.1', 1), ('lvl-4.1', 3), ('input', 'lvl-1')]
G = nx.OrderedDiGraph()
G.add_edges_from(edges)

# Tree
root = "input"
subtrees = {node:ete3.Tree(name=node) for node in G.nodes()}
[*map(lambda edge:subtrees[edge[0]].add_child(subtrees[edge[1]]), G.edges())]
tree = subtrees[root]
print(tree.get_ascii())
#                                /-1
#                         /lvl-4.1
#                  /lvl-3.1      \-3
#                 |      |
#           /lvl-2.1      \-5
#          |      |
# -inputlvl-1      \-2
#          |
#          |       /-4
#           \lvl-2.2
#                  \-6