我在networkx
中创建了一个图表并获取了它的bfs树。
G = nx.Graph()
# build a graph
tree = nx.bfs_tree(G, '1')
现在我想将newick format中的树保存到文件中。最好的方法是什么?
答案 0 :(得分:1)
受this answer的启发,我做了类似的事情:
import networkx as nx
import matplotlib.pyplot as plt
def recursive_search(dict, key):
if key in dict:
return dict[key]
for k, v in dict.items():
item = recursive_search(v, key)
if item is not None:
return item
def bfs_edge_lst(graph, n):
return list(nx.bfs_edges(graph, n))
def load_graph(filename):
G = nx.Graph()
# build the graph
return G
def tree_from_edge_lst(elst):
tree = {'1': {}}
for src, dst in elst:
subt = recursive_search(tree, src)
subt[dst] = {}
return tree
def tree_to_newick(tree):
items = []
for k in tree.keys():
s = ''
if len(tree[k].keys()) > 0:
subt = tree_to_newick(tree[k])
if subt != '':
s += '(' + subt + ')'
s += k
items.append(s)
return ','.join(items)
g = load_graph('dataset.txt')
elst = bfs_edge_lst(g, '1')
tree = tree_from_edge_lst(elst)
newick = tree_to_newick(tree) + ';'
答案 1 :(得分:0)
受@hklel的回答启发,我编写了以下代码:
import networkx as nx
def tree_to_newick(g, root=None):
if root is None:
roots = list(filter(lambda p: p[1] == 0, g.in_degree()))
assert 1 == len(roots)
root = roots[0][0]
subgs = []
for child in g[root]:
if len(g[child]) > 0:
subgs.append(tree_to_newick(g, root=child))
else:
subgs.append(child)
return "(" + ','.join(subgs) + ")"
tree_to_newick(nx.Graph(), None)
似乎可行...