[英]DFS on a graph specified with edges as list of tuples
我有一個圖,其邊指定為元組列表。 比如說,我們有:
edges = [('human', 'mammal'), ('mammal', 'vertebrate'), ('mouse', 'mammal'), ('vertebrate', 'animal')]
我們如何編寫一個遞歸迭代所有可以從上圖中構造的節點的方法,以在 Python 中執行深度優先搜索遍歷?
任何幫助表示贊賞,謝謝!
這是一個相當有趣的問題,您可能仍然應該嘗試自己解決,但這里有一個實現示例:
from typing import Generator, Union
def construct(edges: list[tuple[str, str]]) -> dict:
root = {}
index = {}
for x, isa in edges:
if x in root:
root[isa] = {x: root[x]}
del root[x]
index[isa] = root[isa]
else:
if x in index:
raise SyntaxError(f'Invalid tree structure')
if isa in index:
index[isa][x] = (d := {})
else:
root[isa] = {x: (d := {})}
index[isa] = root[isa]
index[x] = d
return root
def traverse(tree: Union[list[tuple[str, str]], dict]) -> Generator[str, None, None]:
# this assumes that, if tree is a dict, it is a well-constructed tree, as construct() would return it
if not isinstance(tree, dict):
tree = construct(tree)
for node in tree:
yield node
yield from traverse(tree[node])
def main():
edges = [('human', 'mammal'), ('mammal', 'vertebrate'), ('mouse', 'mammal'), ('vertebrate', 'animal')]
for node in traverse(edges):
print(node)
if __name__ == '__main__':
main()
這會在線性時間內構造一棵樹,然后深度優先遍歷它,產生訪問過的節點。 它拒絕具有重復葉或節點值或循環的樹。
輸出:
animal
vertebrate
mammal
human
mouse
我的建議是你試試這個例子,試着理解它,然后用你學到的東西從頭開始編寫你自己的例子。
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