如何统计列表元组表示的无向边中边数最多的节点及对应边数

核心思路 #

无向边的每个节点在元组中出现的次数，就等于该节点关联的边数，我们只需要统计所有节点的出现次数，再取最大值即可。

代码实现 #

依赖标准库的最简实现 #

from collections import Counter

# 你的边列表，注意不要用list作为变量名，会覆盖Python内置关键字
edges = [("a","b"), ("a","c"), ("a","d"), ("b","d"), ("c","d")]

# 展开所有边的节点为一维列表
all_nodes = [node for edge in edges for node in edge]
# 统计每个节点的出现次数（即边数）
node_edge_count = Counter(all_nodes)
# 取出边数最多的节点和对应边数
max_node, max_edge_num = max(node_edge_count.items(), key=lambda item: item[1])

print(f"边数最多的节点是{max_node}，边数为{max_edge_num}")

运行示例输出：

边数最多的节点是a，边数为3

处理多节点并列最大值的场景 #

如果存在多个节点边数相同且均为最大值，可以用以下方式获取所有并列节点：

max_edge_num = max(node_edge_count.values())
max_nodes = [node for node, count in node_edge_count.items() if count == max_edge_num]

print(f"边数最多的节点有{max_nodes}，边数均为{max_edge_num}")

纯原生实现（无需导入标准库） #

如果不想导入collections，也可以用字典手动统计：

edges = [("a","b"), ("a","c"), ("a","d"), ("b","d"), ("c","d")]
node_edge_count = {}

for u, v in edges:
    node_edge_count[u] = node_edge_count.get(u, 0) + 1
    node_edge_count[v] = node_edge_count.get(v, 0) + 1

max_node, max_edge_num = max(node_edge_count.items(), key=lambda x: x[1])
print(f"边数最多的节点是{max_node}，边数为{max_edge_num}")

内容的提问来源于stack exchange，提问作者reallycool123