如何从字典各键中获取多个最大值（或Top N值）

需求背景 #

已基于NetworkX和community库完成网络社区聚类，生成了社区编号到节点列表的communities_dict，原代码仅能获取每个社区的单个最高度节点，现需修改为获取每个社区的Top3最高度节点。

原代码实现（获取单个最高度节点） #

communities_dict = {1: ["A","B","C","D","E","F","G","CB"],
                    2: ["H","I","J","K","L","M","N","XY"],
                    3: ["O","P","Q","R","S","T","U","GT"],
                    4: ["V","W","X","Y","Z","AB","HH","FF"],
                    5: ["CD","EF","GH","IJ","KL","HG","FR","RR"]}
# 加载网络图
G = nx.read_adjlist("./data/networkgraph.csv", delimiter = ',')
# 聚类网络图
partition = community.best_partition(G)
# 获取社区集合
communities = set(partition.values())
# 创建社区编号到对应节点的字典
communities_dict = {c: [k for k, v in partition.items() if v == c] for c in communities}
# 获取每个社区最高度节点的代码
highest_degree = {k: max(v, key=lambda x: G.degree(x)) for k, v in communities_dict.items()}

原代码输出结果：

1: "A"
2: "I"
3: "Q"
4: "AB" 
5: "CD" 

修改后代码（获取Top3最高度节点） #

只需替换获取最高度节点的代码行，实现排序取前3的逻辑：

# 获取每个社区Top3最高度节点
top3_degree = {
    k: sorted(v, key=lambda x: G.degree(x), reverse=True)[:3]
    for k, v in communities_dict.items()
}

修改说明 #

• 用sorted()替代max()，通过reverse=True将社区内节点按度数降序排列
• 利用切片[:3]截取排序后的前3个节点，即为该社区的Top3最高度节点

预期输出示例 #

1: ["A","B","C"]
2: ["I", "J","K"]
3: ["Q","P","O"]
4: ["V","W","X"]
5: ["CD", "EF","GH"]

内容的提问来源于stack exchange，提问作者Jeff