给定一个(父，子)的平面列表，创建一个分层的字典树 [英] Given a flat list of (parent,child), create a hierarchical dictionary tree

问题描述

我有一个包含元组的列表，其中包含名称.这些名称是父名称和子名称，我想用它们的名称创建一个分层的字典树. 例如，我有以下列表:

I have a list with tuples that containing names. These names are parent and child names and I want to create a hierarchical dictionary tree with their names. For example I have the following list:

[('john','marry'),('mike','john'),('mike','hellen'),('john','elisa')]

我想创建这个:

{
    'mike':{
        'john':{
            'marry':{}
            'elisa':{}
         }
         'hellen':{}
        }
}

推荐答案

假设它行为良好(没有循环，没有重复的名称，也没有一个孩子的多个父母)，您可以简单地使用有向图"进行遍历.为了找到根，我还使用了一个包含布尔值的字典，该布尔值指示名称是否存在任何父项:

Assuming it's well-behaved (no cycles, no duplicate names or multiple parents for one child) you could simply use a "directed graph" and traverse it. To find the root(s) I also used a dictionary containing a boolean that indicates if there is any parent for the name:

lst = [('john','marry'), ('mike','john'), ('mike','hellen'), ('john','elisa')]

# Build a directed graph and a list of all names that have no parent
graph = {name: set() for tup in lst for name in tup}
has_parent = {name: False for tup in lst for name in tup}
for parent, child in lst:
    graph[parent].add(child)
    has_parent[child] = True

# All names that have absolutely no parent:
roots = [name for name, parents in has_parent.items() if not parents]

# traversal of the graph (doesn't care about duplicates and cycles)
def traverse(hierarchy, graph, names):
    for name in names:
        hierarchy[name] = traverse({}, graph, graph[name])
    return hierarchy

traverse({}, graph, roots)
# {'mike': {'hellen': {}, 'john': {'elisa': {}, 'marry': {}}}}

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