给定(父,子)的平面列表,创建分层字典树

时间:2017-08-02 12:16:07

标签: python python-3.x dictionary

我有一个包含名称的元组列表。这些名称是父名和子名,我想创建一个带有名称的分层字典树。 例如,我有以下列表:

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

我想创建这个:

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

3 个答案:

答案 0 :(得分:8)

假设它表现良好(没有周期,没有重复的名字或一个孩子的多个父母),你可以简单地使用“有向图”并遍历它。为了找到根,我还使用了一个包含布尔值的字典,该字符表示该名称是否有任何父级:

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': {}}}}

答案 1 :(得分:1)

def get_children(parent, relations):
    children = (r[1] for r in relations if r[0] == parent)
    return {c: get_children(c, relations) for c in children}

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

parents, children = map(set, zip(*the_list))
the_tree = {p: get_children(p, the_list) for p in (parents - children)}

print(the_tree)

答案 2 :(得分:0)

替代方案:

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

roots = set()
mapping = {}
for parent,child in data:
    childitem = mapping.get(child,None)
    if childitem is None:
        childitem =  {}
        mapping[child] = childitem
    else:
        roots.discard(child)
    parentitem = mapping.get(parent,None)
    if parentitem is None:
        mapping[parent] = {child:childitem}
        roots.add(parent)
    else:
        parentitem[child] = childitem

tree = {id : mapping[id] for id in roots}

print (tree)

结果:

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

归功于@WillemVanOnsem Recursively creating a tree hierarchy without using class/object