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':{}
}
}
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