我有一个dict列表,其中包含以下结构:
[{
"parent": "com.company.object.kind.type.subtype.family.Feline",
"class": "com.company.object.kind.type.subtype.family.species.Cat"
}, {
"parent": "com.company.object.kind.type.subtype.Mammal",
"class": "com.company.object.kind.type.subtype.family.Feline"
}, {
"parent": "com.company.object.Being",
"class": "com.company.object.kind.LivingBeing"
}, {
"parent": "com.company.object.kind.type.subtype.family.Canine",
"class": "com.company.object.kind.type.subtype.family.species.Wolf"
}, {
"parent": "com.company.object.kind.type.subtype.Mammal",
"class": "com.company.object.kind.type.subtype.family.Canine"
}, {
"parent": "com.company.object.kind.type.Animal",
"class": "com.company.object.kind.type.subtype.Mammal"
}, {
"parent": "com.company.object.kind.LivingBeing",
"class": "com.company.object.kind.type.Animal"
}, {
"parent": "com.company.object.kind.type.Animal",
"class": "com.company.object.kind.type.subtype.Fish"
}, {
"parent": "com.company.object.kind.StaticBeing",
"class": "com.company.object.kind.type.Solid"
}, {
"parent": "com.company.object.Being",
"class": "com.company.object.kind.StaticBeing"
}, {
"parent": "com.company.object.kind.type.subtype.family.Feline",
"class": "com.company.object.kind.type.subtype.family.species.Lion"
}, {
"parent": "com.company.object.kind.type.subtype.family.Canine",
"class": "com.company.object.kind.type.subtype.family.species.Hyena"
}, {
"parent": "com.company.object.kind.StaticBeing",
"class": "com.company.object.kind.type.Liquid"
}]
需要通过以下方式从中构建层次结构树:
[
"com.company.object.Being" : [
"com.company.object.kind.StaticBeing": [
"com.company.object.kind.type.Solid",
"com.company.object.kind.type.Liquid"
],
"com.company.object.kind.LivingBeing": [
"com.company.object.kind.type.Animal": [
"com.company.object.kind.type.subtype.Fish",
"com.company.object.kind.type.subtype.Mammal": [
"com.company.object.kind.type.subtype.family.Canine": [
"com.company.object.kind.type.subtype.family.species.Wolf",
"com.company.object.kind.type.subtype.family.species.Hyena"
],
"com.company.object.kind.type.subtype.family.Feline": [
"com.company.object.kind.type.subtype.family.species.Lion",
"com.company.object.kind.type.subtype.family.species.Cat"
]
]
]
]
]
]
包可以是不同的并且具有任何类型的深度,它只需要从父子关系构造树。
答案 0 :(得分:1)
这是一种非复杂的方法,循环遍历对象列表三次,将树节点放在字典(treenodes)中,将根节点放在root_node中。
lst是问题中提供的列表。
def display_node(node, indent=0):
print ('.'*indent, node['class'])
indent += 3
for child in node['children']:
display_node(child, indent)
# Create list of classes
classes = []
for item in lst:
name = item['class']
if name not in classes:
classes.append(name)
treenodes = {}
root_node = None
for item in lst: # Create tree nodes
item['children'] = []
name = item['class']
treenodes[name] = item
parent = item['parent']
if parent not in classes: # parent is root node, create
if parent not in treenodes:
node = {}
node['parent'] = None
node['children'] = []
node['class'] = parent
root_node = node
treenodes[parent] = node
# Connect parents and children
for item in lst: # Create tree nodes
parent = item['parent']
parent_node = treenodes[parent]
parent_node['children'].append(item)
display_node(root_node)
将节点创建为对象并省去treenodes字典可能更好。 这个过程可以在一个循环中实现,但可能非常复杂。
答案 1 :(得分:0)
请注意,您在结果中混合使用词典和列表。
假设您想要一个包含密钥id
和children
的字典,那么这是一种递归方式...
def build_tree(elems):
elem_with_children = {}
def _build_children_sub_tree(parent):
cur_dict = {
'id': parent,
# put whatever attributes here
}
if parent in elem_with_children.keys():
cur_dict["children"] = [_build_children_sub_tree(cid) for cid in elem_with_children[parent]]
return cur_dict
for item in elems:
cid = item['id']
pid = item['parent']
elem_with_children.setdefault(pid, []).append(cid)
res = _build_children_sub_tree("com.company.object.Being")
return res