考虑一下，我们有一堆看起来像这样的子标题：
subtree1 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": []
},
{
"id": "edit",
"caption": "Edit",
"children": []
},
{
"id": "tools",
"caption": "Tools",
"children": [
{
"id": "packages",
"caption": "Packages",
"children": []
}
]
},
{
"id": "help",
"caption": "Help",
"children": []
},
]
}
subtree2 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": [
{"caption": "New"},
{"caption": "Exit"},
]
}
]
}
subtree3 = {
"id": "root",
"children": [
{
"id": "edit",
"children": [
{"caption": "Copy"},
{"caption": "Cut"},
{"caption": "Paste"},
]
},
{
"id": "help",
"children": [
{"caption": "About"},
]
}
]
}
subtree4 = {
"id": "root",
"children": [
{
"id": "edit",
"children": [
{
"id": "text",
"caption": "Text",
"children": [
{ "caption": "Insert line before" },
{ "caption": "Insert line after" }
]
}
]
}
]
}
我试图弄清楚如何编写merge函数的代码，例如执行以下操作：
tree0 = merge(subtree1, subtree2)
tree0 = merge(tree0, subtree3)
tree0 = merge(tree0, subtree4)
将产生：
tree0 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": [
{"caption": "New"},
{"caption": "Exit"},
]
},
{
"id": "edit",
"caption": "Edit",
"children": [
{"caption": "Copy"},
{"caption": "Cut"},
{"caption": "Paste"},
{
"id": "text",
"caption": "Text",
"children": [
{ "caption": "Insert line before" },
{ "caption": "Insert line after" }
]
}
]
},
{
"id": "tools",
"caption": "Tools",
"children": [
{
"id": "packages",
"caption": "Packages",
"children": []
}
]
},
{
"id": "help",
"caption": "Help",
"children": [
{"caption": "About"},
]
},
]
}
但是要这样做：
tree1 = merge(subtree1, subtree2)
tree1 = merge(tree1, subtree4)
tree1 = merge(tree1, subtree3)
会产生：
tree1 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": [
{"caption": "New"},
{"caption": "Exit"},
]
},
{
"id": "edit",
"caption": "Edit",
"children": [
{
"id": "text",
"caption": "Text",
"children": [
{ "caption": "Insert line before" },
{ "caption": "Insert line after" }
]
},
{"caption": "Copy"},
{"caption": "Cut"},
{"caption": "Paste"},
]
},
{
"id": "tools",
"caption": "Tools",
"children": [
{
"id": "packages",
"caption": "Packages",
"children": []
}
]
},
{
"id": "help",
"caption": "Help",
"children": [
{"caption": "About"},
]
},
]
}
否则，以相同的顺序加载子树将始终生成相同的树，但是如果您以不同的顺序使用相同的子树列表，则不能保证生成相同的树（因为子列表可以以不同的方式扩展订单）。
我已经尝试对此进行编码，但是我不知道merge算法的行为如何，这就是我的问题。谁能提供代码/伪代码/解释以便我实现？
PS：在下面，您会发现一些随机尝试，我认为这可能会导致我获得胜利
if __name__ == '__main__':
from collections import defaultdict
subtree1 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": []
},
{
"id": "edit",
"caption": "Edit",
"children": []
},
{
"id": "tools",
"caption": "Tools",
"children": [
{
"id": "packages",
"caption": "Packages",
"children": []
}
]
},
{
"id": "help",
"caption": "Help",
"children": []
},
]
}
subtree2 = {
"id": "root",
"children": [
{
"id": "file",
"caption": "File",
"children": [
{"caption": "New"},
{"caption": "Exit"},
]
}
]
}
subtree3 = {
"id": "root",
"children": [
{
"id": "edit",
"children": [
{"caption": "Copy"},
{"caption": "Cut"},
{"caption": "Paste"},
]
},
{
"id": "help",
"children": [
{"caption": "About"},
]
}
]
}
subtree4 = {
"id": "root",
"children": [
{
"id": "edit",
"children": [
{
"id": "text",
"caption": "Text",
"children": [
{"caption": "Insert line before"},
{"caption": "Insert line after"}
]
}
]
}
]
}
lst = [
subtree1,
subtree2,
subtree3,
subtree4
]
def traverse(node, path=[]):
yield node, tuple(path)
for c in node.get("children", []):
path.append(c.get("id", None))
yield from traverse(c)
path.pop()
# Levels & Hooks
dct_levels = defaultdict(list)
dct_hooks = defaultdict(list)
for subtree in lst:
for n, p in traverse(subtree):
if p not in dct_levels[len(p)]:
dct_levels[len(p)].append(p)
dct_hooks[p].append(n)
print(dct_levels)
print(dct_hooks[("file",)])
# Merge should happen here
tree = {
"id": "root",
"children": []
}
for level in range(1, max(dct_levels.keys()) + 1):
print("populating level", level, dct_levels[level])
但是不确定我是否在这里创建正确的结构/助手，因为目前尚不清楚整个算法如何工作……这就是这个问题的全部内容
答案 0 :(得分：7)
使用您的示例在Python 3.5上进行了测试。
from copy import deepcopy
def merge(x: dict, y: dict) -> dict:
'Merge subtrees x y, and return the results as a new tree.'
return merge_inplace(deepcopy(x), y)
def merge_inplace(dest: dict, src: dict) -> dict:
'Merge subtree src into dest, and return dest.'
# perform sanity checks to make the code more rock solid
# feel free to remove those lines if you don't need
assert dest.get('id'), 'Cannot merge anonymous subtrees!'
assert dest.get('id') == src.get('id'), 'Identity mismatch!'
# merge attributes
dest.update((k, v) for k, v in src.items() if k != 'children')
# merge children
if not src.get('children'): # nothing to do, so just exit
return dest
elif not dest.get('children'): # if the children list didn't exist
dest['children'] = [] # then create an empty list for it
named_dest_children = {
child['id']: child
for child in dest['children']
if 'id' in child
}
for child in src['children']:
if 'id' not in child: # anonymous child, just append it
dest['children'].append(child)
elif child['id'] in named_dest_children: # override a named subtree
merge_inplace(named_dest_children[child['id']], child)
else: # create a new subtree
dest['children'].append(child)
named_dest_children[child['id']] = child
return dest
答案 1 :(得分：1)
您可以将itertools.groupby用于递归：
from itertools import groupby
def merge(*args):
if len(args) < 2 or any('id' not in i for i in args):
return list(args)
_d = [(a, list(b)) for a, b in groupby(sorted(args, key=lambda x:x['id']), key=lambda x:x['id'])]
return [{**{j:k for h in b for j, k in h.items()}, 'id':a, 'children':merge(*[i for c in b for i in c['children']])} for a, b in _d]
通过args，此解决方案将每个传递的字典视为children列表的成员。这是为了解决将两个或更多具有不同merge即id和{'id':'root', 'children':[...]}的字典传递给{'id':'root2', 'children':[...]}的可能性。这样，该解决方案将返回[{'id':'root', 'children':[...]}, {'id':'root2', 'children':[...]}]的列表，因为不同的id不提供匹配的途径。因此，在当前问题的上下文中，您需要使用索引来访问结果列表的单个返回元素：合并的dict与id 'root'：
import json
tree0 = merge(subtree1, subtree2)[0]
tree0 = merge(tree0, subtree3)[0]
tree0 = merge(tree0, subtree4)[0]
print(json.dumps(tree0, indent=4))
输出：
{
"id": "root",
"children": [
{
"id": "edit",
"caption": "Edit",
"children": [
{
"caption": "Copy"
},
{
"caption": "Cut"
},
{
"caption": "Paste"
},
{
"id": "text",
"caption": "Text",
"children": [
{
"caption": "Insert line before"
},
{
"caption": "Insert line after"
}
]
}
]
},
{
"id": "file",
"caption": "File",
"children": [
{
"caption": "New"
},
{
"caption": "Exit"
}
]
},
{
"id": "help",
"caption": "Help",
"children": [
{
"caption": "About"
}
]
},
{
"id": "tools",
"caption": "Tools",
"children": [
{
"id": "packages",
"caption": "Packages",
"children": []
}
]
}
]
}
答案 2 :(得分：0)
用于合并JSON文档/对象的手工编码可能不是最佳解决方案。干！
我在这里使用了genson，jsonschema和jsonmerge软件包进行合并。
genson从JSON实例文档生成JSON模式。
jsonschema使用JSON模式验证JSON实例文档。
jsonmerge通过扩展JSON模式来合并对象/ JSON文档。
首先让我们从JSON实例生成JSON模式。
trees = (subtree1, subtree2, subtree3, subtree4)
schema_builder = genson.SchemaBuilder()
for tree in trees:
schema_builder.add_object(tree)
schema = schema_builder.to_schema()
现在指定合并策略。
schema['properties']['children']['mergeStrategy'] = 'arrayMergeById'
schema['properties']['children']['items']['properties']['children']['mergeStrategy'] = 'append'
arrayMergeById策略通过对象的id属性合并对象。
append策略收集数组中的对象。
这是完整的代码；
import genson
import jsonmerge
import jsonschema
subtree1 = {
"id":
"root",
"children": [
{
"id": "file",
"caption": "File",
"children": []
},
{
"id": "edit",
"caption": "Edit",
"children": []
},
{
"id": "tools",
"caption": "Tools",
"children": [{
"id": "packages",
"caption": "Packages",
"children": []
}]
},
{
"id": "help",
"caption": "Help",
"children": []
},
]
}
subtree2 = {
"id":
"root",
"children": [{
"id": "file",
"caption": "File",
"children": [
{
"caption": "New"
},
{
"caption": "Exit"
},
]
}]
}
subtree3 = {
"id":
"root",
"children": [{
"id":
"edit",
"children": [
{
"caption": "Copy"
},
{
"caption": "Cut"
},
{
"caption": "Paste"
},
]
}, {
"id": "help",
"children": [
{
"caption": "About"
},
]
}]
}
subtree4 = {
"id":
"root",
"children": [{
"id":
"edit",
"children": [{
"id":
"text",
"caption":
"Text",
"children": [{
"caption": "Insert line before"
}, {
"caption": "Insert line after"
}]
}]
}]
}
trees = (subtree1, subtree2, subtree3, subtree4)
schema_builder = genson.SchemaBuilder()
for tree in trees:
schema_builder.add_object(tree)
schema = schema_builder.to_schema()
print("Validating schema...", end='')
for tree in trees:
jsonschema.validate(tree, schema)
print(' done')
schema['properties']['children']['mergeStrategy'] = 'arrayMergeById'
schema['properties']['children']['items']['properties']['children']['mergeStrategy'] = 'append'
merger = jsonmerge.Merger(schema=schema)
tree = merger.merge(subtree1, subtree2)
tree = merger.merge(tree, subtree3)
tree = merger.merge(tree, subtree4)
print(tree)