[英]Python recursion to sort list of tuples into tree structure
我是編碼新手,並且在嘗試對元組列表進行遞歸排序時陷入困境。 這是我的代碼:
# table consists on [(code, parent), (code, parent), ...]
table = [('1', ''),
('1.1', '1'),
('2', ''),
('2.1','2'),
('2.1.1','2.1'),
('3',''),
('3.1','3'),
('4',''),
('4.1','4'),
('4.1.1','4.1'),
('4.1.2','4.1')]
content = {}
def rec(table, parent=None):
while True:
try:
_code = table[0][0]
_parent = table [0][1]
if _parent == '':
content[_code] = _parent
return rec(table[1:])
else:
if _parent in content:
if content[_parent] == '':
content[_parent] = table[0]
else:
content[_parent] = content[_parent], table[0]
return rec(table[1:], parent=_parent)
else:
content[_parent] = table[0]
return rec(table[1:], parent=_parent)
except IndexError:
break
return content
print(rec(table))
我得到的輸出:
{'1': ('1.1', '1'), '2': ('2.1', '2'), '2.1': ('2.1.1', '2.1'), '3':('3.1', '3'), '4': ('4.1', '4'), '4.1': (('4.1.1', '4.1'), ('4.1.2','4.1'))}
但所需的輸出是:
{'1': ('1.1', '1'), '2': ('2.1', '2'), {'2.1': ('2.1.1', '2.1')}, '3': ('3.1','3'), '4': ('4.1', '4'), {'4.1': ('4.1.1', '4.1'), ('4.1.2', '4.1')}
我需要類似的東西:
{'node_id': '1', 'name':'somename', 'children': [{'node_id': '1.1' ,'name':'somename', 'children': [{'node_id': '1.1.1', 'name':'somename', 'children': [{'node_id': '1.1.1.1', 'name':'somename', 'children': []}]}, {'node_id': '1.1.2', 'name':'somename', 'children': []}, {'node_id': '1.1.3', 'name':'somename', 'children': []}]}, {'node_id': '1.2', 'name':'somename', 'children': []}]}
關於如何實現我的目標的任何想法?
為了使您的輸出成為嵌套字典樹,它需要有一個常規結構,其中每個節點都是一個字典,其中的值代表一個子節點的字典,直到葉節點,葉節點將為它們的子節點提供一個空字典。
這是一個將構建樹的簡單循環:
table = [('1', ''),
('1.1', '1'),
('2', ''),
('2.1','2'),
('2.1.1','2.1'),
('3',''),
('3.1','3'),
('4',''),
('4.1','4'),
('4.1.1','4.1'),
('4.1.2','4.1')]
tree = { node:dict() for link in table for node in link }
for child,parent in table:
tree[parent].update({child:tree[child]})
輸出:
print(tree[""]) # "" is te root
{
'1': { '1.1': {}},
'2': {
'2.1': { '2.1.1': {}}
},
'3': { '3.1': {}},
'4': {
'4.1': {
'4.1.1': {},
'4.1.2': {}
}
}
}
作為附帶好處,此結構為您提供了樹中所有節點的索引
對於屬性字典(其中之一是子列表),可以使用相同的方法:
tree = { node:{"id":node,"children":[]} for link in table for node in link }
for child,parent in table:
tree[parent]["children"].append(tree[child])
輸出:
print(tree[""]["children"]) # children of root
[ { 'id': '1',
'children': [ { 'id': '1.1', 'children': []} ]
},
{ 'id': '2',
'children': [
{ 'id': '2.1',
'children': [ {'id': '2.1.1', 'children': []} ]
}
]
},
{ 'id': '3',
'children': [ { 'id': '3.1','children': []} ]
},
{ 'id': '4',
'children': [
{ 'id': '4.1',
'children': [
{ 'id': '4.1.1', 'children': []},
{ 'id': '4.1.2', 'children': []}
]
}
]
}
]
遞歸方法很好,但執行速度會慢得多,並且不會生成索引以通過其 Id 訪問節點:
def tree(links,node=""):
return {"id":node, "children":[tree(links,child) for child,parent in links if parent==node] }
root = tree(table)
您可以使用遞歸:
table = [('1', ''), ('1.1', '1'), ('2', ''), ('2.1', '2'), ('2.1.1', '2.1'), ('3', ''), ('3.1', '3'), ('4', ''), ('4.1', '4'), ('4.1.1', '4.1'), ('4.1.2', '4.1')]
def to_dict(d):
return {'node_id':d, 'children':[*map(to_dict, [a for a, b in table if b == d])]}
result = [to_dict(a) for a, b in table if not b]
輸出:
[{'node_id': '1', 'children': [{'node_id': '1.1', 'children': []}]}, {'node_id': '2', 'children': [{'node_id': '2.1', 'children': [{'node_id': '2.1.1', 'children': []}]}]}, {'node_id': '3', 'children': [{'node_id': '3.1', 'children': []}]}, {'node_id': '4', 'children': [{'node_id': '4.1', 'children': [{'node_id': '4.1.1', 'children': []}, {'node_id': '4.1.2', 'children': []}]}]}]
編輯:假設table
的元組有其他信息:
table = [('1', '', 'someval0'), ('1.1', '1', 'someval1'), ('2', '', 'someval2'), ('2.1', '2', 'someval3'), ('2.1.1', '2.1', 'someval4'), ('3', '', 'someval5'), ('3.1', '3', 'someval6'), ('4', '', 'someval7'), ('4.1', '4', 'someval8'), ('4.1.1', '4.1', 'someval9'), ('4.1.2', '4.1', 'someval10')]
def to_dict(d):
return {**(dict(zip(['node_id', 'name'], d))), 'children':[*map(to_dict, [(a, *c) for a, b, *c in table if b == d[0]])]}
result = [to_dict((a, *c)) for a, b, *c in table if not b]
輸出:
[{'node_id': '1', 'name': 'someval0', 'children': [{'node_id': '1.1', 'name': 'someval1', 'children': []}]}, {'node_id': '2', 'name': 'someval2', 'children': [{'node_id': '2.1', 'name': 'someval3', 'children': [{'node_id': '2.1.1', 'name': 'someval4', 'children': []}]}]}, {'node_id': '3', 'name': 'someval5', 'children': [{'node_id': '3.1', 'name': 'someval6', 'children': []}]}, {'node_id': '4', 'name': 'someval7', 'children': [{'node_id': '4.1', 'name': 'someval8', 'children': [{'node_id': '4.1.1', 'name': 'someval9', 'children': []}, {'node_id': '4.1.2', 'name': 'someval10', 'children': []}]}]}]
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