递归错误:获取对象的 str 时超出最大递归深度

Question

在这里寻找了很多关于这个问题的例子，但我无法为我的例子整理出来。

任何建议将不胜感激，我已经对这种递归感到头疼了。

tree = {}
def populate_node(account):
node = '%(LOGIN)s,%(server_id)s' % account
tree[node]['login'] = account['LOGIN']
tree[node]['email'] = account['EMAIL'].lower()
tree[node]['server_id'] = account['server_id']
for account in accounts:
node = '%(LOGIN)s,%(server_id)s' % account
parent = None
if account['AGENT_ACCOUNT']:
parent = '%(AGENT_ACCOUNT)s,%(server_id)s' % account
if node not in tree:
tree[node] = {}
populate_node(account)
if parent:
tree[node]['parent'] = parent
if parent not in tree:
tree[parent] = {
'login': parent,
'server_id': account['server_id'],
'children': [node],
}
else:
if 'children' not in tree[parent]:
tree[parent]['children'] = [node]
else:
tree[parent]['children'].append(node)
def get_path(node, tree):
parent = node.get('parent')
node_login = str(str(node.get('login')) + ',' + str(node.get('server_id')))
if not parent:
return []
elif parent == node_login:
return [parent]
path = get_path(tree[parent], tree)
return [parent] + path

for k, v in tree.items():
v['path'] = get_path(v, tree)
v['level'] = len(v['path']) + (1 if v['login'] != v.get('parent') else 0)

默认情况下：

tree = {}

节点是tree的项目。

示例树：

tree = {
'1987,mt4-demo-0': {
'login': 1987,
'email': 'email_1',
'server_id': 'mt4-demo-0'
},
'16044,mt4-demo-0': {
'login': 16044,
'email': 'email_2',
'server_id': 'mt4-demo-0'
},
'160877748,mt4-demo-0': {
'login': 160877748,
'email': 'email_3',
'server_id': 'mt4-demo-0'
}
}

而且我每次都收到这个递归错误

RecursionError: maximum recursion depth exceeded while getting the str of an object

Answer 1

您的代码假设您始终在处理无环有向图，但您的输入中至少有一个有向循环，其中一个AGENT_ACCOUNT引用直接或间接指向另一个帐户，而另一个帐户又具有指向第一个帐户的AGENT_ACCOUNT值。

例如，如果accounts设置为：

accounts = [
{'LOGIN': 'foo', 'EMAIL': '[e-mail]', 'server_id': 'server 1',
'AGENT_ACCOUNT': 'bar'},
{'LOGIN': 'bar', 'EMAIL': '[e-mail]', 'server_id': 'server 1',
'AGENT_ACCOUNT': 'foo'}]

然后tree变成：

{'bar,server 1': {'children': ['foo,server 1'],
'email': '[e-mail]',
'login': 'bar',
'parent': 'foo,server 1',
'server_id': 'server 1'},
'foo,server 1': {'children': ['bar,server 1'],
'email': '[e-mail]',
'login': 'foo',
'parent': 'bar,server 1',
'server_id': 'server 1'}}

注意foo有一个指向bar的AGENT_ACCOUNT，bar指向foo，形成一个循环。

这将在这两个条目中的任何一个上产生无限递归错误：

>>> get_path(tree['bar,server 1'], tree)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 8, in get_path
File "<stdin>", line 8, in get_path
File "<stdin>", line 8, in get_path
[Previous line repeated 994 more times]
File "<stdin>", line 2, in get_path
RecursionError: maximum recursion depth exceeded while calling a Python object

您可以尽早检测到此类周期，并以更清晰的错误消息退出：

def get_path(node, tree, seen=None):
if seen is None:
seen = set()
parent = node.get('parent')
if parent:
if  parent in seen:
raise ValueError(
'Already handled {!r}, cycle detected. '
'Check all of {}'.format(
parent, sorted(seen)))
seen.add(parent)
node_login = '{0[login]},{0[server_id]}'.format(node)  # cleaner method to generate the key
if not parent:
return []
elif parent == node_login:
return [parent]
path = get_path(tree[parent], tree, seen)  # pass seen along to recursive calls
return [parent] + path

在同一tree上运行此更新版本现在会产生：

>>> get_path(tree['bar,server 1'], tree)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 16, in get_path
File "<stdin>", line 16, in get_path
File "<stdin>", line 9, in get_path
ValueError: Already handled 'foo,server 1', cycle detected. Check all of ['bar,server 1', 'foo,server 1']fted by one space

我假设这样的循环是错误的。如果不是，只需使用if parent in seen: return []返回到该点的路径(因此忽略循环)，但您将为循环的每个成员提供一个路径版本，每条路径都是下一个路径的旋转版本。

您应该真正修复您的帐户信息，并删除此类循环。如果您需要查找所有此类循环，可以使用：

from collections import deque
def find_all_cycles(tree):
visited, cycles, path = set(), [], []
queue = deque(sorted(tree))
while queue:
key = queue.pop()
if key in visited:
continue
visited.add(key)
path.append(key)
parent = tree[key].get('parent')
if not parent:
path = []
elif parent in visited:
# cycle detected!
cycles.append(path + [parent])
path = []
else:
queue.append(parent)
return cycles

Answer 2

我有相同的错误消息，为我修复它是传递模型而不是返回它。

我不得不使用这个：

class User(AbstractUser):
pass

下面的代码不允许将模型传递给其他模型。

class User(AbstractUser):
def __str__(self):
return f"{self}"