python之递归错误 : maximum recursion depth exceeded while getting the str of an object

在这里查看了很多关于这个问题的例子，但我无法为我的例子解决这个问题。

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:

tree = {} 

节点是树的一个项目。

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' 
    } 
} 

我每次都会遇到这个递归错误

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

请您参考如下方法:

您的代码假定您始终处理 acyclic directed graph , 但您的输入至少有一个 directed cycle其中，一个 AGENT_ACCOUNT 引用直接或间接指向另一个帐户，而该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': 'foo@bar.com', 'server_id': 'server 1', 
     'AGENT_ACCOUNT': 'bar'}, 
    {'LOGIN': 'bar', 'EMAIL': 'bar@bar.com', 'server_id': 'server 1', 
     'AGENT_ACCOUNT': 'foo'}] 

然后树变成:

{'bar,server 1': {'children': ['foo,server 1'], 
                  'email': 'bar@bar.com', 
                  'login': 'bar', 
                  'parent': 'foo,server 1', 
                  'server_id': 'server 1'}, 
 'foo,server 1': {'children': ['bar,server 1'], 
                  'email': 'foo@bar.com', 
                  'login': 'foo', 
                  'parent': 'bar,server 1', 
                  'server_id': 'server 1'}} 

注意 foo 有一个 AGENT_ACCOUNT 指向 bar，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 

您可以及早检测到此类循环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 上运行这个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)，但是您将拥有每个成员的路径版本循环，每条路径都是下一条路径的旋转版本。

您真的应该修复您的帐户信息，并消除此类循环。如果你需要找到所有这样的循环，你可以使用:

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