Question

我有一个数据库，用于建模与n嵌套级别的文件夹关系。对于任何给定的文件夹，我想生成所有子文件夹的列表。

假设我有一个名为getChildFolders()的函数，那么调用这种递归循环的最有效方法是什么？

以下代码适用于4个嵌套级别，但我希望在指定递归深度方面更灵活，或者在没有更多子级要遵循的情况下智能地停止循环。

folder_ids = []
folder_ids.append(folder.id)
for entry in child_folders:
    folder_ids.append(entry.id)
    child_folders_1 = getChildFolders(entry.id)
    for entry_1 in child_folders_1:
        folder_ids.append(entry_1.id)
        child_folders_2 = getChildFolders(entry_1.id)
        for entry_2 in child_folders_2:
            folder_ids.append(entry_2.id)
            child_folders_3 = getChildFolders(entry_2.id)
            for entry_3 in child_folders_3:
                folder_ids.append(entry_3.id)

Answer 1

递归函数是一种很好的方法：

def collect_folders(start, depth=-1)
    """ negative depths means unlimited recursion """
    folder_ids = []

    # recursive function that collects all the ids in `acc`
    def recurse(current, depth):
        folder_ids.append(current.id)
        if depth != 0:
            for folder in getChildFolders(current.id):
                # recursive call for each subfolder
                recurse(folder, depth-1)

    recurse(start, depth) # starts the recursion
    return folder_ids

Answer 2

我通常避免像python中的瘟疫一样递归，因为它很慢并且因为整个堆栈溢出错误的事情。

def collect_folders(start):
    stack = [start.id]
    folder_ids = []
    while stack:
        cur_id = stack.pop()
        folder_ids.append(cur_id)
        stack.extend(folder.id for folder in getChildFolders(cur_id))
    return folder_ids

这假定getChildFolders在没有子节点时返回空列表。如果它做了其他事情，比如返回一个标记值或引发异常，则必须进行修改。

Answer 3

def my_recursive_function(x, y, depth=0, MAX_DEPTH=20):
    if depth > MAX_DEPTH:
        return exhausted()
    elif something(x):
        my_recursive_function(frob(x), frob(y), depth + 1)
    elif query(y):
        my_recursive_function(mangle(x), munge(y), depth + 1)
    else:
        process(x, y)

# A normal call looks like this.
my_recursive_function(a, b)

# If you're in a hurry,
my_recursive_function(a, b, MAX_DEPTH=5)
# Or have a lot of time,
my_recursive_function(a, b, MAX_DEPTH=1e9)

Answer 4

这是最接近你的代码，非常unpythonic：

def recurse(folder_ids, count):
  folder_ids.append(folder.id)
  for entry in child_folders:
    folder_ids.append(entry.id)
    child_folders_1 = getChildFolders(entry.id)
    if count > 0:
        recurse(folder_ids, count-1)

folder_ids = []
recurse(folder_ids, 4)

您应该寻找os.walk，并采用类似的方法迭代地遍历树。

Answer 5

我需要类似的东西来检查一个分层树。你可以尝试：

def get_children_folders(self,mother_folder):
    '''
    For a given mother folder, returns all children, grand children
    (and so on) folders of this mother folder.
    '''
    folders_list=[]
    folders_list.append(mother_folder)
    for folder in folders_list:
        if folder not in folders_list: folders_list.append(folder)
        new_children = getChildFolders(folder.id)
        for child in new_children:
            if child not in folders_list: folders_list.append(child)
    return folders_list

Python中的递归循环函数

5 个答案: