递归地迭代嵌套的字典并返回第一个匹配键的值 [英] Recursively iterate through a nested dict and return value of the first matching key

问题描述

我有一个深层嵌套的dict，需要遍历它，并返回与 key 参数（函数的第二个参数）相对应的值。

I have a deeply nested dict and need to iterate through it and return the value corresponding to the key argument, second argument of my function.

例如，使用

tree = {"a": 12, "g":{ "b": 2, "c": 4}, "d":5}

tree_traverse（tree， d）应该返回5

这是我的代码：

def tree_traverse(tree, key):
    for k,v  in tree.items():
        if isinstance(v, dict):
            tree_traverse(v, key)

        elif k == key:
            return v

我遇到的问题是，一旦完成对最深层嵌套dict的迭代，该函数如果找不到匹配的键，则返回None。
在找到匹配密钥之前，我不希望它返回任何内容。

The problem I have is that this function returns None if it doesnt find the matching key once it's done iterating through the deepest nested dict. I don't want it to return anything before the matching key is found.

我没有在另一个线程中找到解决方案，其中大多数使用print语句并且不返回任何内容，所以我想它避免了这个问题。

I didn't find a solution in another thread, most of them use print statements and don't return anything so I guess it avoids this issue.

推荐答案

您必须检查递归调用是否确实找到了某些东西，以便您可以继续循环。例如。尝试以下操作：

You have to check whether the recursive call actually found something so you can continue the loop. E.g. try the following:

def tree_traverse(tree, key):
    for k, v  in tree.items():
        if k == key:
            return v
        elif isinstance(v, dict):
            found = tree_traverse(v, key) 
            if found is not None:  # check if recursive call found it
                return found

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