python字典的递归深度 [英] Recursive depth of python dictionary

问题描述

G'day，

我正在尝试查找拖曳字典的函数的递归深度，我有点失落...
目前我有这样的东西：

I am trying to find the recursive depth of a function that trawls a dictionary and I'm a bit lost... Currently I have something like:

myDict = {'leve1_key1': {'level2_key1': {'level3_key1': {'level4_key_1': {'level5_key1':   'level5_value1'}}}}}

我想知道只是嵌套最嵌套的字典是...所以我做以下...

And I want to know just how nested the most nested dictionary is... so I do the following...

def dict_depth(d, depth):

    for i in d.keys():
        if type(d[i]) is dict:
            newDict = d[i]
            dict_depth(newDict, depth+1)
    return depth

print dict_depth(myDict, 0)

只有问题是，递归循环只返回最终值（0）的返回值。
如果我在d.keys（）：中放入一个打印声明
，那么我至少可以打印最高的递归值，但是返回价值是一个不同的事情...

Only problem is, the recursive loop only returns the return of the final value (0). if I put in a print statement for i in d.keys(): then I can at least print the highest value of recursion, but returning the value is a different matter...

我确信这是直截了当的 - 我刚刚得到jellybrain。

I'm sure this is straightforward - I've just got jellybrain.

干杯

推荐答案

请确保将递归调用的结果分配给深度 。另外，正如@amit所说，请考虑使用 max ，以便您可以处理具有多个键值对（treelike结构）的列表。

Be sure to assign the result of the recursive call to depth. Also, as @amit says, consider using max so that you can handle dicts with multiple key value pairs (a treelike structure).

def dict_depth(d, depth=0):
    if not isinstance(d, dict) or not d:
        return depth
    return max(dict_depth(v, depth+1) for k, v in d.iteritems())

>>> myDict = {'leve1_key1': {'level2_key1': 
               {'level3_key1': {'level4_key_1': 
                  {'level5_key1':   'level5_value1'}}}}}
>>> dict_depth(myDict)
5

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