按路径（字符串）获取字典键 [英] get dictionary key by path (string)

问题描述

我有一条可以随时更改的路径：

I have this path that can change from time to time:

'#/path/to/key'

路径的部分未定义，因此此值也很好

The parts of the path aren't defined, so this value is also fine

'#/this/is/a/longer/path'

我将此键分割为'/'，所以我得到

I'm splitting this key at '/' so I get

['#', 'path', 'to', 'key']

，我需要输入键在这个路径中，假设我的字典是exp，所以我需要去这里：

and I need to get to the key in this path, let's say my dict is exp, so I need to get to here:

exp['path']['to']['key']

我怎么可能知道如何获取此密钥？

how could I possibly know how to get to this key?

推荐答案

使用递归，路加...

Use the recursion, Luke ...

def deref_multi(data, keys):
    return deref_multi(data[keys[0]], keys[1:]) \
        if keys else data

last = deref_multi(exp, ['path','to','key'])

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