使用点符号字符串"a.b.c.d.e"检查嵌套字典，会自动创建缺失级别 [英] Checking a nested dictionary using a dot notation string "a.b.c.d.e", automatically create missing levels

问题描述

给出以下字典:

d = {"a":{"b":{"c":"winning!"}}}

我有这个字符串(来自外部来源，我无法更改此隐喻).

I have this string (from an external source, and I can't change this metaphor).

k = "a.b.c"

我需要确定字典是否具有键 'c'，所以如果没有，我可以添加它.

I need to determine if the dictionary has the key 'c', so I can add it if it doesn't.

这可轻松检索点表示法值:

This works swimmingly for retrieving a dot notation value:

reduce(dict.get, key.split("."), d)

但是我不知道如何减少" has_key支票或类似的东西.

but I can't figure out how to 'reduce' a has_key check or anything like that.

我的最终问题是:给定"a.b.c.d.e"，我需要在字典中创建所有必需的元素，但是如果它们已经存在，则不要踩踏它们.

My ultimate problem is this: given "a.b.c.d.e", I need to create all the elements necessary in the dictionary, but not stomp them if they already exist.

推荐答案

...或使用递归:

def put(d, keys, item):
    if "." in keys:
        key, rest = keys.split(".", 1)
        if key not in d:
            d[key] = {}
        put(d[key], rest, item)
    else:
        d[keys] = item

def get(d, keys):
    if "." in keys:
        key, rest = keys.split(".", 1)
        return get(d[key], rest)
    else:
        return d[keys]

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