Question

如果我有嵌套字典d = {'a':{'b':{}}}和字符串'a.b.c'以及值'X'

我需要根据键字符串将值放在dict中。

我想要实现的内容可以硬编码为d['a']['b']['c'] = 'X'但我需要动态执行。键串可以是任意长度。

对于奖励积分：我还需要创建密钥，如果它们不存在，如'a.b.z'，但我相信如果我能解决它们已经存在的情况，我可以弄明白。

Answer 1

def set(d, key, value):
    dd = d
    keys = key.split('.')
    latest = keys.pop()
    for k in keys:
        dd = dd.setdefault(k, {})
    dd.setdefault(latest, value)

d = {}
set(d, 'a.b.c', 'X')
set(d, 'a.b.d', 'Y')
print(d)

结果：

{'a': {'b': {'c': 'X', 'd': 'Y'}}}

Answer 2

def recursedict(d,keylist,value=None):
    key = keylist.pop(0) # removes and returns the first key
    if len(keylist): # True if there are more levels to go down
        try: assert type(d[key]) is dict
        except KeyError: d[key] = dict()
        except AssertionError: raise ValueError("d[{}] is a {}".format(key,type(d[key])))
        # if d[key] doesn't exist, make it a dict()
        # if d[key] DOES exist, and isn't a dict, raise a KeyError

        recursedict(d[key],keylist,value)
        # recurse
    else:
        if value is None:
            return d[key]
        else:
            d[keylist[0]] = value
            return value

def setdeepdict(d,attributestr,value): # double entendre intentional
    keys = attributestr.split('.')
    recursedict(d,keys,value)

def getdeepdict(d,attributestr): # double entendre SUPER intentional
    keys = attributestr.split('.')
    recursedict(d,keys)

动态地在dict中设置深度值

2 个答案: