Question

我有一份标准名称清单。

standard = ["Richard","Robert","Nicolas"]

and(此处为nick)

aliases = {standard[0]:["Richard","Rick","Dick","Rich"],
           standard[1]:["Robert","Roberto","Bob"],
           standard[2]:["Nicolas","Nick","Nic"]}

我想提出一个新的字典,即我可以把任何名字作为钥匙,并将把标准名称AKA转换为关键和价值。

我迄今唯一的猜测就是这一点。

t = {}
aliases = [t.update(zip(v,[k]*len(v))) for k,v in aliases.items()]
aliases = t

这样做有了一个更严厉或更可读的方法(更倾向于没有临时字典)。

Answer 1

我认为,这更可读:

rev_aliases = {}
for name, nick_list in aliases.iteritems():
    for nick in nick_list:
        rev_aliases[nick] = name

如果你喜欢某种形式的发电机表达,你可以使用:

页: 1

rev_aliases = {nick: name 
               for name, nick_list in aliases.viewitems()
               for nick in nick_list}

甲型六氯环己烷

rev_aliases = dict((nick, name)
                   for name, nick_list in aliases.iteritems()
                   for nick in nick_list)

Answer 2

>>> standard = ["Richard","Robert","Nicolas"] 
>>> aliases = {standard[0]:["Richard","Rick","Dick","Rich"], 
           standard[1]:["Robert","Roberto","Bob"] ,
           standard[2]:["Nicolas","Nick","Nic"] }
>>> def name(nickname):
        return [n for n in aliases if nickname in aliases[n]]

>>> name( Bob )
[ Robert ]
>>>

清单内容繁多。

Answer 3

http://www.ohchr.org。

Answer 4

滥用惯性的手段:

>>> from itertools import izip, repeat, chain
>>> dict(chain.from_iterable(
        izip(iter(b), repeat(a, len(b)))
        for a, b in aliases.iteritems()))
{ Nicolas :  Nicolas ,
  Richard :  Richard ,
  Nic :  Nicolas ,
  Robert :  Robert ,
  Dick :  Richard ,
  Roberto :  Robert ,
  Nick :  Nicolas ,
  Rick :  Richard ,
  Rich :  Richard ,
  Bob :  Robert }

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