Question

我收到一个字典作为输入，并希望返回字典值在该字典范围内唯一的键列表。

我将用一个例子来澄清。假设我的输入是字典a，构造如下：

a = dict()
a['cat'] =      1
a['fish'] =     1
a['dog'] =      2  # <-- unique
a['bat'] =      3
a['aardvark'] = 3
a['snake'] =    4  # <-- unique
a['wallaby'] =  5
a['badger'] =   5

我期望的结果是['dog', 'snake']。

有很明显的蛮力方法来实现这一目标，但我想知道是否有一种巧妙的Python方式来完成工作。

Answer 1

我认为如果dict太大会有效的方式

countMap = {}
for v in a.itervalues():
    countMap[v] = countMap.get(v,0) + 1
uni = [ k for k, v in a.iteritems() if countMap[v] == 1]

Answer 2

请注意，这实际上是一个暴力：

l = a.values()
b = [x for x in a if l.count(a[x]) == 1]

Answer 3

这是一个只需要遍历dict的解决方案：

def unique_values(d):
    seen = {} # dict (value, key)
    result = set() # keys with unique values
    for k,v in d.iteritems():
        if v in seen:
            result.discard(seen[v])
        else:
            seen[v] = k
            result.add(k)
    return list(result)

Answer 4

>>> b = []
>>> import collections
>>> bag = collections.defaultdict(lambda: 0)
>>> for v in a.itervalues():
...     bag[v] += 1
...
>>> b = [k for (k, v) in a.iteritems() if bag[v] == 1]
>>> b.sort() # optional
>>> print b
['dog', 'snake']
>>>

Answer 5

稍微冗长一点，但只需要一次通过：

revDict = {}
for k, v in a.iteritems():
  if v in revDict:
     revDict[v] = None
  else:
     revDict[v] = k

[ x for x in revDict.itervalues() if x != None ]

（我希望它有效，因为我不能在这里测试）

Answer 6

子类化怎么样？

class UniqueValuesDict(dict):

    def __init__(self, *args):
        dict.__init__(self, *args)
        self._inverse = {}

    def __setitem__(self, key, value):
        if value in self.values():
            if value in self._inverse:
                del self._inverse[value]
        else:
            self._inverse[value] = key
        dict.__setitem__(self, key, value)

    def unique_values(self):
        return self._inverse.values()

a = UniqueValuesDict()

a['cat'] =      1
a['fish'] =     1
a[None] =       1
a['duck'] =     1
a['dog'] =      2  # <-- unique
a['bat'] =      3
a['aardvark'] = 3
a['snake'] =    4  # <-- unique
a['wallaby'] =  5
a['badger'] =   5

assert a.unique_values() == ['dog', 'snake']

Answer 7

这是另一种变体。

>>> import collections
>>> inverse= collections.defaultdict(list)
>>> for k,v in a.items():
...     inverse[v].append(k)
... 
>>> [ v[0] for v in inverse.values() if len(v) == 1 ]
['dog', 'snake']

我偏爱这个，因为倒置字典是一种常见的设计模式。

Answer 8

你可以这样做（只计算每个值的出现次数）：

def unique(a):
    from collections import defaultdict
    count = defaultdict(lambda: 0)
    for k, v in a.iteritems():
        count[v] += 1
    for v, c in count.iteritems():
        if c <= 1:
            yield v

Answer 9

使用嵌套列表推导！

print [v[0] for v in 
           dict([(v, [k for k in a.keys() if a[k] == v])
                     for v in set(a.values())]).values()
       if len(v) == 1]

Python：在字典中查找具有唯一值的键？

9 个答案: