Python：在字典中查找具有唯一值的键？
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Python: finding keys with unique values in a dictionary?
提问by Roee Adler
I receive a dictionary as input, and want to return a list of keys for which the dictionary values are unique in the scope of that dictionary.
我收到一个字典作为输入，并希望返回一个键列表，其中的字典值在该字典的范围内是唯一的。
I will clarify with an example. Say my input is dictionary a, constructed as follows:
我会用一个例子来澄清。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
The result I expect is ['dog', 'snake'].
我期望的结果是['dog', 'snake']。
There are obvious brute force ways to achieve this, however I wondered if there's a neat Pythonian way to get the job done.
有明显的蛮力方法可以实现这一点，但是我想知道是否有一种简洁的 Pythonian 方法来完成工作。
采纳答案by Anurag Uniyal
I think efficient way if dict is too large would be
如果 dict 太大，我认为有效的方法是
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]
回答by Bartosz Radaczyński
Note that this actually is a bruteforce:
请注意，这实际上是一种蛮力：
l = a.values()
b = [x for x in a if l.count(a[x]) == 1]
回答by Rick Copeland
Here is a solution that only requires traversing the dict once:
这是一个只需要遍历 dict 一次的解决方案：
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)
回答by John Machin
>>> 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']
>>>
回答by Juergen
A little more verbose, but does need only one pass over a:
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 ]
( I hope it works, since I can't test it here )
（我希望它有效，因为我无法在这里测试）
回答by fabrizioM
What about subclassing?
子类化呢？
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']
回答by S.Lott
Here's another variation.
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']
I'm partial to this because the inverted dictionary is such a common design pattern.
我偏爱这个，因为倒排字典是一种很常见的设计模式。
回答by Alex Morega
You could do something like this (just count the number of occurrences for each value):
你可以做这样的事情（只需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
回答by Greg Bacon
Use nested list comprehensions!
使用嵌套列表理解！
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]
