Checking if all elements in a list are unique


Question

What is the best way (best as in the conventional way) of checking whether all elements in a list are unique?

My current approach using a Counter is:

>>> x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
>>> counter = Counter(x)
>>> for values in counter.itervalues():
        if values > 1: 
            # do something

Can I do better?

1
86
3/11/2011 11:14:41 PM

Accepted Answer

Not the most efficient, but straight forward and concise:

if len(x) > len(set(x)):
   pass # do something

Probably won't make much of a difference for short lists.

142
3/11/2011 8:47:21 PM

Here is a two-liner that will also do early exit:

>>> def allUnique(x):
...     seen = set()
...     return not any(i in seen or seen.add(i) for i in x)
...
>>> allUnique("ABCDEF")
True
>>> allUnique("ABACDEF")
False

If the elements of x aren't hashable, then you'll have to resort to using a list for seen:

>>> def allUnique(x):
...     seen = list()
...     return not any(i in seen or seen.append(i) for i in x)
...
>>> allUnique([list("ABC"), list("DEF")])
True
>>> allUnique([list("ABC"), list("DEF"), list("ABC")])
False

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