Checking if all elements in a list are unique

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

An early-exit solution could be

def unique_values(g):
    s = set()
    for x in g:
        if x in s: return False
        s.add(x)
    return True

however for small cases or if early-exiting is not the common case then I would expect len(x) != len(set(x)) being the fastest method.

for speed:

import numpy as np
x = [1, 1, 1, 2, 3, 4, 5, 6, 2]
np.unique(x).size == len(x)

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.