Python Unique List – Checking if All Elements in a List Are Unique

algorithmlistpythonunique

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?

Best 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.

Related Solutions

Python – Check if All Elements in a List Are Identical

Use itertools.groupby (see the itertools recipes):

from itertools import groupby

def all_equal(iterable):
    g = groupby(iterable)
    return next(g, True) and not next(g, False)

or without groupby:

def all_equal(iterator):
    iterator = iter(iterator)
    try:
        first = next(iterator)
    except StopIteration:
        return True
    return all(first == x for x in iterator)

There are a number of alternative one-liners you might consider:

Converting the input to a set and checking that it only has one or zero (in case the input is empty) items
```
def all_equal2(iterator):
    return len(set(iterator)) <= 1
```
Comparing against the input list without the first item
```
def all_equal3(lst):
    return lst[:-1] == lst[1:]
```

Counting how many times the first item appears in the list

def all_equal_ivo(lst):
    return not lst or lst.count(lst[0]) == len(lst)

Comparing against a list of the first element repeated

def all_equal_6502(lst):
    return not lst or [lst[0]]*len(lst) == lst

But they have some downsides, namely:

all_equal and all_equal2 can use any iterators, but the others must take a sequence input, typically concrete containers like a list or tuple.
all_equal and all_equal3 stop as soon as a difference is found (what is called "short circuit"), whereas all the alternatives require iterating over the entire list, even if you can tell that the answer is False just by looking at the first two elements.
In all_equal2 the content must be hashable. A list of lists will raise a TypeError for example.
all_equal2 (in the worst case) and all_equal_6502 create a copy of the list, meaning you need to use double the memory.

On Python 3.9, using perfplot, we get these timings (lower Runtime [s] is better):

Best Answer

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Python – Check if All Elements in a List Are Identical

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