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You can try making sets.

a, b, c = 1, 2, 3
if len(a,b,c) == 3:
 # Do something

If your variables are kept as a list, it becomes even more simple:

a = [1,2,3,4,4]
if len(set(a)) == len(a):
 # Do something

Here is the official documentation of python sets.

This works only for hashable objects such as integers, as given in the question. For non-hashable objects, see @chepner's more general solution.

This is definitely the way you should go for hashable objects, since it takes O(n) time for the number of objects n. The combinatorial method for non-hashable objects take O(n^2) time.

Assuming hashing is not an option, use itertools.combinations and all.

from itertools import combinations

if all(x != y for x, y in combinations([a,b,c], 2)):
 # All values are unique

It depends a bit on the kind of values that you have.

If they are well-behaved and hashable then you can (as others already pointed out) simply use a set to find out how many unique values you have and if that doesn't equal the number of total values you have at least two values that are equal.

def all_distinct(*values):
 return len(set(values)) == len(values)

all_distinct(1, 2, 3) # True
all_distinct(1, 2, 2) # False

Hashable values and lazy

In case you really have lots of values and want to abort as soon as one match is found you could also lazily create the set. It's more complicated and probably slower if all values are distinct but it provides short-circuiting in case a duplicate is found:

def all_distinct(*values):
 seen = set()
 seen_add = seen.add
 last_count = 0
 for item in values:
 seen_add(item)
 new_count = len(seen)
 if new_count == last_count:
 return False
 last_count = new_count
 return True

all_distinct(1, 2, 3) # True
all_distinct(1, 2, 2) # False

However if the values are not hashable this will not work because set requires hashable values.

Unhashable values

In case you don't have hashable values you could use a plain list to store the already processed values and just check if each new item is already in the list:

def all_distinct(*values):
 seen = []
 for item in values:
 if item in seen:
 return False
 seen.append(item)
 return True

all_distinct(1, 2, 3) # True
all_distinct(1, 2, 2) # False

all_distinct([1, 2], [2, 3], [3, 4]) # True
all_distinct([1, 2], [2, 3], [1, 2]) # False

This will be slower because checking if a value is in a list requires to compare it to each item in the list.

A (3rd-party) library solution

In case you don't mind an additional dependency you could also use one of my libraries (available on PyPi and conda-forge) for this task iteration_utilities.all_distinct. This function can handle both hashable and unhashable values (and a mix of these):

from iteration_utilities import all_distinct

all_distinct([1, 2, 3]) # True
all_distinct([1, 2, 2]) # False

all_distinct([[1, 2], [2, 3], [3, 4]]) # True
all_distinct([[1, 2], [2, 3], [1, 2]]) # False

General comments

Note that all of the above mentioned approaches rely on the fact that equality means "not not-equal" which is the case for (almost) all built-in types but doesn't necessarily be the case!

However I want to point out chepners answers which doesn't require hashability of the values and doesn't rely on "equality means not not-equal" by explicitly checking for !=. It's also short-circuiting so it behaves like your original and approach.

Performance

To get a rough idea about the performance I'm using another of my libraries (simple_benchmark)

I used distinct hashable inputs (left) and unhashable inputs (right). For hashable inputs the set-approaches performed best, while for unhashable inputs the list-approaches performed better. The combinations-based approach seemed slowest in both cases:

I also tested the performance in case there are duplicates, for convenience I regarded the case when the first two elements were equal (otherwise the setup was identical to the previous case):

from iteration_utilities import all_distinct
from itertools import combinations
from simple_benchmark import BenchmarkBuilder

# First benchmark
b1 = BenchmarkBuilder()

@b1.add_function()
def all_distinct_set(values):
 return len(set(values)) == len(values)

@b1.add_function()
def all_distinct_set_sc(values):
 seen = set()
 seen_add = seen.add
 last_count = 0
 for item in values:
 seen_add(item)
 new_count = len(seen)
 if new_count == last_count:
 return False
 last_count = new_count
 return True

@b1.add_function()
def all_distinct_list(values):
 seen = []
 for item in values:
 if item in seen:
 return False
 seen.append(item)
 return True

b1.add_function(alias='all_distinct_iu')(all_distinct)

@b1.add_function()
def all_distinct_combinations(values):
 return all(x != y for x, y in combinations(values, 2))

@b1.add_arguments('number of hashable inputs')
def argument_provider():
 for exp in range(1, 12):
 size = 2**exp
 yield size, range(size)

r1 = b1.run()
r1.plot()

# Second benchmark

b2 = BenchmarkBuilder()
b2.add_function(alias='all_distinct_iu')(all_distinct)
b2.add_functions([all_distinct_combinations, all_distinct_list])

@b2.add_arguments('number of unhashable inputs')
def argument_provider():
 for exp in range(1, 12):
 size = 2**exp
 yield size, [[i] for i in range(size)]

r2 = b2.run()
r2.plot()

# Third benchmark
b3 = BenchmarkBuilder()
b3.add_function(alias='all_distinct_iu')(all_distinct)
b3.add_functions([all_distinct_set, all_distinct_set_sc, all_distinct_combinations, all_distinct_list])

@b3.add_arguments('number of hashable inputs')
def argument_provider():
 for exp in range(1, 12):
 size = 2**exp
 yield size, [0, *range(size)]

r3 = b3.run()
r3.plot()

# Fourth benchmark
b4 = BenchmarkBuilder()
b4.add_function(alias='all_distinct_iu')(all_distinct)
b4.add_functions([all_distinct_combinations, all_distinct_list])

@b4.add_arguments('number of hashable inputs')
def argument_provider():
 for exp in range(1, 12):
 size = 2**exp
 yield size, [[0], *[[i] for i in range(size)]]

r4 = b4.run()
r4.plot()

The slightly neater way is to stick all the variables in a list, then create a new set from the list. If the list and the set aren't the same length, some of the variables were equal, since sets can't contain duplicates:

vars = [a, b, c]
no_dupes = set(vars)

if len(vars) != len(no_dupes):
 # Some of them had the same value

This assumes the values are hashable; which they are in your example.

You can use all with list.count as well, it is reasonable, may not be the best, but worth to answer:

>>> a, b, c = 1, 2, 3
>>> l = [a, b, c]
>>> all(l.count(i) < 2 for i in l)
True
>>> a, b, c = 1, 2, 1
>>> l = [a, b, c]
>>> all(l.count(i) < 2 for i in l)
False
>>> 

Also this solution works with unhashable objects in the list.

A way that only works with hashable objects in the list:

>>> a, b, c = 1, 2, 3
>>> l = [a, b, c]
>>> len(*l) == len(l)
True
>>> 

Actually:

>>> from timeit import timeit
>>> timeit(lambda: *l, number=1000000)
0.5163292075532642
>>> timeit(lambda: set(l), number=1000000)
0.7005311807841572
>>> 

*l is faster than set(l), more info here.