I need a function which takes in a list
and outputs True
if all elements in the input list evaluate as equal to each other using the standard equal
A solution faster than using set() that works on sequences (not iterables) is to simply count the first element. This assumes the list is non-empty (but that's trivial to check, and decide yourself what the outcome should be on an empty list)
x.count(x[0]) == len(x)
some simple benchmarks:
>>> timeit.timeit('len(set(s1))<=1', 's1=[1]*5000', number=10000)
1.4383411407470703
>>> timeit.timeit('len(set(s1))<=1', 's1=[1]*4999+[2]', number=10000)
1.4765670299530029
>>> timeit.timeit('s1.count(s1[0])==len(s1)', 's1=[1]*5000', number=10000)
0.26274609565734863
>>> timeit.timeit('s1.count(s1[0])==len(s1)', 's1=[1]*4999+[2]', number=10000)
0.25654196739196777