Better way to check if all lists in a list are the same length? [duplicate]

Question

Currently running with:

l1 = [i for i in range(0,10)]
l2 = [i for i in range(0,10)]
l3 = [i for i in range(0,10)]
lists = [l1, l2, l3]
length = len(lists[0])
for l in lists:
    if length != len(l):
        raise ValueErrorr('not all lists have same length!')

Is there a prettier way to test for this than a for loop? Is there a faster/better way which isn't O(n) ?

Answer 1

I'd do it with a generator expression and all:

it = iter(lists)
the_len = len(next(it))
if not all(len(l) == the_len for l in it):
     raise ValueError('not all lists have same length!')

This avoids checking the length of the first list twice and does not build throwaway list/set datastructures.

all also evaluates lazily, which means it will stop and return False as soon as the first list which differs in length is yielded by the generator.

Answer 2

You can use a set comprehension in order to preserve the unique lengths, then check if you only have one item in set:

if len({len(i) for i in lists}) == 1:
    # do stuff

Or as a more efficient way you could use a generator expression within any or all.

def check_size_eq(lst):
    # returns true if there's any list with unequal length to the first one
    return not any(len(lst[0])!= len(i) for i in lst)
    # or you could do:
    # return all(len(lst[0])== len(i) for i in lst)

demo :

>>> a = {1}
>>> 
>>> a.pop() and not a
True
>>> a = {1,3}
>>> a.pop() and not a
False