CODEWITHSUNDEEP
python-3.x
scf
scf

Reputation: 406

Does Python3 short-circuit on list comparison?

Given to same-length but different lists a and b, when performing a == b will Python iterate through both entire lists, or will it stop on first inequality?

Upvotes: 0

Views: 30

Answers (1)

Patrick Artner
Patrick Artner

Reputation: 51683

If you got this question, you can simply test it:

a = [9] + list(range(2000))
b = [8] + list(range(2000))

def diff_at_start(): 
    return a == b

aa = list(range(2000)) + [9]
bb = list(range(2000)) + [8]

def diff_at_end(): 
    return aa == bb

if __name__ == '__main__':
    import timeit

    print(timeit.timeit("diff_at_start()", 
          setup="from __main__ import diff_at_start",
          number=1000))

    print(timeit.timeit("diff_at_end()", 
          setup="from __main__ import diff_at_end", 
          number=1000))

Output:

0.000169038772583  # diff_at_start
0.0287721157074    # diff_at_end

showing that in fact, a difference at the begin of a list, shortens the execution time considerably.

See

Upvotes: 1

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