python's timeit is not initialising list in setup code

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I'm timing element removal from a python set vs. a list. My timeit code for the list is raising a ValueError: ... x not in list, but only when I run more than one iteration with timeit !??

It appears that for the list, the variable created in the setup code is re-used in the subsequent iteration (as if the setup code is not run the second time??).

Here's my code:

In [1]: import timeit

In [2]: timeit.timeit(stmt='a.discard(10**5)', setup='a = set(range(10**6))', number=100000)
Out[2]: 0.02187999989837408

In [3]: timeit.timeit(stmt='a.remove(10**5)', setup='a = list(range(10**6))', number=1)
Out[3]: 0.023419374600052834

In [4]: timeit.timeit(stmt='a.remove(10**5)', setup='a = list(range(10**6))', number=2)
ValueError                                Traceback (most recent call last)
ValueError: list.remove(x): x not in list

What's going on??

They key point is that setup is only performed once even when number is >1 (hence the ValueError when you try call to list.remove with a unique value that has already been removed from the list). From the docs (emphasis mine):

Time number executions of the main statement. This executes the setup statement once, and then returns the time it takes to execute the main statement a number of times, measured in seconds as a float. The argument is the number of times through the loop, defaulting to one million. The main statement, the setup statement and the timer function to be used are passed to the constructor.

So, if you want to perform multiple timings of a code snippet like this (say to get more accurate timings) then you have to use number=1, but you could use the repeat argument with timeit.repeat() instead:

>>> timeit.repeat(stmt='a.remove(10**5)', setup='a = list(range(10**6))', number=1, repeat=2)
[0.002321417909115553, 0.0023121219128370285]

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In [4]: timeit.timeit(stmt='a.remove(10**5)', setup='a = list(range(10**6))', number=2)

a.remove(10**5) is executed twice but 10**5 can only be removed once. The setup is only called ONCE so you always work on the same list or set.

This is equivalent to

a = list(range(10**6))
a.remove(10**5)  # works
a.remove(10**5)  # fails with ValueError: list.remove(x): x not in list

This doesn't fail for the set since set discard and list remove behave differently. If you use set remove method you will also end up getting an error (KeyError)

a = set(range(10**6))
a.discard(10**5)  # works
a.discard(10**5)  # works


a = set(range(10**6))
a.remove(10**5)  # works
a.remove(10**5)  # KeyError 10**5

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The discard() operation on sets does nothing if the element being discarded is not in the set. The remove() operation on lists, however, raises a ValueError when the element does not exist. When you initialize your list with range(10**6), the value 10**5 only appears once; in the set this isn't a problem, but in the list it can only be removed once and will give an error on future attempts to remove the same value.

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  • @Laurent H. Happy to use US/UK spelling with -ize, but I'm Australian and -ise is acceptable spelling here
  • I really apologize to Australian pepole. You have learnt me something, thanks !
  • fwiw, removing from sets is always faster, and especially for long lists, where it's much faster (for 10k elements, 10x to remove first element, 200x for last element).
  • @drevicko Right, that makes sense in terms of time complexity since set.remove is normally an O(1) operation (just calculating a hash) while list.remove is an O(n) operation (potentially examining every item in the whole list)
  • also, removing the first item has a hit since the remaining items need to be shifted in memory afaik (though that's pretty efficient actually)