Delete all items in the list

Steven D'Aprano steve at pearwood.info
Sat Feb 28 10:59:18 EST 2009


Chris Rebert wrote:

> Obviously that equivalence is true, but in this case I'm emphasizing
> that it's even worse than that when constant factors are taken into
> account. Big-O is nice in the abstract, but in the real-world those
> constant factors can matter.
> 
> In pure big-O, it is indeed O(M*N) vs. O(N)
> Including constant factors, the performance is roughly 2*M*N*X [X =
> overhead of remove()] vs. N, which makes the badness of the algorithm
> all the more apparent.


So, what you're saying is that if you include a constant factor on one side
of the comparison, and neglect the constant factors on the other, the side
with the constant factor is worse. Well, duh :)

I forget which specific O(N) algorithm you're referring to, but I think it
was probably a list comp:

L[:] = [x for x in L if x != 'a']

As a wise man once said *wink*, "Big-O is nice in the abstract, but in the
real-world those constant factors can matter". This is very true... in this
case, the list comp requires creating a new list, potentially resizing it
an arbitrary number of times, then possibly garbage collecting the previous
contents of L. These operations are not free, and for truly huge lists
requiring paging, it could get very expensive.

Big Oh notation is good for estimating asymptotic behaviour, which means it
is good for predicting how an algorithm will scale as the size of the input
increases. It is useless for predicting how fast that algorithm will run,
since the actual speed depends on those constant factors that Big Oh
neglects. That's not a criticism of Big Oh -- predicting execution speed is
not what it is for.

For what it's worth, for very small lists, the while...remove algorithm is
actually faster that using a list comprehension, despite being O(M*N)
versus O(N), at least according to my tests. It's *trivially* faster, but
if you're interested in (premature) micro-optimisation, you might save one
or two microseconds by using a while loop for short lists (say, around
N<=12 or so according to my tests), and swapping to a list comp only for
larger input.

Now, this sounds silly, and in fact it is silly for the specific problem
we're discussing, but as a general technique it is very powerful. For
instance, until Python 2.3, list.sort() used a low-overhead but O(N**2)
insertion sort for small lists, only kicking off a high-overhead but O(N)
sample sort above a certain size. The current timsort does something
similar, only faster and more complicated. If I understand correctly, it
uses insertion sort to make up short runs of increasing or decreasing
values, and then efficiently merges them.


-- 
Steven




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