Optimizing list processing

Peter Otten __peter__ at web.de
Thu Dec 12 16:08:33 CET 2013


Steven D'Aprano wrote:

> I have some code which produces a list from an iterable using at least
> one temporary list, using a Decorate-Sort-Undecorate idiom. The algorithm
> looks something like this (simplified):
> 
> table = sorted([(x, i) for i,x in enumerate(iterable)])
> table = [i for x,i in table]
> 
> The problem here is that for large iterables, say 10 million items or so,
> this is *painfully* slow, as my system has to page memory like mad to fit
> two large lists into memory at once. So I came up with an in-place
> version that saves (approximately) two-thirds of the memory needed.
> 
> table = [(x, i) for i,x in enumerate(iterable)]
> table.sort()
> for x, i in table:
>     table[i] = x
> 
> For giant iterables (ten million items), this version is a big
> improvement, about three times faster than the list comp version. Since
> we're talking about the difference between 4 seconds and 12 seconds (plus
> an additional 40-80 seconds of general slow-down as the computer pages
> memory into and out of virtual memory), this is a good, solid
> optimization.
> 
> Except that for more reasonably sized iterables, it's a pessimization.
> With one million items, the ratio is the other way around: the list comp
> version is 2-3 times faster than the in-place version. For smaller lists,
> the ratio varies, but the list comp version is typically around twice as
> fast. A good example of trading memory for time.
> 
> So, ideally I'd like to write my code like this:
> 
> 
> table = [(x, i) for i,x in enumerate(iterable)]
> table.sort()
> if len(table) < ?????:
>     table = [i for x,i in table]
> else:
>     for x, i in table:
>         table[i] = x
> 
> where ????? no doubt will depend on how much memory is available in one
> contiguous chunk.
> 
> Is there any way to determine which branch I should run, apart from hard-
> coding some arbitrary and constant cut-off value?

How about using two lists?

keys = list(iterable)
values = range(len(keys))
values.sort(key=keys.__getitem__)
del keys

The intention is to save memory used for the 2-tuples; I don't know if they 
pop up elsewhere.




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