[Tutor] Why do sets use 6 times as much memory as lists?

[wesley chun]

Following Oscar's comment, It's also O(n) vs. O(1) tradeoff.

--wesley


On Tue, May 7, 2013 at 12:09 PM, Oscar Benjamin
<oscar.j.benjamin at gmail.com>wrote:

> On 7 May 2013 19:09, Bjorn Madsen <bjorn.h.madsen at googlemail.com> wrote:
> > Hi, Thanks for the quick response.
> >
> > "Being curious" I actually expected something like this:
> >>>> L={x:None for x in range(10000)}
> >>>> sys.getsizeof(L)
> > 196660
> >>>>
> >
> > That is why I wondered why 6 times is a lot given that a dict can do the
> > same at 3/4 of the mem-footprint.
> > Just got surprised about the overhead as "sets" some are more lightweight
> > than dictionaries.
>
> I think that the 3/4 is pretty much random. sets and dicts resize
> peroidically depending on how they are used.
>
> > additional overhead must have something to do with set-specific
> operations,
> > I imagine such as:
> > set1 - set2
> > set | set2
> > set1 & set2
> > set1 <= set2
> > ....
>
> I don't think so. Here's the results on a 64-bit Linux machine (Python
> 2.7):
> >>> import sys
> >>> sys.getsizeof([x for x in range(10000)])
> 87632
> >>> sys.getsizeof({x for x in range(10000)})
> 524520
> >>> sys.getsizeof({x: None for x in range(10000)})
> 786712
>
> So in this case the set ends up being 2/3 the size of the dict. As I
> said before these results will not be consistent from one computer to
> another.
>
>
> Oscar
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    +wesley chun : wescpy at gmail : @wescpy
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