Why I think range is a wart.

[Peter Dobcsanyi]

Bengt Richter <bokr at oz.net> wrote:
> I wonder how the proportions would change if len(ls)
> were 1,000,000 instead of 1,000
> and/or if you used xrange in place of range.

Good question, as it turned out.
I timed only the following methods:

    def foo(ls, q):
        for i in range(len(ls)):
            if ls[i] == q:
                return i

    def xfoo(ls, q):
        for i in xrange(len(ls)):
            if ls[i] == q:
                return i

    def bar(ls, q):
        i = 0   
        for x in ls:
            if x == q:
                return i
            i += 1

If you have to go over (almost) the entire list then the proportions
don't really change.

    len(ls) = 1000
    search for "998"
    repeat 3000 times

    i in range(len(ls)):   2.02
    i in xrange(len(ls)):  2.25
    x in ls; i+=1 :        2.24


    ----------------------------------------------------------

    len(ls) = 10000
    search for "9998"
    repeat 3000 times

    i in range(len(ls)):   23.71
    i in xrange(len(ls)):  25.04
    x in ls; i+=1 :        25.44

    ----------------------------------------------------------

    len(ls) = 100000
    search for "99998"
    repeat 3000 times

    i in range(len(ls)):   212.83
    i in xrange(len(ls)):  220.57
    x in ls; i+=1 :        257.96

    ----------------------------------------------------------

But when the number of iterations over the list is small compared to the
length of the list then xrange and the counting methods obviously win.
In all cases below I have searched for '998'.

    len(ls) = 1000
    search for "998"
    repeat 3000 times

    i in range(len(ls)):   2.02
    i in xrange(len(ls)):  2.26
    x in ls; i+=1 :        2.28

    ----------------------------------------------------------

    len(ls) = 10000
    search for "998"
    repeat 3000 times

    i in range(len(ls)):   3.36
    i in xrange(len(ls)):  2.19
    x in ls; i+=1 :        2.27

    ----------------------------------------------------------

    len(ls) = 1000000
    search for "998"
    repeat 3000 times

    i in range(len(ls)):   22.40
    i in xrange(len(ls)):  2.22
    x in ls; i+=1 :        2.57


So it depends :-)

    Peter