istep() addition to itertool? (Was: Re: Printing n elements per line in a list)

[Neil Cerutti]

On 2006-08-19, Rhamphoryncus <rhamph at gmail.com> wrote:
> unexpected wrote:
>> If have a list from 1 to 100, what's the easiest, most elegant
>> way to print them out, so that there are only n elements per
>> line.
>
> I've run into this problem a few times, and although many
> solutions have been presented specifically for printing I would
> like to present a more general alternative.
>
> from itertools import chain
> def istepline(step, iterator):
>     i = 0
>     while i < step:
>         yield iterator.next()
>         i += 1
>
> def istep(iterable, step):
>     iterator = iter(iterable)  # Make sure we won't restart iteration
>     while True:
>         # We rely on istepline()'s side-effect of progressing the
>         # iterator.
>         start = iterator.next()
>         rest = istepline(step - 1, iterator)
>         yield chain((start,), rest)
>         for i in rest:
>             pass  # Exhaust rest to make sure the iterator has
>                   # progressed properly.
>
> Would anybody else find this useful?  Maybe worth adding it to
> itertool?

Your note me curious enough to re-read the itertools
documentation, and I found the following in 5.16.3 Recipes:

def grouper(n, iterable, padvalue=None):
    "grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

Wish I'd found that yesterday. ;)

-- 
Neil Cerutti