the python way?

[Andrew Dalke]

Reinhold Birkenfeld wrote:
> To make it short, my version is:
> 
> import random
> def reinterpolate2(word, vocals='aeiouy'):
>     wlist = list(word)
>     random.shuffle(wlist)
>     vees = [c for c in wlist[::-1] if c in vocals]
>     cons = [c for c in wlist[::-1] if c not in vocals]

Why the [::-1]?  If it's randomly shuffled the order isn't important.

>     short, long = sorted((cons, vees), key=len)
>     return ''.join(long[i] + short[i] for i in range(len(short))) + ''.join(long[len(short):])

All the cool kids are using 2.4 these days.  :)

Another way to write this is (assuming the order of characters
can be swapped)

     N = min(len(short), len(long))
     return (''.join( [c1+c2 for (c1, c2) in zip(cons, vees)] +
             cons[N:] + vees[N:])

The main change here is that zip() stops when the first iterator finishes
so there's no need to write the 'for i in range(len(short))'

If the order is important then the older way is

    if len(cons) >= len(vees):
        short, long = vees, cons
    else:
        short, long = cons, vees
    return (''.join( [c1+c2 for (c1, c2) in zip(short, long)] +
             long[len(short):])


'Course to be one of the cool kids, another solution is to use the
roundrobin() implementation found from http://www.python.org/sf/756253

from collections import deque
def roundrobin(*iterables):
    pending = deque(iter(i) for i in iterables)
    while pending:
        task = pending.popleft()
        try:
            yield task.next()
        except StopIteration:
            continue
        pending.append(task)



With it the last line becomes

     return ''.join(roundrobin(short, long))

Anyone know if/when roundrobin() will be part of the std. lib?
The sf tracker implies that it won't be.

				Andrew
				dalke at dalkescientific.com