Arithmetic with Boolean values

[Steven D'Aprano]

On Sun, 12 Aug 2012 07:40:30 -0400, Roy Smith wrote:

> In article <502791ea$0$29978$c3e8da3$5496439d at news.astraweb.com>,
>  Steven D'Aprano <steve+comp.lang.python at pearwood.info> wrote:
> 
>> for x in (0,) if len(L)%2 else (0, 1):
>>     ...
>> 
>> which is even more explicit and simpler to read even though it is
>> longer.
> 
> Ugh.
> 
> do_stuff()
> if len(L) % 2 == 0:
>    do_stuff()  # reprocess even-length list
> 
> Sure, it's 3 lines instead of one, but dead-obvious what the intention
> is.

Well, sure, for that specific case that would work too. Using a for-loop 
to do something once is a little icky. But only a little.

Also, it scales to situations like "repeat 37 times when even, or 82 
times when odd" (or any other values).


> There's two problems with all the looping suggestions people have given.
> One is that the computation of whether you need to do it once or twice
> is messy.  But, but bigger issue is you're trying to warp what's
> fundamentally a boolean value into a looping construct.  That's a
> cognitive mismatch.

Really? You've never used a while loop?

while not finished:
    do_something()



There's nothing wrong with having a for-loop which iterates over a 
computed set of values:

if len(L)%2:
    items = range(len(L)//2 + 1)
else:
    items = range(len(L)//2)
for x in items:
    ...


which can be simplified to:

for x in range(len(L)//2 + len(L)%2):
    ...


-- 
Steven