[Edu-sig] RE: [Tutor] style question

[Seth David Schoen]

alan.gauld@bt.com writes:

> > some feedbakc.  Let's say you have a task of finding out how many of 5
> > positive integer numbers (a-e) are even.  Which (if either) of the two
> > codes would be preferred:
> > 
> > # {option A}
> > odd=a%2+b%2+c%2+d%2+e%2
> > even=5-odd
> 
> Yech! I don't like this under any circumstance.
> If I really wanted to do it in two lines I'd try something like:
> 
> >>> count = 0
> >>> L = [a,b,c,d,e]
> >>> for n in L:
>       if n%2: count += 1
> 
> > # {option B}
> > if a%2==0:	
> > 	even = even + 1
> > if b%2==0:	
> > 	even = even + 1
> > if c%2==0:	
> > 	even = even + 1
> > if d%2==0:	
> > 	even = even + 1
> > if e%2==0:	
> > 	even = even + 1

I like lambda and filter a lot:

L = [a,b,c,d,e]
even = len(L) - len(filter(lambda x: x%2, L))

Some people prefer "x&1" instead of "x%2".  "~x&1" is probably the
shortest Python expression for "x is even", so you could write

even = len(filter(lambda x: ~x&1, L))

-- 
Seth David Schoen <schoen@loyalty.org> | Reading is a right, not a feature!
     http://www.loyalty.org/~schoen/   |                 -- Kathryn Myronuk
     http://vitanuova.loyalty.org/     |