[Tutor] Re: Can you modify every nth item in a list with a single assignment?

[Jeff Shannon]

Andrei wrote:

> >>> [ (item in range(0, len(b), 3) and [b[item]] or [2+b[item]])[0] 
> for item in range(0, len(b))]
> [0, 3, 4, 3, 6, 7, 6, 9, 10]
>
> That's better. 


Well, "better" being a matter of opinion. ;)  This works, sure, but I'd 
rather explicitly build a list of indices, and then explicitly loop 
through that list and modify the primary list -- or else use that list 
to exclude items from processing.

 >>> indices = range(0,9,3)
 >>> b = range(9)
 >>> for n in range(len(b)):
...     if n in indices:
...         continue
...     b[n] += 2
...    
 >>> b
[0, 3, 4, 3, 6, 7, 6, 9, 10]
 >>>

This is more than one line of code, sure... but I can actually *read* 
it!  To quote 'import this', "Readability counts."  As someone who 
frequently maintains old code, I can guarantee you that I'd much rather 
run across the explicit loops than the convoluted list comprehensions. 
 I can figure out the effect of the explicit loop in a matter of 
seconds; the list comprehensions and the and/or tricks would take me *at 
least* several minutes to painstakingly work out.  This strikes me as a 
very poor tradeoff.

Jeff Shannon
Technician/Programmer
Credit International