Simple newbie question

[Robert Brewer]

Or, simply iterate through the list backwards.

FuManChu

> -----Original Message-----
> From: Duncan Booth [mailto:duncan at NOSPAMrcp.co.uk] 
> Sent: Thursday, December 04, 2003 7:51 AM
> To: python-list at python.org
> Subject: Re: Simple newbie question
> 
> 
> Angelo Secchi <secchi at sssup.it> wrote in 
> news:mailman.107.1070547952.16879.python-list at python.org:
> 
> > The code i'm using is the following:
> > 
> > for line in lista:
> >      if sum(line == 0) > 0:
> >      lista.remove(line)
> > 
> > 
> > The problem is that the loop stops at the first line with 
> zeros and it
> > doesn't remove the other lines with zeros (at least if I do 
> not re-run
> > the loop). Probably I do not understand how the loop works.
> 
> The loop will assign the first element of lista to line, then 
> the second, 
> then the third, and so on.
> 
> The problem is that, if you remove an element everything 
> after it shifts up 
> one place. Say you are looking at the first element and 
> remove it, what was 
> the second element now becomes the first element, but the 
> next time round 
> the loop you will look at the new second element and the one that was 
> originally second (and is now first) will be skipped.
> 
> The solution is never to work on the list you are modifying. 
> Just make a 
> copy of the list and iterate over the copy, that means you can safely 
> modify the original. To copy a list, use the 'list' builtin.
> 
>     for line in list(lista):
>         if condition:
>             lista.remove(line)
> 
> Alternatively turn your brain around and instead of deleting specific 
> elements from the list just build a new list with the 
> elements you want to 
> keep. In this case you could do that using a list comprehension:
> 
>    lista = [ line for line in lista if not condition ]
> 
> (Obviously in both these cases replace condition with the correct 
> expression, which as Gerhard pointed out may not be the 
> condition in your 
> original posting.)
> 
> The list comprehension rebinds the lista variable, whereas 
> your original 
> attempt was modifying the list inplace. This might be 
> significant if there 
> are other references to the same list elsewhere in your code. 
> If it does 
> matter, then you can still use a list comprehension to mutate 
> the original 
> list:
> 
>    lista[:] = [ line for line in lista if not condition ]
> 
> Unlike the original 'for' loop, this works by building a complete 
> replacement list then mutating the original list in one go.
>  
> -- 
> Duncan Booth                                             
> duncan at rcp.co.uk
> int month(char 
> *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
> "\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?
> -- 
> http://mail.python.org/mailman/listinfo/python-list
>