[Tutor] list copy and remove question

[Lloyd Kvam]

On Thu, 2004-09-09 at 19:44, Brian van den Broek wrote:
> Hi all,
> 
> I'm back to Python after long enough of an interruption to make me feel a 
> bit like a fresh-faced newbie again.
> 
> I've read around in the docs and done a bit of google, but haven't managed 
> to understand:
> 
> Python 2.3.4 (#53, May 25 2004, 21:17:02)
>  >>> r = [1, 2, 3]
>  >>> s = r[:].remove(2)
>  >>> print s
> None
>  >>> t = r[:]
>  >>> t.remove(2)
>  >>> print t
> [1, 3]
>  >>>
> 
> Both times I expected the second procedure's result. I get that remove 
> modifies in place, but it seems to me like the condensed version should 
> work. My thinking is that it first creates a copy, then modifies the copy 
> in place, then assigns the modified copy to s. 

That would happen if remove returned self.  However, it returns None. 
So s is set to None.  2 is removed from r[:], but that object is no
longer accessible since it is not bound to any name.

> But it would seem that 
> that's not what happens.
> 
> The 2 questions then are: why doesn't the first way work?, and, in the 
> first way, from what (if anything) is 2 being removed?
> 
> Pointers to the doc's explanation that I must have overlooked are welcome.
> 
> Thanks and best to all,
> 
> Brian vdB
> 
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-- 
Lloyd Kvam
Venix Corp