question about list extension

[Bruno Desthuilliers]

J a écrit :
> Ok... I know pretty much how .extend works on a list... basically it
> just tacks the second list to the first list... like so:
> 
>>>> lista=[1]
>>>> listb=[2,3]
>>>> lista.extend(listb)
>>>> print lista;
> [1, 2, 3]
> 
> what I'm confused on is why this returns None:

> So why the None? Is this because what's really happening is that
> extend() and append() are directly manipulating the lista object and
> thus not actuall returning a new object?

Exactly.

> Even if that assumption of mine is correct, I would have expected
> something like this to work:
> 
>>>> lista=[1]
>>>> listb=[2,3]
>>>> print (lista.extend(listb))
> None

So what ? It JustWork(tm). list.extend returns None, so "None" is 
printed !-)


(snip)

> So changing debugger.printout() to:
> 
>    def printout(self,*info):
> 
> lets me pass in multiple lists... and I can do this:
> 
>    for tlist in info:
>        for item in tlist:
>             print item
> 
> which works well...
> 
> So, what I'm curious about, is there a list comprehension or other
> means to reduce that to a single line?

Why do you think the above code needs to be "reduced to a single line" ? 
  What's wrong with this snippet ???


> It's more of a curiosity thing at this point...  and not a huge
> difference in code... I was just curious about how to make that work.

Uh, ok.

What about:

from itertools import chain

def printout(*infos):
    print "\n".join("%s" % item for item in chain(*infos))


HTH