String building using join

[gervaz]

On 2 Gen, 22:37, gervaz <ger... at gmail.com> wrote:
> On 2 Gen, 19:14, Emile van Sebille <em... at fenx.com> wrote:
>
>
>
>
>
> > On 1/2/2011 9:43 AM gervaz said...
>
> > > On 31 Dic 2010, 16:43, Emile van Sebille<em... at fenx.com>  wrote:
> > >> On 12/31/2010 7:22 AM gervaz said...
>
> > >>> Hi all, I would like to ask you how I can use the more efficient join
> > >>> operation in a code like this:
>
> > >>>>>> class Test:
> > >>> ...     def __init__(self, v1, v2):
> > >>> ...         self.v1 = v1
> > >>> ...         self.v2 = v2
> > >>> ...
> > >>>>>> def prg(l):
> > >>> ...     txt = ""
> > >>> ...     for x in l:
> > >>> ...         if x.v1 is not None:
> > >>> ...             txt += x.v1 + "\n"
> > >>> ...         if x.v2 is not None:
> > >>> ...             txt += x.v2 + "\n"
> > >>> ...     return txt
> > >>> ...
> > >>>>>> t1 = Test("hello", None)
> > >>>>>> t2 = Test(None, "ciao")
> > >>>>>> t3 = Test("salut", "hallo")
> > >>>>>> t = [t1, t2, t3]
>
> > >>>>>> prg(t)
> > >>> 'hello\nciao\nsalut\nhallo\n'
>
> > >>> The idea would be create a new list with the values not None and then
> > >>> use the join function... but I don't know if it is really worth it.
> > >>> Any hint?
>
> > >>>>>> def prg2(l):
>
> > >>               return "\n".join([x for x in l if x])
>
> > >> Emile
>
> > >>> ...     e = []
> > >>> ...     for x in l:
> > >>> ...         if x.v1 is not None:
> > >>> ...             e.append(x.v1)
> > >>> ...         if x.v2 is not None:
> > >>> ...             e.append(x.v2)
> > >>> ...     return "\n".join(e)
> > >>> ...
> > >>>>>> prg2(t)
> > >>> 'hello\nciao\nsalut\nhallo'
>
> > >>> Thanks, Mattia- Nascondi testo citato
>
> > >> - Mostra testo citato -- Nascondi testo citato
>
> > >> - Mostra testo citato -
>
> > > Sorry, but it does not work
>
> > Oh -- you want a working solution, not a hint?  OK.
>
> > class Test:
> >       def __init__(self, v1, v2):
> >           self.v1 = v1
> >           self.v2 = v2
>
> > t1 = Test("hello", None)
> > t2 = Test(None, "ciao")
> > t3 = Test("salut", "hallo")
> > t = [t1, t2, t3]
>
> > "\n".join([y for x in t for y in [x.v1,x.v2] if y])
>
> > Emile
>
> > >>>> def prg3(l):
> > > ...     return "\n".join([x for x in l if x])
> > > ...
> > >>>> prg3(t)
> > > Traceback (most recent call last):
> > >    File "<stdin>", line 1, in<module>
> > >    File "<stdin>", line 2, in prg3
> > > TypeError: sequence item 0: expected str instance, Test found- Nascondi testo citato
>
> > - Mostra testo citato -- Nascondi testo citato
>
> > - Mostra testo citato -
>
> Thanks Emile, despite that now the solution runs in quadratic time I
> guess. I could also provide a __str__(self) representation, but in my
> real code I don't have access to the class. Also using str() on an
> empty object (i.e. None), the representation is 'None'.
>
> Ciao,
>
> Mattia- Nascondi testo citato
>
> - Mostra testo citato -

And this one is another working solution...

>>> def prg4(l):
...     s = []
...     for x in l:
...         s.extend(set([x.v1, x.v2]) - set([None]))
...     return "\n".join(s)
...
>>> prg4(t)
'hello\nciao\nhallo\nsalut'

My original question was just related to the fact that I read that the
string concatenation in expensive and it sould be used the join()
function but now probably it is better to stick with the first
implementation, the simplest one.

Ciao,

Mattia