On 08/09/2010 17.50, Roelof Wobben wrote:
> Subject: Re: [Tutor] sort problem
> From: evert....@gmail.com
> Date: Wed, 8 Sep 2010 17:26:58 +0200
> CC: tutor@python.org
> To: rwob...@hotmail.com
...
> > seq2 = list(seq)
> > seq2.sort()
> > print seq2
> > seq.join(seq2)
> > return seq
> >
> > The problem is that if I want to sort the characters in a string,
the list exist of the sorted characters but as soon as I convert them to
a string I get the old string.
Are you sure that you really get your old string? I would expect
something like:
seq = "cba"
seq2 = ["a", "b", "c"]
seq.join(seq2) => "acbabcbac"
that is, all the characters from seq2 separated by copies of seq.
Evert gave you a good advice:
> Carefully read the documentation for str.join:
http://docs.python.org/library/stdtypes.html#str.join
>
> How does it work, what does it return, etc. Then fix the
corresponding line in your code.
> As a hint: str.join does work quite different than list.sort; I
assume you're confusing their syntaxes.
>
> Good luck,
>
> Evert
>
str.join(/iterable/)ΒΆ <#str.join>
How it works.
It puts all the elements of iterable into one string named str.
So it returns a string.
Str is here seq and the iterable is the list made by list.sort so seq2
So I don't see the error in that line.
What does join use as a separator between the elements it joins?
Roelof
Francesco
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