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comp.lang.ruby

Troubles with blocks

Holden Holden

1/3/2008 11:50:00 PM

Hi,

I am a Python programmer and I've just started learning Ruby. Until now,
I've enjoyed it, but I am having troubles with some methods that require
blocks.

This is a simple example that (I expect) summarizes my problems - how to
obtain a list of pairs [index, value] (or [value, index], it doesn't
matter) from an array:

Python:

list(enumerate(['a','b','c']))
=> [(0, 'a'), (1, 'b'), (2, 'c')]

Ruby:

t = []; ['a','b','c'].each_with_index { |*x| t << x }; t
=> [["a", 0], ["b", 1], ["c", 2]]

Which is a very ugly solution. How could I do this on Ruby without a
temporal variable?

thanks
--
Posted via http://www.ruby-....
6 Answers

Jason Roelofs

1/4/2008 12:06:00 AM

0

[Note: parts of this message were removed to make it a legal post.]

On Jan 3, 2008 6:49 PM, Holden Holden <pyarnau@gmail.com> wrote:

> Hi,
>
> I am a Python programmer and I've just started learning Ruby. Until now,
> I've enjoyed it, but I am having troubles with some methods that require
> blocks.
>
> This is a simple example that (I expect) summarizes my problems - how to
> obtain a list of pairs [index, value] (or [value, index], it doesn't
> matter) from an array:
>
> Python:
>
> list(enumerate(['a','b','c']))
> => [(0, 'a'), (1, 'b'), (2, 'c')]
>
> Ruby:
>
> t = []; ['a','b','c'].each_with_index { |*x| t << x }; t
> => [["a", 0], ["b", 1], ["c", 2]]
>
> Which is a very ugly solution. How could I do this on Ruby without a
> temporal variable?
>
> thanks
> --
> Posted via http://www.ruby-....
>
>
Hmm, I can't find a shorter way than what you've got. Why are you trying to
get this kind of information? Maybe there's something earlier in your code
that can be made more Ruby-esque to make this easier for you.

Jason

Gary Wright

1/4/2008 12:09:00 AM

0


On Jan 3, 2008, at 6:49 PM, Holden Holden wrote:
> Ruby:
>
> t = []; ['a','b','c'].each_with_index { |*x| t << x }; t
> => [["a", 0], ["b", 1], ["c", 2]]
>
> Which is a very ugly solution. How could I do this on Ruby without a
> temporal variable?

In Ruby 1.8:

require 'enumerator'
%w{a b c}.enum_for(:each_with_index).to_a


In Ruby 1.9:

%w{a b c}.each_with_index.to_a

Gary Wright

Denis Hennessy

1/4/2008 12:11:00 AM

0



MenTaLguY

1/4/2008 12:13:00 AM

0

On Fri, 4 Jan 2008 08:49:54 +0900, Holden Holden <pyarnau@gmail.com> wrote:
> Python:
>
> list(enumerate(['a','b','c']))
> => [(0, 'a'), (1, 'b'), (2, 'c')]
>
> Ruby:
>
> t = []; ['a','b','c'].each_with_index { |*x| t << x }; t
> => [["a", 0], ["b", 1], ["c", 2]]
>
> Which is a very ugly solution. How could I do this on Ruby without a
> temporal variable?

Try this (you will need to require 'enumerator'):

['a','b','c'].enum_with_index.to_a

-mental

Jason Roelofs

1/4/2008 12:18:00 AM

0

[Note: parts of this message were removed to make it a legal post.]

On Jan 3, 2008 7:13 PM, MenTaLguY <mental@rydia.net> wrote:

> On Fri, 4 Jan 2008 08:49:54 +0900, Holden Holden <pyarnau@gmail.com>
> wrote:
> > Python:
> >
> > list(enumerate(['a','b','c']))
> > => [(0, 'a'), (1, 'b'), (2, 'c')]
> >
> > Ruby:
> >
> > t = []; ['a','b','c'].each_with_index { |*x| t << x }; t
> > => [["a", 0], ["b", 1], ["c", 2]]
> >
> > Which is a very ugly solution. How could I do this on Ruby without a
> > temporal variable?
>
> Try this (you will need to require 'enumerator'):
>
> ['a','b','c'].enum_with_index.to_a
>
> -mental
>
>
Ah, that's where enum_with_index is. I was wondering why it wasn't working
for me.

Jason

Holden Holden

1/4/2008 12:18:00 AM

0

Gary Wright wrote:

> In Ruby 1.8:
>
> require 'enumerator'
> %w{a b c}.enum_for(:each_with_index).to_a
>
> In Ruby 1.9:
>
> %w{a b c}.each_with_index.to_a
>

Thanks! that's exactly what I was looking for. I had already tried the
last solution but I got a "no block given" error (I'm using Ruby 1.8).
I'm glad to hear that versions 1.9 and higher support this construct.
--
Posted via http://www.ruby-....