Assume I have:

for user in @Users
put user.name
put user.name #I want this to be next object in @Users
end

I want the second user.name to be the next object in the list. Is
there a way to do this?

Thanks in advance,
LP

On Wed, 5 Dec 2007 05:35:01 +0900, "lpgauth@gmail.com" <lpgauth@gmail.com> wrote:
> Assume I have:
>
> for user in @Users
> put user.name
> put user.name #I want this to be next object in @Users
> end
>
> I want the second user.name to be the next object in the list. Is
> there a way to do this?

Of course. But what do you want to happen when you're at the end
of the list and there is no longer a next object?

-mental

On Wed, 5 Dec 2007 05:42:21 +0900, MenTaLguY <mental@rydia.net> wrote:
> On Wed, 5 Dec 2007 05:35:01 +0900, "lpgauth@gmail.com" <lpgauth@gmail.com>
> wrote:
>> Assume I have:
>>
>> for user in @Users
>> put user.name
>> put user.name #I want this to be next object in @Users
>> end
>>
>> I want the second user.name to be the next object in the list. Is
>> there a way to do this?
>
> Of course. But what do you want to happen when you're at the end
> of the list and there is no longer a next object?

That is, if you only care about pairs of users, then you can do this:

prior = nil
for user in @Users
if prior
put prior.name
put user.name
end
prior = user
end

Obviously that won't print anything if there is only one entry in
@Users, though.

-mental

On Dec 4, 4:19 pm, MenTaLguY <men...@rydia.net> wrote:
> On Wed, 5 Dec 2007 05:42:21 +0900, MenTaLguY <men...@rydia.net> wrote:
> > On Wed, 5 Dec 2007 05:35:01 +0900, "lpga...@gmail.com" <lpga...@gmail.com>
> > wrote:
> >> Assume I have:
>
> >> for user in @Users
> >> put user.name
> >> put user.name #I want this to be next object in @Users
> >> end
>
> >> I want the second user.name to be the next object in the list. Is
> >> there a way to do this?
>
> > Of course. But what do you want to happen when you're at the end
> > of the list and there is no longer a next object?
>
> That is, if you only care about pairs of users, then you can do this:
>
> prior = nil
> for user in @Users
> if prior
> put prior.name
> put user.name
> end
> prior = user
> end
>
> Obviously that won't print anything if there is only one entry in
> @Users, though.
>
> -mental

I figured it out... You can use in_group_of

On Dec 4, 2007 4:55 PM, lpgauth@gmail.com <lpgauth@gmail.com> wrote:
> On Dec 4, 4:19 pm, MenTaLguY <men...@rydia.net> wrote:
> > On Wed, 5 Dec 2007 05:42:21 +0900, MenTaLguY <men...@rydia.net> wrote:
> > > On Wed, 5 Dec 2007 05:35:01 +0900, "lpga...@gmail.com" <lpga...@gmail.com>
> > > wrote:
> > >> Assume I have:
> >
> > >> for user in @Users
> > >> put user.name
> > >> put user.name #I want this to be next object in @Users
> > >> end
> >
> > >> I want the second user.name to be the next object in the list. Is
> > >> there a way to do this?
> >
> > > Of course. But what do you want to happen when you're at the end
> > > of the list and there is no longer a next object?
> >
> > That is, if you only care about pairs of users, then you can do this:
> >
> > prior = nil
> > for user in @Users
> > if prior
> > put prior.name
> > put user.name
> > end
> > prior = user
> > end
> >
> > Obviously that won't print anything if there is only one entry in
> > @Users, though.
> >
> > -mental
>
> I figured it out... You can use in_group_of
>
>


Cool! That's from Rails (not part of Ruby), but there's also each_slice(n):

require 'enumerable'

a = [1, 2, 3, 4]

a.each_slice(2) {|slice| puts slice[0] + slice[1]}

=>

3
7

Christian von Kleist wrote:
> Cool! That's from Rails (not part of Ruby), but there's also
> each_slice(n):
>
> require 'enumerable'
>
> a = [1, 2, 3, 4]
>
> a.each_slice(2) {|slice| puts slice[0] + slice[1]}

You can also use each_with_index:

users.each_with_index do |user, index|
puts user
if index < users.length
puts users[index + 1]
end
end
--
Posted via http://www.ruby-forum.com/.

On Dec 4, 2007 9:34 PM, Chris Fisher <fourfifty@gmail.com> wrote:
> Christian von Kleist wrote:
> > Cool! That's from Rails (not part of Ruby), but there's also
> > each_slice(n):
> >
> > require 'enumerable'
> >
> > a = [1, 2, 3, 4]
> >
> > a.each_slice(2) {|slice| puts slice[0] + slice[1]}
>
> You can also use each_with_index:
>
> users.each_with_index do |user, index|
> puts user
> if index < users.length
> puts users[index + 1]
> end
> end
> --
> Posted via http://www.ruby-forum.com/.
>
>

I think using each_with_index that way will give behavior like each_cons.

On 04.12.2007 23:18, Christian von Kleist wrote:
> On Dec 4, 2007 4:55 PM, lpgauth@gmail.com <lpgauth@gmail.com> wrote:
>> On Dec 4, 4:19 pm, MenTaLguY <men...@rydia.net> wrote:
>>> On Wed, 5 Dec 2007 05:42:21 +0900, MenTaLguY <men...@rydia.net> wrote:
>>>> On Wed, 5 Dec 2007 05:35:01 +0900, "lpga...@gmail.com" <lpga...@gmail.com>
>>>> wrote:
>>>>> Assume I have:
>>>>> for user in @Users
>>>>> put user.name
>>>>> put user.name #I want this to be next object in @Users
>>>>> end
>>>>> I want the second user.name to be the next object in the list. Is
>>>>> there a way to do this?
>>>> Of course. But what do you want to happen when you're at the end
>>>> of the list and there is no longer a next object?
>>> That is, if you only care about pairs of users, then you can do this:
>>>
>>> prior = nil
>>> for user in @Users
>>> if prior
>>> put prior.name
>>> put user.name
>>> end
>>> prior = user
>>> end
>>>
>>> Obviously that won't print anything if there is only one entry in
>>> @Users, though.
>>>
>>> -mental
>> I figured it out... You can use in_group_of
>>
>>
>
>
> Cool! That's from Rails (not part of Ruby), but there's also each_slice(n):
>
> require 'enumerable'
>
> a = [1, 2, 3, 4]
>
> a.each_slice(2) {|slice| puts slice[0] + slice[1]}

I'd probably rather use #each_cons:

irb(main):001:0> %w{a b c}.each_cons(2) {|*a| p a}
[["a", "b"]]
[["b", "c"]]
=> nil

Cheers

robert