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find previous & next in array

John Griffiths

5/14/2009 3:45:00 AM

hi, anyone know how to do this?

basically i've got an array of items

a = ['a','b','c','d']

now given i can find the position of say 'c' using .index('c'), how do i
get the positions of the ones previous & next to it.

a recursive routine would probably do it, but looking for something more
flash

any ideas?
--
Posted via http://www.ruby-....
6 Answers

Boby Selamet Hartono

5/14/2009 3:48:00 AM

0

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

2009/5/14 John Griffiths <indiehead@gmail.com>

> hi, anyone know how to do this?
>
> basically i've got an array of items
>
> a = ['a','b','c','d']
>
> now given i can find the position of say 'c' using .index('c'), how do i
> get the positions of the ones previous & next to it.
>
> a recursive routine would probably do it, but looking for something more
> flash
>
> any ideas?
> --
> Posted via http://www.ruby-....
>
>
a[a.index(:c)-1] or a[a.index(:c)+1]?

--
Tidak ada yang lebih baik dari kembali ke asal
Nothing can be better than back to the roots

Joshua Collins

5/14/2009 6:55:00 AM

0

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

Boby had it almost right!

instead of:

a[a.index(:c)+1]

it is:

a[a.index('c')+1]

:-)

On Wed, May 13, 2009 at 11:48 PM, Boby Selamet Hartono
<vecciora@gmail.com>wrote:

> 2009/5/14 John Griffiths <indiehead@gmail.com>
>
> > hi, anyone know how to do this?
> >
> > basically i've got an array of items
> >
> > a = ['a','b','c','d']
> >
> > now given i can find the position of say 'c' using .index('c'), how do i
> > get the positions of the ones previous & next to it.
> >
> > a recursive routine would probably do it, but looking for something more
> > flash
> >
> > any ideas?
> > --
> > Posted via http://www.ruby-....
> >
> >
> a[a.index(:c)-1] or a[a.index(:c)+1]?
>
> --
> Tidak ada yang lebih baik dari kembali ke asal
> Nothing can be better than back to the roots
>

Boby Selamet Hartono

5/14/2009 7:34:00 AM

0

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

2009/5/14 Joshua Collins <kidguko@gmail.com>

> Boby had it almost right!
>
> instead of:
>
> a[a.index(:c)+1]
>
> it is:
>
> a[a.index('c')+1]
>
> :-)
>
> On Wed, May 13, 2009 at 11:48 PM, Boby Selamet Hartono
> <vecciora@gmail.com>wrote:
>
> > 2009/5/14 John Griffiths <indiehead@gmail.com>
> >
> > > hi, anyone know how to do this?
> > >
> > > basically i've got an array of items
> > >
> > > a = ['a','b','c','d']
> > >
> > > now given i can find the position of say 'c' using .index('c'), how do
> i
> > > get the positions of the ones previous & next to it.
> > >
> > > a recursive routine would probably do it, but looking for something
> more
> > > flash
> > >
> > > any ideas?
> > > --
> > > Posted via http://www.ruby-....
> > >
> > >
> > a[a.index(:c)-1] or a[a.index(:c)+1]?
> >
> > --
> > Tidak ada yang lebih baik dari kembali ke asal
> > Nothing can be better than back to the roots
> >
>

It was a copy from my irb. I'm use symbols as objects in my array. I've
never use index method but since I figure out that method returns an Integer
I think it would be something like that (thanks to irb). I'm a newbie and
just joining this ML yesterday :-).

--
Tidak ada yang lebih baik dari kembali ke asal
Nothing can be better than back to the roots

Robert Klemme

5/14/2009 7:46:00 AM

0

2009/5/14 John Griffiths <indiehead@gmail.com>:
> hi, anyone know how to do this?
>
> basically i've got an array of items
>
> a = ['a','b','c','d']
>
> now given i can find the position of say 'c' using .index('c'), how do i
> get the positions of the ones previous & next to it.
>
> a recursive routine would probably do it, but looking for something more
> flash
>
> any ideas?

Depends on what you need to do. Do you need those indexes or do you
only need access to those elements?

There are a few variants, but the nice ones aren't really safe against
edge cases (first and last element):

irb(main):004:0> a = ['a','b','c','d']
=> ["a", "b", "c", "d"]
irb(main):005:0> i = a.index('c') and a[i-1 .. i+1]
=> ["b", "c", "d"]
irb(main):006:0> a[a.index('c') - 1, 3] rescue nil
=> ["b", "c", "d"]
irb(main):007:0> i = a.index('a') and a[i-1 .. i+1]
=> []
irb(main):008:0> a[a.index('a') - 1, 3] rescue nil
=> ["d"]

The safest is probably

i = a.index('c')

if i
predecessor = i > 0 ? a[i -1] : nil
successor = a[i + 1] # safe!
end

Or as a variant

i = a.index('c') and begin
predecessor = i > 0 ? a[i -1] : nil
successor = a[i + 1] # safe!
end

Kind regards

robert

--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestprac...

John Griffiths

5/14/2009 3:38:00 PM

0

> The safest is probably
>
> i = a.index('c')
>
> if i
> predecessor = i > 0 ? a[i -1] : nil
> successor = a[i + 1] # safe!
> end
>
> Or as a variant
>
> i = a.index('c') and begin
> predecessor = i > 0 ? a[i -1] : nil
> successor = a[i + 1] # safe!
> end
>
> Kind regards
>
> robert

thanks Robert, i was trying to build up a little function to find the
prev/next elements in an array, thanks to your help it's sorted;
couldn't find a solution online. appreciate it all.


John.
--
Posted via http://www.ruby-....

Joshua Collins

5/14/2009 5:35:00 PM

0

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

Bobby, I am new too, so I might be missing something as well.

I tried your method first, and when I tried to do the symbol in irb it gave
an error. Maybe you have settings I do not?

On Thu, May 14, 2009 at 11:38 AM, John Griffiths <indiehead@gmail.com>wrote:

> > The safest is probably
> >
> > i = a.index('c')
> >
> > if i
> > predecessor = i > 0 ? a[i -1] : nil
> > successor = a[i + 1] # safe!
> > end
> >
> > Or as a variant
> >
> > i = a.index('c') and begin
> > predecessor = i > 0 ? a[i -1] : nil
> > successor = a[i + 1] # safe!
> > end
> >
> > Kind regards
> >
> > robert
>
> thanks Robert, i was trying to build up a little function to find the
> prev/next elements in an array, thanks to your help it's sorted;
> couldn't find a solution online. appreciate it all.
>
>
> John.
> --
> Posted via http://www.ruby-....
>
>