# Initializing an NxN array

Discussion in 'Ruby' started by Ruby Student, Mar 19, 2009.

1. ### Ruby StudentGuest

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

Team,

What is the easiest way to initialize a square matrix?
For example, to initialize a 3x3 array elements to 0, I am doing what you
see below. But I am not sure how to proceed if, for instance, I want a NxN
array where N > 10 or a huge value?
I played a bit on IRB but could not find the way to do it easily.

irb(main):008:0> ary = Array.new(9,[0,0,0])
=> [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0,
0], [0, 0, 0], [0, 0, 0]]

Thank you

--
Ruby Student

Ruby Student, Mar 19, 2009

2. ### matt neuburgGuest

matt neuburg, Mar 19, 2009

Ruby Student wrote:
> Team,
>
> What is the easiest way to initialize a square matrix?
> For example, to initialize a 3x3 array elements to 0, I am doing what
> you
> see below. But I am not sure how to proceed if, for instance, I want a
> NxN
> array where N > 10 or a huge value?
> I played a bit on IRB but could not find the way to do it easily.
>
> irb(main):008:0> ary = Array.new(9,[0,0,0])
> => [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0],
> [0, 0,
> 0], [0, 0, 0], [0, 0, 0]]
>
> Thank you

Wouldn't the best solution be:

require 'Matrix'
a = Matrix.zero(3)

?

Assuming you actually want to use it for Matrix math and not as just a
2D array.

If you want a 2D array, I'd do ary = Array.new(3) {|row| Array.new(3)
{|col| 0}}
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4. ### MarceloGuest

On Thu, Mar 19, 2009 at 10:05, Ruby Student <> wrote:

> For example, to initialize a 3x3 array elements to 0, I am doing what you
> see below. But I am not sure how to proceed if, for instance, I want a NxN
> array where N > 10 or a huge value?

N=3
Array.new(N) { Array.new(N,0) }

-m.

Marcelo, Mar 19, 2009
5. ### Evan FarrarGuest

Since * is defined on array, you can start with a 1x1 array and multiply
to the size you need.

n = 3;
ary = [[0]*n]*n

results in:
ary => [[0, 0, 0], [0, 0, 0], [0, 0, 0]]
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Evan Farrar, Mar 19, 2009
6. ### Evan FarrarGuest

Evan Farrar wrote:
> Since * is defined on array, you can start with a 1x1 array and multiply
> to the size you need.
>
>
> n = 3;
> ary = [[0]*n]*n
>
> results in:
> ary => [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

whoops! that has the same problem mentioned above:
irb(main):026:0> ary[0][0] = :foo; ary
=> [[:foo, 0, 0], [:foo, 0, 0], [:foo, 0, 0]]

I retract!
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Evan Farrar, Mar 19, 2009
7. ### Ruby StudentGuest

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

On Thu, Mar 19, 2009 at 1:36 PM, Evan Farrar <> wrote:

> Evan Farrar wrote:
> > Since * is defined on array, you can start with a 1x1 array and multiply
> > to the size you need.
> >
> >
> > n = 3;
> > ary = [[0]*n]*n
> >
> > results in:
> > ary => [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

>
> whoops! that has the same problem mentioned above:
> irb(main):026:0> ary[0][0] = :foo; ary
> => [[:foo, 0, 0], [:foo, 0, 0], [:foo, 0, 0]]
>
> I retract!
> --
> Posted via http://www.ruby-forum.com/.
>
>

Thanks to all, your help is appreicated!

--
Ruby Student

Ruby Student, Mar 19, 2009