Producing 2D array from a 1D array in MATLAB

Question

Does anyone know if there is a way to produce a 2D array from a 1D array, where the rows in the 2D are generated by repeating the corresponding elements in the 1D array.

I.e.:

1D array      2D array

  |1|       |1 1 1 1 1|
  |2|       |2 2 2 2 2|
  |3|  ->   |3 3 3 3 3|
  |4|       |4 4 4 4 4|
  |5|       |5 5 5 5 5|

Amro · Accepted Answer · 2010-02-05 20:19:55Z

9

In the spirit of bonus answers, here are some of my own:

Let A = (1:5)'

Using indices [faster than repmat]:
```
B = A(:, ones(5,1))
```
Using matrix outer product:
```
B = A*ones(1,5)
```

Using bsxfun() [not the best way of doing it]

B = bsxfun(@plus, A, zeros(1,5))
%# or
B = bsxfun(@times, A, ones(1,5))

edited Feb 5, 2010 at 20:19

answered Feb 5, 2010 at 20:11

Amro

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1 Comment

Amro Over a year ago

@merv: you can read more about these "Techniques for Improving Performance" at mathworks.com/access/helpdesk/help/techdoc/matlab_prog/… @woodchips: the first example you gave is the same as the one given by gnovice. The second is not really a general repmat alternative, only a special case for this particular A..

gnovice · Accepted Answer · 2010-02-05 18:39:26Z

8

You can do this using the REPMAT function:

>> A = (1:5).'

A =

     1
     2
     3
     4
     5

>> B = repmat(A,1,5)

B =

     1     1     1     1     1
     2     2     2     2     2
     3     3     3     3     3
     4     4     4     4     4
     5     5     5     5     5

EDIT: BONUS ANSWER! ;)

For your example, REPMAT is the most straight-forward function to use. However, another cool function to be aware of is KRON, which you could also use as a solution in the following way:

B = kron(A,ones(1,5));

For small vectors and matrices KRON may be slightly faster, but it is quite a bit slower for larger matrices.

edited Feb 5, 2010 at 18:39

answered Feb 5, 2010 at 18:05

gnovice

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Comments

Ofri Raviv · Accepted Answer · 2010-02-05 18:07:34Z

1

repmat(a, [1 n]), but you should also take a look at meshgrid.

answered Feb 5, 2010 at 18:07

Ofri Raviv

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Comments

zdav · Accepted Answer · 2010-02-05 18:08:37Z

0

You could try something like:

a = [1 2 3 4 5]'
l = size(a)
for i=2:5
    a(1:5, i) = a(1:5)

The loop just keeps appending columns to the end.

answered Feb 5, 2010 at 18:08

zdav

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2 Comments

Doresoom Over a year ago

For small matrices this solution is faster, but for larger sizes repmat is a much better solution. (With a 1000x1000 matrix, repmat is 500+ times faster!)

zdav Over a year ago

I agree, my solution is a pretty naive on, using repmat is a much better/more elegant solution in general.

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