select entries and put in a matrix without loop

Question

I have the following:

   b = [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]

whose dimensions are variable.

b{1}

ans =

 0     0     0     0

I want to put the first entry of each of the 10 vectors as the first column of matrix A

2nd column of matrix A will be as v the 1st entry of each of the 10 vectors of r:

r = 

[1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]    [1x4 double]

r{1} --> ans = 10 10 10 10

This is what i need to get:

A = 

    v{1}(1)   r{1}(1)
    v{2}(1)   r{2}(1)
    v{3}(1)   r{3}(1)

How to do that without a loop is there a way?

I'm not sure I understand this question but look into the Matlab function, cellfun (http://www.mathworks.com/help/techdoc/ref/cellfun.html) — Dan, Apr 25 '12 at 06:15
The title says "select entries and put in a **cell** without loop", but I assume that you want a put the entries in a **matrix**, as you explain so in your question. Should `A` be a **matrix**, not a **cell**? — nrz, Apr 25 '12 at 07:49

nrz · Accepted Answer · 2012-04-25T07:40:48.077

Some example data:

b = {[ 101:104 ], [ 201:204 ], [ 301:304 ], [ 401:404 ], [ 501:504 ], [ 601:604 ], [ 701:704 ], [ 801:804 ], [ 901:904 ], [ 1001:1004 ]};

r = {[ 2101:2104 ], [ 2201:2204 ], [ 2301:2304 ], [ 2401:2404 ], [ 2501:2504 ], [ 2601:2604 ], [ 2701:2704 ], [ 2801:2804 ], [ 2901:2904 ], [ 3001:3004 ]};

Edit: a lot faster solution without looping by using vertcat. Edit: corrected a typo in code.

bMatrix = vertcat(b{:});
rMatrix = vertcat(r{:});
A = [ bMatrix(:,1), rMatrix(:,1) ];

A lot slower solution by using cellfun (cellfun does loop) :

A = [ cellfun(@(x) x(1), b)', cellfun(@(x) x(1), r)' ];

Or in parts:

ColumnOneOfMatrixA = cellfun(@(x) x(1), b)';
ColumnTwoOfMatrixA = cellfun(@(x) x(1), r)';
A = [ ColumnOneOfMatrixA, ColumnTwoOfMatrixA ];

Both ways give the same result.

A =
     101        2101
     201        2201
     301        2301
     401        2401
     501        2501
     601        2601
     701        2701
     801        2801
     901        2901
    1001        3001

score 0 · Answer 2 · answered Apr 25 '12 at 07:15

As Dan notes, cellfun is the trick to avoiding this loop.

%Setup test data
for ix = 1:10
    b{ix} = ones(1,4)*(ix-1);
    r{ix} = ones(1,4)*(ix+9);
end

%Cellfun based definition of the "A" matrix
A = [...
    cellfun(  @(x)x(1),  b(1:10)  ); ...
    cellfun(  @(x)x(1),  r(1:10)  ); ...
    ]';

Here the cellfun calls have been set up to return a numeric array containikng the first element of each numeric array in the cell array. The anonymous function @(x)x(1) serves as the core, just returning the first element, and cellfun takes care of implementing the appropriate looping without bothering you with the details.

Note that cellfun is usually not any faster than the loop that it replaces. It simply requires less typing, and is arguably easier to read after you learn to work with it.

select entries and put in a matrix without loop

2 Answers2