Assigning values to an array of arbitrary dimensions in MATLAB.
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Michael Ransom
on 16 Mar 2019
Answered: Christine Tobler
on 17 Jul 2020
I have a problem that requires assigning values to an array of arbitrary dimensions.
For a given integer j, I pre-allocate a regular j-dimensional array such that
where are integers for . Such an array in MATLAB looks like:
Array = NaN.*ones( repmat( d , 1 , j ) ) ;
The problem is assigning values for elements in this array. Specifically, given are vectors of length , I require all combinations of element-wise multiplication for each of these vectors, where each result is assigned to .
For the simplest case , assuming vectors are stored in cell-arrays, I can do this quite easily using loops:
for m = 1:d
for n = 1:d
Array( m , n ) = v{ 1 }( m ).*v{ 2 }( n ) ;
end
end
But, a "psudeo-code" implementation with j loops would look something like:
for m( 1 ) = 1:d
...
for m( j ) = 1:d
Array( m( 1 ) , ... , m( j ) ) = v{ 1 }( m( j ) ).* ... .*v{ j }( m( j ) ) ;
end
...
end
The indexing of now becomes variable, which I have had no success in implementing.
Does there exist a way of assigning values to an array of arbitrary dimensions in MATLAB in this manner, or perhaps a neater method?
Accepted Answer
per isakson
on 17 Mar 2019
Edited: per isakson
on 18 Mar 2019
"The problem I have is to be able to this with an arbitrary value of j." I think value2arrayND does that. It takes a cell array, {dim1,dim2,dim3,...}, and returns an array, the size of which is [dim1,dim2,dim3,...]. value2array2D and value2array3D are based on the code of your comment. They are used to produce expected values.
>> a2 = value2array2D( );
>> aN2 = value2arrayND( {5,5} );
>> aN2-a2
ans =
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
>> a3 = value2array3D( );
>> aN3 = value2arrayND( {5,5,5} );
>> aN3(:,:,3)-a3(:,:,3)
ans =
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
where
function array = value2arrayND( d )
rng('default'); % for reproducing rand;
N = length(d);
v = cell( 1, N );
for jj = 1 : N
v{ jj } = rand( d{jj}, 1 );
end
array = nan(d{:});
nel = numel( array );
sz = [d{:}];
dim = cell( 1, N );
for ii = 1 : nel
[dim{:}] = ind2sub( sz, ii );
array(ii) = prod( arrayfun( @(jj) v{jj}(dim{jj}), 1:N, 'uni',true ) );
end
end
and
function array = value2array3D( )
rng('default'); % for reproducing rand;
d = { 5, 5, 5 }; % vector length; arbitrary
v{ 1 } = rand( d{1}, 1 ); % vector 1
v{ 2 } = rand( d{2}, 1 ); % vector 2
v{ 3 } = rand( d{3}, 1 ); % vector 3
array = nan(d{:});
for p = 1 : d{1}
for q = 1 : d{2}
for r = 1 : d{3}
array( p, q, r ) = v{ 1 }( p ).*v{ 2 }( q ).*v{ 3 }( r );
end
end
end
end
and
function array = value2array2D( )
rng('default'); % for reproducing rand;
d = { 5, 5 }; % vector length; arbitrary
v{ 1 } = rand( d{1}, 1 ); % vector 1
v{ 2 } = rand( d{2}, 1 ); % vector 2
array = nan(d{:});
for m = 1 : d{1}
for n = 1 : d{2}
array( m, n ) = v{ 1 }( m ).*v{ 2 }( n );
end
end
end
2 Comments
dpb
on 17 Mar 2019
+1 Per...was thinking on those lines when made initial comment; it came to me overnight precisely how and I was all excited this AM to post my clever solution! Then, you beat me to it! :)
More Answers (1)
Christine Tobler
on 17 Jul 2020
Using the somewhat recent implicit expansion, this can also be done without indexing into every element of the array:
function Array = value2arrayND(v)
Array = 1;
for ii=1:length(v)
% Turn d-by-1 vector into 1-by-...-by-1-by-d vector, where d is in dimension ii
viiPermuted = permute(v{ii}(:), [2:ii 1 ii+1:length(v)]);
Array = Array .* viiPermuted;
end
For the 2D case, this corresponds to
Array = v{1}(:) .* v{2}(:).'; % (d-by-1 matrix) .* (1-by-d matrix)
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