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matlabmatrix
Maddy
Maddy

Reputation: 2570

repeat values in a row 'N' number of times and then increment the values - Matlab (varying N)

I have a matrix:

mat = [  2009 3 ;
         2010 2 ] ;

I need to repeat the Col1 as per Col2. The solution at Repeat copies of array elements: Run-length decoding in MATLAB is helpful. However, my main problem is to then increment the years according, as in:

Ansmat = [  2009 3
            2010 3
            2011 3
            2010 2
            2011 2 ] ;

I want to avoid a for loop here. Thanks for your help! This would be a great help!

Upvotes: 2

Views: 3091

Answers (3)

gnovice
gnovice

Reputation: 125874

You can first replicate your matrix using my vectorized answer to the previous question:

>> mat = [2009 3; 2010 2];
>> index = zeros(1, sum(mat(:, 2)));
>> index([1; cumsum(mat(1:end-1, 2))+1]) = 1;
>> Ansmat = mat(cumsum(index), :)

Ansmat =

        2009           3
        2009           3
        2009           3
        2010           2
        2010           2

Next, you can create a column vector of offsets to add to the dates in the first column. Here's how you can do this in a vectorized way.

>> offset = ones(size(Ansmat, 1), 1);
>> offset([1; cumsum(mat(1:end-1, 2))+1]) = [0; 1-mat(1:end-1, 2)];
>> Ansmat(:, 1) = Ansmat(:, 1)+cumsum(offset)

Ansmat =

        2009           3
        2010           3
        2011           3
        2010           2
        2011           2

Upvotes: 2

Serg
Serg

Reputation: 14118

incr = (0 : max(mat(:, 2)))';
incr = [incr, 0 * incr];
Ansmat = [];
for k = 1 : size(mat, 1)
    Ansmat = cat(1, Ansmat, repmat(mat(k, :), mat(k, 2), 1) + incr(1 : mat(k, 2), :));
end

But if the expected size of matrix Ansmat is large, the first solution of GummiV will be faster.

Upvotes: 0

GummiV
GummiV

Reputation: 236

I'm not sure if there is a way to do this without a loop since this is a somewhat obscure operation we're performing. If I understand your algorithm correctly here are two methods that use a single for loop:

B = zeros(sum(A(:,2)), 2);
counter = 1;
for i = 1:size(A,2)
    n = A(i,2);
    B(counter:counter+n-1,1) = A(i,1)+(0:n-1)';
    B(counter:counter+n-1,2) = n;
    counter = counter+n;
end

You could do away with the preallocate. If the counter variable is confusing then you could check out this one which appends the matrix B every iteration but needs an inital starting-case instead.

n = A(1,2);
B = [A(1,1)+(0:n-1)', n*ones(n,1)];
for i = 2:size(A,2)
    n = A(i,2);
    B = [B; A(i,1)+(0:n-1)', n*ones(n,1)];
end

Upvotes: 0

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