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Asked by Michael
on 19 Nov 2012

Hello, I'd like to replace a percentage of each existing value in a matrix or vector with another value? For example, I have matrix B which has values 1, 2, 3, and 4 and I want to replace 25% of each of these extant values with '5,' is there a sweet way to do this? The result would be matrix C, that is, 25% would mean one element of each extant value changed to / replaced with '5'. I'm afraid I am not sure where to start.

B = [1 1 1 1; 2 2 2 2; 3 3 3 3; 4 4 4 4];

C = [1 1 1 5; 2 2 5 2; 3 5 3 3; 4 4 5 4];

Thank you,

Mike

Answer by Richard Zapor
on 20 Nov 2012

Accepted answer

function replace25 m = randi(4, 9, 9) newvalue = 5; % determine qty of values 1 thru 4, a(1) qty 1, a(4) number 4s a=hist(m(:),4)

for i=1:4 % find first 25 % of a value and give it new value m(find(m==i,round(a(i)/4)))=newvalue; end m end

Answer by Matt Fig
on 19 Nov 2012

B(randperm(16,4)) = 5

Show 2 older comments

Matt Fig
on 20 Nov 2012

Further questions: Are your matrices all made of such a nice pattern as the example matrix (row of 1s, row of 2s, etc.) or are they more like:

B = randi(5,5,5)

In either case, what if there are a number of, say, 1s that is not evenly divisible by 4? Say there are three 1s, how many to replace? What if there is only one 1?

For the particular example you give, this would work:

idx = (1:4) + (randi(4,1,4)-1)*4; B(idx) = 5

Michael
on 20 Nov 2012

Matt,
I agree, *unique* is better than *extant* for this arena, thanks for bringing that to my attention.

All of the matrices I will be using will have 5 initial unique values. They will all be ordered similarly to the above. However, as in the example below, sometimes there will be more than 1 row per unique value. As far as the divisibility issue, I'd say round up to the next integer. I think I understand how the code works that you posted most recently, thanks.

Thanks again,

Mike

mat = zeros(10,10); [M N] = size(mat); vals = [0 1 2 3 4]; % 5 initial unique values vL = numel(mat);

nVals = length(vals); idx = floor(linspace(1,nVals+1-2*eps(nVals),vL)); out_vec = vals(idx); out_mat = vec2mat(out_vec,M); % orderly unique values, aggregated by row

Matt Fig
on 20 Nov 2012

O.k. Here is a more generalized code:

N = 4; % This controls the code: a positive integer. vals = 0:N; nVals = N + 1; M = nVals*randi(4); % Up to size nVals*4 idx = floor(linspace(1,nVals+1-2*eps(nVals),M^2)); out_vec = vals(idx); % orderly unique values, aggregated by row: out_mat = vec2mat(out_vec,M) % Now, we have nVals unique values in out_mat. % So we want 1/4 of the set of each of % those (M*N) values set to another value, % say -1 for example....

CV = -1; % The change value, set as needed... PE = floor(M^2/nVals/4); % 1/4 of the number of each

for ii = 0:M/nVals:M-1 cnt = 0; while cnt<PE R = randi(M/nVals) + ii; C = randi(M);

if out_mat(R,C)~=CV out_mat(R,C) = CV; cnt = cnt + 1; end end end

out_mat

Answer by Image Analyst
on 20 Nov 2012

Michael, I think this code does what you want in a pretty flexible and robust way:

m = int32(randi(4, 9, 9)) replacementValue = 5;

% Get the unique values in m uniqueValues = unique(m) % Make a new output array so that the replacement value % can be one of the input values. This allows % maximum flexibility and robustness. % For example what if the replacment value = 3 % without having a copy you'd replace values already replaced. mOut = m;

for k = 1 : length(uniqueValues) thisValue = uniqueValues(k) % Count the number originally at this value. count = sum(m(:) == thisValue) % Get the number to replace, rouding up. numberToReplace = int32(ceil(0.25 * count)) % Pick them from random locations linearIndexes = randperm(numel(m)); indexesToReplace = linearIndexes(1:numberToReplace); % Do the assignments. mOut(indexesToReplace) = replacementValue; end % Do the replacements m = mOut; % Display final m. m

Michael
on 20 Nov 2012

Image Analyst,

Thank you for the reply. I didn't know about the unique command, thanks. If, for example, I replace the *m* in your code, with *B* above -- I have the same issue as I did with the first code suggested by Matt above. That is, whilst your code is much more adaptable, it still is replacing 25% of the entire matrix, and not 25% of each value. Is is it possible to replace *x*% of each unique value?

Thank you,

Mike

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