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Replace percentage of existing values with new value

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

0 Comments

Michael

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3 Answers

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

1 Comment

Michael on 20 Nov 2012

Richard,

Thanks this works and I was able to adapt it to the matrices (described above) that I will be using.

Cheers,

Michael

Richard Zapor
Answer by Matt Fig on 19 Nov 2012
B(randperm(16,4)) = 5

5 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
Matt Fig
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

1 Comment

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

Image Analyst

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