The usage of find in matlab for finding the length of repeated element

Question

yuese zheng 2015-7-31

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编辑： Andrei Bobrov 2015-7-31

Let' say I have a vector [1 1 1 2 2]. I want to know what elements are repeated, and repeated for how many times. Basically, I want the program to tell me 1s and 2s are repeated, and 1s are repeated 3 times, and 2s are repeated 2 times. So far,

index=[find(x(1:end-1) ~= x(2:end))]; 
len=diff([0 index]);

gives me a value of 3 and 2 respectively, which is what I want for the number of times each element is repeated. However, I am not sure how the "find" function is doing its job here. Could anyone explain to me? I checked the find function docs but no luck. Thank you.

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per isakson 2015-7-31

编辑：per isakson 2015-7-31

See RunLength by Jan Simon and test your code with a few different vectors, x

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Answer 1

Azzi Abdelmalek 2015-7-31

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在 MATLAB Online 中打开

In your case find is not doing the job, your code doesn't give the expected result

x=[1 1 1 2 2];
index=[find(x(1:end-1) ~= x(2:end))]; 
len=diff([0 index]);

The result

index =
  3
len =
  3

This is one way to do it:

x=[1 1 1  2 2];
[~,~,kk]=unique(x);
out=accumarray(kk,1)

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Answer 2

Andrei Bobrov 2015-7-31

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编辑：Andrei Bobrov 2015-7-31

在 MATLAB Online 中打开

a = unique(x(:));
out = [a,histcounts(x,[a,a(end)+1])'];

for v = [1, 1, 1, 2, 2, 1, 1, 7, 7, 4, 4, 4, 2]';

i0 = [true;diff(v)~=0];
ii = cumsum(i0);
t = [v(i0),accumarray(ii,1)];
[a,~,c] = unique(t(:,1));
out = [num2cell(a),num2cell(accumarray(c,1)),accumarray(c,t(:,2),[],@(x){x})];

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Answer 3

Image Analyst 2015-7-31

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编辑：Image Analyst 2015-7-31

在 MATLAB Online 中打开

Try this:

vector = [1, 1, 1, 2, 2, 1, 1, 7, 7, 4, 4, 4, 2]
for k = unique(vector)
  measurements = regionprops(vector == k, 'Area');
  numberOfRegions = length(measurements);
  fprintf('%d occurs in %d regions with lengths: ', k, numberOfRegions);
  allLengths = [measurements.Area];
  fprintf('%d, ', allLengths);
  fprintf('\n'); % Go to the next line.
end

Output in the command window:

vector =
   1     1     2     2     1     1     7     7     4     4     4     2
occurs in 2 regions with lengths: 3, 2, 
occurs in 2 regions with lengths: 2, 1, 
occurs in 1 regions with lengths: 3, 
occurs in 1 regions with lengths: 2,

It has the advantage over the other methods in that it will work where the same number is separated into 2 or more regions (like in the vector example I show), whereas the other methods don't. However you need the Image Processing Toolbox to use this method, but that's a very common toolbox.