How do I find the duration of a number in an array?

Given the array

c=[3 3 3 5 5 6 7 8 1 1 3 3 3 0 0 0 4 4 4 5 6 7 8 3 3 3 9 10 3];

I am trying to find the average duration of each number in this array. For example, I am looking to find the average duration for 3 in this array. I know that 3 occurs 10 times but is grouped into 4 separate occurrences. I want the output to be the average duration for 3...which is 2.5. please help!

Answer 1

[Andrei Bobrov]

c = c(:);
ii = [true;diff(c)~=0];
i2 = diff([find(ii);numel(ii)+1]);

[a,b,c0] = unique(c(ii),'first'); % in new version of MATLAB:
[~,jj] = sort(b);                 %
[~,j1] = sort(jj);                % [a1,~,c1] = unique(c(ii),'stable');         
a1 = a(jj);                       %
c1 = j1(c0);                      %

out = [a1,accumarray(c1,i2,[],@mean)];

or with sorting

ii = [true;diff(c)~=0];
i2 = diff([find(ii);numel(ii)+1]);

[a,~,c0] = unique(c(ii));

out1 = [a,accumarray(c0,i2,[],@mean)]

[Joe]

It worked thanks!

[Joe]

I am applying this to my data set and the inputs range from 1-10. However, if there are no occurrences of an input (example= there is no 2 in the data set above how can I make sure the output array will have a zero in the place of the 2 or any other number in the series? Is there a way to output a c(2,10) array with c(1,10) being equal to 1:10 and the c(2,10) array being filled with the output values for average duration. if there are none I would want an empty place value to be filled with a zero.

Answer 2

[Azzi Abdelmalek]

c=[3 3 3 5 5 6 7 8 1 1 3 3 3 0 0 0 4 4 4 5 6 7 8 3 3 3 9 10 3];
[ii,jj,kk]=unique(c)
d=accumarray(kk,1)
for k=1:numel(ii)
  a=ii(k)
  res(k)=d(k)/numel(strfind([0 c==a 0],[0,1]))'
end
out=[ii' res']