Insert certain value after occurence of a set of n equal values - matlab

Question

Example:
input = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 1]; % array of numbers (uint8)
output = [1 255 0 0 0 255 9 9 9 255 1 6 6 6 255 6 6 6 255 1];
% output must have 255 inserted at positions 6, 10, 15, 19
% because 0, 9, 6, 6 have occurred three times respectively
outputIndex = [6 10 15 19];
% outputIndex must indicate the positions where 255 was inserted

Answer 1

This could be one vectorized approach to get things done efficiently -
%// Input
A = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 1]
%// Input paramter (how many times a value must be repeated for detection)
search_count = 3;
%// Find difference between consecutive elemnts and set all non zero
%// differences as ones, otherwise as zeros in a binary array
diffA = diff(A)~=0
%// Find start and end indices of "islands" of same value
starts = strfind([1 diffA],[1 zeros(1,search_count-1)])
ends = strfind([diffA 1],[zeros(1,search_count-1) 1])+search_count
%// For each island of same valued elements, find out where first group ends
firstgrp = starts + search_count
%// Find how many times a group of that same value of search_count times repeats
%// within each "island" of same valued elements. Also get the max repeats.
pattern_repeats = floor((ends - starts)./search_count)
max_repeat = max(pattern_repeats)
%// All possible repeat indices within all islands
all_repeats = bsxfun(#plus,firstgrp,[0:max_repeat-1]'*(search_count)) %//'
%// Use a binary mask to select only those repeats allowed with pattern_repeat
out_idx = all_repeats(bsxfun(#lt,[0:max_repeat-1]',pattern_repeats)) %//'
out_idx = out_idx + [0:numel(out_idx)-1]' %//'
%// Create output arary, insert 255 at out_idx locations and put values
%// from input array into rest of the locations
out = zeros(1,numel(A)+numel(out_idx));
out(out_idx) = 255
out(out==0) = A
Code run -
>> A
A =
Columns 1 through 13
1 255 0 0 0 9 9 9 1 6 6 6 6
Columns 14 through 16
6 6 1
>> out_idx
out_idx =
6
10
15
19
>> out
out =
Columns 1 through 13
1 255 0 0 0 255 9 9 9 255 1 6 6
Columns 14 through 20
6 255 6 6 6 255 1

Answer 2

I don't understand the downvotes, it's actually an interesting question.
Here the long answer:
n = 3;
subst = 255;
input = [1 255 0 0 0 9 9 9 1 6 6 6 6 6 6 61];
%// mask
X = NaN(1,numel(input));
%// something complicated (see below)
X(filter(ones(1,n-1),1,~([0. diff(input)])) == n-1) = 1;
%// loop to split multiple occurences of n-tuples
for ii = 1:numel(input)
if X(ii) == 1 && ii < numel(X)-n+1
X(ii+1:ii+n-1) = NaN(1,n-1);
end
end
%// output vector
D = [input; X.*subst];
E = D(:);
output = E(isfinite(E))
%// indices of inserted 255
D = [input.*0; X.*subst];
E = D(:);
outputIndex = find(E(isfinite(E)))
Explanation of the complicated part:
%// finite differences of input
A = [0 diff(input)];
%// conversion to logical
B = ~A;
%// mask consecutive values
mask = filter(ones(1,n-1),1,B) == n-1;
%// set masked values to 1
X(mask) = 1;
If you have the image processing toolbox you can save the loop with this fancy oneliner for getting the mask:
mask = accumarray(bwlabel(filter(ones(1,n-1),1,~([0. diff(input)])) == n-1).'+1,1:numel(input),[],#(x) {getfield(sort(x),{find(mod(cumsum(1:numel(x)),n) == 1)})});
X = NaN(1,numel(input));
X(vertcat(mask{2:end})) = subst;
%// output vector
D = [input; X];
E = D(:);
output = E(isfinite(E))
%// indices of inserted 255
D = [input.*0; X];
E = D(:);
outputIndex = find(E(isfinite(E)))