Change a value after a maximum five-consecutive column of zero

I have a matrix [3000,1000] which has only 0, 1, and -1. If there is a -1 in a row followed by a (maximum) consecutive five-columns of zeros then followed by 1, this 1 value should change to 2.

I am looking for the fastest way to do it in Matlab. Is there anyway without a loop? if not, what is the best way?

For example

A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1;
     0 -1 0 0 0 0 0 0 1 -1 0 1 0 0 -1];

% I would like to change matrix A as follow

A = [1 0 0 -1 0 0 0 0 2 0 0 -1 0 0 2;
     0 -1 0 0 0 0 0 0 1 -1 0 2 0 0 -1 ];

Thanks in advance,

[Jan]

Why is the output not:

A = [1 0 0 -1 0 0 0 0 2 0 0 -1 0 0 2;
     0 -1 0 0 0 0 0 2 -1 0 2 0 0 -1 0];
                    ^

There is a -1 at A(2, 2), 5 consecutive zeros and then a 1. A(2,8) should be a 2 in consequence.

Please post the relevant sizes. It matters if you have a [1e6, 10] or [10 x 1e6] array.

[Jalu Naradi]

Sorry, it was my mistake. I edit the example so that there are 6 consecutive zeros now before 1

[Jalu Naradi]

@Jan, the matrix size is [3000,1000]. Thank you

Answer 1

[Guillaume]

I don't think this can be done more efficiently than with a loop over the rows:

for row = 1:size(A, 1)
   col = find(A(row, :));
   tochange = diff(col) <= 5 & A(row, col(1:end-1)) == -1 &  A(row, col(2:end)) == 1;
   A(row, col([false, tochange])) = 2;
end

[Jalu Naradi]

Thank you! It works like a charm!

[Jalu Naradi]

I tried to find a reason for using a "false" function

 A(row, col([false, tochange])) = 2;

But I still couldn't find it. Could you please elaborate more the logic behind it?

Thank you

[Guillaume]

The result of diff is a vector with one less element than col. tochange(1) actually corresponds to col(2|, so I just prepend false to vector (since the 1st element is never going to have to be changed anyway).

[Jalu Naradi]

Thank you for your explanation

Answer 2

[Jan]

A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1; ...
     0 -1 0 0 0 0 0 1 -1 0 1 0 0 1 0];

[s1, s2] = size(A);
Av = reshape(A.', 1, []);                  % Convert it to 1 vector
for k = 0:5
  index = strfind(Av, [-1, zeros(1, k), 1]);
  index = index(mod(index, s2) < s2 - k);  % Omit matches at end of row
  Av(index + k + 1) = 2;        
end
B = reshape(Av, s2, []).';

[Jalu Naradi]

Thank you Jan, it works!

Answer 3

[Ameer Hamza]

This is an approach to modify an identify an arbitrary pattern and modify a value, Using for loop and comparing them by converting them to char array.

A = [1 0 0 -1 0 0 0 0 1 0 0 -1 0 0 1;
   0 -1 0 0 0 0 0 1 -1 0 1 0 0 1 0];

pattern = [-1 0 0 0 0 1];

strA = num2str(A+1); % +1 is added to remove negative signs for easy manipulation as strings
strPattern = strrep(num2str(pattern+1), ' ', '');

for i=1:size(A,1)
  index = strfind( strrep(strA(i,:), ' ', ''), strPattern) + length(pattern) -1;
  if ~isempty(index)
      A(i, index) = 2;
  end
end

[Guillaume]

Note that you don't need to convert numbers to char to use strfind. Despite not being actually documented, strfind works just as well for detecting patterns of numbers

strfind(A(i, :), [-1 0 0 0 0 0 1])

The problem here is that several patterns are acceptable, [-1 1], [-1 0 1], ..., [-1 0 0 0 0 0 1], so a pattern search is not particularly useful. I guess a regexp would work (which does requires a conversion to char) but I'm not convinced the extra complexity and time taken by the regex would be better than the simple for loop I've detailed.

[Jalu Naradi]

Thank you Ameer for your answer. As Guillaume mentioned above, there are several patterns that acceptable here.