Matlab replace consecutive zero value with others value - matlab

Question

I have this matrix:
A = [92 92 92 91 91 91 146 146 146 0
0 0 112 112 112 127 127 127 35 35
16 16 121 121 121 55 55 55 148 148
0 0 0 96 96 0 0 0 0 0
0 16 16 16 140 140 140 0 0 0]
How can I replace consecutive zero value with shuffled consecutive value from matrix B?
B = [3 3 3 5 5 6 6 2 2 2 7 7 7]
The required result is some matrix like this:
A = [92 92 92 91 91 91 146 146 146 0
6 6 112 112 112 127 127 127 35 35
16 16 121 121 121 55 55 55 148 148
7 7 7 96 96 5 5 3 3 3
0 16 16 16 140 140 140 2 2 2]

Answer 1

You simply can do it like this:
[M,N]=size(A);
for i=1:M
for j=1:N
if A(i,j)==0
A(i,j)=B(i+j);
end
end
end

Answer 2

If I understand it correctly from what you've described, your solution is going to need the following steps:
Loop over the rows of your matrix, e.g. for row = 1:size(A, 1)
Loop over the elements of each row, identify where each run of zeroes starts and store the indices and the length of the run. For example you might end up with a matrix like: consecutiveZeroes = [ 2 1 2 ; 4 1 3 ; 4 6 5 ; 5 8 3 ] indicating that you have a run at (2, 1) of length 2, a run at (4, 1) of length 3, a run at (4, 6) of length 5, and a run at (5, 8) of length 3.
Now loop over the elements of B counting up how many elements there are of each value. For example you might store this as replacementValues = [ 3 3 ; 2 5 ; 2 6 ; 3 2 ; 3 7 ] meaning three 3's, two 5's, two 6's etc.
Now take a row from your consecutiveZeroes matrix and randomly choose a row of replacementValues that specifies the same number of elements, replace the zeroes in A with the values from replacementValues, and delete the row from replacementValues to show that you've used it.
If there isn't a row in replacementValues that describes a long enough run of values to replace one of your runs of zeroes, find a combination of two or more rows from replacementValues that will work.
You can't do this with just a single pass through the matrix, because presumably you could have a matrix A like [ 15 7 0 0 0 0 0 0 3 ; 2 0 0 0 5 0 0 0 9 ] and a vector B like [ 2 2 2 3 3 3 7 7 5 5 5 5 ], where you can only achieve what you want if you use the four 5's and two 7's and not the three 2's and three 3's to substitute for the run of six zeroes, because you have to leave the 2's and 3's for the two runs of three zeroes in the next row. The easiest approach if efficiency is not critical would probably be to run the algorithm multiple times trying different random combinations until you get one that works - but you'll need to decide how many times to try before giving up in case the input data actually has no solution.
If you get stuck on any of these steps I suggest asking a new, more specific question.