How to check whether a 2d matrix is gradually increasing in values in row direction.

Lets say u have a matrix A=[2 4 7;3 4 6;] So we can see the A(4)==3 in row 2 has increased from A(1)==2 progression,

And the 6th element,A(6)==6 has reduced from being A(3)==7 to 6.

So the A(6) needs to be replaced by Nan

This is basically the thing. Needs to be done in a large matrix. Any ideas on doing it faster than for loops.

Answer 1

[Stephen23]

Using cummax is simple:

>> A = [2,4,7;3,4,6]
A =
   2   4   7
   3   4   6
>> A(A<cummax(A,1)) = NaN
A =
     2     4     7
     3     4   NaN

EDIT: to also ignore adjacent repeated values:

>> A = [2,4,7;3,4,6]
A =
   2   4   7
   3   4   6
>> idx = A<cummax(A,1) | 0==diff([NaN*A(1,:);A],1,1);
>> A(idx) = NaN
A =
     2     4     7
     3   NaN   NaN

[MSP]

One thing though that I forgot to mention is the A(5) is same as A(2),so it should be also replaced by Nan.Thanks in advance for doing this

[Stephen23]

"...I forgot to mention is the A(5) is same as A(2)..."

>> A(5)
ans =  7
>> A(2)
ans =  3

How is A(5) the "same" as A(2) ?

EDIT: aaah, I see how: your understanding of linear indices is incorrect. When I read your question it is clear that "A(3)==7" is not correct, because in fact the value 7 has linear index 5. You can check this yourself quite easily:

>> A(5) % the correct linear index
ans =  7
>> A(3) % what you think it is
ans =  4

Linear indices work along the columns, not along the rows.

So when I fix your incorrect linear indices it turns out you are talking about column 2, with linear indices 3 and 4. Now your point is clear. And the fix is easy too, see my edited answer.

[Andrei Bobrov]

+1

[MSP]

Yeah sorry guys I had posted the question rather casually.

Answer 2

[Andrei Bobrov]

B = cummax(A);
A([false(1,size(A,2));diff(B)==0]) = nan;

[MSP]

This one is much faster tbh.

Answer 3

[Cedric]

 >> flags = [false( 1, size( A, 2 )); diff( A ) < 0]
 flags =
  2×3 logical array
   0   0   0
   0   0   1

 >> A(flags) = NaN
 A =
     2     4     7
     3     4   NaN

EDIT 5:40pm EST:

 >> A = [5, 3, 4, 6; 2, 4, 2, 3].'
 A =
     5     2
     3     4
     4     2
     6     3

 >> select = any((A-permute(A,[3,2,1])) .* permute(tril(ones(size(A,1)*[1,1]),-1),[1,3,2]) < 0, 3)
 select =
  4×2 logical array
   0   0
   1   0
   1   1
   0   1

 >> A(select) = NaN
 A =
     5     2
   NaN     4
   NaN   NaN
     6   NaN

and if you have an old version of MATLAB, the expansions must be performed using BSXFUN:

 select = any(bsxfun(@times, bsxfun(@minus, A, permute(A, [3,2,1])), ...
     permute(tril(ones(size(A, 1) * [1,1]), -1), [1,3,2])) < 0, 3) ;

[MSP]

No,would you check out my previous comment on Image Analyst answer

[Cedric]

See my edited solution.

Answer 4

[Image Analyst]

Of course, simply use conv2():

A=randi(9, 10, 3) 
zeroRow = zeros(1, size(A, 2))
m = [zeroRow; conv2(A, [1;-1], 'valid')]
A(m<0) = nan

[MSP]

But the code isnt doing what I wanted though.From the image I posted you can view what I mean,like the 3 should be replaced by Nan,otherwise its not maintaining the sequence of gradual increasing value.which in case of interpolation of missing values will lead to the same scenario again

[Image Analyst]

Well whatever was after the 9 was less than a 9, let's say it was a 1. So then the 1 goes to a NAN, but 3 is more than the 1 so it gets kept.

What you want is a moving peak detector. I don't think MATLAB has a movpeak() function but I think I saw someone make one in effect through some trick. Of course you could just to a for loop which should be fast as long as your array doesn't have millions of rows.