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

Question

MSP 2017-9-21

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https://ww2.mathworks.cn/matlabcentral/answers/357758-how-to-check-whether-a-2d-matrix-is-gradually-increasing-in-values-in-row-direction

评论： MSP 2017-9-22

采纳的回答： Stephen23

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.

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Answer 1

Stephen23 2017-9-22

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编辑：Stephen23 2017-9-22

在 MATLAB Online 中打开

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

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Andrei Bobrov 2017-9-22

+1

MSP 2017-9-22

Yeah sorry guys I had posted the question rather casually.

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Answer 2

Andrei Bobrov 2017-9-22

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在 MATLAB Online 中打开

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

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MSP 2017-9-22

This one is much faster tbh.

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Answer 3

Cedric 2017-9-21

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编辑：Cedric 2017-9-21

在 MATLAB Online 中打开

 >> 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) ;

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MSP 2017-9-21

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

Cedric 2017-9-21

See my edited solution.

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Answer 4

Image Analyst 2017-9-21

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在 MATLAB Online 中打开

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

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MSP 2017-9-21

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 2017-9-22

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.

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