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comparing relevant elemnts of two matrix

Asked by som on 30 Nov 2012

Hi all,

I have two matrics like below:

      a=[9 7 NaN; 6 3	8; 15 NaN 5; NaN 4 2];
      b= [10,14,NaN;10,10,13;10,NaN,10;NaN,10,10;] ;

I want to see if each element of "a" is less than or equal to its corresponding element in "b"?

 how can I write this program .
 Thanks in advance. 

1 Comment

Muruganandham Subramanian on 30 Nov 2012

Can you mention the range?

som

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4 Answers

Answer by Image Analyst on 30 Nov 2012
Accepted answer

Maybe this is what you want:

a=[9 7 NaN; 6 3	8; 15 NaN 5; NaN 4 2]
b= [10,14,NaN;10,10,13;10,NaN,10;NaN,10,10;]
% Subtract the two. 
% The value will be negative when a(r,c) < b(r,c).
differenceMatrix = a-b
% Make that into a boolean map of where a<b.
% It will be true (1) where a<b.
mapOfWhereAisSmallerThanB = differenceMatrix<0
% Just for fun, do the converse.
% Make that into a boolean map of where b<a.
% It will be true (1) where b<a.
mapOfWhereBisSmallerThanA = differenceMatrix > 0
% See if all non-Nan values have a<b
nonNanIndexes = ~isnan(a) & ~isnan(b)
if all(differenceMatrix(nonNanIndexes) < 0)
	msgbox('All of a < b');
else
	msgbox('Not all of a is < b');
end

In the command window:

a =
     9     7   NaN
     6     3     8
    15   NaN     5
   NaN     4     2
b =
    10    14   NaN
    10    10    13
    10   NaN    10
   NaN    10    10
differenceMatrix =
    -1    -7   NaN
    -4    -7    -5
     5   NaN    -5
   NaN    -6    -8
mapOfWhereAisSmallerThanB =
     1     1     0
     1     1     1
     0     0     1
     0     1     1
mapOfWhereBisSmallerThanA =
     0     0     0
     0     0     0
     1     0     0
     0     0     0
nonNanIndexes =
     1     1     0
     1     1     1
     1     0     1
     0     1     1

0 Comments

Image Analyst
Answer by Harshit on 30 Nov 2012

size((A-B)>0)

0 Comments

Harshit
Answer by Vishal Rane on 30 Nov 2012
Edited by Vishal Rane on 30 Nov 2012

By 'little-equal' I assume you mean 'less than or equal'.

Use

 a <= b

assuming a,b are of same dimensions.

See Relational Operators for more info.

0 Comments

Vishal Rane
Answer by Wayne King on 30 Nov 2012
Edited by Wayne King on 30 Nov 2012

I'll assume that " littel-equal " means "less than or equal to"

 a=[9 7 NaN; 6 3	8; 15 NaN 5; NaN 4 2];
 b= [10,14,NaN;10,10,13;10,NaN,10;NaN,10,10;] ;
 indices = find(a<=b);
 a(indices)

The above gives the elements of a that are less than or equal to the corresponding element of b.

or

C = a<=b;

The matrix C has a 1 where the element of a is less than or equal to the element of b and a 0 otherwise

1 Comment

som on 30 Nov 2012

Thanks for your answer, but please pay attention that both matrics have "NaN" elements. so the problem is still remaind.

Any help would be appreciated.

Wayne King

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