Counting values above a threshold within a matrix.

Hello,

Im looking at finding a way of counting values that are >0 from a matrix, for example:-

   if true %if statement used to display matrix correctly
00100
01110
11111
01110
00100
  end

Would provide the output 1,3,5,3,1. Any help would light up my day.

Cheers

Gareth

[Azzi Abdelmalek]

is this your matrix?

0 0 2 0 0
0 3 4 5 0
1 2 3 4 5
0 6 5 4 0 
0 0 3 0 0

Answer 1

[Greg]

sum(myMat>0, 2)

the result is the number of values in each row > 0.

[Greg]

to get the number of values in each row > 0 use:

 sum(myMat > 0, 2)

to get the number of values in each column > 0 use:

 sum(myMat > 0, 1)

the default value is 1 so the following is another way to get number of values in each column > 0:

 sum(myMat > 0)

[Gareth Evans]

A little closer to the solution, for which i really appreciate.

Say the matrix is a 24X365 where 24 is equal to hours and 365 to days. This matrix contains values which relate to the intensity of sunlight hourly per day for 365 days.

How can i create an output from the matrix that will provide a matrix (1X365) with each cell in the matrix (1x365) containing values that exceed 0. or in other words how many hours a day sunlight per day?

I hope this makes the problem clearer as the code given above is outputting the wrong dimensions. and i cant seem to get the right solution by playing with it.

cheers guys.

[Gareth Evans]

Wonderful. Thankyou..(-:

Answer 2

[Babak]

you can run the below function for each of your numbers:

function num_of_1s = no1(n)

if n==0
    num_of_1s = 0;

else

      length = floor(log10(n))+1;
      num_of_1s = 0;

      for j=length:-1:1

          if floor(log10(n)) == j-1
              num_of_1s = num_of_1s +1;
              n = n-10^(j-1);
          end

      end
  end
  end

Then say for example:

 no1(1010111)

which returns

 5

[Gareth Evans]

I do apologise, I believe that the way i used to display the matrix using this forum editor might of caused a little confusion. I used the code tab to enable me to display the matrix as you would write down on a piece of paper.

00200 03450 12345 06540 00300 would be a 5x5 matrix.

Could you solve this problem with out an if loop?

with thanks.

Answer 3

[nah]

A simple way is:

myMat =

     0     0     2     0     0
     0     3     4     5     0
     1     2     3     4     5
     0     6     5     4     0
     0     0     3     0     0
[nrows,ncols] = size(myMat);
counts_above_0 = zeros(nrows,1);

for rx = 1:nrows                                   
counts_above_0(rx,1) = length(find(myMat(rx,:)>0));
end

[Emmagnio Desir]

Really good