Finding sorted rows in a matrix

Hi,

I need to know which rows of a matrix are sorted in ascending without using the "for" cycle. For instance, if i have:

outsx = [

1 1

2 1

3 1

1 2

2 2

3 2

1 3

2 3

3 3]

i want a vector containing indexes of which rows are sorted in ascending order. In this case.

ii= [1 4 5 7 8 9]

Thank you in advance

Answer 1

[the cyclist]

For my current understanding of what you want, I believe this works:

ii = find(diff(outsx,[],2)>=0)

[Jan]

I assume an all() is wanted:

ii = find(all(diff(outsx,[],2) >= 0, 2))

[the cyclist]

Yes, thanks for catching that

Answer 2

[the cyclist]

outsx = [
     1     1
     2     1
     3     1
     1     2
     2     2
     3     2
     1     3
     2     3
     3     3];

ii = find(outsx(:,2)>=outsx(:,1))

ii = 6×1

1 4 5 7 8 9

[Marco Gaetani d'Aragona]

Hi,

I would need a more general formulation, cosidering matrixes with different number of rows and columns

Answer 3

[Star Strider]

Two approaches:

 for k = 1:size(outsx,1)
     TF(k,:) = issorted(outsx(k,:),2,'ascend')
 end
 Result = find(TF)
 
 TF = outsx(:,2) >= outsx(:,1);
 Result = find(TF)

both producing:

Result =
     1
     4
     5
     7
     8
     9

.

[Marco Gaetani d'Aragona]

Hi,

I need to perform this check for matrixes nxm, so I need a faster procedure without employing for cycles.

[the cyclist]

FYI, just because code has a for loop does not necessarily mean it is slower than other solutions.

Finding a solution that works is the first step (partly to verify that the solution does what you want), and then thinking about readability and optimization follows.