finding last non-zero value from column

Given a matrix looking something like e.g. M = [ 1 1 1 1 1 1 ; 1 1 1 0 0 0 ; 1 1 0 0 0 0 ; 1 0 0 0 0 0; 0 0 0 0 0 0 ];

How can I find the coordinate (for plotting these values as a line) for which states the last non-zero element in each column?

For this example I want the coordinates for the column number [ 4, 3, 2, 1, 1, 1]

Would the same code work for 3D-matrix?

[Sean de Wolski]

We could absolutely get the code to work for a 3d matrix, but you have to define what you want. Would you want a two d plane through the third dimension with each column's contribution, or would you like it reshaped?

[Lizan]

For example, I have a matrix M = (x,y,z).

I want to plot a line separating the 1-region with the 0-region for each z.

for example

z1 = d1; M1 = [ 1 1 1 1 1 1 ; 1 1 1 0 0 0 ; 1 1 0 0 0 0 ; 1 0 0 0 0 0; 0 0 0 0 0 0 ];

z2 = d2; M2 = [ 1 1 1 1 1 1 ; 1 1 1 1 1 1 ; 1 1 1 1 1 0 ; 1 1 1 1 0 0; 1 1 1 0 0 0 ];

z3 = d3; M3 = [ 1 1 1 1 1 1 ; 1 1 1 1 1 0 ; 1 1 1 1 1 0 ; 1 1 1 1 0 0; 1 1 1 1 0 0 ];

z4 = d4; M4 = [ 1 0 0 0 0 0 ; 1 0 0 0 0 0 ; 1 1 0 0 0 0 ; 1 1 0 0 0 0; 1 1 0 0 0 0 ];

where d1,d2,d3,d4 is some parameter.

Answer 1

[Fangjun Jiang]

M = [ 1 1 1 1 1 1 ; 1 1 1 0 0 0 ; 1 1 0 0 0 0 ; 1 0 0 0 0 0; 0 0 0 0 0 0 ];
n=M~=0;
[dummy,Index]=sort(n);
Index=Index(end,:).*any(n)

Use M=randint(x,y) to generate testing data, I am thinking my solution has an edge.

for 3D matrix:

clear M;clc;
M(1,:,:) = [ 1 1 1 1 1 1 ; 1 1 1 0 0 0 ; 1 1 0 0 0 0 ; 1 0 0 0 0 0; 0 0 0 0 0 0 ]; 
M(2,:,:) = [ 1 1 1 1 1 1 ; 1 1 1 1 1 1 ; 1 1 1 1 1 0 ; 1 1 1 1 0 0; 1 1 1 0 0 0 ]; 
M(3,:,:) = [ 1 1 1 1 1 1 ; 1 1 1 1 1 0 ; 1 1 1 1 1 0 ; 1 1 1 1 0 0; 1 1 1 1 0 0 ]; 
M(4,:,:) = [ 1 0 0 0 0 0 ; 1 0 0 0 0 0 ; 1 1 0 0 0 0 ; 1 1 0 0 0 0; 1 1 0 0 0 0 ];

n=M~=0;
[dummy,Index]=sort(n,2);
Index=Index(:,end,:).*any(n,2);
Index=reshape(Index,size(M,1),size(M,3))

[Lizan]

About the 3D Matrix.

I get;

ndex =

4 3 3 3 3 3

4 3 3 3 3 2

4 4 3 3 3 4

4 4 3 3 4 4

4 4 3 3 4 4

Does that really agree with above?

Looking at it manually I get

Index = 4 3 2 1 1 1

5 5 5 4 3 2

5 5 5 5 2 1

5 2 0 0 0 0 % this last one is opposite to the others

Am I wrong?

[Fangjun Jiang]

Yeah. I need to work on the 3D Matrix. My brain is flat now!

[Fangjun Jiang]

All right! See updated version! Considered all-zero columns for both 2D and 3D!

Answer 2

[Andrei Bobrov]

in your case

sum(M)

ADD

sum(cumsum(flipud(M~=0))~=0)

[Sean de Wolski]

Nice, no reason for the ~=0 tests

sum(logical(cumsum(flipud(M))))

[Lizan]

That is effective.. how would this work for matrix like M4 (which the zeros is on the top instead of below)?

[Fangjun Jiang]

sum() won't work for cases like [0 0;1 1]

Answer 3

[Sean de Wolski]

[junk, idx] = max(flipud(M),[],1); %flip it and find first maximizer
idx = size(M,1)-idx+1

for 3d:

[junk, idx] = max(flipdim(rand(10,10,10)>.5,1),[],1);
idx = size(M,1)-idx+1