finding last non-zero value from column

2011 8 15
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0 个投票

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?

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Sean de Wolski
Sean de Wolski 2011-8-15
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
Lizan 2011-8-15
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.

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 采纳的回答

Fangjun Jiang
Fangjun Jiang 2011-8-15

1 个投票

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))

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Lizan
Lizan 2011-8-15
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
Fangjun Jiang 2011-8-15
Yeah. I need to work on the 3D Matrix. My brain is flat now!
Fangjun Jiang
Fangjun Jiang 2011-8-15
All right! See updated version! Considered all-zero columns for both 2D and 3D!

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更多回答（2 个）

Andrei Bobrov
Andrei Bobrov 2011-8-15

1 个投票

in your case
sum(M)
ADD
sum(cumsum(flipud(M~=0))~=0)

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Sean de Wolski
Sean de Wolski 2011-8-15
Nice, no reason for the ~=0 tests
sum(logical(cumsum(flipud(M))))
Lizan
Lizan 2011-8-15
That is effective.. how would this work for matrix like M4 (which the zeros is on the top instead of below)?
Fangjun Jiang
Fangjun Jiang 2011-8-15
sum() won't work for cases like [0 0;1 1]

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Sean de Wolski
Sean de Wolski 2011-8-15

0 个投票

[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

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