# Need algorithm to create the list

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Fayyaz on 6 Jul 2014
Commented: Fayyaz on 6 Jul 2014
Hello,
I need to create a matrix of order 1614006*3, The sequence should be
1st Col (origin) , 2nd Col (Destination) , 3rd Col (Survey_Location)
Origin (1) Destination(1) Survey_Location(1)
Origin (1) Destination(1) Survey_Location(2)
.......
Origin (1) Destination(1) Survey_Location(246)
Origin (1) Destination(2) Survey_Location(1)
...........
Origin (1) Destination(81) Survey_Location(246)
......
Origin (2) Destination(1) Survey_Location(1)
.........
Origin (81) Destination(81) Survey_Location(246)
For this case rows should be 81*81*246=1614006
Star Strider on 6 Jul 2014
How does this differ from your earlier question: Survey data Matrix arrangement?
Fayyaz on 6 Jul 2014
Hello, it is not different from the previous question. I just posted this question in a more clear way. I am going to delete the previous question.
Thanks.

Image Analyst on 6 Jul 2014
Edited: Image Analyst on 6 Jul 2014
I'm sure there's a more clever vectorized way - Andrei can probably do it in a single line! - but there's always the brute force way:
Origin = randi(9, [81, 1]);
Destination = randi(9, [81, 1]);
Survey_Location = randi(9, [246, 1]);
numOutputRows = size(Origin, 1) * size(Destination, 1) * size(Survey_Location, 1)
output = zeros(numOutputRows, 3);
tic;
row = 1;
for r1 = 1 : size(Origin, 1)
for r2 = 1 : size(Destination, 1)
for r3 = 1 : size(Survey_Location, 1)
output(row, :) = [Origin(r1), Destination(r2), Survey_Location(r3)];
row = row + 1;
end
end
end
toc
Takes 1.4 second on my computer.
Image Analyst on 6 Jul 2014
Oh, I thought you already had some arrays so I just used sample arrays. If you want sequential arrays, you can do
Origin = (1:81)';
Destination = (1:81)';
Survey_Location = (1:246)';
Fayyaz on 6 Jul 2014
Perfect! Actually I am not that much familiar with MATLAB, but I hope I will be able to understand quite soon. Thanks again.

Andrei Bobrov on 6 Jul 2014
[z,y,x] = ndgrid(Survey_Location,Destination,Origin);
out = [x(:),y(:),z(:)];
##### 3 CommentsShow 1 older commentHide 1 older comment
Image Analyst on 6 Jul 2014
Aha! I knew Andrei could do it in a line or two. He's the master of that.
Fayyaz on 6 Jul 2014
Ya I've realized when Andrei did this.
@Andrei, if possible, kindly look into this question too, you've almost solved it.

Cedric on 6 Jul 2014
Edited: Cedric on 6 Jul 2014
Here is an example, with 3 elements Origin and Destination, and a 4 elements Survey_Location..
% - Dummy test case.
Origin = 10 : 12 ;
Destination = 100 : 102 ;
Survey_Location = [500, 600, 700, 800] ;
% - Define columns of final array and concatenate.
c1 = reshape( repmat( Origin(:)', 3*4, 1 ), [], 1 ) ;
c2 = repmat( reshape( repmat( Destination(:)', 4, 1 ), [], 1 ), 3, 1 ) ;
c3 = repmat( Survey_Location(:), 3*3, 1 ) ;
X = [c1, c2, c3] ;
Running this, you get
>> X
X =
10 100 500
10 100 600
10 100 700
10 100 800
10 101 500
10 101 600
10 101 700
10 101 800
10 102 500
10 102 600
10 102 700
10 102 800
11 100 500
11 100 600
11 100 700
11 100 800
11 101 500
11 101 600
11 101 700
11 101 800
11 102 500
11 102 600
11 102 700
11 102 800
12 100 500
12 100 600
12 100 700
12 100 800
12 101 500
12 101 600
12 101 700
12 101 800
12 102 500
12 102 600
12 102 700
12 102 800
EDIT : for your case, you want to replace 3 by 81, and 4 by 246.