Need help using colon operator with multiple matrices - I'm really close to being loopless!

Simplified problem is: a=[1 2 3]; b=[4 5 6]; How can I get to: c=[1 2 3 4; 2 3 4 5; 3 4 5 6];

I've tried the obvious, a:b will return only c=[1 2 3 4]; Any ideas?

Many thanks, - Matt

Answer 1

[Teja Muppirala]

If you know that it's going to end up being a rectangular matrix like in your example, then the (b-a) all have to be the same:

if all( (b-a) == (b(1)-a(1)) ) 
  bsxfun(@plus,a',0:(b(1)-a(1)))
end

Answer 2

[Andrei Bobrov]

more variant

a=[1 2 3];
b=[4 5 6]; 
c = [a b];
c = c(cumsum([1:4; ones(2,4)]));

or in this case, just

c = cumsum([1:4; ones(2,4)])

more

c1 = repmat(min(a):max(b),length(a),1)';
c = reshape(c(ones(size(c1,1),1)*a<=c1&c1<=ones(size(c1,1),1)*b),[],length(a))';

Answer 3

[Sean de Wolski]

Bruno's mcolon on the FEX:

http://www.mathworks.com/matlabcentral/fileexchange/29854-multiple-colon

[Matt H]

Not sure I like this option, I feel like I'd be better off (crunch time wise) with a for loop.

[Sean de Wolski]

Then go with a for loop.

I think this is the thread that started mcolon:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/298813

Answer 4

[Matt Fig]

I am not sure if this is in the above thread or not. But perhaps the simplest FOR loop version is this:

a=[1 2 3]; 
b=[4 5 6];
C = cell(1,length(a));

for ii = 1:length(a)
   C{ii} = a(ii):b(ii); 
end

C = [C{:}]

But, like Sean de, I would use the MEX version if speed due to very large a and b is at all a concern.

EDIT In response to Oleg's comment.

Oleg makes the point that my code above is not the simplest FOR loop implementation. I guess the urge to pre-allocate is too strong in me! This is simpler, though slower on my machine for larger a and b:

a=[1 2 3]; 
b=[4 5 6];
C = []

for ii = 1:length(a)
   C = [C a(ii):b(ii)]; 
end

[Oleg Komarov]

Why the cell Matt?

[Oleg Komarov]

Assuming b(i)-a(i) = k, for all i:

C = zeros(numel(a),b(1)-a(1))

Answer 5

[Shravan Chandrasekaran]

Hi Matt,

 This works

clear all;
clc;
a=[1 2 3]
b=[4 5 6]
A=[a b]
for i=1:1:3
    for j=1:1:4
        AA(i,j)=A(1,i+(j-1));
    end
end
AA

Regards, Shravan.

[Sean de Wolski]

DON'T CLEAR ALL!!!!! (five exclamation points should be enough)