Assign values to elements of a matrix with indices less than those specified in an array

Hi,

I'm not really sure how to word my question (hence the confusing title!). What I'm trying to do is take an array of values, like A=[2,4,1,3]

and use it to make a matrix B that looks like;

1 1 1 1
1 1 0 1
0 1 0 1
0 1 0 0

i.e. there are ones for rows lower than the value in the column from A. I can see how I'd do this using a for loop, but we'd like to speed up our (currently very slow) program, so I wondered if anybody had an idea how I could make this vectorised?

Cheers

[Azzi Abdelmalek]

What is the relation between A and B?

[dpb]

Nothing "magic" comes to mind otomh...sometimes a loop is best. What are you now doing to generate it? The crude would be

B=zeros(max(A));  % preallocate
for i=1:length(B)
  B(:,i)=ones(A(i),1);
end

Unless A is huge or you've no preallocated I'd not think this would take long at all. Have you actually profile your code to find out what is the bottleneck or are you guessing about what might be?

[Mike]

within each column in B, the value of B is 1 if the row number is less than or equal to the value in A

e.g. the first column of A contains a 2

the first column of B contains 2 1's, then zeros

the last column of A contains a 3

the last column of B contains 3 1's, then zeros

I apologise about my inability to describe this clearly!

[dpb]

Just, forgot the length argument in B...

B=zeros(max(A));  % preallocate
for i=1:length(B)
  B(1:A(i),i)=ones(A(i),1);
end

Answer 1

[Azzi Abdelmalek]

Not the best solution, but it should work

A=[2,4,1,3]
B=zeros(max(A),numel(A))
idx=cell2mat(arrayfun(@(x) sub2ind(size(B),(1:A(x)),ones(1,A(x))*x),1:numel(A),'un',0))
B(idx)=1

[Mike]

perfect!

Thanks so much!

I had been messing with sub2ind, but not quite worked out what to do with it!

[Azzi Abdelmalek]

Or simply

A=[2,4,1,3];
n=max(A);
m=numel(A);
B=zeros(n,m);
for k=1:m
  B(1:A(k),k)=1;
end
disp(B)

[dpb]

Yep...simpler is often better. I just forgot to write the explicit length on the fly in original response.

Answer 2

[Andrei Bobrov]

n = max(A);
nn = numel(A);
p = zeros(n,nn);
p(A + n*(0:nn-1))=1;
out = flipud(cumsum(flipud(p)));