matlab: Sparse Function for any matrix

Question

I am trying to write a sparse function code for any matrix size. For example if i have a matrix:

A = [ 0   0   0   5
      0   2   0   0
      1   3   0   0
      0   0   4   0];
  a=size(A);
  b=size(A);
  c=0;
  position=0;
  for i=1:a for j=1:b if A(i,j) ~=0
              c=c+1;
              position=position+1;
              S(c,:)=[position,i,j,A(i,j)];
          end
      end
  end
S

S --> is the storage matrix for all the non zero elements for A matrix. In addition to this information(Index,Row Number, Column Number, Value) how do I include two more columns in the matrix which shows the next element in row & the next element in column.

Any reason why the built-in sparse matrix functions are inadequate? — Adam, Oct 24 '13 at 05:02
I am trying to write code without using find function and sparse function — user2913990, Oct 24 '13 at 05:04
@user2913990 abd why would you like to do that??? I'm working with `sparse` and `find` for quite a while - they rock! What **exactly** is the problem you are trying to solve by re-writing the sparse class? — Shai, Oct 24 '13 at 05:18
@user2913990 - one more thing - pre-allocate `S`. Your current code re-allocates `S` at each iteration, and this kind of thing **kills** Matlab's performance... — Shai, Oct 24 '13 at 06:26
Just one observation - you're storing way more information than Matlab does for a sparse matrix. So in terms of memory you'll be much better off using the builtin `sparse`, and since you're implementing things in matlab-code the builtin surely is faster as well. So if it's not only for academic reasons, it's difficult to see the point — sebastian, Oct 24 '13 at 06:41

Rody Oldenhuis · Answer 1 · 2013-10-24T12:29:27.300

Here's the fruits of some academic masturbation :)

1) It uses an array-growth scheme that is a lot more efficient than growing the array at each iteration 2) It optimizes memory footprint/usage by using the minimum necessary data class for the indices

K      = 1.1;
inds   = zeros(10,2, 'uint8');
values = zeros(10,1, class(A));

count = 1;
for ii = 1:numel(A)    
    if A(ii)

        %// grow arrays and/or change type when necessary
        if count > size(inds,1)

            %// Re-cast datatype of inds
            if count > intmax(class(inds))
                switch class(inds)
                    case 'uint8'
                        inds = uint8(inds);
                    case 'uint16'
                        inds = uint32(inds);
                    case 'uint32'
                        inds = uint64(inds);
                    case 'uint64'
                        error('There are too many non-zero elements.');
                    otherwise
                        error('...huh?!');
                end
            end

            %// now grow the arrays
            inds   = [inds;   zeros(ceil((K-1)*count), 2, class(inds))  ]; %#ok<AGROW>
            values = [values; zeros(ceil((K-1)*count), 1, class(values))]; %#ok<AGROW>

        end

        %// assign the indices and values
        inds(count,:) = [mod(ii-1,size(A,1))+1; ceil(ii/size(A,1))];
        values(count) = A(ii);
        count = count + 1;

    end
end

%// chop-off any remaining zeros
inds   = inds  (1:count-1,:);
values = values(1:count-1,:);

Aside from some relatively minor optimizations you could still do, this is about as efficient as you can make it in MATLAB.

Testing this implementation vs. the built-in sparse for A = rand(500); A(A<0.5) = 0; gives:

Elapsed time is 1.802105 seconds.  %// implementation above
Elapsed time is 0.006877 seconds.  %// sparse

So about 250 times slower. All of this just proves one thing:

DON'T DO THIS SORT OF THING IN MATLAB!

@DennisJaheruddin: see [here](http://www.urbandictionary.com/define.php?term=mental%20masturbation). — Rody Oldenhuis, Oct 24 '13 at 12:31
I don't see how this invalidates my suggestion, but if it is just me I won't make a point out of it. My vote stays as the example is very educational. --- Sidenote: 50% empty cells is not a very nice sparse example (doesn't influence the timings though) — Dennis Jaheruddin, Oct 24 '13 at 12:40

score 0 · Answer 2 · answered Oct 28 '13 at 18:18

A = [ -1   0  -2   0   0
       2   8   0   1   0
       0   0   3   0  -2
       0  -3   2   0   0
       1   2   0   0  -4];

  a=size(A);
  b=size(A);
  c=0;
  position=0;
  for i=1:a 
      for j=1:b 
          if A(i,j) ~=0
              c=c+1;
              position=position+1;
              S(c,:)=[position,i,j,A(i,j)];
          end
      end
  end

  index=size(S,1);
  NIR = [];
for i = 1:index
    for j = 1:index
        if i ~= j
            if S(i,2) == S(j,2)if S(i,3) < S(j,3)if length(NIR) < i
                        NIR(i,1) = S(j,1);
                    end
                end
            end
        end
    end
    if length(NIR) < i
        NIR(i,1) = 0;
    end
end

%finding NIC
NIC = [];
for i = 1:index
    for j = 1:index
        if i ~= j
            if S(i,3) == S(j,3)if S(i,2) < S(j,2)if length(NIC) < i
                        NIC(i,1) = S(j,1);
                    end
                end
            end
        end
    end
    if length(NIC) < i
        NIC(i,1) = 0;
    end
end

Sprs=[S,NIR,NIC]

n=nnz(A);
r=size(A);
m=r(1,1);
position=zeros(n,1);
t=1;
FIR= zeros(m,1);
FIC= zeros(m,1);
for i= 1: n
    position(i,1)= i;
end

for i= 1: m
    for j= 1: m
        if (A(i,j)~= 0)
            value(var,1)= A(i,j);
            row(var,1)= i;
            col(var,1)= j;
            var= var+ 1;
        end
    end
end
for j= 1: m
    for i= t: n
        if row(i,1)== j
            FIR(j,1)= position(i,1);
            t= i+1;
            break;
        end
    end
end
for k= 1: m
for l= 1: n
    if (col(l,1)== k)
        FIC(k,1)= position(l,1);
        break;
    end
end
end

S1=[FIR, FIC]

matlab: Sparse Function for any matrix

2 Answers2

DON'T DO THIS SORT OF THING IN MATLAB!