Updating a matrix based on conditions in other matrixes

I have coded the attached loop to update matrix Lr but I would ideally like to use code that can achieve what the shown loop is doing in one sweep (i.e. without needing to loop); the actual matrixes I have can be extremely large.

The code is also shown below, but the layout is off:

k = 1;
pos_neg = [1;1;0];
E = [round(rand(1,3));round(rand(1,3));round(rand(1,3))];
Lr = zeros(numel(pos_neg(:,1)),numel(E(:,1)));
L = 0.5;

for i = 1:numel(E(:,1)) 
    for ii = 1:numel(E(:,1)) 
        if  sum(E(i,:)) <= k
            if  E(i,ii) == 0
                Lr(i,ii) = L;
            end
            if  E(i,ii) == 1
                Lr(i,ii) = -L;
            end    
        end
        if  sum(E(i,:)) >= k
            if  E(i,ii) == 1
                Lr(i,ii) = L;
            end
            if  E(i,ii) == 0
                Lr(i,ii) = -L;
            end    
        end
    end
end

[Ngonidzashe Nedziwe]

hi i would like to help but im confused about what u exactly mean or are trying to achieve when u say in one sweep

[Ulrik William Nash]

Thank you, Ngonidzashe, what I mean is without looping.

Answer 1

[Walter Roberson]

If I work through the logic correctly,

Lr = L*(1 - 2 * bsxfun( @eq, sum(E,2) <= k, E));

Or if you have R2016b or later,

Lr = L*(1 - 2 * ((sum(E,2) <= k) == E));

[Ulrik William Nash]

Hi Walter,

If I set

E =

     1     1     1
     0     0     1
     1     0     0

then I get

Lr =

    0.5000    0.5000    0.5000
   -0.5000   -0.5000    0.5000
    0.5000   -0.5000   -0.5000

but your code yields:

Lr =

    0.5000    0.5000    0.5000
    0.5000    0.5000   -0.5000
   -0.5000    0.5000    0.5000

[Walter Roberson]

Sorry, I would have to look again later when I am less tired.

[Ulrik William Nash]

Thank you Walter, no problem. In the mean time I will look at your approach. Maybe I can figure it out.

[Ulrik William Nash]

Hi Walter,

I think your mistake was very small. As far as I can tell, all that is required is the following change:

From: Lr = L*(1 - 2 * bsxfun( @eq, sum(E,2) <= k, E))

To:

Lr = L*(2 * bsxfun( @eq, sum(E,2) >= k, E)-1)

Answer 2

[Jan]

Start with cleaning the code by avoiding all repeated calculations inside the loop:

k = 1;
pos_neg = [1;1;0];
E  = [round(rand(1,3));round(rand(1,3));round(rand(1,3))];
Lr = zeros(numel(pos_neg(:,1)),numel(E(:,1)));
L = 0.5;

sE = size(E, 1);
for i = 1:sE
    sumEi = sum(E(i,:));
    for ii = 1:sE
        if sumEi <= k
            if  E(i,ii) == 0
                Lr(i,ii) = L;
            end
            if  E(i,ii) == 1
                Lr(i,ii) = -L;
            end    
        end
        if sumEi >= k
            if  E(i,ii) == 1
                Lr(i,ii) = L;
            end
            if  E(i,ii) == 0
                Lr(i,ii) = -L;
            end    
        end
    end
end

Now you see that the "if sumEi <= k" test does not depend on the "for ii" loop. Then move it outside this loop:

sE = size(E, 1);
for i = 1:sE
    sumEi = sum(E(i,:));
    if sumEi <= k
      for ii = 1:sE
        if  E(i,ii) == 0
           Lr(i,ii) = L;
        elseif E(i,ii) == 1
           Lr(i,ii) = -L;
        end    
      end
    elseif sumEi >= k
      for ii = 1:sE
        if  E(i,ii) == 1
           Lr(i,ii) = L;
        elseif E(i,ii) == 0
           Lr(i,ii) = -L;
        end    
      end
    end
end

Now see, shat you want to achieve actually (this should be contained as a clear and exhaustive comment in the code): Lr is set to L or -L. So start with set all elements to L and swap the sign for the specific elements:

Lr = repmat(L size(pos_neg, 1), size(E,1));

size(E, 1) is more efficient than numel(E(:,1)), because the latter creates a temporary vector only to measure its length.

sumEi = sum(E, 1);
index = ((sumEi <= k) & (E == 1)) | ...
         (sumEi >= k) & (E == 0));    % >= R2016b

Note that sumEi == k belongs to both cases. I assume one of them should be a > or < .

If the values of E are either 1 or 0, this can be expressed in one logical comparison only. I cannot test this currently, but it is something like:#

Lr    = repmat(L, size(pos_neg, 1), size(E,1));
index = xor(sum(E, 1) <= k, E == 1);  % Or E==0? Or < instead of <= ? Try this by your own
Lr(index) = -L;

[Ulrik William Nash]

Hi Jan,

This was very instructive! I have been making some inefficiencies in my coding.

Regarding the specific solution you find, perhaps it is dependent on the Matlab version (I have 2015), because I obtain the following error:

xor(sum(E,1) <= k, E == 1) Error using xor Matrix dimensions must agree.

[Andrei Bobrov]

index = bsxfun(@xor,sum(E,1) <= k,E == 1);