i need some help on matrix operations!

if i have the index matrix a

a=[0 1 0 1 1 0 1]

and matrix b contains the actual values

v=[2 3 4 2 6 1 8]

here i'm going to check if a(i)=1 then i'm going to do the following:

a(2)=1 then sum=v(4)+v(5)+v(7)

and this will be done again to each one alone..

how to do that in an optimal way?

Answer 1

[sixwwwwww]

do you need something like this:

a=[0 1 0 1 1 0 1];
v=[2 3 4 2 6 1 8];
for i = 1:numel(a)
    sum = 0;
    for j = i:numel(a)
        if a(j) == 1
            sum = sum + v(j);
        end
    end
    sumArray(i) = sum;
end

[mary]

no for example if i have

a=[1 0 1 1]

b=[2 3 4 5]

here a(1)=1 then sum(1) will be equal to 4+5=9

a(3)=1 then sum(2)=2+5=7

[sixwwwwww]

mary try this:

a = [1 0 1 1];
v = [2 3 4 5];
sumArray = zeros(1, numel(a));
for i = 1:numel(a)
    if a(i) ~= 0
        for j = 1:numel(a)
            if a(j) == 1 && j ~= i
                sumArray(i) = sumArray(i) + v(j);
            end
        end
    end
end

[mary]

yea that worked thanks

Answer 2

[Azzi Abdelmalek]

a=[1 0 1 1];
v=[2 3 4 5];
idx=find(a);
n=numel(idx);
ii=cell2mat(arrayfun(@(x) circshift(idx,[0 -x]),(1:n)','un',0));
s=sum(v(ii(:,1:n-1)),2)

Answer 3

[Image Analyst]

Mary, a vectorized, more "MATLAB-ish" way of doing it is:

% Make logical matrix.
a= logical([0 1 0 1 1 0 1]) 
% The "v" matix.
v = [2 3 4 2 6 1 8]

%------------------------------------------------
% Initialize
partialSum = a .* (sum(v(a)) * ones(1, length(a)))
% Subtract the v value
partialSum(a) = partialSum(a)-v(a)

In the command window, you'll see:

a =
     0     1     0     1     1     0     1
v =
     2     3     4     2     6     1     8
partialSum =
     0    19     0    19    19     0    19
partialSum =
     0    16     0    17    13     0    11