take line sum of part of matrix

Hi all!

I'm trying to use (as always) vectorized commands but came to a problem which looks quite simple at the beginning. Let me give you an example.

A = [1, 2, 54, 4, 5, 6; ...
1, 2, 88, 4, 5, 6; ...
1, 2, 0, 4, 5, 6; ]; % identical rows except for column 3

b = [5;2;6];

The values in b give the last column in A I want to build the line-sum for: My result should give me

c = ??? = [1+2+54+4+5; 
1+2; 
1+2+0+4+5+6];

Thank you for your help!

Answer 1

[Daniel Shub]

There are a number of ways to do this. A fully vectorized way is with cumsum, but it has a lot of extra calculations so may not be best, especially since MATLAB loops are no longer slow.

A = [1, 2, 54, 4, 5, 6; ...
  1, 2, 88, 4, 5, 6; ...
  1, 2, 0, 4, 5, 6; ];
b = [5;2;6];

Then get the cumulative sum of each row from the beginning to the end.

x = cumsum(A')';

Then extract the values you want.

x(sub2ind(size(x), 1:3, b'))

Solution 2

Give the rows are all identical except for the third column, you can cheat a little bit. If we leave out the 3rd column, then we only need to sum the first row up to the max of b.

maxb = max(b);
n = 3;
x = cumsum(A(1, [1:(n-1), (n+1):maxb]))';

If every element of b was less than 3, we could do

x(b)

If the every element of b was greater than or equal to 3, we could do something like

x(b-1)+*A(:, 3)

where we need to subtract the 1 from the index to account for the fact that we skipped the third column. These two statements can be combined into a single statement with a weight vector that tells us if b is greater or less than 3 (n).

w = (b >= n);

Then we adjust the indices

c = b;
c(w) = b(w)-1;

and finally get the answer

x(c)+w.*A(:, 3)

[Mat]

Thanks, Solution 2 seems a bit overkilling - beside it may not be the third column... Thanks again - didn't know there was the function cumsum!

thanks also to all other answerers! Array function was what I didn't want acutally, but could you tell me (in general) if array/cell functions are faster/slower than loops?!

Thanks!

[Jan]

cumsum(X, 2) is more efficient than cumsum(X')'.

[Daniel Shub]

Of course it is. I am curious as to what lead me to that strange construct and why I didn't notice.

[Matt Fig]

Mat,

I find ARRAYFUN almost always slower than loops and CELLFUN almost always faster. Strange but true. CELLFUN with the string args is very fast indeed.

Answer 2

[Matt Fig]

Another:

% Given data
A = [1, 2, 54, 4, 5, 6;...
     1, 2, 88, 4, 5, 6;...
     1, 2, 0, 4, 5, 6];  
b = [5;2;6];

% The method
R = size(A,1);           % Just to be general.
S = cumsum(A,2);         % The cumulative sum.
S = S((1:R).' + (b-1)*R) % The desired values.

Answer 3

[Thomas]

A = [1, 2, 54, 4, 5, 6;  1, 2, 88, 4, 5, 6; 1, 2, 0, 4, 5, 6; ];
b = [5;2;6];
for ii=1:length(b)
sumA(ii)=sum(A(ii,1:b(ii)));
end

sumA

Answer 4

[Andrei Bobrov]

A = [1, 2, 54, 4, 5, 6; ...
1, 2, 88, 4, 5, 6; ...
1, 2, 0, 4, 5, 6; ];
b = [5;2;6];

s = size(A);
q = zeros(s);
q((b - 1)*s(1) + (1:s(1))') = 1;
i1 = bsxfun(@times,fliplr(cumsum(fliplr(q),2)),(1:s(1))');
t = i1>0;
out = accumarray(i1(t),A(t));