Can you split a vector according to a pre-defined sequence?

Can anyone suggest a way in which it is possible to perform operations on a vector according to a predetermined sequence - for example I have a vector of different values, M, which is < 8760x1 > in size. I have another vector with a sequence of numbers, P, (size < 300x1 >) and this sequence sums to 8760. I would like to use these P values to index the vector M and find the product of each index.

An example to make this clearer:

M = [1,2,4,2,3,4,5,3,4,2];

P = [2,2,4,2];

Result = [3,6,15,6]

Any help here would be greatly appreciated.

Peter.S.

[Azzi Abdelmalek]

how did you find 6 ?

[Image Analyst]

Can you explain how you got Result? Because I don't understand. Using P as indexes of M gives [M(2), M(2), M(4), M(2)], and the product of all those i = 2*2*2*2 which equals 16, not 3 (which is Result(1)). Why do you have 4 Result elements and not just 1? On the other hand "the product of each index." would be 2*2*4*2 which is the product of all the indexes (the numbers in P), and that is 32 - again, not 3 and not [3,6,15,6].

[Pete]

Sorry I'm not sure which 6 you are referring to -

Result gives the value of M(1,1)+M(1,2) =3, then M(1,3)+M(1,4) =6, then M(1,5)+M(1,6)+M(1,7)+M(1,8)=15 and M(1,9)+ M(1,10) =6

Answer 1

[Azzi Abdelmalek]

M = [1,2,4,2,3,4,5,3,4,2];
P = [2,2,4,2];
id2=cumsum(P);
id1=[1 id2(1:end-1)+1];
for k=1:numel(id1)
  Result(k)=sum(M(id1(k):id2(k)));
end

%or

id2=cumsum(P);
id1=[1 id2(1:end-1)+1];
Result=arrayfun(@(x,y) sum(M(x:y)),id1,id2)

[Pete]

Thankyou that does work perfectly. I also need the product of these values i.e. so the Result would now be [2,8,180,8]. Is there a similar method to achieve this?

Thanks again for your help.

[Pete]

Sorry that was not a good question, simply swap the sum for prod! Thanks again!

Answer 2

[Cedric]

 id = zeros(numel(M), 1) ;
 id(1+cumsum(P)) = 1 ;
 id = 1 + cumsum(id(1:end-1)) ;
 result = accumarray(id, M(:)).' ;

With the vectors M and P from your statement, this code leads to:

 result =
     3     6    15     6

Note that for small sizes of M and P, Azzi's solution is much more efficient than mine, and my solution becomes more efficient than his when sizes become significant.

[Azzi Abdelmalek]

Cedric, I've tried your code for

M = randi(101,1,10001)
P = randi(80,1,100)

It seems there is a problem

[Cedric]

Hello Azzi, yes, in your test sum(P) is not equal to length(M).

I performed my tests on larger arrays with

 n = 1000 ;    % 1 for original setup.
 M = repmat([1,2,4,2,3,4,5,3,4,2], 1, n) ;
 P = repmat([2,2,4,2], 1, n) ;

 tic ;
 result_AA1 = zeros(size(P)) ;
 idx=cumsum(P);  ii=1;  jj=P(1);
 for k=1:numel(idx)
    jj=P(k)+ii-1;
    result_AA1(k)=sum(M(ii:jj));
    ii=jj+1;
 end
 toc

 tic ;
 id2=cumsum(P);
 id1=[1 id2(1:end-1)+1];
 result_AA2 = arrayfun(@(x,y) sum(M(x:y)),id1,id2) ;
 toc

 tic ;
 id = zeros(numel(M), 1) ;
 id(1+cumsum(P)) = 1 ;
 id = 1 + cumsum(id(1:end-1)) ;
 result_CW = accumarray(id, M(:)).' ;
 toc

 all(result_AA2 == result_AA1 & result_CW == result_AA1)

And I obtained

 Elapsed time is 0.010770 seconds.
 Elapsed time is 0.031326 seconds.
 Elapsed time is 0.000788 seconds.
 ans =
     1

EDIT: for n=1, however:

 Elapsed time is 0.000036 seconds.
 Elapsed time is 0.001322 seconds.
 Elapsed time is 0.000582 seconds.
 ans =
     1

[Azzi Abdelmalek]

Exact, I was wondering what was wrong.