# Checking parts of different sizes of a vector for ones

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Hello,
I am trying to check if parts of different sizes of a vector are equal to ones' vector of the part's size. Let me be more clear on what I mean:
Suppose that we are given a matrix like the following:
a = [0 0 0 0 1;
0 1 0 0 0;
0 0 0 1 0;
0 0 1 0 0;
1 0 0 0 0]
and also a vector with elements whose sum is equal to the number of the columns in the a matrix. For example:
b =[3 2];
As you can see 3+2 = 5, where 5 is the number of columns in the a matrix.
Now, we are summing the rows of the a matrix dynamically in a loop, and we store this sum in a vector called c. In the example I am using c would be like that in the loop:
first iteration:
c = [0 1 0 0 1]
second iteration:
c = [0 1 0 1 1];
third iteration:
c = [0 1 1 1 1];
fourth iteration:
c = [1 1 1 1 1];
For each iteration what I want to do is to check the following in the example I am using:
I want to check if c from 1 to 3 is equal to ones(1,3) and if c from 4 to 5 is equal to ones(1,2).
I want this check to be done dynamically in the loop and print something when this happens.
I have written a following sample of code which works for some examples I have used:
ni = 5;
noperations = [3 2];
sumoperations = [0 0 0 0 1;0 1 0 0 0;0 0 0 1 0;0 0 1 0 0;1 0 0 0 0]
k=2;
oper = zeros(1,ni);
i=1;
paraggelies = zeros(1,length(noperations));
index = 1;
while (k<=ni)
sumoperations(k,:) = sumoperations(k-1,:) + sumoperations(k,:);
oper(1,:) = sumoperations(k,:)
for i=1:length(noperations)
if(i == 1)
if(oper(1:noperations(i)) == ones(1,noperations(i)))
% disp('Finished Order number')
% disp(i)
paraggelies(index) = i;
index = index+1;
end
else
if(oper(noperations(i-1)+1:noperations(i-1)+noperations(i)) == ones(1,noperations(i)))
% disp('Finished Order number')
% disp(i)
paraggelies(index) = i;
index = index+1;
end
end
end
k = k+1;
end
paraggeliesnew = [];
index = 1;
for i = 2:length(paraggelies)
if(paraggelies(i) ~= paraggelies(i-1))
paraggeliesnew(index) = paraggelies(i-1);
index = index+1;
end
end
but it is pretty slow for the purpose I want to use it and maybe it is buggy.
Is there another way (more vectorized/more "matlab") to achieve efficiency with speed of the problem I described you?
I would greatly appreciate your help.
Chris

Image Analyst on 6 Nov 2016
% Get sums of columns:
c = sum(sumoperations, 1);
Then check c :
indexes1 = 1:b(1);
indexes2 = (b(1)+1) : (b(1) + b(2));
if all(c(indexes1) == 1)
% Then do something.....
end
if all(c(indexes2) == 1)
% Then do something.....
end
I'm using all() to check that every one of the c elements is exactly 1. This can be extended to any number of elements in b of course -- just follow the pattern.
Christos Patrinos on 6 Nov 2016
Thank you for answering... I applied what you suggested in a more general input function and it worked perfectly for some examples I used. Thank you!