Select every Nth row from number groups

I have a vector that looks like

a = [1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 1....]

I would like to find the vector location where I would find the first 1, second 1, third 1 and so on, in the trains of 1's in a quickish way.

As an output... I would like new vectors containing only the first 1 in every sequence of 1's

eg

b = [1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0....]
c = [0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0....]

OR

a Matrix containing the locations of each number in a sequence...

b = [1 13 19]
c = [2 14 20]
d = [3 15 21].... and so forth

[Ondrej]

Can you please clarify your question (maybe some example of what you expect)? Do you want the indices of all ones in the vector?, e.g.

a = [1 1 0 1], result = [1 2 4]

?

Answer 1

[Mat]

I eventually solved this, and am posting my solution... The actual example had some strings containing a mixture of A and B's also, so that's included in this answer.

a='A'
b='B'
aab='AAB'
asandbs=[a;a;a;a;a;a;aab;b;b;b;b;b;b;b;b;a;a;a;a;a;b;b;b;b;a;a;a;a;a;a;a]
str2cell(asandbs)

Blocs=strcmp(asandbs,'B')==1;
Alocs=strcmp(asandbs,'A')==1;
changeovers=~(Alocs+Blocs);

Borders=zeros(size(Blocs,1),1);
for i=2:size(Blocs,1)
    if Blocs(i,:)==1
        Borders(i,1)=1+Borders(i-1,1)
    else
        Borders(i,1)=0
    end
end

Then repeat the loop for A's

Answer 2

[Ondrej]

If the n "trains" of ones have the same length always, then this probably what you want:

result = repmat(find(diff([0 a])==1),n,1)+(0:n-1)'*ones(1,n)

For example:

a = [1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1];
n = 3;
result = repmat(find(diff([0 a])==1),n,1)+(0:n-1)'*ones(1,n)

>> result =

       1    13    21
       2    14    22
       3    15    23

UPDATE:

If the number of your outputs should be equal to the shortest length of the train of ones (in your case 3), then you have to find 'n', e.g.,

tmp = find(diff([0 a])==1);
n = min(find(diff([a 0])==-1) - tmp)+1;
m = length(tmp);
result = repmat(tmp,n,1)+(0:n-1)'*ones(1,m)

[Mat]

Unfortunately, they are not always the same length (as example)!!

[Ondrej]

If they are not, what would be the next output for your case?

e = [4 NaN 22] ?

Answer 3

[Elias Gule]

a = [1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1];

isTake = false;

newMatrix = nan*ones(size(a)); count = 0;

for k = 1 : length(a) val = a(k);

    if((val == 1) && ~isTake)
        count = count + 1;
        newMatrix(count) = k;
        isTake = true;
    else
        isTake = (val == 1);
    end

end

newMatrix = newMatrix(~isnan(newMatrix));

Answer 4

[Jos (10584)]

a = [1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 1]
k = 1 ;
while any(a)
  i0 = strfind([0 a],[0 1]) ;
  OUT{k} = i0 ;
  a(i0) = 0 ;
  k = k + 1 ;
end

% OUT{X} holds those indices of A where a 1 occurs as the X-th element in the series of 1's