# Select every Nth row from number groups

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Mat on 20 Jan 2015
Answered: Jos (10584) on 17 Feb 2015
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 on 20 Jan 2015
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]
?

Mat on 17 Feb 2015
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

Ondrej on 20 Jan 2015
Edited: Ondrej on 20 Jan 2015
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 on 20 Jan 2015
Unfortunately, they are not always the same length (as example)!!
Ondrej on 20 Jan 2015
Edited: Ondrej on 20 Jan 2015
If they are not, what would be the next output for your case?
e = [4 NaN 22] ?

Elias Gule on 17 Feb 2015
Edited: Elias Gule on 17 Feb 2015
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));

Jos (10584) on 17 Feb 2015
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