detecting 6 ones in a vector

i need a code to detect 6 consecutive ones in a vector and their places in the vector

[jgg]

Does this do what you want: http://www.mathworks.com/matlabcentral/fileexchange/23998-findsubmat

[Image Analyst]

What if there are 10 ones in a row? You could fit 6 in there in a bunch of places. Would you want the only starting index of the run of 10 elements? Would you want all elements that are part of the 10? Or do you just want the 5 starting indices of where a segment of 6 could fit? You need to clarify because these would be three different algorithms.

[shimaa ali]

@jgg (Findsubmat) geives me an error message(undefined function for variable double ) @Imag Analyst i have a very big vector consists of zeros and ones...i need to find every 6 consecutive ones and add 0 after them...every 6 consecutive ones put 0...that what i want

Answer 1

[John D'Errico]

Trivially, in one line, just us strfind, a tool that works on numeric vectors as well as strings.

% A test case:
V = rand(1,1000) > 0.5;

ind = strfind(V,ones(1,6))
ind =
   323   371   411   412   413   533   576   577   578   600   650   651   652   865   894   955   956   957   958

The vector ind contains the location of the start point of any string of 6 consecutive ones. For example, it found 411, 412, 413, so there was actually a string of 8 consecutive ones, and at the end, a strong of 9 consecutive ones.

If your goal is to find EXACTLY a string of only 6 ones, then use the pattern [0 1 1 1 1 1 1 0]. You would also need to test the first 6 and the last 6 elements then.

Or, you could append a zero to the start and end of the string of elements. Then no special test at the ends would be needed.

So this test will identify strings of EXACTLY 6 ones:

ind = strfind([0,V,0],[0 1 1 1 1 1 1 0])
ind =
   323   371   533   600   865   894

[shimaa ali]

isn't there any other finction that returns the last indices of the occurrences ?

[John D'Errico]

Why do you need another tool?

If the string of interest is known to be 6 elements long, just add 5 to the start point to get the index of the end point.

[shimaa ali]

ok that works with me .. thank u alot :)) i have another question how can i insert an element in a row verctor between 2 elements without replacing any of the matrix elements

Answer 2

[Andrei Bobrov]

b = A == 1; % A - your array
t = [true;diff(b)~=0];
n = find(t);
p = [n,diff([n;numel(A)+1])];
out1 = p(A(n)==1,:);
out = out1(out1(:,2)==6,:);

or with Image Processing Toolbox

c = regionprops(A(:) == 1,'BoundingBox');
k = cat(1,c.BoundingBox);
out = ceil(k(k(:,4)==6,[2,4]));