find sequence in a matrix

Hi, it possibile to velocize it and avoid loop? (I will be a matrix .,not a single array)

(if is possible to use vectorization)

I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0]

I = 1×18

0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0

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 k=3;
 
 for i=1:numel(I)
     if i>=k+1
         if I(i-1) && ~I(i) %%i want to find the first "1 0"
            saved=i; 
            break;
         end
     end    
 end   

 saved

saved = 7

Answer 1

[Torsten]

I = [0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0];
k = 3;
saved = k + find(diff(I(k:end))==-1,1)

saved = 7

If I is a matrix, you will have to tell us your search direction and the role of k.

[shamal]

can you do an example using matrix?

I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0]

[Torsten]

Using @Matt J 's idea:

I = [0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0];
k = 3;
[~,loc]=min(diff(I(:,k:end),1,2),[],2);
k + loc

ans = 3×1

7 5 5

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[Image Analyst]

Where is the search pattern defined in that code? Where does it say it's looking for the pattern [1, 0]?

Also can it find a 2-D pattern in a 2-D matrix. Like for the given matrix

I = [0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0]

I = 3×18

0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0

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Can it find the pattern

1 0 1

1 0 0

?

[Torsten]

Where is the search pattern defined in that code? Where does it say it's looking for the pattern [1, 0]?

Let x be a vector consisting only of 0 and 1 and assume that the pattern [1 0] somewhere appears in the vector x (if it's not present in x, the method will give a wrong answer).

diff(x) can give values -1, 0 and 1 - so -1 is the minimum value you can get when you compute diff(x). On the other hand, getting -1 for diff(x) in position i means x(i+1) - x(i) = -1 which is only possible if x(i+1) = 0 and x(i) = 1. Thus a position i where diff(x) attains its minimum (namely -1) is a position where the pattern [1 0] (x(i) = 1, x(i+1) = 0) appears in the vector x.

Can it find the pattern

1 0 1

1 0 0

?

diff([0 0 0],2)

ans = 0

diff([1 1 1],2)

ans = 0

diff([1 0 0],2)

ans = 1

diff([0 1 0],2)

ans = -2

diff([0 0 1],2)

ans = 1

diff([1 1 0],2)

ans = -1

diff([0 1 1],2)

ans = -1

diff([1 0 1],2)

ans = 2

Only pattern [0 1 0] gives answer -2 and pattern [1 0 1] gives answer 2. Thus these two patterns could be found by a similar method (min and max) as the one used for pattern [1 0].

Answer 2

[Image Analyst]

The first [1, 0] shows up at index 2, not 7. It also appears at index 6 and others.

Probably the simplest way (a single line of code) is to use strfind. (Yes it works with numbers as well as character strings).

% Set up parameters:
I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0];
searchPattern = [1, 0];

% Now find all locations where the vector has values [1, 0]
indexes = strfind(I, searchPattern) 

indexes = 1×5

2 6 12 15 17

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If you want the first location, just take indexes(1).

If you have a 2-D array, you can use normxcorr2. See attached demo.

[shamal]

if you read my post with attention i want to begin from k=3..look my code

[Image Analyst]

You didn't explicitly say that. And by looking at your code, it only starts looking at values for i>=k+1, or 4, not 3. And then it looks like it finds the trailing 0, not the leading 1, so it finds index 7 rather than 6 which is where the [1,0] pattern starts. So it was kind of confusing to me. Not sure what you want exactly.

If you want to find the index of the trailing 0 after index 3, you can do

% Set up parameters:
I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0];
k=3;
searchPattern = [1, 0];
% Now find locations where the array has values [1, 0]
indexes = strfind(I(k:end), searchPattern) + k

indexes = 1×4

7 13 16 18

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If you want to find the index of the leading 1 after index 3, you can do

% Set up parameters:
I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0];
k=3;
searchPattern = [1, 0];
% Now find locations where the array has values [1, 0]
indexes = strfind(I(k:end), searchPattern) + k - 1

indexes = 1×4

6 12 15 17

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Either way, it's still simpler than the other solutions.

[Matt J]

Simpler, but not as efficient. strfind() will not operate row-wise on a matrix. You will have to loop. Additionally, it will search the entire row, unlike the other solutions which will stop at the first occurence.

Answer 3

[Matt J]

I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0]

I = 3×18

0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0

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k=3;


[minval,loc]=min(diff(I(:,k:end),1,2),[],2);
loc(minval>-1)=nan;
loc=loc+(k-1)

loc = 3×1

6 4 4

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Answer 4

[Matt J]

I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0];
k=3;

Ik=string(char(I(:,k:end)+'0'));

loc=strlength(extractBefore( Ik , '10'))+k

loc = 3×1

6 4 4

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Answer 5

[Matt J]

I=[0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0;
    0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1;
    1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0]

I = 3×18

0 1 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 1 1 0 1 0 1 0 0 0 1 0 1 0 0 1 1 1 0

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k=3;

C=conv2(I,[0,1],'valid')./conv2(I,[1,1],'valid')==1

C = 3x18 logical array
   0   1   0   0   0   1   0   0   0   0   0   1   0   0   1   0   1   0
   0   1   0   1   0   1   0   1   0   1   0   1   0   0   1   0   0   1
   0   1   0   1   0   1   0   0   0   1   0   1   0   0   0   0   1   0

C(:,1:k-1)=0;
[~,loc]=max(C,[],2)

loc = 3×1

6 4 4

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