find sequence in a matrix

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shamal
shamal 2025-5-11
编辑： Torsten 2025-5-13
Ran in:
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

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采纳的回答

Torsten
Torsten 2025-5-11
编辑：Torsten 2025-5-11
Ran in:
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.
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Image Analyst
Image Analyst 2025-5-13
Ran in:
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
Torsten 2025-5-13
编辑：Torsten 2025-5-13
Ran in:
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].

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更多回答（4 个）

Image Analyst
Image Analyst 2025-5-11
Ran in:
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.
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Image Analyst
Image Analyst 2025-5-12
编辑：Image Analyst 2025-5-12
Ran in:
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
Matt J 2025-5-12
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.

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Matt J
Matt J 2025-5-11
编辑：Matt J 2025-5-13
Ran in:
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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Matt J
Matt J 2025-5-12
编辑：Matt J 2025-5-13
Ran in:
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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Matt J
Matt J 2025-5-13
Ran in:
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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