How can I find the coordinates of a known vector, within a larger vector?

3 次查看(过去 30 天)
I'd like to take a group of linear vectors, and find where a smaller known vector is placed within them.
For example if I have:
Vector1 = 3 2 4 1 3 2 4 1 3 2 4 2 4 3 4 3 1 4 2 2 3 4 1 4 2
Vector2 = 1 4 2 3 1 4 2 4 2 3 4 3 1 4 2 2 4 1 3 2 4 1 2 3 1 1 1
Vector3 = 1 4 2 3 4 1 4 2 3 1 4 3 4 3 1 4 2 1 4 2 3 1 4 1 3
Vector4 = 1 3 2 4 2 4 1 4 3 2 3 4 3 1 4 2 1 1 3 2 4 2 4 1 2 3
Vector5 = 1 3 2 4 2 1 1 1 3 4 3 2 3 4 3 1 4 2 1 1 3 2 4 2 4 1 2 3
ConstantVector = 3 4 3 1 4 2
I know that each of the vectors 1-5 have the ConstantVector in them somewhere. I would like to be able to quickly output the position of the vector in each case.
The end aim is to be able to put Vectors 1-5 into a matrix of zeros such the sections denoted by ConstantVector are all aligned in columns. For example the above vectors would become:
3 2 4 1 3 2 4 1 3 2 4 2 4|3 4 3 1 4 2|2 3 4 1 4 2 0 0 0 0 0 0
0 0 0 0 1 4 2 3 1 4 2 4 2|3 4 3 1 4 2|2 4 1 3 2 4 1 2 3 1 1 1
ConstantOverLaying = 0 0 1 4 2 3 4 1 4 2 3 1 4|3 4 3 1 4 2|1 4 2 3 1 4 1 3 0 0 0 0
0 0 0 1 3 2 4 2 4 1 4 3 2|3 4 3 1 4 2|1 1 3 2 4 2 4 1 2 3 0 0
0 1 3 2 4 2 1 1 1 3 4 3 2|3 4 3 1 4 2|1 1 3 2 4 2 4 1 2 3 0 0
I've put lines on either side of the ConstantVector just to show that they overlap.
My current thinking is that once I have the coordinates I can use them to automatically shift them left or right as needed when adding them to the matrix of zeroes. If anyone can think of any better ways to do this it is of course welcome.

采纳的回答

Michael Haderlein
You can use strfind:
Vector1 = [3 2 4 1 3 2 4 1 3 2 4 2 4 3 4 3 1 4 2 2 3 4 1 4 2];
Vector2 = [1 4 2 3 1 4 2 4 2 3 4 3 1 4 2 2 4 1 3 2 4 1 2 3 1 1 1];
Vector3 = [1 4 2 3 4 1 4 2 3 1 4 3 4 3 1 4 2 1 4 2 3 1 4 1 3];
Vector4 = [1 3 2 4 2 4 1 4 3 2 3 4 3 1 4 2 1 1 3 2 4 2 4 1 2 3];
Vector5 = [1 3 2 4 2 1 1 1 3 4 3 2 3 4 3 1 4 2 1 1 3 2 4 2 4 1 2 3];
ConstantVector = [3 4 3 1 4 2];
strfind(Vector1,ConstantVector)
if you want all the vectors 1...5 together:
V={Vector1;Vector2;Vector3;Vector4;Vector5};
cellfun(@(x) strfind(x, ConstantVector),V)
Are you able to do the rest on your own? If you need further help, just ask here.
  6 个评论
Michael Haderlein
You've been close to the solution:
FrameStart = max(Positions)
FrameShift = FrameStart-Positions
MaxLength=max(FrameShift+Lengths)
Empty=zeros(length(V),MaxLength);
for cnt=1:length(V)
Empty(cnt,1+FrameShift(cnt):FrameShift(cnt)+Lengths(cnt))=V{cnt};
end
Instead of an arbitrary starting point, I choose the one goes most left. That simplifies further code. The FrameShift must be defined the other way around - big numbers must be shifted far to the right. Finally, it's just simple replacing.
Arthur Radley
Arthur Radley 2015-2-3
Thanks both of you! :)
Michael I understand your version and think I could do it again from scratch.
Stephen your version has things I've never seen before such as num2cell, but I will definitely spend some time "reverse engineering" it to learn the new commands.
Again thanks

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