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Determining new x,y location of an object

Asked by Sara on 28 Nov 2012

If I know the original x,y location of an object, then move it randomly to a different point on an image, how can I determine it's new x,y location?

2 Comments

John Petersen on 29 Nov 2012

If you moved it, just remember where you moved it to.

Jan Simon on 30 Nov 2012

@Sara: Most likely the problem gets clear, when you post the relevant part of the code.

Of course the information where the parts of the scrambled image are located afterwards are available in your program already. But we cannot guess where or in which format.

Sara

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3 Answers

Answer by Image Analyst on 29 Nov 2012

1 Comment

Image Analyst on 29 Nov 2012

If you move the object by deltaX and deltaY, then obviously the new location is

newX = startingX + deltaX;
newY = startingY + deltaY;

Needless to say, it's new position is (newX, newY). But this is way way too obvious (you wouldn't have even asked about something so trivial and simple) so we must have misunderstood your question. Can you try to describe it in more detail this time?

Image Analyst
Answer by Sara on 29 Nov 2012

The object is moved via code. It is moved to a random location on the image.

5 Comments

Image Analyst on 30 Nov 2012

Let's take it one step at a time. You said "I know the original x,y location of an object" so let's assume it's at (50,50). Now let's say that your image is 1024 pixels wide by 960 pixels high and you chop it up into 2 tiles high by 4 tiles wide so that each tile is 480 pixels high and 256 pixels wide. Now let's say your scrambled the image (in code) so you know that you sent the upper left tile where your object was to the second rows, third panel over. So now the (50,50) point will be sent to the (50+480) row and the (50+256*2) column, or in other words, it's now at (row, column) = (520, 562). Make sense?

Sara on 30 Nov 2012

Here is my code. I am not sending one tile to the same new location each time. What you say makes sense, I'm just not sure how that works with this:

A=mat2cell(image,[384,384],[256,256,256,256],[3]);

B=randperm(8);

C=reshape(B,2,4);

D=mat2cell(C,[1,1],[1,1,1,1]);

[row,col]=find(C==1)

D(row,col)=A(1,1)

[row,col]=find(C==2)

D(row,col)=A(1,2)

[row,col]=find(C==3)

D(row,col)=A(1,3)

[row,col]=find(C==4)

D(row,col)=A(1,4)

[row,col]=find(C==5)

D(row,col)=A(2,1)

[row,col]=find(C==6)

D(row,col)=A(2,2)

[row,col]=find(C==7)

D(row,col)=A(2,3)

[row,col]=find(C==8)

D(row,col)=A(2,4)

scram=cell2mat(D); fig3=figure('Visible','on'); imshow(scram)

Sara on 30 Nov 2012

the intial location of the pedestrian is located in an excel file that i upload, and I am running this for about 500 images.

Sara
Answer by John Petersen on 30 Nov 2012

You know which cell in A that the pedestrian is located, right? Say it's the A(r,c) cell. Then,

ped = (c-1)*2 + r; % location of the pedestrian in A strung out, 
                   % i.e. A(ped) = A(r,c)
B=randperm(8);
C=reshape(B,2,4);
D=mat2cell(C,[1,1],[1,1,1,1]);
for i=1:8
   [row,col]=find(C==i) % move tile i to row,col
   D(row,col)=A(i);  % Build new image 
   if (C(i)==ped)
        Row_newped = row;  %FOUND PEDESTRIAN in New image
        Col_newped = col;
   end
end

There's probably a more efficient code that doesn't require a loop, but this should be clear to you so you understand the idea.

1 Comment

Sara on 3 Dec 2012

I apologize to everyone for being unclear. I am teaching myself Matlab and I appreciate everyone's help with this.

@John - Thank you for your post, I understand now what I need to do.

John Petersen

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