CODEWITHSUNDEEP
c#imageimage-processingemgucvimage-scaling
Khalil Khalaf
Khalil Khalaf

Reputation: 9407

How to transform an image to another coordinates?

I am trying to find duplicate videos in my database, and in order to do so I grab two frames from a videos pair, resize them to the same width and height and then compare both images pixel by pixel.

I have a case where images from the a videos pair are like below:

enter image description here ----- enter image description here

These are actually the same videos (images), but because of the aspect ratio of the videos (16:9, 4:3 .. etc) the result is negative when comparing pixel by pixel (no match).

If my standard is 50x50, how can I transform any Region Of Interest to 50x50?

For the above example:

Pixel [5,0] shall be [0,0]

Pixel [45,0] shall be [50,0]

Pixel [5,50] shall be [0,50]

Pixel [45,50] shall be [50,50]

and all other pixels are transformed

Upvotes: 1

Views: 1226

Answers (1)

javaLover
javaLover

Reputation: 6425

Encouraged by OP that pseudo-code can be helpful ....
I have no knowledge about "emgucv", so I will answer in pseudo-code.

Definition

Let SRC be a source image - to be read.
Let DST be a destination image - to be written.

Both SRC and DST are 2D-array, can be accessed as ARRAY[int pixelX,int pixelY].

Here is the pseudo-code :-

input : int srcMinX,srcMinY,srcMaxX,srcMaxY;
float linearGra(float dst1,float dst2,float src1,float src2,float dst3){
    return ( (dst3-dst1)*src2+  (dst2-dst3)*src1) / (dst2-dst1);
};
for(int y=0;y<50;y++){     //y of DST
    for(int x=0;x<50;x++){ //x of DST
        float xSRC=linearGra(0,50,srcMinX,srcMaxX,x);
        float ySRC=linearGra(0,50,srcMinY,srcMaxY,y);
        DST[x,y]=SRC[round(xSRC),round(ySRC)];  //find nearest pixel
    }
}

Description

The main idea is to use linear-interpolation.

The function linearGra takes two points in a 2D graph (dst1,src1) and (dst2,src2) .

Assuming that it is a linear function (it is true because scaling+moving is linear function between SRC and DST coordinate), it will find the point (dst3,?) that lying in the graph.

I used this function to calculate pixel coordinate in SRC that match a certain pixel in DST.

Further work

If you are a perfectionist, you may want to :-

  • bounded the index (xSRC,ySRC) - so it will not index-out-of-bound
  • improve the accurary :-
    I currently ignore some pixels (I use Nearest-neighbor w/o interpolation).
    The better approach is to integrate all involved SRC's pixel, but you will get a-bit-blurry image in some cases.

You may also be interested in this opencv link (not emgucv).

Upvotes: 1

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