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Attention please:
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If you want to reprint my article, please mark the original address and author(刘书裴).
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If you are puzzled about a certain part or have some better suggestions, you can contact me: 3017218062@tju.edu.cn/1005968086@qq.com
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If my blog has mistakes, I'm so sorry!
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I think and write everything. Please don't copy.
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Thanks for watching!
| item | tool |
|---|---|
| image | mspaint |
| formula | Online LaTeX Equation Editor |
| language | python3.7 |
| date | 2019.11.10 |
| author | 刘书裴 |
Original image:
- Here is the pixel distribution of the two pictures(src:3x3 and dst:5x5):
- Then overlap two pictures:
- Choose the white pixel and calculate it:
1. Take the boundary pixel of the upper left corner as the origin (0,0).
For the destination matrix, the pixel I choose is (2,1).
realPixel1 = (2, 1)
Calculate the ratio between src and dst:
heightRatio = srcHeight / dstHeight = 0.6
widthRatio = srcWidth / dstWidth = 0.6
2. Take the boundary point of the upper left corner as the origin (0,0).
For the white pixel, its coordinate in destination matrix is (2.5,1.5):
realPixel2 = (2.5, 1.5) => _realPixel2 = realPixel1 + (0.5, 0.5)
3. Converts coordinates from the dst coordinate system to the src coordinate system:
virtualPixel2 = realPixel2 * (heightRatio, widthRatio) = (1.5, 0.9)
virtualPixel1 = (1.0, 0.4) => virtualPixel1 = virtualPixel2 - (0.5, 0.5)
4. In summary:
virtualPixel = (realPixel + 0.5) * (heightRatio, WidthRatio) - 0.5
For nearest, we only need to take the rounding value of the corresponding coordinates.
nearestPixel = int(virtualPixel + 0.5)
If we choose single coordinate system like pixel coordinate, the virtualPixel(realPixel*ratio) may be out of bounds when the radtio is greater than or equal to 0.5. For point coordinate, it can't represents the pixel between two points.
As an example:
for the realPixel (src:3x3,dst:6x6)
int((5,5)*0.5)=int((2.5,2.5))=(3,3) is out of range(0,3).
For many programming languages, they can remove data after decimal point when they change the type of float to int.
And Adding 0.5 can a number greater than or equal to 0.5 to add 1 to its integer.
Calculate the nearest coordinate matrix, then replace with the pixel corresponding to the coordinate.
For a virtualPixel O(x,y), its neighbors are
- A(int(x),int(y))
- B(int(x),int(y+1))
- C(int(x+1),int(y))
- D(int(x+1),int(y+1)).
Specially, for boundary pixels, I delete the neighbors out of bounds.
At first, we should know that the order of interpolation has no effect on the result.
- Assume that A(x1,y1), B(x1,y2), C(x2,y1), D(x2,y2), O(x,y).
- For A and B,
- For C and D,
- For AB and CD,
Obviously:
xUp<=xDown and yLeft<=yRight
From virtualPixel = (realPixel + 0.5) * (heightRatio, WidthRatio) - 0.5, we can know:
-0.5<x<height-0.5 and -0.5<y<width-0.5
If (xUp<0 or yLeft<0):
Look at the int(x), int(y), int(x)>=0 and int(y)>=0.
So (xUp<0 or yLeft<0) is False, and we don't need to deal with it.
If (xDown>height-1 or yRight>width-1):
There is a x or a y, int(x+1)>height-1 and int(y+1)>width-1.
So we need to substract 1 from xDown or yRight out of bounds.
Thanks for 展希希鸿's blog. On this basis, I improve the algorithm and improve its generalization ability.
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Assume that A(0,0), B(0,1), C(1,0), D(2,2), O(x,y).
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We can get the formula:
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Then we replace it with matrix:
The Basic algorithm is for special cases. So we can't directly use it.
However, the coordinates can be translated. So we can translate A/B/C/D/O to coordinates between 0 and 1.
To be convenient, we only need to change O with x%1 and y%1.
- For RGB, we should repeat f(O) three times.
for i in range(3):
fr(O)|fg(O)|fb(O)
- But we can simplify it:
For single channel, the double loop can be replaced with A, B, C, D according to the formula.
The we can calculate a channel with the formula.
