如何使用PIL将矩形图像映射到四边形? [英] How to map rectangle image to quadrilateral with PIL?

问题描述

Python PIL库允许我使用

Python PIL library allows me to map any quadrilateral in an image to rectangle using

im.transform(size, QUAD, data)

我需要的是一个相反的功能，即将矩形图像映射到指定的四边形.

What I need is a function that does the opposite, i.e. map a rectangular image to specified quadrilateral.

我认为这可以通过上述功能来实现:

I figured this might be achieved with the above mentioned function like this:

即我会找到这样的四边形(图像中的红色)，可以使用函数im.transform(size，QUAD，data)将图像转换为我想要的四边形.问题是我不知道如何找到红色四边形.

I.e. I would find such quad (the red one in the image) that would, using the function im.transform(size, QUAD, data) transform the image to quad I want. The problem is I don't know how to find the red quad.

对于任何关于如何找到红色四边形的想法，或任何其他将rect图像映射到四边形的方法，我将不胜感激.

I would appreciate any idea on how to find the red quad or any other way to map a rect image to quad, only with PIL if possible.

推荐答案

因此，我通过简单的前向映射而不是逆向映射解决了这个问题，通常情况下，逆向映射会更好，但是在我的应用程序中，我只将矩形映射到小于矩形的四边形，因此变换后的图像中通常没有孔.代码如下:

So I solved the issue with a simple forward mapping, rather than inverse mapping, which is usually better, but in my application I only ever map the rectangle to a quad that is smaller than the rectangle, so there are usually no holes in the transformed image. The code is as follows:

def reverse_quad_transform(image, quad_to_map_to, alpha):
  # forward mapping, for simplicity

  result = Image.new("RGBA",image.size)
  result_pixels = result.load()

  width, height = result.size

  for y in range(height):
    for x in range(width):
      result_pixels[x,y] = (0,0,0,0)

  p1 = (quad_to_map_to[0],quad_to_map_to[1])
  p2 = (quad_to_map_to[2],quad_to_map_to[3])
  p3 = (quad_to_map_to[4],quad_to_map_to[5])
  p4 = (quad_to_map_to[6],quad_to_map_to[7])

  p1_p2_vec = (p2[0] - p1[0],p2[1] - p1[1])
  p4_p3_vec = (p3[0] - p4[0],p3[1] - p4[1])

  for y in range(height):
    for x in range(width):
      pixel = image.getpixel((x,y))

      y_percentage = y / float(height)
      x_percentage = x / float(width)

      # interpolate vertically
      pa = (p1[0] + p1_p2_vec[0] * y_percentage, p1[1] + p1_p2_vec[1] * y_percentage) 
      pb = (p4[0] + p4_p3_vec[0] * y_percentage, p4[1] + p4_p3_vec[1] * y_percentage)

      pa_to_pb_vec = (pb[0] - pa[0],pb[1] - pa[1])

      # interpolate horizontally
      p = (pa[0] + pa_to_pb_vec[0] * x_percentage, pa[1] + pa_to_pb_vec[1] * x_percentage)

      try:
        result_pixels[p[0],p[1]] = (pixel[0],pixel[1],pixel[2],min(int(alpha * 255),pixel[3]))
      except Exception:
        pass

  return result

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