PIL 的Image.transform透视模式需要 8 个数据元组，但我不知道如何将 30 度的右倾斜转换为该元组。
谁能解释一下？
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4 回答
93
要应用透视变换，您首先必须知道平面 A 中的四个点将映射到平面 B 中的四个点。使用这些点，您可以推导出单应变换。通过这样做，您可以获得 8 个系数，并且可以进行转换。
站点http://xenia.media.mit.edu/~cwren/interpolator/（镜像：WebArchive）以及许多其他文本描述了如何确定这些系数。为方便起见，这里是根据上述链接的直接实现：
import numpy
def find_coeffs(pa, pb):
matrix = []
for p1, p2 in zip(pa, pb):
matrix.append([p1[0], p1[1], 1, 0, 0, 0, -p2[0]*p1[0], -p2[0]*p1[1]])
matrix.append([0, 0, 0, p1[0], p1[1], 1, -p2[1]*p1[0], -p2[1]*p1[1]])
A = numpy.matrix(matrix, dtype=numpy.float)
B = numpy.array(pb).reshape(8)
res = numpy.dot(numpy.linalg.inv(A.T * A) * A.T, B)
return numpy.array(res).reshape(8)
其中pb是当前平面中的pa四个顶点，并且包含结果平面中的四个顶点。
因此，假设我们将图像转换为：
import sys
from PIL import Image
img = Image.open(sys.argv[1])
width, height = img.size
m = -0.5
xshift = abs(m) * width
new_width = width + int(round(xshift))
img = img.transform((new_width, height), Image.AFFINE,
(1, m, -xshift if m > 0 else 0, 0, 1, 0), Image.BICUBIC)
img.save(sys.argv[2])
这是上面代码的示例输入和输出：
我们可以继续上一个代码并执行透视变换以恢复剪切：
coeffs = find_coeffs(
[(0, 0), (256, 0), (256, 256), (0, 256)],
[(0, 0), (256, 0), (new_width, height), (xshift, height)])
img.transform((width, height), Image.PERSPECTIVE, coeffs,
Image.BICUBIC).save(sys.argv[3])
导致：
您还可以通过目的地点获得一些乐趣：
于 2013-01-06T02:35:35.790 回答
12
我将稍微劫持这个问题，因为它是谷歌上唯一与 Python 中的透视转换有关的东西。下面是一些基于上述代码的更通用的代码，它创建一个透视变换矩阵并生成一个函数，该函数将在任意点上运行该变换：
import numpy as np
def create_perspective_transform_matrix(src, dst):
""" Creates a perspective transformation matrix which transforms points
in quadrilateral ``src`` to the corresponding points on quadrilateral
``dst``.
Will raise a ``np.linalg.LinAlgError`` on invalid input.
"""
# See:
# * http://xenia.media.mit.edu/~cwren/interpolator/
# * http://stackoverflow.com/a/14178717/71522
in_matrix = []
for (x, y), (X, Y) in zip(src, dst):
in_matrix.extend([
[x, y, 1, 0, 0, 0, -X * x, -X * y],
[0, 0, 0, x, y, 1, -Y * x, -Y * y],
])
A = np.matrix(in_matrix, dtype=np.float)
B = np.array(dst).reshape(8)
af = np.dot(np.linalg.inv(A.T * A) * A.T, B)
return np.append(np.array(af).reshape(8), 1).reshape((3, 3))
def create_perspective_transform(src, dst, round=False, splat_args=False):
""" Returns a function which will transform points in quadrilateral
``src`` to the corresponding points on quadrilateral ``dst``::
>>> transform = create_perspective_transform(
... [(0, 0), (10, 0), (10, 10), (0, 10)],
... [(50, 50), (100, 50), (100, 100), (50, 100)],
... )
>>> transform((5, 5))
(74.99999999999639, 74.999999999999957)
If ``round`` is ``True`` then points will be rounded to the nearest
integer and integer values will be returned.
>>> transform = create_perspective_transform(
... [(0, 0), (10, 0), (10, 10), (0, 10)],
... [(50, 50), (100, 50), (100, 100), (50, 100)],
... round=True,
... )
>>> transform((5, 5))
(75, 75)
If ``splat_args`` is ``True`` the function will accept two arguments
instead of a tuple.
>>> transform = create_perspective_transform(
... [(0, 0), (10, 0), (10, 10), (0, 10)],
... [(50, 50), (100, 50), (100, 100), (50, 100)],
... splat_args=True,
... )
>>> transform(5, 5)
(74.99999999999639, 74.999999999999957)
If the input values yield an invalid transformation matrix an identity
function will be returned and the ``error`` attribute will be set to a
description of the error::
>>> tranform = create_perspective_transform(
... np.zeros((4, 2)),
... np.zeros((4, 2)),
... )
>>> transform((5, 5))
(5.0, 5.0)
>>> transform.error
'invalid input quads (...): Singular matrix
"""
try:
transform_matrix = create_perspective_transform_matrix(src, dst)
error = None
except np.linalg.LinAlgError as e:
transform_matrix = np.identity(3, dtype=np.float)
error = "invalid input quads (%s and %s): %s" %(src, dst, e)
error = error.replace("\n", "")
to_eval = "def perspective_transform(%s):\n" %(
splat_args and "*pt" or "pt",
)
to_eval += " res = np.dot(transform_matrix, ((pt[0], ), (pt[1], ), (1, )))\n"
to_eval += " res = res / res[2]\n"
if round:
to_eval += " return (int(round(res[0][0])), int(round(res[1][0])))\n"
else:
to_eval += " return (res[0][0], res[1][0])\n"
locals = {
"transform_matrix": transform_matrix,
}
locals.update(globals())
exec to_eval in locals, locals
res = locals["perspective_transform"]
res.matrix = transform_matrix
res.error = error
return res
于 2014-06-06T18:33:00.933 回答
9
8个变换系数（a、b、c、d、e、f、g、h）对应如下变换：
x' = (a x + b y + c) / (g x + h y + 1)
y' = (d x + e y + f) / (g x + h y + 1)
这 8 个系数通常可以通过求解 8 个（线性）方程来找到，这些方程定义平面上的 4 个点如何变换（2D 中的 4 个点 -> 8 个方程），请参阅 mmgp 的答案以获取解决此问题的代码，尽管您可能发现改线更准确一点
res = numpy.dot(numpy.linalg.inv(A.T * A) * A.T, B)
至
res = numpy.linalg.solve(A, B)
即，没有真正的理由实际反转 A 矩阵，或将其乘以它的转置并失去一点准确性，以求解方程。
至于您的问题，对于（x0，y0）周围的θ度的简单倾斜，您正在寻找的系数是：
def find_rotation_coeffs(theta, x0, y0):
ct = cos(theta)
st = sin(theta)
return np.array([ct, -st, x0*(1-ct) + y0*st, st, ct, y0*(1-ct)-x0*st,0,0])
通常，任何仿射变换都必须使 (g, h) 等于零。希望有帮助！
于 2017-06-01T10:45:22.383 回答
7
这是生成变换系数的纯 Python 版本（正如我已经看到几个人要求的那样）。我制作并使用它来制作PyDraw纯 Python 图像绘图包。
如果将它用于您自己的项目，请注意计算需要几个高级矩阵运算，这意味着该函数需要另一个，幸运的是，纯 Python 矩阵库，matfunc最初由 Raymond Hettinger 编写，您可以在此处或此处下载。
import matfunc as mt
def perspective_coefficients(self, oldplane, newplane):
"""
Calculates and returns the transform coefficients needed for a perspective
transform, ie tilting an image in 3D.
Note: it is not very obvious how to set the oldplane and newplane arguments
in order to tilt an image the way one wants. Need to make the arguments more
user-friendly and handle the oldplane/newplane behind the scenes.
Some hints on how to do that at http://www.cs.utexas.edu/~fussell/courses/cs384g/lectures/lecture20-Z_buffer_pipeline.pdf
| **option** | **description**
| --- | ---
| oldplane | a list of four old xy coordinate pairs
| newplane | four points in the new plane corresponding to the old points
"""
# first find the transform coefficients, thanks to http://stackoverflow.com/questions/14177744/how-does-perspective-transformation-work-in-pil
pb,pa = oldplane,newplane
grid = []
for p1,p2 in zip(pa, pb):
grid.append([p1[0], p1[1], 1, 0, 0, 0, -p2[0]*p1[0], -p2[0]*p1[1]])
grid.append([0, 0, 0, p1[0], p1[1], 1, -p2[1]*p1[0], -p2[1]*p1[1]])
# then do some matrix magic
A = mt.Matrix(grid)
B = mt.Vec([xory for xy in pb for xory in xy])
AT = A.tr()
ATA = AT.mmul(A)
gridinv = ATA.inverse()
invAT = gridinv.mmul(AT)
res = invAT.mmul(B)
a,b,c,d,e,f,g,h = res.flatten()
# finito
return a,b,c,d,e,f,g,h
于 2014-09-23T21:05:48.180 回答