Question

我的目标是以三个源点映射到空数组中的三个目标点的方式转换图像。我已经解决了正确的仿射矩阵的发现，但我无法在彩色图像上应用仿射变换。

更具体地说，我正在努力正确使用scipy.ndimage.interpolation.affine_transform方法。正如这个question和它的结论指出的那样，affine_transform方法可能有点不直观（特别是关于偏移计算），但是，用户timday显示了如何在图像上应用旋转和剪切并将其放置在另一个数组中，而用户地理数据则提供更多背景信息。

我的问题是将那里显示的方法（1）概括为彩色图像，（2）我自己计算的任意变换。

这是我的代码（应该在您的计算机上运行）：

import numpy as np
from scipy import ndimage
import matplotlib.pyplot as plt


def calcAffineMatrix(sourcePoints, targetPoints):
    # For three source- and three target points, find the affine transformation
    # Function works correctly, not part of the question
    A = []
    b = []
    for sp, trg in zip(sourcePoints, targetPoints):
        A.append([sp[0], 0, sp[1], 0, 1, 0])
        A.append([0, sp[0], 0, sp[1], 0, 1])
        b.append(trg[0])
        b.append(trg[1])
    result, resids, rank, s = np.linalg.lstsq(np.array(A), np.array(b))

    a0, a1, a2, a3, a4, a5 = result
    # Ignoring offset here, later use timday's suggested offset calculation
    affineTrafo = np.array([[a0, a1, 0], [a2, a3, 0], [0, 0, 1]], 'd')

    # Testing the correctness of transformation matrix
    for i, _ in enumerate(sourcePoints):
        src = sourcePoints[i]
        src.append(1.)
        trg = targetPoints[i]
        trg.append(1.)
        at = affineTrafo.copy()
        at[2, 0:2] = [a4, a5]
        assert(np.array_equal(np.round(np.array(src).dot(at)), np.array(trg)))
    return affineTrafo


# Prepare source image
sourcePoints = [[162., 112.], [130., 112.], [162., 240.]]
targetPoints = [[180., 102.], [101., 101.], [190., 200.]]
image = np.empty((300, 300, 3), dtype='uint8')
image[:] = 255
# Mark border for better visibility
image[0:2, :] = 0
image[-3:-1, :] = 0
image[:, 0:2] = 0
image[:, -3:-1] = 0
# Mark source points in red
for sp in sourcePoints:
    sp = [int(u) for u in sp]
    image[sp[1] - 5:sp[1] + 5, sp[0] - 5:sp[0] + 5, :] = np.array([255, 0, 0])

# Show image
plt.subplot(3, 1, 1)
plt.imshow(image)

# Prepare array in which the image is placed
array = np.empty((400, 300, 3), dtype='uint8')
array[:] = 255
a2 = array.copy()
# Mark target points in blue
for tp in targetPoints:
    tp = [int(u) for u in tp]
    a2[tp[1] - 2:tp[1] + 2, tp[0] - 2:tp[0] + 2] = [0, 0, 255]

# Show array
plt.subplot(3, 1, 2)
plt.imshow(a2)

# Next 5 program lines are actually relevant for question:

# Calculate affine matrix
affineTrafo = calcAffineMatrix(sourcePoints, targetPoints)

# This follows the c_in-c_out method proposed in linked stackoverflow issue
# extended for color channel (no translation here)
c_in = np.array([sourcePoints[0][0], sourcePoints[0][1], 0])
c_out = np.array([targetPoints[0][0], targetPoints[0][1], 0])
offset = (c_in - np.dot(c_out, affineTrafo))

# Affine transform!
ndimage.interpolation.affine_transform(image, affineTrafo, order=2, offset=offset,
                                       output=array, output_shape=array.shape,
                                       cval=255)
# Mark blue target points in array, expected to be above red source points
for tp in targetPoints:
    tp = [int(u) for u in tp]
    array[tp[1] - 2:tp[1] + 2, tp[0] - 2:tp[0] + 2] = [0, 0, 255]

plt.subplot(3, 1, 3)
plt.imshow(array)

plt.show()

我尝试的其他方法包括使用反转，转置或两者都使用affineTrafo：

affineTrafo = np.linalg.inv(affineTrafo)
affineTrafo = affineTrafo.T
affineTrafo = np.linalg.inv(affineTrafo.T)
affineTrafo = np.linalg.inv(affineTrafo).T

在他的回答中，地理数据显示了如何计算affine_trafo进行缩放和旋转所需的矩阵：

如果首先想要缩放S然后想要一个旋转R它保持T=R*S并因此T.inv=S.inv*R.inv（请注意相反的顺序）。

我尝试使用矩阵分解进行复制（将我的仿射变换分解为旋转，剪切和另一次旋转）：

u, s, v = np.linalg.svd(affineTrafo[:2,:2])
uInv = np.linalg.inv(u)
sInv = np.linalg.inv(np.diag((s)))
vInv = np.linalg.inv(v)
affineTrafo[:2, :2] = uInv.dot(sInv).dot(vInv)

再次，没有成功。

对于我的所有结果，它不是（仅）偏移问题。从图片中可以清楚地看出，源点和目标点的相对位置不对应。

我搜索了网络和stackoverflow，但没有找到我的问题的答案。请帮我！ :)

Answer 1

由于AlexanderReynolds暗示使用另一个图书馆，我终于得到了它。这当然是一种解决方法;我使用scipy的affine_transform无法使用它，所以我使用了OpenCV cv2.warpAffine。如果这对其他人有帮助，这是我的代码：

import numpy as np
import matplotlib.pyplot as plt
import cv2

# Prepare source image
sourcePoints = [[162., 112.], [130., 112.], [162., 240.]]
targetPoints = [[180., 102.], [101., 101.], [190., 200.]]
image = np.empty((300, 300, 3), dtype='uint8')
image[:] = 255
# Mark border for better visibility
image[0:2, :] = 0
image[-3:-1, :] = 0
image[:, 0:2] = 0
image[:, -3:-1] = 0
# Mark source points in red
for sp in sourcePoints:
    sp = [int(u) for u in sp]
    image[sp[1] - 5:sp[1] + 5, sp[0] - 5:sp[0] + 5, :] = np.array([255, 0, 0])

# Show image
plt.subplot(3, 1, 1)
plt.imshow(image)

# Prepare array in which the image is placed
array = np.empty((400, 300, 3), dtype='uint8')
array[:] = 255
a2 = array.copy()
# Mark target points in blue
for tp in targetPoints:
    tp = [int(u) for u in tp]
    a2[tp[1] - 2:tp[1] + 2, tp[0] - 2:tp[0] + 2] = [0, 0, 255]

# Show array
plt.subplot(3, 1, 2)
plt.imshow(a2)

# Calculate affine matrix and transform image
M = cv2.getAffineTransform(np.float32(sourcePoints), np.float32(targetPoints))
array = cv2.warpAffine(image, M, array.shape[:2], borderValue=[255, 255, 255])

# Mark blue target points in array, expected to be above red source points
for tp in targetPoints:
    tp = [int(u) for u in tp]
    array[tp[1] - 2:tp[1] + 2, tp[0] - 2:tp[0] + 2] = [0, 0, 255]

plt.subplot(3, 1, 3)
plt.imshow(array)

plt.show()

评论：

有趣的是它在更改库后几乎立即起作用。花了一天多的时间试图让它与scipy一起工作后，这对我自己更快地更换库来说是一个教训。
如果有人想要根据超过三个点找到仿射变换的（最小二乘）近似值，这就是你得到与cv2.warpAffine一起使用的矩阵的方法：

代码：

def calcAffineMatrix(sourcePoints, targetPoints):
    # For three or more source and target points, find the affine transformation
    A = []
    b = []
    for sp, trg in zip(sourcePoints, targetPoints):
        A.append([sp[0], 0, sp[1], 0, 1, 0])
        A.append([0, sp[0], 0, sp[1], 0, 1])
        b.append(trg[0])
        b.append(trg[1])
    result, resids, rank, s = np.linalg.lstsq(np.array(A), np.array(b))

    a0, a1, a2, a3, a4, a5 = result
    affineTrafo = np.float32([[a0, a2, a4], [a1, a3, a5]])
    return affineTrafo

如何使用scipy的affine_transform对彩色图像进行任意仿射变换？

1 个答案: