裁剪后如何检测和对齐倾斜的图像

Question

我已经在我的解决方案上实现了一个效果很好的裁剪算法。问题是当图像倾斜时，裁剪会起作用，但它会像图像一样显示背景空间。

种植流程：

第一步：

第二步：

最终结果：

我已经搜索/尝试了多种解决方案，但无法获得不错的结果，或者我的想法不正确。

预期的结果是这样的：

编辑[最终结果]：

import cv2
import numpy as np

def order_corner_points(corners):
    # Separate corners into individual points
    # Index 0 - top-right
    #       1 - top-left
    #       2 - bottom-left
    #       3 - bottom-right
    corners = [(corner[0][0], corner[0][1]) for corner in corners]
    top_r, top_l, bottom_l, bottom_r = corners[0], corners[1], corners[2], corners[3]
    return (top_l, top_r, bottom_r, bottom_l)

def perspective_transform(image, corners):
    # Order points in clockwise order
    ordered_corners = order_corner_points(corners)
    top_l, top_r, bottom_r, bottom_l = ordered_corners

    # Determine width of new image which is the max distance between 
    # (bottom right and bottom left) or (top right and top left) x-coordinates
    width_A = np.sqrt(((bottom_r[0] - bottom_l[0]) ** 2) + ((bottom_r[1] - bottom_l[1]) ** 2))
    width_B = np.sqrt(((top_r[0] - top_l[0]) ** 2) + ((top_r[1] - top_l[1]) ** 2))
    width = max(int(width_A), int(width_B))

    # Determine height of new image which is the max distance between 
    # (top right and bottom right) or (top left and bottom left) y-coordinates
    height_A = np.sqrt(((top_r[0] - bottom_r[0]) ** 2) + ((top_r[1] - bottom_r[1]) ** 2))
    height_B = np.sqrt(((top_l[0] - bottom_l[0]) ** 2) + ((top_l[1] - bottom_l[1]) ** 2))
    height = max(int(height_A), int(height_B))

    # Construct new points to obtain top-down view of image in 
    # top_r, top_l, bottom_l, bottom_r order
    dimensions = np.array([[0, 0], [width - 1, 0], [width - 1, height - 1], 
                    [0, height - 1]], dtype = "float32")

    # Convert to Numpy format
    ordered_corners = np.array(ordered_corners, dtype="float32")

    # Find perspective transform matrix
    matrix = cv2.getPerspectiveTransform(ordered_corners, dimensions)

    # Return the transformed image
    return cv2.warpPerspective(image, matrix, (width, height))

def get_image_width_height(image):
    image_width = image.shape[1]  # current image's width
    image_height = image.shape[0]  # current image's height
    return image_width, image_height

def calculate_scaled_dimension(scale, image):
    image_width, image_height = get_image_width_height(image)
    ratio_of_new_with_to_old = scale / image_width


    dimension = (scale, int(image_height * ratio_of_new_with_to_old))
    return dimension

def scale_image(image, size):
    image_resized_scaled = cv2.resize(
        image,
        calculate_scaled_dimension(
            size,
            image
        ),
        interpolation=cv2.INTER_AREA
    )
    return image_resized_scaled

def rotate_image(image, angle):
    # Grab the dimensions of the image and then determine the center
    (h, w) = image.shape[:2]
    (cX, cY) = (w / 2, h / 2)

    # grab the rotation matrix (applying the negative of the
    # angle to rotate clockwise), then grab the sine and cosine
    # (i.e., the rotation components of the matrix)
    M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
    cos = np.abs(M[0, 0])
    sin = np.abs(M[0, 1])

    # Compute the new bounding dimensions of the image
    nW = int((h * sin) + (w * cos))
    nH = int((h * cos) + (w * sin))

    # Adjust the rotation matrix to take into account translation
    M[0, 2] += (nW / 2) - cX
    M[1, 2] += (nH / 2) - cY

    # Perform the actual rotation and return the image
    return cv2.warpAffine(image, M, (nW, nH))

image = cv2.imread('images/damina_cc_back.jpg')
original_image = image.copy()

image = scale_image(image, 500)

# convert the image to grayscale, blur it, and find edges in the image
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = cv2.bilateralFilter(gray, 11, 17, 17)
edged = cv2.Canny(gray, 30, 200)

cnts = cv2.findContours(edged.copy(), cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
cnts = sorted(cnts, key = cv2.contourArea, reverse = True)[:10]
screen_cnt = None

# loop over our contours
for c in cnts:
    # approximate the contour
    peri = cv2.arcLength(c, True)
    approx = cv2.approxPolyDP(c, 0.015 * peri, True)

    if len(approx) == 4:
        screen_cnt = approx
        transformed = perspective_transform(image, screen_cnt)
        break

# Draw ROI
cv2.drawContours(image, [screen_cnt], -1, (0, 255, 0), 1)

(h, w) = transformed.shape[:2]

if (h > w):
    rotated = rotate_image(transformed, 90)
else:
    rotated = transformed

cv2.imshow("image", original_image)
cv2.imshow("ROI", image)
cv2.imshow("transformed", transformed)
cv2.imshow("rotated", rotated)
cv2.waitKey(0)

Answer 1

要在裁剪后对齐图像，我们可以使用perspective transformation。首先，我们将矩形的四个角分成cv2.approxPolyDP() 给我们的各个点。我们使用此函数将点重新排列为顺时针方向（左上、右上、右下、左下）：

def order_corner_points(corners):
    # Separate corners into individual points
    # Index 0 - top-right
    #       1 - top-left
    #       2 - bottom-left
    #       3 - bottom-right
    corners = [(corner[0][0], corner[0][1]) for corner in corners]
    top_r, top_l, bottom_l, bottom_r = corners[0], corners[1], corners[2], corners[3]
    return (top_l, top_r, bottom_r, bottom_l)

这个函数为我们提供了 ROI 的边界框坐标

现在有了孤立的角点，我们可以使用cv2.getPerspectiveTransform()获得变换矩阵，并使用cv2.warpPerspective()实际获得变换后的图像。

def perspective_transform(image, corners):
    # Order points in clockwise order
    ordered_corners = order_corner_points(corners)
    top_l, top_r, bottom_r, bottom_l = ordered_corners

    # Determine width of new image which is the max distance between 
    # (bottom right and bottom left) or (top right and top left) x-coordinates
    width_A = np.sqrt(((bottom_r[0] - bottom_l[0]) ** 2) + ((bottom_r[1] - bottom_l[1]) ** 2))
    width_B = np.sqrt(((top_r[0] - top_l[0]) ** 2) + ((top_r[1] - top_l[1]) ** 2))
    width = max(int(width_A), int(width_B))

    # Determine height of new image which is the max distance between 
    # (top right and bottom right) or (top left and bottom left) y-coordinates
    height_A = np.sqrt(((top_r[0] - bottom_r[0]) ** 2) + ((top_r[1] - bottom_r[1]) ** 2))
    height_B = np.sqrt(((top_l[0] - bottom_l[0]) ** 2) + ((top_l[1] - bottom_l[1]) ** 2))
    height = max(int(height_A), int(height_B))

    # Construct new points to obtain top-down view of image in 
    # top_r, top_l, bottom_l, bottom_r order
    dimensions = np.array([[0, 0], [width - 1, 0], [width - 1, height - 1], 
                    [0, height - 1]], dtype = "float32")

    # Convert to Numpy format
    ordered_corners = np.array(ordered_corners, dtype="float32")

    # Find perspective transform matrix
    matrix = cv2.getPerspectiveTransform(ordered_corners, dimensions)

    # Return the transformed image
    return cv2.warpPerspective(image, matrix, (width, height))

这是结果

我们可以用这个函数来旋转图像

def rotate_image(image, angle):
    # Grab the dimensions of the image and then determine the center
    (h, w) = image.shape[:2]
    (cX, cY) = (w / 2, h / 2)

    # grab the rotation matrix (applying the negative of the
    # angle to rotate clockwise), then grab the sine and cosine
    # (i.e., the rotation components of the matrix)
    M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
    cos = np.abs(M[0, 0])
    sin = np.abs(M[0, 1])

    # Compute the new bounding dimensions of the image
    nW = int((h * sin) + (w * cos))
    nH = int((h * cos) + (w * sin))

    # Adjust the rotation matrix to take into account translation
    M[0, 2] += (nW / 2) - cX
    M[1, 2] += (nH / 2) - cY

    # Perform the actual rotation and return the image
    return cv2.warpAffine(image, M, (nW, nH))

旋转后的最终结果：

完整代码

import cv2
import numpy as np

def order_corner_points(corners):
    # Separate corners into individual points
    # Index 0 - top-right
    #       1 - top-left
    #       2 - bottom-left
    #       3 - bottom-right
    corners = [(corner[0][0], corner[0][1]) for corner in corners]
    top_r, top_l, bottom_l, bottom_r = corners[0], corners[1], corners[2], corners[3]
    return (top_l, top_r, bottom_r, bottom_l)

def perspective_transform(image, corners):
    # Order points in clockwise order
    ordered_corners = order_corner_points(corners)
    top_l, top_r, bottom_r, bottom_l = ordered_corners

    # Determine width of new image which is the max distance between 
    # (bottom right and bottom left) or (top right and top left) x-coordinates
    width_A = np.sqrt(((bottom_r[0] - bottom_l[0]) ** 2) + ((bottom_r[1] - bottom_l[1]) ** 2))
    width_B = np.sqrt(((top_r[0] - top_l[0]) ** 2) + ((top_r[1] - top_l[1]) ** 2))
    width = max(int(width_A), int(width_B))

    # Determine height of new image which is the max distance between 
    # (top right and bottom right) or (top left and bottom left) y-coordinates
    height_A = np.sqrt(((top_r[0] - bottom_r[0]) ** 2) + ((top_r[1] - bottom_r[1]) ** 2))
    height_B = np.sqrt(((top_l[0] - bottom_l[0]) ** 2) + ((top_l[1] - bottom_l[1]) ** 2))
    height = max(int(height_A), int(height_B))

    # Construct new points to obtain top-down view of image in 
    # top_r, top_l, bottom_l, bottom_r order
    dimensions = np.array([[0, 0], [width - 1, 0], [width - 1, height - 1], 
                    [0, height - 1]], dtype = "float32")

    # Convert to Numpy format
    ordered_corners = np.array(ordered_corners, dtype="float32")

    # Find perspective transform matrix
    matrix = cv2.getPerspectiveTransform(ordered_corners, dimensions)

    # Return the transformed image
    return cv2.warpPerspective(image, matrix, (width, height))

def rotate_image(image, angle):
    # Grab the dimensions of the image and then determine the center
    (h, w) = image.shape[:2]
    (cX, cY) = (w / 2, h / 2)

    # grab the rotation matrix (applying the negative of the
    # angle to rotate clockwise), then grab the sine and cosine
    # (i.e., the rotation components of the matrix)
    M = cv2.getRotationMatrix2D((cX, cY), -angle, 1.0)
    cos = np.abs(M[0, 0])
    sin = np.abs(M[0, 1])

    # Compute the new bounding dimensions of the image
    nW = int((h * sin) + (w * cos))
    nH = int((h * cos) + (w * sin))

    # Adjust the rotation matrix to take into account translation
    M[0, 2] += (nW / 2) - cX
    M[1, 2] += (nH / 2) - cY

    # Perform the actual rotation and return the image
    return cv2.warpAffine(image, M, (nW, nH))

image = cv2.imread('1.PNG')
original_image = image.copy()

# convert the image to grayscale, blur it, and find edges in the image
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
gray = cv2.bilateralFilter(gray, 11, 17, 17)
edged = cv2.Canny(gray, 30, 200)

cnts = cv2.findContours(edged.copy(), cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
cnts = sorted(cnts, key = cv2.contourArea, reverse = True)[:10]
screen_cnt = None

# loop over our contours
for c in cnts:
    # approximate the contour
    peri = cv2.arcLength(c, True)
    approx = cv2.approxPolyDP(c, 0.015 * peri, True)

    if len(approx) == 4:
        screen_cnt = approx
        transformed = perspective_transform(original_image, screen_cnt)
        break

# Draw ROI
cv2.drawContours(image, [screen_cnt], -1, (0, 255, 0), 3)

# Rotate image
rotated = rotate_image(transformed, -90)

cv2.imshow("image", original_image)
cv2.imshow("ROI", image)
cv2.imshow("transformed", transformed)
cv2.imshow("rotated", rotated)
cv2.waitKey(0)

Answer 2

我假设您正在寻找找到边缘（或者可能是某些分位数）的最小和最大 u 和 v 位置，以找到裁剪的矩形。即遍历所有标记为边缘的图像像素并更新 u/v/ min/max 值。

如果计算时间对您来说不是问题，您可以简单地保持算法不变，另外循环多次旋转并更新每个旋转的特殊值。伪代码：

for v
  for u
    if (u,v) is edge
      for rotation_matrix
        (ur, vr) = rotation_matrix * (u,v)
        update boundary for given rotation matrix

最后你可以选择最小的旋转矩阵的边界框。

如果上述算法对您的用例来说太慢，您也可以尝试使用 opencv HoughLinesP 函数提取主轴。这当然不适用于所有类型的图像，但对于身份证来说可能就足够了。

最后，要应用旋转校正，请参阅this 教程。