Opencv：从许可证中裁剪文本区域

我有一张驾驶执照的下图，我想提取有关驾驶执照、姓名、出生日期等的信息。我的思考过程是找到一种方法将它们逐行分组，并裁剪出包含姓名的单个矩形、 eng 和 ara 的许可证等。但我却惨遭失败。

import cv2
import os
import numpy as np

scan_dir = os.path.dirname(__file__)
image_dir = os.path.join(scan_dir, '../../images')


class Loader(object):
    def __init__(self, filename, gray=True):
        self.filename = filename
        self.gray = gray
        self.image = None

    def _read(self, filename):
        rgba = cv2.imread(os.path.join(image_dir, filename))

        if rgba is None:
            raise Exception("Image not found")

        if self.gray:
            gray = cv2.cvtColor(rgba, cv2.COLOR_BGR2GRAY)

        return gray, rgba


    def __call__(self):
        return self._read(self.filename)


class ImageScaler(object):

    def __call__(self, gray, rgba, scale_factor = 2):
        img_small_gray = cv2.resize(gray, None, fx=scale_factor, fy=scale_factor, interpolation=cv2.INTER_AREA)
        img_small_rgba = cv2.resize(rgba, None, fx=scale_factor, fy=scale_factor, interpolation=cv2.INTER_AREA)


        return img_small_gray, img_small_rgba



class BoxLocator(object):
    def __call__(self, gray, rgba):
        # image_blur = cv2.medianBlur(gray, 1)
        ret, image_binary = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
        image_not = cv2.bitwise_not(image_binary)

        erode_kernel = np.ones((3, 1), np.uint8)
        image_erode = cv2.erode(image_not, erode_kernel, iterations = 5)

        dilate_kernel = np.ones((5,5), np.uint8)
        image_dilate = cv2.dilate(image_erode, dilate_kernel, iterations=5)


        kernel = np.ones((3, 3), np.uint8)
        image_closed = cv2.morphologyEx(image_dilate, cv2.MORPH_CLOSE, kernel)
        image_open = cv2.morphologyEx(image_closed, cv2.MORPH_OPEN, kernel)

        image_not = cv2.bitwise_not(image_open)
        image_not = cv2.adaptiveThreshold(image_not, 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 15, -2)

        image_dilate = cv2.dilate(image_not, np.ones((2, 1)), iterations=1)
        image_dilate = cv2.dilate(image_dilate, np.ones((2, 10)), iterations=1)

        image, contours, heirarchy = cv2.findContours(image_dilate, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)

        for contour in contours:
            x, y, w, h = cv2.boundingRect(contour)
            # if w > 30 and h > 10:
            cv2.rectangle(rgba, (x, y), (x + w, y + h), (0, 0, 255), 2)

        return image_dilate, rgba



def entry():
    loader = Loader('sample-004.jpg')
    # loader = Loader('sample-004.jpg')
    gray, rgba = loader()

    imageScaler = ImageScaler()
    image_scaled_gray, image_scaled_rgba = imageScaler(gray, rgba, 1)

    box_locator = BoxLocator()
    gray, rgba = box_locator(image_scaled_gray, image_scaled_rgba)

    cv2.namedWindow('Image', cv2.WINDOW_NORMAL)
    cv2.namedWindow('Image2', cv2.WINDOW_NORMAL)

    cv2.resizeWindow('Image', 600, 600)
    cv2.resizeWindow('Image2', 600, 600)

    cv2.imshow("Image2", rgba)
    cv2.imshow("Image", gray)

    cv2.moveWindow('Image', 0, 0)
    cv2.moveWindow('Image2', 600, 0)

    cv2.waitKey()
    cv2.destroyAllWindows()

当我运行上面的代码时，我得到下面的分段。这与我想要的不接近

But below is what I want to achieve, for all input license

从我的头脑中，我可以想到两种方法：

方法1.正如评论中提到的，您可以裁剪鹰符号在左上角和flag在右上角，使用它们作为模板，并根据找到的模板的位置找到您感兴趣的两个框，左下（小框）和中心（大框）。首先，您可以使用以下命令：

模板1

模板2

Code:

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

image = cv2.imread("ID_card.jpg")

template_1 = cv2.imread("template_1.jpg", 0)
w_1, h_1 = template_1.shape[::-1]

template_2 = cv2.imread("template_2.jpg", 0)
w_2, h_2 = template_2.shape[::-1]

res_1 = cv2.matchTemplate(image=image, templ=template_1, method=cv2.TM_CCOEFF)
min_val_1, max_val_1, min_loc_1, max_loc_1 = cv2.minMaxLoc(res_1)

res_2 = cv2.matchTemplate(image=image, templ=template_2, method=cv2.TM_CCOEFF)
min_val_2, max_val_2, min_loc_2, max_loc_2 = cv2.minMaxLoc(res_2)

cv2.rectangle(image, max_loc_1, (max_loc_1[0] + w_1, max_loc_1[1] + h_1), 255, 2)
cv2.rectangle(image, max_loc_2, (max_loc_2[0] + w_2, max_loc_2[1] + h_2), 255, 2)

Result:

您可以使用找到的模板的中心来获取所需框（小和大）的相对位置。

方法 2. 与基于轮廓所做的类似，基本思想是使用形态学在更大的框中获得明确的线条。

Code:

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

image = cv2.imread("ID_card.jpg")
imgray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)

ret, thresh = cv2.threshold(imgray, 150, 255, 0)
# cv2.imwrite("thresh.jpg", thresh)

# Morphological operation
thresh = cv2.morphologyEx(thresh, cv2.MORPH_OPEN, 
cv2.getStructuringElement(cv2.MORPH_RECT, (7, 7)))

im2, contours, heirarchy = cv2.findContours(thresh, cv2.RETR_TREE, 
cv2.CHAIN_APPROX_SIMPLE)

# Sort the contours based on area
cntsSorted = sorted(contours, key=lambda x: cv2.contourArea(x), reverse=True)

approxes = []

for cnt in cntsSorted[1:10]:
    peri = cv2.arcLength(cnt, True)
    # approximate the contour shape
    approx = cv2.approxPolyDP(cnt, 0.04 * peri, True)
    approxes.append(approx)
    if len(approx) == 4:
    # length of 4 means 4 vertices so it should be a quadrilateral
        cv2.drawContours(image, approx, -1, (0, 255, 0), 10)

cv2.imwrite("ID_card_contours.jpg", image)
print(approxes)

Results:

阈值图像

形态学开放后

最终图像，两个预期框的各自角标有绿色

因此，这种方法非常简单，我相信您可以完成剩下的工作，从大盒子中找到较小的子集。如果没有，请给我留言，我很乐意提供帮助（基本上从图像中裁剪该区域，使用霍夫林斯P https://docs.opencv.org/3.4.0/d9/db0/tutorial_hough_lines.html你应该没问题。或者，我可以看到较小的子集具有相同的宽度，因此您可以根据 y 坐标裁剪它们）

附言。希望“更大”、“更小”的盒子能被很好地理解，为我的懒惰而没有在图像中显示它们的内容表示歉意。

Note：仅给出一张图像，我无法确定它是否适用于数据集中的所有图像。你可能需要调整临界点 and 变形开放参数。如果您可以上传更多图片，我可以尝试一下。

礼貌：OpenCV 形状检测 https://www.pyimagesearch.com/2016/02/08/opencv-shape-detection/用于检测轮廓中的形状。