我正在关注这个问题：
如何从左到右、从上到下对轮廓进行排序？ https://stackoverflow.com/questions/38654302/how-can-i-sort-contours-from-left-to-right-and-top-to-bottom
从左到右和从上到下对轮廓进行排序。然而，我的轮廓是使用这个（OpenCV 3）找到的：
im2, contours, hierarchy = cv2.findContours(threshold,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
它们的格式如下：
array([[[ 1, 1]],
[[ 1, 36]],
[[63, 36]],
[[64, 35]],
[[88, 35]],
[[89, 34]],
[[94, 34]],
[[94, 1]]], dtype=int32)]
当我运行代码时
max_width = max(contours, key=lambda r: r[0] + r[2])[0]
max_height = max(contours, key=lambda r: r[3])[3]
nearest = max_height * 1.4
contours.sort(key=lambda r: (int(nearest * round(float(r[1])/nearest)) * max_width + r[0]))
我收到错误
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
所以我把它改成这样：
max_width = max(contours, key=lambda r: np.max(r[0] + r[2]))[0]
max_height = max(contours, key=lambda r: np.max(r[3]))[3]
nearest = max_height * 1.4
contours.sort(key=lambda r: (int(nearest * round(float(r[1])/nearest)) * max_width + r[0]))
但现在我收到错误：
TypeError: only length-1 arrays can be converted to Python scalars
EDIT:
阅读下面的答案后我修改了我的代码：
EDIT 2
这是我用来“放大”字符并找到轮廓的代码
kernel = cv2.getStructuringElement(cv2.MORPH_RECT,(35,35))
# dilate the image to get text
# binaryContour is just the black and white image shown below
dilation = cv2.dilate(binaryContour,kernel,iterations = 2)
编辑 2 结束
im2, contours, hierarchy = cv2.findContours(dilation,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
myContours = []
# Process the raw contours to get bounding rectangles
for cnt in reversed(contours):
epsilon = 0.1*cv2.arcLength(cnt,True)
approx = cv2.approxPolyDP(cnt,epsilon,True)
if len(approx == 4):
rectangle = cv2.boundingRect(cnt)
myContours.append(rectangle)
max_width = max(myContours, key=lambda r: r[0] + r[2])[0]
max_height = max(myContours, key=lambda r: r[3])[3]
nearest = max_height * 1.4
myContours.sort(key=lambda r: (int(nearest * round(float(r[1])/nearest)) * max_width + r[0]))
i=0
for x,y,w,h in myContours:
letter = binaryContour[y:y+h, x:x+w]
cv2.rectangle(binaryContour,(x,y),(x+w,y+h),(255,255,255),2)
cv2.imwrite("pictures/"+str(i)+'.png', letter) # save contour to file
i+=1
排序前的轮廓：
[(1, 1, 94, 36), (460, 223, 914, 427), (888, 722, 739, 239), (35,723, 522, 228),
(889, 1027, 242, 417), (70, 1028, 693, 423), (1138, 1028, 567, 643),
(781, 1030, 98, 413), (497, 1527, 303, 132), (892, 1527, 168, 130),
(37, 1719, 592, 130), (676, 1721, 413, 129), (1181, 1723, 206, 128),
(30, 1925, 997, 236), (1038, 1929, 170, 129), (140, 2232, 1285, 436)]
排序后的轮廓：
(NOTE:这不是我想要的轮廓排序顺序。请参阅底部的图像）
[(1, 1, 94, 36), (460, 223, 914, 427), (35, 723, 522, 228), (70,1028, 693, 423),
(781, 1030, 98, 413), (888, 722, 739, 239), (889, 1027, 242, 417),
(1138, 1028, 567, 643), (30, 1925, 997, 236), (37, 1719, 592, 130),
(140, 2232, 1285, 436), (497, 1527, 303, 132), (676, 1721, 413, 129),
(892, 1527, 168, 130), (1038, 1929, 170, 129), (1181, 1723, 206, 128)]
我正在使用的图片
I want to find the contours in the following order:
Dilation image used for finding contours
您实际需要的是设计一个公式将轮廓信息转换为排名并使用该排名对轮廓进行排序，因为您需要从上到下、从左到右对轮廓进行排序，因此您的公式必须涉及origin给定轮廓来计算其等级。例如我们可以使用这个简单的方法：
def get_contour_precedence(contour, cols):
origin = cv2.boundingRect(contour)
return origin[1] * cols + origin[0]
它根据轮廓的原点为每个轮廓提供排名。当两个连续的轮廓垂直放置时，它的变化很大，但当轮廓水平堆叠时，它的变化很小。因此，通过这种方式，首先将从上到下对轮廓进行分组，并且在发生冲突的情况下，将使用水平铺设轮廓中变化较小的值。
import cv2
def get_contour_precedence(contour, cols):
tolerance_factor = 10
origin = cv2.boundingRect(contour)
return ((origin[1] // tolerance_factor) * tolerance_factor) * cols + origin[0]
img = cv2.imread("/Users/anmoluppal/Downloads/9VayB.png", 0)
_, img = cv2.threshold(img, 70, 255, cv2.THRESH_BINARY)
im, contours, h = cv2.findContours(img.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
contours.sort(key=lambda x:get_contour_precedence(x, img.shape[1]))
# For debugging purposes.
for i in xrange(len(contours)):
img = cv2.putText(img, str(i), cv2.boundingRect(contours[i])[:2], cv2.FONT_HERSHEY_COMPLEX, 1, [125])
如果你仔细看的话，第三行是3, 4, 5, 6轮廓放置在6介于 3 和 5 之间，原因是6轮廓线略低于3, 4, 5轮廓。
告诉我您是否想要以其他方式输出，我们可以调整get_contour_precedence to get 3, 4, 5, 6轮廓的等级已校正。