Python opencv排序轮廓

Question

我正在关注这个问题：

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。但是，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]))

我收到了错误

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

所以我把它改成了这样：

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]))

但现在我得到了错误：

TypeError: only length-1 arrays can be converted to Python scalars

编辑：

阅读下面的答案后，我修改了我的代码：

编辑 2

这是我用来“扩张”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)

编辑结束 2

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：

（注意：这不是我希望对轮廓进行排序的顺序。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)]

我正在使用的图像

我想按以下顺序查找轮廓：

用于寻找轮廓的膨胀图像

Answer 1

您链接的question 似乎不适用于原始轮廓，而是首先使用cv2.boundingRect 获得一个边界矩形。只有这样计算max_width 和max_height 才有意义。您发布的代码表明您正在尝试对原始轮廓进行排序，而不是对矩形进行排序。如果不是这样，您能否提供一段更完整的代码，包括您尝试排序的多个轮廓的列表？

Answer 2

您实际需要的是设计一个公式将您的轮廓信息转换为等级并使用该等级对轮廓进行排序，因为您需要从上到下和从左到右对轮廓进行排序，因此您的公式必须涉及给定轮廓的origin 以计算其排名。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]

它根据轮廓的原点对每个轮廓进行排名。当两个连续的轮廓垂直放置时变化很大，但当轮廓水平堆叠时变化很小。因此，以这种方式，首先轮廓将从上到下分组，并且在发生冲突的情况下，将使用水平放置的轮廓中较小的变量值。

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])

如果你仔细观察，3, 4, 5, 6 轮廓线所在的第三行 6 介于 3 和 5 之间，原因是 6th 轮廓线略低于 3, 4, 5 轮廓线。

告诉我您是否希望以其他方式输出，我们可以调整get_contour_precedence 以校正3, 4, 5, 6 的轮廓等级。

Answer 3

这是由 Adrian Rosebrock 提供的，用于根据位置 link 对轮廓进行排序：

# import the necessary packages
import numpy as np
import argparse
import imutils
import cv2


def sort_contours(cnts, method="left-to-right"):
    # initialize the reverse flag and sort index
    reverse = False
    i = 0

    # handle if we need to sort in reverse
    if method == "right-to-left" or method == "bottom-to-top":
        reverse = True

    # handle if we are sorting against the y-coordinate rather than
    # the x-coordinate of the bounding box
    if method == "top-to-bottom" or method == "bottom-to-top":
        i = 1

    # construct the list of bounding boxes and sort them from top to
    # bottom
    boundingBoxes = [cv2.boundingRect(c) for c in cnts]
    (cnts, boundingBoxes) = zip(*sorted(zip(cnts, boundingBoxes),
        key=lambda b:b[1][i], reverse=reverse))

    # return the list of sorted contours and bounding boxes
    return (cnts, boundingBoxes)

def draw_contour(image, c, i):
    # compute the center of the contour area and draw a circle
    # representing the center
    M = cv2.moments(c)
    cX = int(M["m10"] / M["m00"])
    cY = int(M["m01"] / M["m00"])

    # draw the countour number on the image
    cv2.putText(image, "#{}".format(i + 1), (cX - 20, cY), cv2.FONT_HERSHEY_SIMPLEX,
        1.0, (255, 255, 255), 2)

    # return the image with the contour number drawn on it
    return image

# construct the argument parser and parse the arguments
ap = argparse.ArgumentParser()
ap.add_argument("-i", "--image", required=True, help="Path to the input image")
ap.add_argument("-m", "--method", required=True, help="Sorting method")
args = vars(ap.parse_args())

# load the image and initialize the accumulated edge image
image = cv2.imread(args["image"])
accumEdged = np.zeros(image.shape[:2], dtype="uint8")

# loop over the blue, green, and red channels, respectively
for chan in cv2.split(image):
    # blur the channel, extract edges from it, and accumulate the set
    # of edges for the image
    chan = cv2.medianBlur(chan, 11)
    edged = cv2.Canny(chan, 50, 200)
    accumEdged = cv2.bitwise_or(accumEdged, edged)

# show the accumulated edge map
cv2.imshow("Edge Map", accumEdged)

# find contours in the accumulated image, keeping only the largest
# ones
cnts = cv2.findContours(accumEdged.copy(), cv2.RETR_EXTERNAL,
    cv2.CHAIN_APPROX_SIMPLE)
cnts = imutils.grab_contours(cnts)
cnts = sorted(cnts, key=cv2.contourArea, reverse=True)[:5]
orig = image.copy()

# loop over the (unsorted) contours and draw them
for (i, c) in enumerate(cnts):
    orig = draw_contour(orig, c, i)

# show the original, unsorted contour image
cv2.imshow("Unsorted", orig)

# sort the contours according to the provided method
(cnts, boundingBoxes) = sort_contours(cnts, method=args["method"])

# loop over the (now sorted) contours and draw them
for (i, c) in enumerate(cnts):
    draw_contour(image, c, i)

# show the output image
cv2.imshow("Sorted", image)
cv2.waitKey(0)

Answer 4

您可以简单地检查轮廓的距离并对其进行排名，听一个例子

def get_distance(x,y):
    return math.sqrt(x*x+y*y)


img = cv2.imread("/image.png", 0)

res, img = cv2.threshold(img, 70, 255, cv2.THRESH_BINARY)

contours, h = cv2.findContours(img.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_A

for i in xrange(len(contours)):
    [x, y, w, h] = cv2.boundingRect(contours[i])
    img = cv2.putText(img, str(get_distance(x,y)), 
    cv2.boundingRect(contours[i])[:2], cv2.FONT_HERSHEY_COMPLEX, 1, [125])