最佳答案
这可以使用 morphological 解决诸如 eroding and dilating 之类的操作.这两个操作将有助于创建封闭的矩形。 之后,您可以使用此 page 中的教程检测矩形等简单形状。
我实现了一个快速演示,它适用于您提供的图像。
main.py:
import cv2
import numpy as np
from shapeDetector import ShapeDetector
import imutils
img = cv2.imread('t.png')
kernel = np.ones((5,5),np.uint8)
erosion = cv2.erode(img,kernel,iterations = 10)
dilate = cv2.dilate(erosion,kernel,iterations = 10)
侵 eclipse 使所有线条变粗,因此要恢复到正常宽度,我们需要在侵 eclipse 后扩大。我建议对扩张操作进行一次评论,以了解侵 eclipse 是如何工作的,反之亦然。 此操作将像这样转换您的图像
我使用的检测算法需要黑色背景上的白线。 这就是我们需要反转图像的原因。
cv2.bitwise_not ( dilate, dilate )
之后,我们就可以使用教程中的代码了。
image = dilate
resized = imutils.resize(image, width=300)
ratio = image.shape[0] / float(resized.shape[0])
# convert the resized image to grayscale, blur it slightly,
# and threshold it
gray = cv2.cvtColor(resized, cv2.COLOR_BGR2GRAY)
blurred = cv2.GaussianBlur(gray, (5, 5), 0)
thresh = cv2.threshold(blurred, 60, 255, cv2.THRESH_BINARY)[1]
#thresh = dilate
# find contours in the thresholded image and initialize the
# shape detector
cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,
cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if imutils.is_cv2() else cnts[1]
sd = ShapeDetector()
# loop over the contours
for c in cnts:
# compute the center of the contour, then detect the name of the
# shape using only the contour
M = cv2.moments(c)
cX = int((M["m10"] / M["m00"]) * ratio)
cY = int((M["m01"] / M["m00"]) * ratio)
shape = sd.detect(c)
# multiply the contour (x, y)-coordinates by the resize ratio,
# then draw the contours and the name of the shape on the image
c = c.astype("float")
c *= ratio
c = c.astype("int")
cv2.drawContours(image, [c], -1, (0, 255, 0), 2)
cv2.putText(image, shape, (cX, cY), cv2.FONT_HERSHEY_SIMPLEX,
0.5, (255, 255, 255), 2)
# show the output image
cv2.imshow("Image", image)
cv2.waitKey(0)
shapeDetector.py:
# import the necessary packages
import cv2
class ShapeDetector:
def __init__(self):
pass
def detect(self, c):
# initialize the shape name and approximate the contour
shape = "unidentified"
peri = cv2.arcLength(c, True)
approx = cv2.approxPolyDP(c, 0.04 * peri, True)
# if the shape is a triangle, it will have 3 vertices
if len(approx) == 3:
shape = "triangle"
# if the shape has 4 vertices, it is either a square or
# a rectangle
elif len(approx) == 4:
# compute the bounding box of the contour and use the
# bounding box to compute the aspect ratio
(x, y, w, h) = cv2.boundingRect(approx)
ar = w / float(h)
# a square will have an aspect ratio that is approximately
# equal to one, otherwise, the shape is a rectangle
shape = "square" if ar >= 0.95 and ar <= 1.05 else "rectangle"
# if the shape is a pentagon, it will have 5 vertices
elif len(approx) == 5:
shape = "pentagon"
# otherwise, we assume the shape is a circle
else:
shape = "circle"
# return the name of the shape
return shape
结果:
关于python - 如何识别openCV中的不完整矩形,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44295099/