OpenCV - 正确使用cv2.approxPolyDP（）

Question

我试图使用cv2.approxPolyDP()提取图像中轮廓的近似值。这是我正在使用的图像：

我的代码试图隔离主岛并定义和绘制轮廓近似和轮廓外壳。我绘制了绿色的轮廓，红色的近似值：

import numpy as np
import cv2

# load image and shrink - it's massive
img = cv2.imread('../data/UK.png')
img = cv2.resize(img, None,fx=0.25, fy=0.25, interpolation = cv2.INTER_CUBIC)

# get a blank canvas for drawing contour on and convert img to grayscale
canvas = np.zeros(img.shape, np.uint8)
img2gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)

# filter out small lines between counties
kernel = np.ones((5,5),np.float32)/25
img2gray = cv2.filter2D(img2gray,-1,kernel)

# threshold the image and extract contours
ret,thresh = cv2.threshold(img2gray,250,255,cv2.THRESH_BINARY_INV)
im2,contours,hierarchy = cv2.findContours(thresh, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)


# find the main island (biggest area)
cnt = contours[0]
max_area = cv2.contourArea(cnt)

for cont in contours:
    if cv2.contourArea(cont) > max_area:
        cnt = cont
        max_area = cv2.contourArea(cont)

# define main island contour approx. and hull
perimeter = cv2.arcLength(cnt,True)
epsilon = 0.01*cv2.arcLength(cnt,True)
approx = cv2.approxPolyDP(cnt,epsilon,True)

hull = cv2.convexHull(cnt)

# cv2.isContourConvex(cnt)

cv2.drawContours(canvas, cnt, -1, (0, 255, 0), 3)
cv2.drawContours(canvas, approx, -1, (0, 0, 255), 3)
## cv2.drawContours(canvas, hull, -1, (0, 0, 255), 3) # only displays a few points as well.

cv2.imshow("Contour", canvas)
k = cv2.waitKey(0)

if k == 27:         # wait for ESC key to exit
    cv2.destroyAllWindows()

以下是生成的图像：

第一张图像以绿色绘制轮廓。第二个用红色绘制近似值 - 如何将此近似值绘制为连续闭合曲线？

documentation并不是非常清晰，tutorial也不是，但我的理解是cv2.approxPolyDP()应该定义一条连续的，闭合的曲线，我应该能够情节cv2.drawContours()。那是对的吗？如果是这样，我做错了什么？

Answer 1

问题仅在于可视化：drawContours需要轮廓的数组（在python的情况下为list），而不仅仅是一个numpy数组（从approxPolyDP返回）。

解决方案如下：替换

cv2.drawContours(canvas, approx, -1, (0, 0, 255), 3)

到

cv2.drawContours(canvas, [approx], -1, (0, 0, 255), 3)

Answer 2

cv2.approxPolyDP()

approx = cv2.approxPolyDP(cnt, 0.03 * cv2.arcLength(cnt, True), True)