我正在处理一个需要查找轮廓内部区域颜色的需求。我们正在使用OpenCv和Python,这是我在Python中的代码:
import imutils
import cv2
import numpy as np
path = "multiple_grains_1.jpeg"
img = cv2.imread(path)
resized = imutils.resize(img, width=900)
ratio = img.shape[0] / float(resized.shape[0])
gray = cv2.cvtColor(resized, cv2.COLOR_BGR2GRAY)
(ret, thresh) = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
edge = cv2.Canny(thresh, 100, 200)
( _,cnts, _) = cv2.findContours(edge.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
for c in cnts:
rect = cv2.minAreaRect(c)
box = cv2.boxPoints(rect)
box = np.int0(box)
area = cv2.contourArea(c)
if area > 1:
cv2.drawContours(resized,[box],0,(0,0,255),2)
cv2.drawContours(resized, [c], -1, (0, 255, 0), 2)
#print("area : "+str(area))
#print('\nContours: ' + str(c[0]))
#img[c[0]]
pixelpoints = np.transpose(np.nonzero(c))
#print('\pixelpoints: ' + str(pixelpoints))
# accessed the center of the contour using the followi
M = cv2.moments(c)
if M["m00"] != 0:
cX = int((M["m10"] / M["m00"]) * ratio)
cY = int((M["m01"] / M["m00"]) * ratio)
#print (cX,cY)
cord = img[int(cX)+3,int(cY)+3]
print(cord)
cv2.imshow("Output", resized)
cv2.waitKey(0)
exit()
当我检查轮廓的质心颜色时,我无法获取正确的颜色。有谁知道如何使用OpenCv和python获取轮廓内的颜色吗?
答案 0 :(得分:1)
我简化了您的代码,无需使用矩便可以获得质心的颜色。
import imutils
import cv2
import numpy as np
import matplotlib.pyplot as plt
img = cv2.imread("multiplegrains.png")
resized = imutils.resize(img, width=900)
ratio = img.shape[0] / float(resized.shape[0])
gray = cv2.cvtColor(resized, cv2.COLOR_BGR2GRAY)
ret, thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)
_, cnts, _ = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
# if you want cv2.contourArea >1, you can just comment line bellow
cnts = np.array(cnts)[[cv2.contourArea(c)>10 for c in cnts]]
grains = [np.int0(cv2.boxPoints(cv2.minAreaRect(c))) for c in cnts]
centroids =[(grain[2][1]-(grain[2][1]-grain[0][1])//2, grain[2][0]-(grain[2][0]-grain[0][0])//2) for grain in grains]
colors = [resized[centroid] for centroid in centroids]