OpenCV：为颜色过滤选择 HSV 阈值

Question

为了从图像中滤除颜色，有必要设置需要检测哪种颜色的边界。我有一种感觉，这主要是一个反复试验的过程。有什么方法可以快速找到特定颜色的正确阈值？在这种特定情况下，我试图检测下图中图形的灰色区域。这当然没有检测到虚线。对于这个例子，我需要非常具体的边界。问题是，我怎样才能轻松找到它们？

hsv = cv2.cvtColor(im, cv2.COLOR_BGR2HSV)

lower = np.array([0, 0, 0], np.uint8)
upper = np.array([180, 255, 200], np.uint8)

mask = cv2.inRange(hsv, lower, upper)

Answer 1

您可以使用 HSV 颜色阈值脚本来隔离所需的颜色范围

import cv2
import sys
import numpy as np

def nothing(x):
    pass

# Create a window
cv2.namedWindow('image')

# create trackbars for color change
cv2.createTrackbar('HMin','image',0,179,nothing) # Hue is from 0-179 for Opencv
cv2.createTrackbar('SMin','image',0,255,nothing)
cv2.createTrackbar('VMin','image',0,255,nothing)
cv2.createTrackbar('HMax','image',0,179,nothing)
cv2.createTrackbar('SMax','image',0,255,nothing)
cv2.createTrackbar('VMax','image',0,255,nothing)

# Set default value for MAX HSV trackbars.
cv2.setTrackbarPos('HMax', 'image', 179)
cv2.setTrackbarPos('SMax', 'image', 255)
cv2.setTrackbarPos('VMax', 'image', 255)

# Initialize to check if HSV min/max value changes
hMin = sMin = vMin = hMax = sMax = vMax = 0
phMin = psMin = pvMin = phMax = psMax = pvMax = 0

img = cv2.imread('1.png')
output = img
waitTime = 33

while(1):

    # get current positions of all trackbars
    hMin = cv2.getTrackbarPos('HMin','image')
    sMin = cv2.getTrackbarPos('SMin','image')
    vMin = cv2.getTrackbarPos('VMin','image')

    hMax = cv2.getTrackbarPos('HMax','image')
    sMax = cv2.getTrackbarPos('SMax','image')
    vMax = cv2.getTrackbarPos('VMax','image')

    # Set minimum and max HSV values to display
    lower = np.array([hMin, sMin, vMin])
    upper = np.array([hMax, sMax, vMax])

    # Create HSV Image and threshold into a range.
    hsv = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)
    mask = cv2.inRange(hsv, lower, upper)
    output = cv2.bitwise_and(img,img, mask= mask)

    # Print if there is a change in HSV value
    if( (phMin != hMin) | (psMin != sMin) | (pvMin != vMin) | (phMax != hMax) | (psMax != sMax) | (pvMax != vMax) ):
        print("(hMin = %d , sMin = %d, vMin = %d), (hMax = %d , sMax = %d, vMax = %d)" % (hMin , sMin , vMin, hMax, sMax , vMax))
        phMin = hMin
        psMin = sMin
        pvMin = vMin
        phMax = hMax
        psMax = sMax
        pvMax = vMax

    # Display output image
    cv2.imshow('image',output)

    # Wait longer to prevent freeze for videos.
    if cv2.waitKey(waitTime) & 0xFF == ord('q'):
        break

cv2.destroyAllWindows()

Answer 2

另一种选择是使用在线图像颜色选择器。您可以上传您的图像，并会获得一些与HSV: 97.5\xc2\xb0 5.1% 61.57%您的情况类似的值。请注意，您需要将它们转换为 H、S 和 V 的 OpenCV 比例。

\n\n

H，OpenCV中的色调从0到180变化，但在外部世界通常以0到360的度数来测量，所以要得到你的颜色的Hh = 97.5\xc2\xb0 / 2 = 48.7

\n\n

S 和 V 是从0 ( = 0% in outer world)到测量的255 ( = 100% in outer world)，所以

\n\n

s = 255 * 5.1% = 13\nv = 255 * 61.57% = 157\n

\n\n

因此，目标 HSV 颜色为(49, 13, 157)。我建议使用 \xc2\xb110 作为范围。或者说更加严格。我认为对于您的情况来说，只选择中心图的像素，没有任何标签，然后根据需要应用形态学操作可能没问题。

\n