也许是这样的?使用 Python 绑定完成,但很容易将方法转换为其他绑定...
#!/usr/local/bin/python
import cv
import colorsys
# get orginal image
orig = cv.LoadImage('car.jpg')
# show orginal
cv.ShowImage("orig", orig)
# get mask image
maskimg = cv.LoadImage('carcontour.jpg')
# split original image into hue and value
hsv = cv.CreateImage(cv.GetSize(orig),8,3)
hue = cv.CreateImage(cv.GetSize(orig),8,1)
val = cv.CreateImage(cv.GetSize(orig),8,1)
cv.CvtColor(maskimg,hsv,cv.CV_BGR2HSV)
cv.Split(hsv, hue, None, val, None)
# build mask from val image, select values NOT black
mask = cv.CreateImage(cv.GetSize(orig),8,1)
cv.Threshold(val,mask,0,255,cv.CV_THRESH_BINARY)
# show the mask
cv.ShowImage("mask", mask)
# calculate colour (hue) histgram of only masked area
hue_bins = 180
hue_range = [0,180]
hist = cv.CreateHist([hue_bins], cv.CV_HIST_ARRAY, [hue_range], 1)
cv.CalcHist([hue],hist,0,mask)
# create the colour histogram
(_, max_value, _, _) = cv.GetMinMaxHistValue(hist)
histimg = cv.CreateImage((hue_bins*2, 200), 8, 3)
for h in range(hue_bins):
bin_val = cv.QueryHistValue_1D(hist,h)
norm_val = cv.Round((bin_val/max_value)*200)
rgb_val = colorsys.hsv_to_rgb(float(h)/180.0,1.0,1.0)
cv.Rectangle(histimg,(h*2,0),
((h+1)*2-1, norm_val),
cv.RGB(rgb_val[0]*255,rgb_val[1]*255,rgb_val[2]*255),
cv.CV_FILLED)
cv.ShowImage("hist",histimg)
# wait for key press
cv.WaitKey(-1)
找到掩码有点笨拙 - 我想知道可能是由于图像中的 JPEG 压缩伪影...如果您有原始轮廓,则很容易将其“渲染”为掩码。
示例直方图渲染函数也有点基本 - 但我认为它显示了这个想法(以及汽车如何主要是红色的!)。请注意 OpenCV 对 Hue 的解释仅在 [0-180] 度范围内。
编辑:如果您想使用蒙版来计算原始图像中的颜色 - 从第 15 行向下编辑:
# split original image into hue
hsv = cv.CreateImage(cv.GetSize(orig),8,3)
hue = cv.CreateImage(cv.GetSize(orig),8,1)
cv.CvtColor(orig,hsv,cv.CV_BGR2HSV)
cv.Split(hsv, hue, None, None, None)
# split mask image into val
val = cv.CreateImage(cv.GetSize(orig),8,1)
cv.CvtColor(maskimg,hsv,cv.CV_BGR2HSV)
cv.Split(hsv, None, None, val, None)
(我认为这更符合预期,因为掩码是单独派生的,并应用于完全不同的图像。两种情况下的直方图大致相同......)