我想在图像上找到主色。为此,我知道我应该使用图像直方图。但我不确定图像格式。应该使用rgb,hsv或灰色图像中的哪一个?
计算直方图后,我应该在直方图上找到最大值。为此,我应该找到hsv图像的最大binVal值以下吗?为什么我的结果图像只包含黑色?
float binVal = hist.at<float>(h, s);
编辑:
我尝试过以下代码。我绘制h-s直方图。我的结果图片就在这里。二进制阈值后我没有找到任何东西。也许我发现最大直方图值不正确。
cvtColor(src, hsv, CV_BGR2HSV);
// Quantize the hue to 30 levels
// and the saturation to 32 levels
int hbins = 20, sbins = 22;
int histSize[] = {hbins, sbins};
// hue varies from 0 to 179, see cvtColor
float hranges[] = { 0, 180 };
// saturation varies from 0 (black-gray-white) to
// 255 (pure spectrum color)
float sranges[] = { 0, 256 };
const float* ranges[] = { hranges, sranges };
MatND hist;
// we compute the histogram from the 0-th and 1-st channels
int channels[] = {0, 1};
calcHist( &hsv, 1, channels, Mat(), // do not use mask
hist, 2, histSize, ranges,
true, // the histogram is uniform
false );
double maxVal=0;
minMaxLoc(hist, 0, &maxVal, 0, 0);
int scale = 10;
Mat histImg = Mat::zeros(sbins*scale, hbins*10, CV_8UC3);
int maxIntensity = -100;
for( int h = 0; h < hbins; h++ ) {
for( int s = 0; s < sbins; s++ )
{
float binVal = hist.at<float>(h, s);
int intensity = cvRound(binVal*255/maxVal);
rectangle( histImg, Point(h*scale, s*scale),
Point( (h+1)*scale - 1, (s+1)*scale - 1),
Scalar::all(intensity),
CV_FILLED );
if(intensity > maxIntensity)
maxIntensity = intensity;
}
}
std::cout << "max Intensity " << maxVal << std::endl;
Mat dst;
cv::threshold(src, dst, maxIntensity, 255, cv::THRESH_BINARY);
namedWindow( "Dest", 1 );
imshow( "Dest", dst );
namedWindow( "Source", 1 );
imshow( "Source", src );
namedWindow( "H-S Histogram", 1 );
imshow( "H-S Histogram", histImg );
答案 0 :(得分:8)
或者您可以尝试k-means方法。 Calculate k
clusters与k ~ 2..5
并以最大群体的质心为主色。
OpenCv的python文档有一个illustrated example,它可以很好地获得主色:
答案 1 :(得分:7)
解决方案
答案 2 :(得分:5)
以下是一些可以帮助您入门的建议。
尝试转换为HSV,然后在H通道上计算直方图。
正如您所说,您希望在直方图中找到最大值。但是:
20-40
,而不只是30
。尝试不同的范围大小。 H=0
和H=360
是相同的。答案 3 :(得分:0)
这是一种使用K-Means Clustering来确定sklearn.cluster.KMeans()
图像中主色的Python方法
输入图片
结果
使用n_clusters=5
,这是最主要的颜色和百分比分布
[14.69488554 34.23074345 41.48107857] 13.67%
[141.44980073 207.52576948 236.30722987] 15.69%
[ 31.75790423 77.52713644 114.33328324] 18.77%
[ 48.41205713 118.34814452 176.43411287] 25.19%
[ 84.04820266 161.6848298 217.14045211] 26.69%
每个颜色簇的可视化
与n_clusters=10
,
[ 55.09073171 113.28271003 74.97528455] 3.25%
[ 85.36889668 145.80759374 174.59846237] 5.24%
[164.17201088 223.34258123 241.81929254] 6.60%
[ 9.97315932 22.79468111 22.01822211] 7.16%
[19.96940211 47.8375841 72.83728002] 9.27%
[ 26.73510467 70.5847759 124.79314278] 10.52%
[118.44741779 190.98204701 230.66728334] 13.55%
[ 51.61750364 130.59930047 198.76335878] 13.82%
[ 41.10232129 104.89923271 160.54431333] 14.53%
[ 81.70930412 161.823664 221.10258949] 16.04%
import cv2, numpy as np
from sklearn.cluster import KMeans
def visualize_colors(cluster, centroids):
# Get the number of different clusters, create histogram, and normalize
labels = np.arange(0, len(np.unique(cluster.labels_)) + 1)
(hist, _) = np.histogram(cluster.labels_, bins = labels)
hist = hist.astype("float")
hist /= hist.sum()
# Create frequency rect and iterate through each cluster's color and percentage
rect = np.zeros((50, 300, 3), dtype=np.uint8)
colors = sorted([(percent, color) for (percent, color) in zip(hist, centroids)])
start = 0
for (percent, color) in colors:
print(color, "{:0.2f}%".format(percent * 100))
end = start + (percent * 300)
cv2.rectangle(rect, (int(start), 0), (int(end), 50), \
color.astype("uint8").tolist(), -1)
start = end
return rect
# Load image and convert to a list of pixels
image = cv2.imread('1.jpg')
image = cv2.cvtColor(image, cv2.COLOR_BGR2RGB)
reshape = image.reshape((image.shape[0] * image.shape[1], 3))
# Find and display most dominant colors
cluster = KMeans(n_clusters=5).fit(reshape)
visualize = visualize_colors(cluster, cluster.cluster_centers_)
visualize = cv2.cvtColor(visualize, cv2.COLOR_RGB2BGR)
cv2.imshow('visualize', visualize)
cv2.waitKey()