AHF*_*AHF 11 c++ opencv image-processing filter computer-vision
我们如何在opencv中制作晕影过滤器?我们需要为它实现任何算法还是仅使用BGR的值?我们如何制作这种类型的过滤器.我在这里看到了它的实现,但我没有清楚地理解它.任何拥有完整算法指导和实施指南的人都非常高兴.
之后Abid rehman K回答我在C试过这种++
int main()
{
Mat v;
Mat img = imread ("D:\\2.jpg");
img.convertTo(v, CV_32F);
Mat a,b,c,d,e;
c.create(img.rows,img.cols,CV_32F);
d.create(img.rows,img.cols,CV_32F);
e.create(img.rows,img.cols,CV_32F);
a = getGaussianKernel(img.cols,300,CV_32F);
b = getGaussianKernel(img.rows,300,CV_32F);
c = b*a.t();
double minVal;
double maxVal;
cv::minMaxLoc(c, &minVal, &maxVal);
d = c/maxVal;
e = v*d ; // This line causing error
imshow ("venyiet" , e);
cvWaitKey();
}
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d显示正确,但e=v*d行导致运行时错误
OpenCV Error: Assertion failed (type == B.type() && (type == CV_32FC1 || type ==
CV_64FC1 || type == CV_32FC2 || type == CV_64FC2)) in unknown function, file ..
\..\..\src\opencv\modules\core\src\matmul.cpp, line 711
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kar*_*lip 18
首先,Abid Rahman K描述了这种过滤器的最简单方法.你应该认真研究他的答案时间和注意力.对于那些从未听说过这个过滤器的人来说,维基百科对Vignetting的看法也很明确.
Browny 对此过滤器的实现要复杂得多.但是,我将他的代码移植到C++ API并简化它,以便您可以自己遵循说明.
#include <math.h>
#include <vector>
#include <cv.hpp>
#include <highgui/highgui.hpp>
// Helper function to calculate the distance between 2 points.
double dist(CvPoint a, CvPoint b)
{
return sqrt(pow((double) (a.x - b.x), 2) + pow((double) (a.y - b.y), 2));
}
// Helper function that computes the longest distance from the edge to the center point.
double getMaxDisFromCorners(const cv::Size& imgSize, const cv::Point& center)
{
// given a rect and a line
// get which corner of rect is farthest from the line
std::vector<cv::Point> corners(4);
corners[0] = cv::Point(0, 0);
corners[1] = cv::Point(imgSize.width, 0);
corners[2] = cv::Point(0, imgSize.height);
corners[3] = cv::Point(imgSize.width, imgSize.height);
double maxDis = 0;
for (int i = 0; i < 4; ++i)
{
double dis = dist(corners[i], center);
if (maxDis < dis)
maxDis = dis;
}
return maxDis;
}
// Helper function that creates a gradient image.
// firstPt, radius and power, are variables that control the artistic effect of the filter.
void generateGradient(cv::Mat& mask)
{
cv::Point firstPt = cv::Point(mask.size().width/2, mask.size().height/2);
double radius = 1.0;
double power = 0.8;
double maxImageRad = radius * getMaxDisFromCorners(mask.size(), firstPt);
mask.setTo(cv::Scalar(1));
for (int i = 0; i < mask.rows; i++)
{
for (int j = 0; j < mask.cols; j++)
{
double temp = dist(firstPt, cv::Point(j, i)) / maxImageRad;
temp = temp * power;
double temp_s = pow(cos(temp), 4);
mask.at<double>(i, j) = temp_s;
}
}
}
// This is where the fun starts!
int main()
{
cv::Mat img = cv::imread("stack-exchange-chefs.jpg");
if (img.empty())
{
std::cout << "!!! Failed imread\n";
return -1;
}
/*
cv::namedWindow("Original", cv::WINDOW_NORMAL);
cv::resizeWindow("Original", img.size().width/2, img.size().height/2);
cv::imshow("Original", img);
*/
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img看起来像什么:

cv::Mat maskImg(img.size(), CV_64F);
generateGradient(maskImg);
/*
cv::Mat gradient;
cv::normalize(maskImg, gradient, 0, 255, CV_MINMAX);
cv::imwrite("gradient.png", gradient);
*/
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maskImg看起来像什么:

cv::Mat labImg(img.size(), CV_8UC3);
cv::cvtColor(img, labImg, CV_BGR2Lab);
for (int row = 0; row < labImg.size().height; row++)
{
for (int col = 0; col < labImg.size().width; col++)
{
cv::Vec3b value = labImg.at<cv::Vec3b>(row, col);
value.val[0] *= maskImg.at<double>(row, col);
labImg.at<cv::Vec3b>(row, col) = value;
}
}
cv::Mat output;
cv::cvtColor(labImg, output, CV_Lab2BGR);
//cv::imwrite("vignette.png", output);
cv::namedWindow("Vignette", cv::WINDOW_NORMAL);
cv::resizeWindow("Vignette", output.size().width/2, output.size().height/2);
cv::imshow("Vignette", output);
cv::waitKey();
return 0;
}
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什么输出看起来像:

如上面的代码说明,通过改变的值firstPt,radius并且power可以实现强/弱艺术效果.
祝好运!
Abi*_*n K 10
您可以使用OpenCV中提供的高斯内核进行简单的实现.
请参阅下面的实现(对于灰度图像):
import cv2
import numpy as np
from matplotlib import pyplot as plt
img = cv2.imread('temp.jpg',0)
row,cols = img.shape
a = cv2.getGaussianKernel(cols,300)
b = cv2.getGaussianKernel(rows,300)
c = b*a.T
d = c/c.max()
e = img*d
cv2.imwrite('vig2.png',e)
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以下是我的结果:

同样对于彩色图像:

注意:当然,它是居中的.您需要进行其他修改才能将焦点移至其他位置.
与阿比德的回答类似。但代码是针对彩色图像的
import cv2
import numpy as np
from matplotlib import pyplot as plt
img = cv2.imread('turtle.jpg',1)
rows,cols = img.shape[:2]
zeros = np.copy(img)
zeros[:,:,:] = 0
a = cv2.getGaussianKernel(cols,900)
b = cv2.getGaussianKernel(rows,900)
c = b*a.T
d = c/c.max()
zeros[:,:,0] = img[:,:,0]*d
zeros[:,:,1] = img[:,:,1]*d
zeros[:,:,2] = img[:,:,2]*d
cv2.imwrite('vig2.png',zeros)
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原始图像(根据 CC0 许可证取自 Pexels)
应用 sigma 为 900 的 Vignette 后(即 `cv2.getGaussianKernel(cols,900))
应用 sigma 为 300 的 Vignette 后(即 `cv2.getGaussianKernel(cols,300))
此外,您可以通过简单地将高斯平均值转移到您的焦点,将晕影效果聚焦到您想要的坐标,如下所示。
import cv2
import numpy as np
img = cv2.imread('turtle.jpg',1)
fx,fy = 1465,180 # Add your Focus cordinates here
fx,fy = 145,1000 # Add your Focus cordinates here
sigma = 300 # Standard Deviation of the Gaussian
rows,cols = img.shape[:2]
fxn = fx - cols//2 # Normalised temperory vars
fyn = fy - rows//2
zeros = np.copy(img)
zeros[:,:,:] = 0
a = cv2.getGaussianKernel(2*cols ,sigma)[cols-fx:2*cols-fx]
b = cv2.getGaussianKernel(2*rows ,sigma)[rows-fy:2*rows-fy]
c = b*a.T
d = c/c.max()
zeros[:,:,0] = img[:,:,0]*d
zeros[:,:,1] = img[:,:,1]*d
zeros[:,:,2] = img[:,:,2]*d
zeros = add_alpha(zeros)
cv2.imwrite('vig4.png',zeros)
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海龟图像的尺寸为 1980x1200(宽x高)。以下是聚焦在坐标 1465,180(即fx,fy = 1465,180)处的示例(请注意,我已减少方差以举例说明焦点的变化)
以下是聚焦于坐标 145,1000(即fx,fy = 145,1000)的示例
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