Question

是否有一种有效的方法可以在不使用任何插值的情况下在OpenCV中调整图像大小？我希望我的图像能够将像素重新映射为更大的图像，而不是传统的“调整大小”，而是用0填充其他所有图像。

e.g。将img1放大到2x以下到img2：

img1 = [ 1, 2, 3,
         4, 5, 6,
         7, 8, 9 ]

cv::resize(img1, img2, cv::Size(6, 6));

img2 = [ 1, 0, 2, 0, 3, 0,
         0, 0, 0, 0, 0, 0,
         4, 0, 5, 0, 6, 0,
         0, 0, 0, 0, 0, 0,
         7, 0, 8, 0, 9, 0,
         0, 0, 0, 0, 0, 0 ]

我知道显而易见的方法是使用for循环，但我想知道是否有更有效的方式使用OpenCV调用？

Answer 1

我想到的一个选项是将cv::resize与INTER_NEAREST一起使用，然后屏蔽不需要的像素。

示例：

#include <opencv2/opencv.hpp> #include <cstdint> #include <iostream> int main() { cv::Mat m1((cv::Mat_<uint8_t>(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9)); std::cout << "Input:\n" << m1 << "\n\n"; cv::Mat mask((cv::Mat_<uint8_t>(2, 2) << 255, 0, 0, 0)); mask = cv::repeat(mask, m1.rows, m1.cols); std::cout << "Mask:\n" << mask << "\n\n"; cv::Mat m2; cv::resize(m1, m2, cv::Size(), 2, 2, cv::INTER_NEAREST); std::cout << "Resized:\n" << m2 << "\n\n"; cv::bitwise_and(m2, mask, m2); std::cout << "Masked:\n" << m2 << "\n\n"; }

控制台输出：

Input: [ 1, 2, 3; 4, 5, 6; 7, 8, 9] Mask: [255, 0, 255, 0, 255, 0; 0, 0, 0, 0, 0, 0; 255, 0, 255, 0, 255, 0; 0, 0, 0, 0, 0, 0; 255, 0, 255, 0, 255, 0; 0, 0, 0, 0, 0, 0] Resized: [ 1, 1, 2, 2, 3, 3; 1, 1, 2, 2, 3, 3; 4, 4, 5, 5, 6, 6; 4, 4, 5, 5, 6, 6; 7, 7, 8, 8, 9, 9; 7, 7, 8, 8, 9, 9] Masked: [ 1, 0, 2, 0, 3, 0; 0, 0, 0, 0, 0, 0; 4, 0, 5, 0, 6, 0; 0, 0, 0, 0, 0, 0; 7, 0, 8, 0, 9, 0; 0, 0, 0, 0, 0, 0]

更新

如果我们删除了对我们特定场景不必要的部分Miki代码，我们几乎将它简化为一个简单的循环。

进行一些快速比较，结果有点快。

#include <opencv2/opencv.hpp> #include <chrono> #include <cstdint> #include <iostream> cv::Mat resize_1(cv::Mat image) { cv::Mat result(cv::Mat::zeros(image.rows * 2, image.cols * 2, CV_8UC1)); for (int ra(0); ra < image.rows; ++ra) { for (int ca = 0; ca < image.cols; ++ca) { result.at<uint8_t>(ra * 2, ca * 2) = image.at<uint8_t>(ra, ca); } } return result; } cv::Mat resize_2(cv::Mat image) { cv::Mat mask((cv::Mat_<uint8_t>(2, 2) << 255, 0, 0, 0)); mask = cv::repeat(mask, image.rows, image.cols); cv::Mat result; cv::resize(image, result, cv::Size(), 2, 2, cv::INTER_NEAREST); cv::bitwise_and(result, mask, result); return result; } template<typename T> void timeit(T f) { using std::chrono::high_resolution_clock; using std::chrono::duration_cast; using std::chrono::microseconds; cv::Mat m1((cv::Mat_<uint8_t>(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9)); m1 = cv::repeat(m1, 1024, 1024); high_resolution_clock::time_point t1 = high_resolution_clock::now(); for (uint32_t i(0); i < 256; ++i) { cv::Mat result = f(m1); } high_resolution_clock::time_point t2 = high_resolution_clock::now(); auto duration = duration_cast<microseconds>(t2 - t1).count(); double t_ms(static_cast<double>(duration) / 1000.0); std::cout << "Total = " << t_ms << " ms\n" << "Iteration = " << (t_ms / 256) << " ms\n" << "FPS = " << (256 / t_ms * 1000.0) << "\n"; } int main() { timeit(&resize_1); timeit(&resize_2); }

<强>定时：

resize_1

Total = 6344.86 ms Iteration = 24.7846 ms FPS = 40.3476

resize_2

Total = 7271.31 ms Iteration = 28.4036 ms FPS = 35.2068

更新2

并行化版本：

class ResizeInvoker : public cv::ParallelLoopBody { public: ResizeInvoker(cv::Mat const& src, cv::Mat& dst) : image(src) , result(dst) { } void operator()(const cv::Range& range) const { for (int y(range.start); y < (range.end); ++y) { for (int x(0); x < image.cols; ++x) { result.at<uint8_t>(y * 2, x * 2) = image.at<uint8_t>(y, x); } } } cv::Mat const& image; cv::Mat& result; }; cv::Mat resize_3(cv::Mat image) { cv::Mat result(cv::Mat::zeros(image.rows * 2, image.cols * 2, CV_8UC1)); ResizeInvoker loop_body(image, result); cv::parallel_for_(cv::Range(0, image.rows) , loop_body , result.total() / (double)(1 << 16)); return result; }

<强>定时：

resize_3

Total = 3876.63 ms Iteration = 15.1431 ms FPS = 66.0367

更新3

如果我们在调用者中使用原始指针，我们可以做得更好：

void operator()(const cv::Range& range) const { for (int y(range.start); y < (range.end); ++y) { uint8_t* D = result.data + result.step * y * 2; uint8_t const* S = image.data + image.step * y; for (int x(0); x < image.cols; ++x) { D[x * 2] = S[x]; } } }

定时：

Total = 3387.87 ms Iteration = 13.2339 ms FPS = 75.5636

Answer 2

您可以使用图片的Kronecker product和类似的图案：

1, 0
0, 0

结果是：

Input:
[1, 2, 3;
 4, 5, 6;
 7, 8, 9]

Output:
[1, 0, 2, 0, 3, 0;
 0, 0, 0, 0, 0, 0;
 4, 0, 5, 0, 6, 0;
 0, 0, 0, 0, 0, 0;
 7, 0, 8, 0, 9, 0;
 0, 0, 0, 0, 0, 0]

代码：

#include <opencv2/opencv.hpp>
#include <iostream>

using namespace std;
using namespace cv;

Mat1b kron(const Mat1b& A, const Mat1b& B)
{
    Mat1b K(A.rows * B.rows, A.cols * B.cols, uchar(0));
    for (int ra = 0; ra < A.rows; ++ra)
    {
        for (int ca = 0; ca < A.cols; ++ca)
        {
            K(Range(ra*B.rows, (ra + 1)*B.rows), Range(ca*B.cols, (ca + 1)*B.cols)) = B.mul(A(ra, ca));
        }
    }
    return K;
}

int main()
{
    Mat1b img = (Mat1b(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9);
    std::cout << "Input:\n" << img << "\n\n";

    // Define the pattern
    Mat1b pattern = (Mat1b(2, 2) << 1, 0, 
                                    0, 0);

    Mat1b out = kron(img, pattern);
    std::cout << "Output:\n" << out << "\n\n";

    return 0;
}

OpenCV没有实现Kronecker产品，因此您需要编写自定义函数。对于适用于所有数据类型（1个通道）的更通用的实现，请查看here。

我发现@Dan Masek方法更快。这是因为我的kron实现未进行优化。我希望这种方法能很好地实现。

Answer 3

考虑分享以下方法，因为它有点不同。我不知道与其他方法相比有多高效。至少你可以使用opencv调用而不需要任何循环，并且可以轻松地为x和y使用任意比例因子。

首先将图像转换为浮点类型，然后使用warpAffine进行缩放（使用线性插值）。使用最近邻方法调整相同图像的大小。比较两个结果图像元素以获得掩码。使用此蒙版复制任何结果图像中的相关元素。

以下是我得到的代码和一些结果：

uchar data[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
Mat im(3, 3, CV_8U, data);
im.convertTo(im, CV_32F);
// x and y scale
int xscale = 2, yscale = 2;
Size size(im.cols * xscale, im.rows * yscale);
float tr[] = {xscale, 0, 0, 0, yscale, 0};
Mat m(2, 3, CV_32F, tr);    // transformation matrix

Mat resized1, resized2;
warpAffine(im, resized1, m, size);  // affine scaling with linear interpolation
resize(im, resized2, size, 0, 0, INTER_NEAREST);    // resize with nearest neighbor
// get the mask
Mat resized = resized1 == resized2;
// copy the pixels
resized1.copyTo(resized, resized);

cout << "image:\n" << im << endl;
cout << "M:\n" << m << endl;
cout << "affine(scaled):\n" << resized1 << endl;
cout << "resized:\n" << resized2 << endl;
cout << "mask:\n" << resized << endl;
cout << "output:\n" << resized << endl;

对于xscale = 2，yscale = 2

image:
[1, 2, 3;
  4, 5, 6;
  7, 8, 9]
M:
[2, 0, 0;
  0, 2, 0]
affine(scaled):
[1, 1.5, 2, 2.5, 3, 1.5;
  2.5, 3, 3.5, 4, 4.5, 2.25;
  4, 4.5, 5, 5.5, 6, 3;
  5.5, 6, 6.5, 7, 7.5, 3.75;
  7, 7.5, 8, 8.5, 9, 4.5;
  3.5, 3.75, 4, 4.25, 4.5, 2.25]
resized:
[1, 1, 2, 2, 3, 3;
  1, 1, 2, 2, 3, 3;
  4, 4, 5, 5, 6, 6;
  4, 4, 5, 5, 6, 6;
  7, 7, 8, 8, 9, 9;
  7, 7, 8, 8, 9, 9]
mask:
[1, 0, 2, 0, 3, 0;
  0, 0, 0, 0, 0, 0;
  4, 0, 5, 0, 6, 0;
  0, 0, 0, 0, 0, 0;
  7, 0, 8, 0, 9, 0;
  0, 0, 0, 0, 0, 0]
output:
[1, 0, 2, 0, 3, 0;
  0, 0, 0, 0, 0, 0;
  4, 0, 5, 0, 6, 0;
  0, 0, 0, 0, 0, 0;
  7, 0, 8, 0, 9, 0;
  0, 0, 0, 0, 0, 0]

对于xscale = 4，yscale = 3

output:
[1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  4, 0, 0, 0, 5, 0, 0, 0, 6, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  7, 0, 0, 0, 8, 0, 0, 0, 9, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

如何在OpenCV中没有插值（零填充）的情况下调整大小？

3 个答案:

更新

更新2

更新3