C++ 在OpenCV的两个矩阵中保留具有最小绝对值的有符号值

在OpenCV中，我有两个大小相同的矩阵1和矩阵2。我想找到在两个矩阵中具有最小绝对值的有符号值，并将其保留在矩阵1中。为此，我使用以下代码：

for (int i = 0; i < One.rows; ++i)
{
    p=One.ptr<float>(i);
    p_two = Two.ptr<float>(i);
    for (int j = 0; j < One.cols; ++j)
    {
        if(fabsf(p_two[j])<fabsf(p[j])) 
             p[j] = p_two[j];
    }
}

for（int i=0；i<1.rows；++i）
{
p=1.ptr（i）；
p_two=2.ptr（i）；
对于（int j=0；j<1.cols；++j）
{
如果（fabsf（p_two[j]）你的代码不是你程序的瓶颈。它确实非常快。你需要分析你的代码，看看实际的瓶颈在哪里


如果矩阵是连续的（这在实践中非常常见），您可以对其进行一些优化，例如：

您可以看到您的方法非常快。我的方法快了一点。@s1h速度较慢，但更简洁易读

代码
您可以使用以下方法在电脑上评估结果：

#include <opencv2/opencv.hpp>
#include <iostream>
using namespace std;
using namespace cv;

int main()
{
    vector<Size> sizes{ Size(3, 3), Size(10, 10), Size(100, 100), Size(1280, 720), Size(1000, 1000), Size(1920, 1080), Size(4096, 3112), Size(5000, 5000) };

    cout << "Size: \t\tYuanhao \ts1h \t\tMiki" << endl;

    for (int is = 0; is < sizes.size(); ++is)
    {
        Size sz = sizes[is];

        cout << sz << "\t";

        Mat1f img1(sz);
        randu(img1, Scalar(-100), Scalar(100));

        Mat1f img2(sz);
        randu(img2, Scalar(-100), Scalar(100));

        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            for (int r = 0; r < one.rows; ++r)
            {
                float* pone = one.ptr<float>(r);
                float* ptwo = two.ptr<float>(r);
                for (int c = 0; c < one.cols; ++c)
                {
                    if (fabs(ptwo[c]) < fabs(pone[c]))
                    {
                        pone[c] = ptwo[c];
                    }
                }
            }

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }


        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            two.copyTo(one, abs(two) < abs(one));

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }


        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            int rows = one.rows;
            int cols = one.cols;
            if (one.isContinuous() && two.isContinuous())
            {
                cols = rows * cols;
                rows = 1;
            }

            for (int r = 0; r < rows; ++r)
            {
                float* pone = one.ptr<float>(r);
                float* ptwo = two.ptr<float>(r);
                for (int c = 0; c < cols; ++c)
                {
                    if (fabs(ptwo[c]) < fabs(pone[c]))
                    {
                        pone[c] = ptwo[c];
                    }
                }
            }

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }

        cout << endl;
    }

    getchar();
    return 0;
}

#包括
#包括
使用名称空间std；
使用名称空间cv；
int main（）
{
向量大小{Size（3,3）、Size（10,10）、Size（100100）、Size（1280720）、Size（10001000）、Size（19201080）、Size（40963112）、Size（50005000）}；
就速度而言，我还没有测试过这一点，但一般来说，您可以用以下行替换您的循环：Two.copyTo（one，cv:：abs（Two）ParallelLoopBody
）（或）使用SSE指令的单矩阵运算，使用SSE加速循环。非常感谢！B）是一个很好的选择。我也考虑保存绝对值，只是做比较。其他选项是合理的，但目前我不能完成它。BTW，我正在尝试优化代码，也许你可以给我一些想法来改进它。表演？提前谢谢！
two.copyTo(one, abs(two) < abs(one));

Size:           Yuanhao         s1h             Miki
[3 x 3]         0.000366543     0.117294        0.000366543
[10 x 10]       0.00109963      0.0157614       0.00109963
[100 x 100]     0.0964009       0.139653        0.112529
[1280 x 720]    8.70577         11.0267         8.65372
[1000 x 1000]   9.66538         13.5068         9.02026
[1920 x 1080]   16.5681         26.9706         15.7412
[4096 x 3112]   104.423         135.629         102.595
[5000 x 5000]   196.124         277.457         187.203

#include <opencv2/opencv.hpp>
#include <iostream>
using namespace std;
using namespace cv;

int main()
{
    vector<Size> sizes{ Size(3, 3), Size(10, 10), Size(100, 100), Size(1280, 720), Size(1000, 1000), Size(1920, 1080), Size(4096, 3112), Size(5000, 5000) };

    cout << "Size: \t\tYuanhao \ts1h \t\tMiki" << endl;

    for (int is = 0; is < sizes.size(); ++is)
    {
        Size sz = sizes[is];

        cout << sz << "\t";

        Mat1f img1(sz);
        randu(img1, Scalar(-100), Scalar(100));

        Mat1f img2(sz);
        randu(img2, Scalar(-100), Scalar(100));

        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            for (int r = 0; r < one.rows; ++r)
            {
                float* pone = one.ptr<float>(r);
                float* ptwo = two.ptr<float>(r);
                for (int c = 0; c < one.cols; ++c)
                {
                    if (fabs(ptwo[c]) < fabs(pone[c]))
                    {
                        pone[c] = ptwo[c];
                    }
                }
            }

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }


        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            two.copyTo(one, abs(two) < abs(one));

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }


        {
            Mat1f one = img1.clone();
            Mat1f two = img2.clone();

            double tic = double(getTickCount());

            int rows = one.rows;
            int cols = one.cols;
            if (one.isContinuous() && two.isContinuous())
            {
                cols = rows * cols;
                rows = 1;
            }

            for (int r = 0; r < rows; ++r)
            {
                float* pone = one.ptr<float>(r);
                float* ptwo = two.ptr<float>(r);
                for (int c = 0; c < cols; ++c)
                {
                    if (fabs(ptwo[c]) < fabs(pone[c]))
                    {
                        pone[c] = ptwo[c];
                    }
                }
            }

            double toc = (double(getTickCount()) - tic) * 1000. / getTickFrequency();
            cout << toc << " \t";
        }

        cout << endl;
    }

    getchar();
    return 0;
}