Question

现在我要使用openCV Mat结构将图像旋转90度，代码在下面。而且我发现了一种现象，当图像宽度为4096时，运行时间约为图像宽度运行时间的两倍，为4097.以下是运行时间输出。有人知道为什么吗？

#include<string>
#include<fstream>
#include<iostream>
#include "opencv.hpp"
using namespace cv;
void main()
{
std::string ImageFileName0 = "F:\\temp\\4096-3000.bmp";
std::string ImageFileName1 = "F:\\temp\\4097-3000.bmp";

Mat image4096 = imread(ImageFileName0);
Mat image4097 = imread(ImageFileName1);

int lWidth4096 = image4096.size().width;
int lHeight4096 = image4096.size().height;
Mat image4096Res(lWidth4096, lHeight4096, CV_8UC1);
for (int i=0; i<lHeight4096; i++)
{
    for (int j=0; j<lWidth4096;j++)
    {
        image4096Res.at<uchar>(j, i) = 0;
    }
}

int lWidth4097 = image4097.size().width;
int lHeight4097 = image4097.size().height;
Mat image4097Res(lWidth4097, lHeight4097, CV_8UC1);
for (int i = 0; i < lHeight4097; i++)
{
    for (int j = 0; j < lWidth4097; j++)
    {
        image4097Res.at<uchar>(j, i) = 0;
    }
}

for (int i = 0; i < 10; i++)
{
    double time0 = static_cast<double>(getTickCount());
    for (int j = 0; j < lWidth4096; ++j)
    {
        for (int i = 0; i<lHeight4096; ++i)
        {
            image4096Res.at<uchar>(j, i) = image4096.at<uchar>(i, j);
        }
    }
    double time4096 = (static_cast<double>(getTickCount()) - time0) / getTickFrequency();

    double time1 = static_cast<double>(getTickCount());
    for (int j = 0; j < lWidth4097; ++j)
    {
        for (int i = 0; i<lHeight4097; ++i)
        {
            //srcTmp = srcimage.PointToRow(i);
            //*(dstTmp + i) = srcimage.GetPixelValue(j, i);
            image4097Res.at<uchar>(j, i) = image4097.at<uchar>(i, j);

        }
    }
    double time4097 = (static_cast<double>(getTickCount()) - time1) / getTickFrequency();
    std::cout << "4096 time:" << time4096*1000 << std::endl;
    std::cout << "4097 time:" << time4097*1000 << std::endl;
    std::cout << std::endl;
}
namedWindow("aa",CV_WINDOW_NORMAL);
imshow("aa", image4096Res);

waitKey();
}

运行时间结果：

4096时间：149.337

4097时间：56.8092

4096时间：143.556

4097时间：67.4758

4096时间：142.07

4097时间：58.2825

4096时间：153.973

4097时间：57.1894

4096时间：145.086

4097时间：58.7944

4096时间：156.33

4097时间：87.9404

4096时间：140.224

4097时间：56.9525

4096时间：144.413

4097时间：57.133

4096时间：141.672

4097时间：54.916

4096时间：148.443

4097时间：55.8449

Time consuming varies with the width of the image is here

很明显，只有当图像宽度为1024,2048,4096时，执行时间才会异常，而整体趋势会线性增加。

Answer 1

您是否可以尝试使用此版本来查看是否存在相同的差异还添加了测试以查看加载的图像是否为CV_8UC1类型。请注意，您和我的解决方案的结果是镜像图像

int lWidth4096 = image4096.size().width;
int lHeight4096 = image4096.size().height;
Mat image4096Res = cv::Mat::zeros(lWidth4096, lHeight4096, CV_8UC1);

int lWidth4097 = image4097.size().width;
int lHeight4097 = image4097.size().height;
Mat image4097Res = cv::Mat::zeros(lWidth4097, lHeight4097, CV_8UC1);

if (image4096.type() != CV_8UC1)
    throw "need 8bit image as input";

if (image4097.type() != CV_8UC1)
    throw "need 8bit image as input";

for (int i = 0; i < 10; i++)
{
    double time0 = static_cast<double>(getTickCount());
    for (int j = 0; j < lWidth4096; ++j)
    {
        uint8_t *buf = image4096Res.ptr(j);
        for (int i = 0; i<lHeight4096; ++i)
        {
            //image4096Res.at<uchar>(j, i) = image4096.at<uchar>(i, j);
            //buf[i] = image4096.at<uchar>(i, j);
            buf[i] = (image4096.data + image4096.step[0] * i)[j];
        }
    }
    double time4096 = (static_cast<double>(getTickCount()) - time0) / getTickFrequency();

    double time1 = static_cast<double>(getTickCount());
    for (int j = 0; j < lWidth4097; ++j)
    {
        uint8_t *buf = image4097Res.ptr(j);
        for (int i = 0; i<lHeight4097; ++i)
        {
            //image4097Res.at<uchar>(j, i) = image4097.at<uchar>(i, j);
            //buf[i] = image4097.at<uchar>(i, j);
            buf[i] = (image4097.data + image4097.step[0] * i)[j];
        }
    }
    double time4097 = (static_cast<double>(getTickCount()) - time1) / getTickFrequency();
    std::cout << "4096 time:" << time4096 * 1000 << std::endl;
    std::cout << "4097 time:" << time4097 * 1000 << std::endl;
    std::cout << std::endl;
}

当灰度图像宽度为2048,4096时，旋转90图像的运行时间比附近

1 个答案: