Opencv遍历像素和imadjust函数实现

最新推荐文章于 2025-09-12 13:20:46 发布

原创最新推荐文章于 2025-09-12 13:20:46 发布 · 1.1k 阅读

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本文介绍了使用Opencv进行像素遍历的三种高效方法，并对比了它们在处理4000*6000大小图像时的性能。通过直接访问、连续内存优化及lambda表达式，实现了快速图像处理。

Opencv遍历像素有四种方式

第一种

利用opencv中

   for (int i = 0; i < nrow; i++)
   {
       for (int j = 0; j < ncol; j++)
       {
           //qDebug() << "original pixel is" << source.at<uchar>(i, j);
           source.at<uchar>(i, j) = uchar(fn(source.at<uchar>(i, j), low_in, high_in, low_out, high_out));
           //qDebug() << "processed pixel is" << source.at<uchar>(i, j);

       }
   }

第二种

   for (int i = 0; i < nrow; i++)
   {
       uchar* data = source.ptr<uchar>(i);
       for (int j = 0; j < ncol; j++)
       {
data[j] = uchar(fn(data[j], low_in, high_in, low_out, high_out));
       }
   }

第三种

   if (source.isContinuous())
   {
    nc = nrow * ncol;
       uchar* inData = source.ptr<uchar>(0);
       for (int i = 0; i < nc; i++)
       {
           *inData = fn(*inData, low_in, high_in, low_out, high_out);
           inData++;
       }
   }

图像IMadjust利用lambda表达式

    auto fn = [](double x, double low_in, double high_in, double low_out, double high_out) {
       if (x < low_in)
           return low_out;
       if (x > high_in)
           return high_out;
       return (high_out - high_in) / (low_out - low_in)*(x - low_in) + high_in;
   };

整体测试结果对于4000*6000大小的图片，三种方法的用时分别为0.873s,0.203s，0.195s