OpenCV的resize函数优化
背景
在使⽤OpenCV做图像处理的时候，最常见的问题是c++版本性能不⾜，以resize函数为例来说明，将size为[864,1323,3]的函数缩⼩⼀半：
Mat img0;
gettimeofday(&t4, NULL);
cv::resize(source, img0, cv::Size(cols_out,rows_out));
gettimeofday(&t5, NULL);
time = 1000*(t5.tv_sec - t4.tv_sec) + (double)(t5.tv_usec - t4.tv_usec)/1000;
printf("frank cv::resize = %lf\n", time);
output:
frank cv::resize = 13.569000(ms) on Macmini(2.6 GHz Intel Core i5)
需要13ms才能完成这个操作，对于⼀半应⽤的实时性要求，要再33ms内处理⼀帧，超过1/3的时间花在resize上，是⾮常不合理的。
优化1
针对上⾯这种1/2⼤⼩的resize，可以这样来思考，既然我已经知道输出mat的长宽是输⼊mat的⼀半，那么输出mat的每个像素值都是从原图的4个像素得到的⼀个映射，事实上OpenCV也是⽤类似的技巧，从原图的点映射到输出图像的像素点，这个映射可以是Nearest Neiborhood，也可以是双线性插值、平均、取左上⾓等等。对于⼀般⾃然图像，我们可以直接使⽤NN映射来获得输出图像的点。
使⽤上⾯的⽅法：
double time = 0.0;
gettimeofday(&t1, NULL);
Mat half_img = resizeByHalf(source, &leftTopInt4);
int wh = ls;
int hh = ws;
printf("half_img = %d, %d\n", hh, wh);
gettimeofday(&t2, NULL);
time = 1000*(t2.tv_sec - t1.tv_sec) + (double)(t2.tv_usec - t1.tv_usec)/1000;
printf("frank resizeByHalf = %lf\n", time);
output:frank resizeByHalf = 10.539000
时间相⽐OpenCV的版本，缩⼩了3ms
优化2 进⼀步借鉴最邻近算法
如果我们不是resize到1/2等特定的长宽，⽽且任意尺⼨呢？上⾯的⽅法就不能直接使⽤，但是这个思想是可以借鉴的，我们可以进⼀步将最临近算法发挥的更好，对outuput的每个点，先根据长宽⽐计算其在原图中最邻近的像素点，然后直接根据最邻近的思想，直接拷贝Channel个字节作为输出图像：
Mat new_img = Mat(rows_out, cols_out, CV_8UC3);
resizeByNN(source.data, new_img.data, ws, ls, source.channels(), ws,
ls);
gettimeofday(&t4, NULL);
time = 1000*(t4.tv_sec - t3.tv_sec) + (double)(t4.tv_usec - t3.tv_usec)/1000;
printf("frank resizeByNN = %lf\n", time);
output: frank resizeByNN = 6.738000
时间已经缩⼩到原先的⼀半！
优化3
还有⼀个杀招，待⽇后更新。。。。
图
原图：
resizeByNN的结果：
上⾯两个函数的源码resizeByHalf
uchar maxInt4(uchar x0, uchar x1, uchar x2, uchar x3){
uchar t0 = (x0 >= x1) ? x0 : x1;
uchar t1 = (x2 >= x3) ? x2 : x3;
return (t0 >= t1) ? t0 : t1;
}
resize函数vbauchar meanInt4(uchar x0, uchar x1, uchar x2, uchar x3){
float t = (x0 + x1 + x2 + x3)/4.0;
return uchar(round(t));  //四舍五⼊
}
uchar leftTopInt4(uchar x0, uchar x1, uchar x2, uchar x3){
return x0;
}
Mat resizeByHalf(Mat input, uchar (*pfun)(uchar x0, uchar x1, uchar x2, uchar x3)){    Mat output;
int h_in = ws;
int w_in = ls;
int c = input.channels();
int h_out = h_in/2;
int w_out = w_in/2;
uchar *data_source = input.data;
uchar *data_half = output.data;
int bpl_source = w_in*3;
int bpl_half = w_out*3;
for (int i = 0; i < h_out; i++) {
for (int j = 0; j < w_out; j++) {
uchar *sr = data_source + i*2*bpl_source + j*2*c;
uchar *hr = data_half + i*bpl_half + j*c;
for (int k = 0; k < c; k++) {
*(hr+k) = pfun(*(sr+k), *(sr+k+c), *(sr+k+bpl_source), *(sr+k+c+bpl_source));            }
}
}
return output;
}
resizeByNN
void resizeByNN(uchar *input, uchar *output, int height_in, int width_in, int channels, int height_out, int width_out){
uchar *data_source = input;
uchar *data_half = output;
int bpl_source = width_in*3;
int bpl_dst = width_out*3;
int pos = 0;
int sep = 0;
uchar *sr = nullptr;
uchar *hr = nullptr;
float step = 0.0;
float step_x = float(width_in)/float(width_out);
float step_y = float(height_in)/float(height_out);
for (int i = 0; i < height_out; i++) {
for (int j = 0; j < width_out; j++) {
sep = int(step_y*i);
step = int(j*step_x);
sr = data_source + sep*bpl_source;
hr = data_half + i*bpl_dst +j*channels;
pos = step*channels;
memcpy(hr, sr+pos, channels);
}
}
return;
}
多谢阅读！

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