我最近使用了这段代码来保存来自网络摄像头的帧数据
#include <opencv\cv.h>
#include <opencv\highgui.h>
#include <opencv2/opencv.hpp>
using namespace cv;
#include <fstream>
using namespace std;
int main(int argc, char** argv)
{
VideoCapture cap(0); // open the default camera
if (!cap.isOpened()) // check if we succeeded
return -1;
cap.set(CV_CAP_PROP_FPS, 15);
Mat edges;
namedWindow("image", 1);
std::vector<cv::Mat> images(100);
for (int i = 0; i < 100; ++i) {
// this is optional, preallocation so there's no allocation
// during capture
images[i].create(480, 640, CV_8UC3);
}
for (int i = 0; i < 100; ++i)
{
Mat frame;
cap >> frame; // get a new frame from camera
frame.copyTo(images[i]);
}
cap.release();
for (int i = 0; i < 100; ++i)
{
imshow("image", images[i]);
if (waitKey(30) >= 0) break;
}在此之后,我想使用imread来分析新拆分的帧。然而,我想不出一种方法来实现这一点。
我试过了:Mat colorImage = imread(images[i]);
但是,它会导致:
error C2664: 'cv::Mat cv::imread(const cv::String &,int)': cannot convert argument 1 from 'std::vector<cv::Mat,std::allocator<_Ty>>' to 'const cv::String &' with [ _Ty=cv::Mat ]
在此之前,非常感谢:)
发布于 2016-09-26 07:18:37
imread函数用于从磁盘打开镜像。
你已经有了图像的向量,所以你只需要做:
Mat colorImage = imagesi;
顺便说一句。不需要这样做:
for (int i = 0; i < 100; ++i) {
// this is optional, preallocation so there's no allocation
// during capture
images[i].create(480, 640, CV_8UC3);
}因为您无论如何都要分配新的空间,除非您像这样直接捕获帧:
cap >> images[i];https://stackoverflow.com/questions/39690566
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