使用 C++/OpenCV 从 100fps GoPro .mp4 视频创建 25fps 慢动作视频
Creating a 25fps slow motion video from a 100fps GoPro .mp4 video with C++/OpenCV
我有一个 100fps .mp4 GoPro 视频,我想用它创建一个 25fps 的慢动作视频。
我已经尝试了两天,但无济于事。我可以播放视频,从 GoPro 的 WiFi 流中保存视频,但是当我尝试读取 100fps 并将其保存在另一个 25fps 的视频文件中时,我得到空文件!
我怀疑用于编码新 mp4 视频的编解码器,但我不确定。
这是代码(我在 Visual Studio 2013 社区 Windows 10 预览版上使用 OpenCV 3.0.0 和 Visual C++)。
#include <iostream>
#include <vector>
#include <random>
#include <functional>
#include <algorithm>
#include <string>
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
using namespace cv;
using namespace std;
int main()
{
VideoCapture inputVideo("GOPR1016.MP4"); // Open the video File
if (!inputVideo.isOpened()) {
cout << "Error opening the video" << endl;
return -1;
}
int frames_num = int(inputVideo.get(CAP_PROP_FRAME_COUNT)); // Get the number of frames in the video
cout << "Num of frames: " << frames_num << endl;
int fps = int(inputVideo.get(CAP_PROP_FPS)); // get the frame rate
cout << "FPS: " << fps << endl;
int frame_width = inputVideo.get(CAP_PROP_FRAME_WIDTH);
int frame_height = inputVideo.get(CAP_PROP_FRAME_HEIGHT);
VideoWriter outputVideo;
string name = "outputVideo.avi";
Size size = Size((int)inputVideo.get(CAP_PROP_FRAME_WIDTH), (int)inputVideo.get(CAP_PROP_FRAME_HEIGHT)); // get the resolution
outputVideo.open(name, CV_FOURCC('3', 'I', 'V', 'X'), 25, size, true); // create a new videoFile with 25fps
Mat src;
for (int i = 0; i < frames_num; i++)
{
inputVideo >> src; // read
if (src.empty()) {
break; // in case ther's nothing to read
}
outputVideo << src; // write
}
waitKey(0); // key press to close window
return 1;
}
结果如下:
如我所料,是 Coded!我用了很多,但后来我发现了这个问题:Create Video from images using VideoCapture (OpenCV)
然后我在以下位置使用了编码的 MJPG:
outputVideo.open(name, CV_FOURCC('M', 'J', 'P', 'G'), 25, size, true); // create a new videoFile with 25fps
成功了!
结果如下:
我有一个 100fps .mp4 GoPro 视频,我想用它创建一个 25fps 的慢动作视频。 我已经尝试了两天,但无济于事。我可以播放视频,从 GoPro 的 WiFi 流中保存视频,但是当我尝试读取 100fps 并将其保存在另一个 25fps 的视频文件中时,我得到空文件! 我怀疑用于编码新 mp4 视频的编解码器,但我不确定。
这是代码(我在 Visual Studio 2013 社区 Windows 10 预览版上使用 OpenCV 3.0.0 和 Visual C++)。
#include <iostream>
#include <vector>
#include <random>
#include <functional>
#include <algorithm>
#include <string>
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/imgproc/imgproc.hpp>
using namespace cv;
using namespace std;
int main()
{
VideoCapture inputVideo("GOPR1016.MP4"); // Open the video File
if (!inputVideo.isOpened()) {
cout << "Error opening the video" << endl;
return -1;
}
int frames_num = int(inputVideo.get(CAP_PROP_FRAME_COUNT)); // Get the number of frames in the video
cout << "Num of frames: " << frames_num << endl;
int fps = int(inputVideo.get(CAP_PROP_FPS)); // get the frame rate
cout << "FPS: " << fps << endl;
int frame_width = inputVideo.get(CAP_PROP_FRAME_WIDTH);
int frame_height = inputVideo.get(CAP_PROP_FRAME_HEIGHT);
VideoWriter outputVideo;
string name = "outputVideo.avi";
Size size = Size((int)inputVideo.get(CAP_PROP_FRAME_WIDTH), (int)inputVideo.get(CAP_PROP_FRAME_HEIGHT)); // get the resolution
outputVideo.open(name, CV_FOURCC('3', 'I', 'V', 'X'), 25, size, true); // create a new videoFile with 25fps
Mat src;
for (int i = 0; i < frames_num; i++)
{
inputVideo >> src; // read
if (src.empty()) {
break; // in case ther's nothing to read
}
outputVideo << src; // write
}
waitKey(0); // key press to close window
return 1;
}
结果如下:
如我所料,是 Coded!我用了很多,但后来我发现了这个问题:Create Video from images using VideoCapture (OpenCV) 然后我在以下位置使用了编码的 MJPG:
outputVideo.open(name, CV_FOURCC('M', 'J', 'P', 'G'), 25, size, true); // create a new videoFile with 25fps
成功了!
结果如下: