如何在 OpenCV 中获取网络摄像头的 fps 速率？

Question

所以我需要在 OpenCV 中获取网络摄像头的 fps 速率。哪个函数可以做这样的事情？

Answer 1

int cvGetCaptureProperty( CvCapture* capture, int property_id);

与 property_id = CV_CAP_PROP_FPS

Answer 2

似乎对于实时网络摄像头捕获，您可以设置任意 fps 并读回相同的 fps，这与网络摄像头的真实 fps 无关。这是一个错误吗？

例如：

cvSetCaptureProperty(capture,CV_CAP_PROP_FPS,500);

稍后

double rates = cvGetCaptureProperty(capture,CV_CAP_PROP_FPS);
printf("%f\n",rates);

会给你 500。

但是如果我使用 网络摄像头 fps 链接，大约是正常的 30fps。

Answer 3

就我而言， fps = video.get(cv2.CAP_PROP_FPS) 不起作用。

所以，我在此链接中找到了此代码：

https://www.learnopencv.com/how-to-find-frame-rate-or-frames-per-second-fps-in-opencv-python-cpp/< /a>

import cv2
import time

if __name__ == '__main__':

    video = cv2.VideoCapture(1)

    # Find OpenCV version
    (major_ver, _, _) = (cv2.__version__).split('.')

    # With webcam get(CV_CAP_PROP_FPS) does not work.
    # Let's see for ourselves.

    if int(major_ver) < 3:
        fps = video.get(cv2.cv.CV_CAP_PROP_FPS)
        print "Frames per second using video.get(cv2.cv.CV_CAP_PROP_FPS): {0}".format(fps)
    else:
        fps = video.get(cv2.CAP_PROP_FPS)
        print "Frames per second using video.get(cv2.CAP_PROP_FPS) : {0}".format(fps)

    # Number of frames to capture
    num_frames = 120

    print "Capturing {0} frames".format(num_frames)

    # Start time
    start = time.time()

    # Grab a few frames
    for i in xrange(0, num_frames):
        ret, frame = video.read()

    # End time
    end = time.time()

    # Time elapsed
    seconds = end - start
    print "Time taken : {0} seconds".format(seconds)

    # Calculate frames per second
    fps = num_frames / seconds
    print "Estimated frames per second : {0}".format(fps);

    # Release video
    video.release()

Answer 4

*OpenCV 2 解决方案：

C++: double VideoCapture::get(int propId)

例如

VideoCapture myvid("video.mpg");
int fps=myvid.get(CV_CAP_PROP_FPS);