CODEWITHSUNDEEP
pythonopencvurllib
DanGoodrick
DanGoodrick

Reputation: 3208

Efficient link handling with OpenCV's VideoCapture method? {edited}

I'm using python with OpenCV. I was passing an MP4 link to VideoCapture but then I found my code runs ~10X faster (including urllib download time) if I download the video file with urllib.request first and then pass the filename to VideoCapture.

My question is if there is a way to download a video clip (subject to my machine's hardware {and software} constraints) into a capture object and then just let that object get disposed of with the normal python garbage collection when I am done with it?

Saving a video file to disk is slow compared to just loading it into memory. Any suggestions on how to do this with an internet file?

Upvotes: 3

Views: 424

Answers (1)

DanGoodrick
DanGoodrick

Reputation: 3208

urllib doesn't have to save the downloaded object to disk so all I needed to do was:

import urllib.request
url = 'https://previews.customer.envatousercontent.com/h264-video-previews/f331148f-3aa1-42ff-ae89-1a67274f2bd6/2009010.mp4'
video, http = urllib.request.urlretrieve(url)
cap = cv2.VideoCapture(video)

Upvotes: 2

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