ffmpeg 使用 python 子进程设置文件名模式

Question

When running on command line, i use this code:

ffmpeg -i video-frame-%06d.png -c:v libx264 -vf "fps=25,format=yuv420p" out.mp4

But im trying to run it inside python script,and Im getting no such file or directory error and I think because of the formatting type for the image filename numerals -%06d

How can I concatenate or use the same file pattern I run in command line to use it in my python script

subprocess.Popen(['ffmpeg -i dir/video-frame-%06d.png -c:v libx264 -vf "fps=25,format=yuv420p" out.mp4', shell=True])

the sample image filename inside the folder are:

1. video-frame-00001.png
2. video-frame-00002.png
3. video-frame-00003.png

Answer 1

I believe if you want to run subprocess you have 2 options:

• one line command and use shell=True which is not recommended.
• or use a proper list.

try this:

cmd = 'ffmpeg -i dir/video-frame-%06d.png -c:v libx264 -vf "fps=25,format=yuv420p" out.mp4'
subprocess.run(cmd.split())

Answer 2

You should use the full path instead of relative path. Eg.: based on path of your Python file:

dir_of_imgs = os.path.join(os.path.realpath(os.path.dirname(__file__)), "dir")

And use this variable to define the path of image:

img_path = os.path.join(dir_of_imgs, "video-frame-%06d.png")

It means your command should be something similar like that:

subprocess.Popen(['ffmpeg -i {} -c:v libx264 -vf "fps=25,format=yuv420p" out.mp4'.format(img_path), shell=True])

NOTE:

Of course, probably your image files are in different folder but you can use the path of your Python file as a reference (as I show above).