ffmpeg in Python subprocess - Unable to find a suitable output format for 'pipe:'
Question: Trying to burn subs into video with ffmpeg via Python. Works fine in the
command line, but when calling from Python subprocess with:

    
    
    p = subprocess.Popen('cd ~/Downloads/yt/; ffmpeg -i ./{video} -vf subtitles=./{subtitles} {out}.mp4'.format(video=vid.replace(' ', '\ '), subtitles=subs, out='out.mp4'), shell=True)
    

I get:

    
    
    Unable to find a suitable output format for 'pipe:'
    

Full traceback:

    
    
    'ffmpeg version 2.7.2 Copyright (c) 2000-2015 the FFmpeg developers
      built with Apple LLVM version 6.1.0 (clang-602.0.53) (based on LLVM 3.6.0svn)
      configuration: --prefix=/usr/local/Cellar/ffmpeg/2.7.2_1 --enable-shared --enable-pthreads --enable-gpl --enable-version3 --enable-hardcoded-tables --enable-avresample --cc=clang --host-cflags= --host-ldflags= --enable-opencl --enable-libx264 --enable-libmp3lame --enable-libvo-aacenc --enable-libxvid --enable-libfreetype --enable-libvpx --enable-libass --enable-libfdk-aac --enable-nonfree --enable-vda
      libavutil      54. 27.100 / 54. 27.100
      libavcodec     56. 41.100 / 56. 41.100
      libavformat    56. 36.100 / 56. 36.100
      libavdevice    56.  4.100 / 56.  4.100
      libavfilter     5. 16.101 /  5. 16.101
      libavresample   2.  1.  0 /  2.  1.  0
      libswscale      3.  1.101 /  3.  1.101
      libswresample   1.  2.100 /  1.  2.100
      libpostproc    53.  3.100 / 53.  3.100
    Input #0, mov,mp4,m4a,3gp,3g2,mj2, from './OnHub - a router for the new way to Wi-Fi-HNnfHP7VDP8.mp4':
      Metadata:
        major_brand     : isom
        minor_version   : 512
        compatible_brands: isomiso2avc1mp41
        encoder         : Lavf56.36.100
      Duration: 00:00:53.94, start: 0.000000, bitrate: 2092 kb/s
        Stream #0:0(und): Video: h264 (High) (avc1 / 0x31637661), yuv420p, 1920x1080 [SAR 1:1 DAR 16:9], 1961 kb/s, 23.98 fps, 23.98 tbr, 90k tbn, 47.95 tbc (default)
        Metadata:
          handler_name    : VideoHandler
        Stream #0:1(und): Audio: aac (LC) (mp4a / 0x6134706D), 44100 Hz, stereo, fltp, 125 kb/s (default)
        Metadata:
          handler_name    : SoundHandler
    [NULL @ 0x7fc07b077600] Unable to find a suitable output format for 'pipe:'
    pipe:: Invalid argument'
    


Answer: I'm guessing the problem was that you had spaces in some argument that you
weren't escaping. You could just escape it, but this is a better way to do
what you're trying to do:

    
    
    import os
    
    
    directory_path = os.path.expanduser('~/Downloads/yt/')
    video_path = 'path/to/video'
    subtitles_path = 'path/to/subtitles'
    outfile_path = 'out.mp4'
    
    args = ['ffmpeg', '-i', video_path, '-vf',
            'subtitles={}'.format(subtitles_path), outfile_path]]
    p = subprocess.Popen(args, cwd=directory_path)
    

The main difference is that you're not using `shell=True`, which is good
practice for security and other reasons, including the fact that you don't
have to worry about quoting arguments with spaces. Because it's not using
`shell=True`, you have to pass in the command line as a list of strings, one
element per argument, instead of one string like before.

