Contiguous frames

You have a couple options:

1. Use vframes (or equivalently frames:v) to read n frames at once:

fs, frames = ffmpegio.video.read(video_path, ss=timestamps[0], vframes=len(timestamps), pix_fmt=image_format)

2. Use to to specify the ending timestamp:

fs, frames = ffmpegio.video.read(video_path, ss=timestamps[0], to=timestamps[-1], pix_fmt=image_format)

Caveat: I'm not 100% sure if this will always include the last frame at to, i.e., whether to is an inclusive or exclusive end point. I recommend using the first option.

Non-contiguous frames

If the # of keeper frames >> # of skipped frames

You can just slice the frames to your liking

fs, frames = ffmpegio.video.read(video_path, ss=timestamps[0], to=timestamps[-1], pix_fmt=image_format)

n = np.round(timestamps*fs).astype(int)
frames = frames[n-n[0], ...]

This could very well be the fastest way regardless of how much you're keeping if you're using CUDA as you implied on your other thread.

If the # of keeper frames << # of skipped frames

You can use select filter to specify which frames to return. Learn the FFmpeg expression from its documentation to formulate your selection logic, but using one of the examples:

vf_expr = ''select='not(mod(n\,100))'" # every 100 frames
fs, frames = ffmpegio.video.read(video_path, ss=timestamps[0], vframes=len(timestamps), pix_fmt=image_format, vf=vf_expr)

Use the vframes option in conjunction so FFmpeg stops immediately after retrieving the last frame.

If your timestamps are randomly chosen (as in no formula can describe your choice) then your expression could get too long. Also, there are several timestamp related options that you may need to set for it to return timestamp accurate frames (e.g., copyts). Nonetheless, give it a try first.