linux - 如何使用 .sh 文件提取帧并设置每秒帧数

Question

我正在尝试使用 shell 脚本文件从视频中提取特征，同时从不知道如何设置每秒帧数的视频中提取特征。

#!/bin/bash
frames_folder_path=./data
videos_folder_path=./videos
ext=mp4

mkdir "${frames_folder_path}"

for video_file_path in "${videos_folder_path}"/*."${ext}"; do
    slash_and_video_file_name="${video_file_path:${#videos_folder_path}}"
    slash_and_video_file_name_without_extension="${slash_and_video_file_name%.${ext}}"
    video_frames_folder_path="${frames_folder_path}${slash_and_video_file_name_without_extension}";
    mkdir "${video_frames_folder_path}"
    ffmpeg -i "${video_file_path}" "${video_frames_folder_path}/%d.jpg"
done

我试过这段代码来提取特征。我只想每秒提取 2 帧，但它以默认帧速率每秒删除 30 帧。
如何使用 shell 脚本文件解决此问题。

Answer 1

添加 -r 2在文件路径和输出路径之间，这里 2 是每秒帧数(FPS)ffmpeg -i "${video_file_path}" "${video_frames_folder_path}/%d.jpg"用这个ffmpeg -i "${video_file_path}" -r 2 "${video_frames_folder_path}/%d.jpg"

#!/bin/bash
    frames_folder_path=./data
    videos_folder_path=./vid
    ext=mp4
    
    mkdir "${frames_folder_path}"
    
    for video_file_path in "${videos_folder_path}"/*."${ext}"; do
        slash_and_video_file_name="${video_file_path:${#videos_folder_path}}"
        slash_and_video_file_name_without_extension="${slash_and_video_file_name%.${ext}}"
        video_frames_folder_path="${frames_folder_path}${slash_and_video_file_name_without_extension}";
        mkdir "${video_frames_folder_path}"
        ffmpeg -i  "${video_file_path}" -r 2 "${video_frames_folder_path}/%d.jpg" 
    done