将电影字幕整理后，便于打印学习

看电影学英语

电影字幕可以从网上下载scr文件，下载后便于学习。但是对于孩子，总看计算机也不好，因此需要将字幕打印下来。如果直接打印scr文件，则文件相当大，没用的换行也非常多。因此对其整理后，可以减少打印数量。手工整理非常费时费力，因此想到使用python写个简单的处理代码。由于很简单，未加注释，对照scr文件，很容易理解。

处理过程中，如果对白之间有超过5秒的空白，则认为切换到另一个对话场景。

python代码如下：

from __future__ import print_function

import sys
import os

def time_interval(start, end):
    int_start = int(start[:2])*3600 + int(start[3:5])*60 + int(start[6:8])
    int_end = int(end[:2])*3600 + int(end[3:5])*60 + int(end[6:8])
    return int_end - int_start
    
def space_print(x, n):
    if len(x)<n:
        for ln in range(n-len(x)):
            print(" ", end='')
    print(x, end='')

def subtitles2screen(scr):
    with open(scr) as f_scr:
        previous_finish_time = ""
        previous_subtitles_index = ""
        subtitles_index = "";
        scene_index = 1
        for line in f_scr.readlines():
            line = line.replace("\n", "").strip()
            if line.isdigit():
                subtitles_index = line
            elif line.find(" --> ") != -1:
                if previous_finish_time.strip()=="":
                    print("scene: " + str(scene_index), end='')
                    scene_index += 1
                elif time_interval(previous_finish_time, line[0:12]) > 5:
                    print("\nscene: " + str(scene_index) + " " + previous_finish_time, end='')
                    scene_index += 1
                previous_finish_time = line[17:29]
            elif line == "":
                pass
            else:
                if subtitles_index == previous_subtitles_index:
                    print(" " + line, end='')
                else:
                    print("")
                    space_print(subtitles_index, 4)
                    print(": " + line, end='')
                previous_subtitles_index = subtitles_index

if __name__ == "__main__" :
    if len(sys.argv) < 2:
        print("usage: python " + sys.argv[0] + " ")
    elif not os.path.exists(sys.argv[1]):
        print("subtitles file does not exist")
    else:
        subtitles2screen(sys.argv[1])