how to convert datetime-like string into milliseconds

Question

I have a user-defined function ( return_times ) that takes json file and returns two datetime-like strings.

time_1, time_2= return_times("file.json")
print(time_1, time_2) # outputs: 00:00:11.352 00:01:51.936

By datetime-like string I mean 00:00:11.352 which suits '%H:%M:%S.%f' formatting. However, when I try to convert them into milliseconds, I get negative values.

from datetime import datetime

dt_obj_1 = datetime.strptime(time_1, '%H:%M:%S.%f')
start_ms = dt_obj_1.timestamp() * 1000

dt_obj_2 = datetime.strptime(time_2, '%H:%M:%S.%f')
end_ms = dt_obj_2.timestamp() * 1000

print(start_ms, end_ms ) # outputs: -2209019260648.0 -2209019160064.0

If I success I would like to trim a video with the following command:

from moviepy.video.io.ffmpeg_tools import ffmpeg_extract_subclip
ffmpeg_extract_subclip("long_video.mp4", start_ms, end_ms, targetname="video_trimmed.mp4"), so just delete ` * 1000` part. 

Note that ffmpeg_extract_subclip requires its t1 and t2 parameters to be in seconds, not in milliseconds as I initially thought.

Because of those negative integers I am not able to successfully run the trimming process. I searched the web that mainly discusses several formats for the year, month and day, but not '%H:%M:%S.%f' .

What may I be overlooking?

Answer 1

What may I be overlooking?

time.strptime docs

The default values used to fill in any missing data when more accurate values cannot be inferred are (1900, 1, 1, 0, 0, 0, 0, 1, -1) .

whilst start of epoch is 1970. You might get what you want by computing delta between what you parsed and default strptime as follows:

import datetime
time1 = "00:00:11.352"
delta = datetime.datetime.strptime(time1, "%H:%M:%S.%f") - datetime.datetime.strptime("", "")
time_s = delta.total_seconds()
print(time_s)

output

11.352

Answer 2

You need to add the year date (year, month, day) to datetime , else this will default to 1 January 1900.

What you do is this:

from datetime import datetime
s = "00:00:11.352" 
f = '%H:%M:%S.%f'
datetime.strptime(s, f) # datetime.datetime(1900, 1, 1, 0, 0, 11, 352000)

One way to do this is to append the date-string to the time-string you receive from return_times

Answer 3

From https://stackoverflow.com/a/59200108/2681662

The year 1900 was before the beginning of the UNIX epoch , which was in 1970, so the number of seconds returned by timestamp must be negative.

What to do?

It's better to use a time object instead of a datetime object.

from datetime import time

time_1 = "00:00:11.352"
hours, minutes, seconds = time_1.split(":")

print(time(hour=int(hours), minute=int(minutes), second=int(float(seconds)),
           microsecond=int(float(seconds) % 1 * 1000000)))

Answer 4

您可以将时间字符串拆分为小时、分钟、秒和毫秒，通过一些简单的数学计算，您可以获得以毫秒为单位的整个时间