Python 以天、小时、分钟、秒表示的经过时间

Question

I would like to measure the execution time of some piece of code in days , hours, minutes and seconds. This is what I have so far:

import time
start_time = time.time()

# some code

elapsed = time.strftime("%H:%M:%S", time.gmtime(time.time() - start_time))
print(f"Took: {elapsed}")

The problem is that if the code that I am measuring takes longer than 24h, the time displayed overflows and starts from zero again. I would like something like this:

# Example: 12 hours and 34 minutes should be printed as 
> Took: 12:34:00

# Example: 26 hours and 3 minutes should be printed as 
> Took: 1:02:03:00

Answer 1

The result of time.gmtime(time.time() - start_time) is not what you seem to think it is. Instead of being a duration of time it is a point in time. Let me explain.

The result of time.time() is the number of seconds since January 1, 1970, 00:00:00 (UTC) at the time of calling. Therefore, the statement time.time() - start_time will produce the number of seconds between the two calls. So far so good. However, the time.gmtime function is interpreting this duration as the number of seconds since January 1, 1970, 00:00:00 (UTC) and formatting the time accordingly. What you are seeing then is the time portion of the date January 1, 1970, 12:34:00 (UTC).

I suggest you either use the datetime.timedelta object and format using that, or as others have suggested, output the duration in seconds or milliseconds.

If you want to format this number yourself, you could use something like this:

def format_duration(duration):
    mapping = [
        ('s', 60),
        ('m', 60),
        ('h', 24),
    ]
    duration = int(duration)
    result = []
    for symbol, max_amount in mapping:
        amount = duration % max_amount
        result.append(f'{amount}{symbol}')
        duration //= max_amount
        if duration == 0:
            break

    if duration:
        result.append(f'{duration}d')

    return ' '.join(reversed(result))

Answer 2

You could use datetime:

from datetime import datetime as dt

start = dt.fromtimestamp(1588432670)
end = dt.now()
elapsed=end-start
print("Took: %02d:%02d:%02d:%02d" % (elapsed.days, elapsed.seconds // 3600, elapsed.seconds // 60 % 60, elapsed.seconds % 60))

Output:

Took: 33:00:21:49

Answer 3

You should try this:

import time
start_time = time.time()
...
elapsed_time = time.time() - start_time
days = 0
if elapsed_time >= 86400:
    days = int(elapsed_time / 86400)
elapsed = time.strftime("%H:%M:%S", time.gmtime(time.time() - start_time))
if days == 0:
    print(f"Took: {elapsed}")
else:
    print(f"Took: {days}:{eplased}")

  

Answer 4

Try using timeit:

import timeit
timeit.timeit(<callable function>, number = 100)

Here timeit will call callable function number times and give you the average runtime.

Answer 5

Time types can only include hours and less units. You should use datetime instead of time as follows:

from datetime import datetime
start_time = datetime.now()

# some code

elapsed = datetime.now() - start_time)
print(f"Took: {elapsed}")

Example usage of Datetime:

from datetime import datetime
d1 = datetime(2013,9,1,5,5,4)
d2 = datetime(2013,1,13,3,2,1)
result1 = d1-d2
print ('{} between {} and {}'.format(result1, d1, d2))

This produces following output:

231 days, 2:03:03 between 2013-09-01 05:05:04 and 2013-01-13 03:02:01