我有一个以秒为单位返回信息的函数,但我需要以小时:分钟:秒为单位存储该信息。

在Python中是否有一种简单的方法将秒转换为这种格式?


当前回答

有点离题,但可能对某人有用

def time_format(seconds: int) -> str:
    if seconds is not None:
        seconds = int(seconds)
        d = seconds // (3600 * 24)
        h = seconds // 3600 % 24
        m = seconds % 3600 // 60
        s = seconds % 3600 % 60
        if d > 0:
            return '{:02d}D {:02d}H {:02d}m {:02d}s'.format(d, h, m, s)
        elif h > 0:
            return '{:02d}H {:02d}m {:02d}s'.format(h, m, s)
        elif m > 0:
            return '{:02d}m {:02d}s'.format(m, s)
        elif s > 0:
            return '{:02d}s'.format(s)
    return '-'

结果:

print(time_format(25*60*60 + 125)) 
>>> 01D 01H 02m 05s
print(time_format(17*60*60 + 35)) 
>>> 17H 00m 35s
print(time_format(3500)) 
>>> 58m 20s
print(time_format(21)) 
>>> 21s

其他回答

dateutil。如果你需要将小时、分钟和秒作为浮点数访问,Relativedelta也很方便。datetime。Timedelta没有提供类似的接口。

from dateutil.relativedelta import relativedelta
rt = relativedelta(seconds=5440)
print(rt.seconds)
print('{:02d}:{:02d}:{:02d}'.format(
    int(rt.hours), int(rt.minutes), int(rt.seconds)))

打印

40.0
01:30:40

我是这样得到它的。

def sec2time(sec, n_msec=3):
    ''' Convert seconds to 'D days, HH:MM:SS.FFF' '''
    if hasattr(sec,'__len__'):
        return [sec2time(s) for s in sec]
    m, s = divmod(sec, 60)
    h, m = divmod(m, 60)
    d, h = divmod(h, 24)
    if n_msec > 0:
        pattern = '%%02d:%%02d:%%0%d.%df' % (n_msec+3, n_msec)
    else:
        pattern = r'%02d:%02d:%02d'
    if d == 0:
        return pattern % (h, m, s)
    return ('%d days, ' + pattern) % (d, h, m, s)

一些例子:

$ sec2time(10, 3)
Out: '00:00:10.000'

$ sec2time(1234567.8910, 0)
Out: '14 days, 06:56:07'

$ sec2time(1234567.8910, 4)
Out: '14 days, 06:56:07.8910'

$ sec2time([12, 345678.9], 3)
Out: ['00:00:12.000', '4 days, 00:01:18.900']

division = 3623 // 3600 #to hours
division2 = 600 // 60 #to minutes
print (division) #write hours
print (division2) #write minutes

PS:我的代码不专业

下面是一个简单的程序,它读取当前时间并将其转换为以小时、分钟和秒为单位的一天时间

import time as tm #import package time
timenow = tm.ctime() #fetch local time in string format

timeinhrs = timenow[11:19]

t=tm.time()#time.time() gives out time in seconds since epoch.

print("Time in HH:MM:SS format is: ",timeinhrs,"\nTime since epoch is : ",t/(3600*24),"days")

输出为

Time in HH:MM:SS format is:  13:32:45 
Time since epoch is :  18793.335252338384 days

通过使用divmod()函数,它只做一个除法就能得到商和余数,你只需要两个数学运算就能很快得到结果:

m, s = divmod(seconds, 60)
h, m = divmod(m, 60)

然后使用字符串格式将结果转换为您想要的输出:

print('{:d}:{:02d}:{:02d}'.format(h, m, s)) # Python 3
print(f'{h:d}:{m:02d}:{s:02d}') # Python 3.6+