我需要将以下格式给出的时间值字符串转换为秒，例如：
1.'00:00:00,000' -> 0 seconds
2.'00:00:10,000' -> 10 seconds
3.'00:01:04,000' -> 64 seconds
4.'01:01:09,000' -> 3669 seconds
我需要使用正则表达式来做到这一点吗？我尝试使用时间模块，但是
time.strptime('00:00:00,000','%I:%M:%S')
抛出：
ValueError: time data '00:00:00,000' does not match format '%I:%M:%S'
编辑：
看起来像这样：
from datetime import datetime
pt = datetime.strptime(timestring,'%H:%M:%S,%f')
total_seconds = pt.second + pt.minute*60 + pt.hour*3600
给出正确的结果。我只是使用了错误的模块。
>>> import datetime
>>> import time
>>> x = time.strptime('00:01:00,000'.split(',')[0],'%H:%M:%S')
>>> datetime.timedelta(hours=x.tm_hour,minutes=x.tm_min,seconds=x.tm_sec).total_seconds()
60.0
我认为更pythonic的方式是：
timestr = '00:04:23'
ftr = [3600,60,1]
sum([a*b for a,b in zip(ftr, map(int,timestr.split(':')))])
输出为 263 秒。
我很想看看是否有人可以进一步简化它。
没有进口
time = "01:34:11"
sum(x * int(t) for x, t in zip([3600, 60, 1], time.split(":")))
要得到timedelta()，你应该减去1900-01-01：
>>> from datetime import datetime
>>> datetime.strptime('01:01:09,000', '%H:%M:%S,%f')
datetime.datetime(1900, 1, 1, 1, 1, 9)
>>> td = datetime.strptime('01:01:09,000', '%H:%M:%S,%f') - datetime(1900,1,1)
>>> td
datetime.timedelta(0, 3669)
>>> td.total_seconds() # 2.7+
3669.0
%H上面暗示输入小于一天，支持一天以上的时差：
>>> import re
>>> from datetime import timedelta
>>> td = timedelta(**dict(zip("hours minutes seconds milliseconds".split(),
... map(int, re.findall('\d+', '31:01:09,000')))))
>>> td
datetime.timedelta(1, 25269)
>>> td.total_seconds()
111669.0
.total_seconds()在 Python 2.6 上进行模拟：
>>> from __future__ import division
>>> ((td.days * 86400 + td.seconds) * 10**6 + td.microseconds) / 10**6
111669.0
看起来你愿意剥夺几分之一秒......问题是你不能使用'00'作为小时%I
>>> time.strptime('00:00:00,000'.split(',')[0],'%H:%M:%S')
time.struct_time(tm_year=1900, tm_mon=1, tm_mday=1, tm_hour=0, tm_min=0, tm_sec=0, tm_wday=0, tm_yday=1, tm_isdst=-1)
>>>
def time_to_sec(t):
h, m, s = map(int, t.split(':'))
return h * 3600 + m * 60 + s
t = '10:40:20'
time_to_sec(t) # 38420
总是有手工解析
>>> import re
>>> ts = ['00:00:00,000', '00:00:10,000', '00:01:04,000', '01:01:09,000']
>>> for t in ts:
... times = map(int, re.split(r"[:,]", t))
... print t, times[0]*3600+times[1]*60+times[2]+times[3]/1000.
...
00:00:00,000 0.0
00:00:10,000 10.0
00:01:04,000 64.0
01:01:09,000 3669.0
>>>
import time
from datetime import datetime
t1 = datetime.now().replace(microsecond=0)
time.sleep(3)
now = datetime.now().replace(microsecond=0)
print((now - t1).total_seconds())
结果：3.0
灵感来自sverrir-sigmundarson 的评论：
def time_to_sec(time_str):
return sum(x * int(t) for x, t in zip([1, 60, 3600], reversed(time_str.split(":"))))
def time_to_sec(time):
sep = ','
rest = time.split(sep, 1)[0]
splitted = rest.split(":")
emel = len(splitted) - 1
i = 0
summa = 0
for numb in splitted:
szor = 60 ** (emel - i)
i += 1
summa += int(numb) * szor
return summa
HH:MM:SS 和 MM:SS 的动态解决方案。如果要处理命令，请使用split(',')除以 1000 或其他内容，然后添加。
_time = 'SS'
_time = 'MM:SS'
_time = 'HH:MM:SS'
seconds = sum(int(x) * 60 ** i for i, x in enumerate(reversed(_time.split(':'))))
# multiple timestamps
_times = ['MM:SS', 'HH:MM:SS', 'SS']
_times = [sum(int(x) * 60 ** i for i, x in enumerate(reversed(_time.split(':')))) for _time in times]
为什么不使用functools.reduce？
from functools import reduce
def str_to_seconds(t):
reduce(lambda prev, next: prev * 60 + next, [float(x) for x in t.replace(',', '.').split(":")], 0)
一个功能，适用10,40于09:12,40或02:08:14,59。如果您使用.而不是,十进制符号，它会更简单：
def str_to_seconds(t):
reduce(lambda prev, next: prev * 60 + next, [float(x) for x in t.split(":")], 0)
.total_seconds()似乎直截了当。
from datetime import datetime
FMT = '%H:%M:%S.%f'
#example
s2 = '11:01:49.897'
s1 = '10:59:26.754'
# calculate difference
pt = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)
# compute seconds number (answer)
total_seconds = pt.total_seconds()
# output: 143.143