如何使用python解析包含毫秒的时间字符串？
如何使用python解析包含毫秒的时间字符串？
提问于 2009-03-30 17:36:53
我可以使用time.strptime解析包含日期/时间的字符串
>>> import time
>>> time.strptime('30/03/09 16:31:32', '%d/%m/%y %H:%M:%S')
(2009, 3, 30, 16, 31, 32, 0, 89, -1)如何解析包含毫秒的时间字符串？
>>> time.strptime('30/03/09 16:31:32.123', '%d/%m/%y %H:%M:%S')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.5/_strptime.py", line 333, in strptime
data_string[found.end():])
ValueError: unconverted data remains: .123回答 7
Stack Overflow用户
回答已采纳
发布于 2009-03-30 17:49:33
Python2.6添加了一个新的strftime/strptime宏%f。这些文档有一点误导，因为它们只提到微秒，但%f实际上可以解析任何小数位的秒数，最多6位，这意味着它也可以在毫秒甚至厘米秒或十分之一秒内工作。
time.strptime('30/03/09 16:31:32.123', '%d/%m/%y %H:%M:%S.%f')然而，time.struct_time实际上并不存储毫秒/微秒。您最好使用datetime，如下所示：
>>> from datetime import datetime
>>> a = datetime.strptime('30/03/09 16:31:32.123', '%d/%m/%y %H:%M:%S.%f')
>>> a.microsecond
123000如您所见，.123被正确地解释为123 000微秒。
Stack Overflow用户
发布于 2010-10-12 23:09:20
我知道这是一个老问题，但我仍然在使用Python 2.4.3，我需要找到一种更好的方法将数据字符串转换为日期时间。
如果datetime不支持%f并且不需要try/except，那么解决方案是：
(dt, mSecs) = row[5].strip().split(".")
dt = datetime.datetime(*time.strptime(dt, "%Y-%m-%d %H:%M:%S")[0:6])
mSeconds = datetime.timedelta(microseconds = int(mSecs))
fullDateTime = dt + mSeconds 这适用于输入字符串"2010-10-06 09:42:52.266000“
Stack Overflow用户
发布于 2009-03-30 17:48:23
给出(来自its source的) nstehr's answer引用的代码：
def timeparse(t, format):
"""Parse a time string that might contain fractions of a second.
Fractional seconds are supported using a fragile, miserable hack.
Given a time string like '02:03:04.234234' and a format string of
'%H:%M:%S', time.strptime() will raise a ValueError with this
message: 'unconverted data remains: .234234'. If %S is in the
format string and the ValueError matches as above, a datetime
object will be created from the part that matches and the
microseconds in the time string.
"""
try:
return datetime.datetime(*time.strptime(t, format)[0:6]).time()
except ValueError, msg:
if "%S" in format:
msg = str(msg)
mat = re.match(r"unconverted data remains:"
" \.([0-9]{1,6})$", msg)
if mat is not None:
# fractional seconds are present - this is the style
# used by datetime's isoformat() method
frac = "." + mat.group(1)
t = t[:-len(frac)]
t = datetime.datetime(*time.strptime(t, format)[0:6])
microsecond = int(float(frac)*1e6)
return t.replace(microsecond=microsecond)
else:
mat = re.match(r"unconverted data remains:"
" \,([0-9]{3,3})$", msg)
if mat is not None:
# fractional seconds are present - this is the style
# used by the logging module
frac = "." + mat.group(1)
t = t[:-len(frac)]
t = datetime.datetime(*time.strptime(t, format)[0:6])
microsecond = int(float(frac)*1e6)
return t.replace(microsecond=microsecond)
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原文链接：
https://stackoverflow.com/questions/698223
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