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我正在尝试使用 datetime.strptime 从两个字符串中获取单个日期时间。

时间很容易（例如晚上 8:53），所以我可以执行以下操作：

theTime = datetime.strptime(givenTime, "%I:%M%p")

但是，字符串不仅仅是一个日期，它是一个格式类似于http://site.com/?year=2011&month=10&day=5&hour=11. 我知道我可以做类似的事情：

theDate = datetime.strptime(givenURL, "http://site.com/?year=%Y&month=%m&day=%d&hour=%H")

但我不想从链接中获得那个小时，因为它正在其他地方检索。有没有办法放置一个虚拟符号（如 %x 或其他东西）作为最后一个变量的灵活空间？

最后，我设想有一行类似于：

theDateTime = datetime.strptime(givenURL + givenTime, ""http://site.com/?year=%Y&month=%m&day=%d&hour=%x%I:%M%p")

（尽管显然不会使用 %x ）。有任何想法吗？

4

3 回答 3

2

认为如果您想从 URL 中简单地跳过时间，您可以使用 split 例如以下方式：

givenURL = 'http://site.com/?year=2011&month=10&day=5&hour=11'
pattern = "http://site.com/?year=%Y&month=%m&day=%d"
theDate = datetime.strptime(givenURL.split('&hour=')[0], pattern)

所以不确定是否正确理解了您，但是：

givenURL = 'http://site.com/?year=2011&month=10&day=5&hour=11'
datePattern = "http://site.com/?year=%Y&month=%m&day=%d"
timePattern = "&time=%I:%M%p"

theDateTime = datetime.strptime(givenURL.split('&hour=')[0] + '&time=' givenTime, datePattern + timePattern)
于 2011-08-15T20:20:16.383 回答
1 
import datetime
import re

givenURL  = 'http://site.com/?year=2011&month=10&day=5&hour=11'
givenTime = '08:53PM'

print ' givenURL == ' + givenURL
print 'givenTime == ' + givenTime

regx = re.compile('year=(\d\d\d\d)&month=(\d\d?)&day=(\d\d?)&hour=\d\d?')
print '\nmap(int,regx.search(givenURL).groups()) ==',map(int,regx.search(givenURL).groups())

theDate = datetime.date(*map(int,regx.search(givenURL).groups()))
theTime = datetime.datetime.strptime(givenTime, "%I:%M%p")

print '\ntheDate ==',theDate,type(theDate)
print '\ntheTime ==',theTime,type(theTime)


theDateTime = theTime.replace(theDate.year,theDate.month,theDate.day)
print '\ntheDateTime ==',theDateTime,type(theDateTime)

结果

 givenURL == http://site.com/?year=2011&month=10&day=5&hour=11
givenTime == 08:53PM

map(int,regx.search(givenURL).groups()) == [2011, 10, 5]

theDate == 2011-10-05 <type 'datetime.date'>

theTime == 1900-01-01 20:53:00 <type 'datetime.datetime'>

theDateTime == 2011-10-05 20:53:00 <type 'datetime.datetime'>

编辑 1

由于strptime()很慢，我改进了代码以消除它

from datetime import datetime
import re
from time import clock


n = 10000

givenURL  = 'http://site.com/?year=2011&month=10&day=5&hour=11'
givenTime = '08:53AM'

# eyquem
regx = re.compile('year=(\d\d\d\d)&month=(\d\d?)&day=(\d\d?)&hour=\d\d? (\d\d?):(\d\d?)(PM|pm)?')
t0 = clock()
for i in xrange(n):
    given = givenURL + ' ' + givenTime
    mat = regx.search(given)
    grps = map(int,mat.group(1,2,3,4,5))
    if mat.group(6):
        grps[3] += 12 # when it is PM/pm, the hour must be augmented with 12
    theDateTime1 = datetime(*grps)
print clock()-t0,"seconds   eyquem's code"
print theDateTime1


print

# Artsiom Rudzenka
dateandtimePattern = "http://site.com/?year=%Y&month=%m&day=%d&time=%I:%M%p"
t0 = clock()
for i in xrange(n):
    theDateTime2 = datetime.strptime(givenURL.split('&hour=')[0] + '&time=' + givenTime, dateandtimePattern)
print clock()-t0,"seconds   Artsiom's code"
print theDateTime2

print
print theDateTime1 == theDateTime2

结果

0.460598763251 seconds   eyquem's code
2011-10-05 08:53:00

2.10386180366 seconds   Artsiom's code
2011-10-05 08:53:00

True

我的代码快了 4.5 倍。如果有很多这样的转换要执行，那可能会很有趣

于 2011-08-15T21:18:30.843 回答
0

格式字符串无法做到这一点。但是，如果时间无关紧要，您可以像在第一个示例中一样从 URL 获取它，然后调用theDateTime.replace(hour=hour_from_a_different_source).

这样您就不必进行任何额外的解析。

于 2011-08-15T20:19:30.620 回答