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如何在Python中导出给定（iso）周数/年的周开始

发布于 2024-10-13 15:33:34 字数 73 浏览 1 评论 0原文

我知道我可以使用 datetime.isocalendar() 来获取给定特定日期的周数。给定周数和年份，如何逆向检索该周的第一天。

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明天过后 2024-10-20 15:33:34

如果您仅限于 stdlib，您可以执行以下操作：

>>> datetime.datetime.strptime('2011, 4, 0', '%Y, %U, %w')
datetime.datetime(2011, 1, 23, 0, 0)

If you're limited to stdlib you could do the following:

>>> datetime.datetime.strptime('2011, 4, 0', '%Y, %U, %w')
datetime.datetime(2011, 1, 23, 0, 0)

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梦里南柯 2024-10-20 15:33:34

找不到标准库，所以不得不推出自己的库。主要困难是找到 ISO 年份的第一天（可能是上一年）。您需要添加一些输入检查等...

import datetime
def isoWeekToGregorian(isoYear, isoWeek, isoDayOfWeek):
    #we have to find the date of the iso year
    t0 = datetime.datetime(isoYear,1,1,0,0,0)
    t0iso = t0.isocalendar()
    if (t0iso[1] != 1):
        t0prime = t0 + datetime.timedelta(7-t0iso[2])
    else:
        t0prime = t0 - datetime.timedelta(t0iso[2])
    #we can add our weeks and days...
    return t0prime + datetime.timedelta(7*(isoWeek-1) + isoDayOfWeek)

我们可以创建一个简单的测试：

#TEST: we know 2004 has 53 weeks...
t0 = datetime.datetime(2004,12,26,0,0,0)
t1 = datetime.datetime(2005,1,10,0,0,0)
ndays = (t1-t0).days + 1
for i in range(ndays):
    d0 = t0 + datetime.timedelta(i)
    d0iso = d0.isocalendar()
    if (d0 != isoWeekToGregorian(d0iso[0],d0iso[1],d0iso[2])):
        print "failed: %s" % d0
    else:
        print d0, d0iso

哪个正确打印：

2004-12-26 00:00:00 (2004, 52, 7)
2004-12-27 00:00:00 (2004, 53, 1)
2004-12-28 00:00:00 (2004, 53, 2)
2004-12-29 00:00:00 (2004, 53, 3)
2004-12-30 00:00:00 (2004, 53, 4)
2004-12-31 00:00:00 (2004, 53, 5)
2005-01-01 00:00:00 (2004, 53, 6)
2005-01-02 00:00:00 (2004, 53, 7)
2005-01-03 00:00:00 (2005, 1, 1)
2005-01-04 00:00:00 (2005, 1, 2)
2005-01-05 00:00:00 (2005, 1, 3)
2005-01-06 00:00:00 (2005, 1, 4)
2005-01-07 00:00:00 (2005, 1, 5)
2005-01-08 00:00:00 (2005, 1, 6)
2005-01-09 00:00:00 (2005, 1, 7)
2005-01-10 00:00:00 (2005, 2, 1)

Couldn't find a standard library, so had to roll my own. Main difficulty is finding the first day of the ISO year (which could be in the previous year). You'll need to add some input checks etc...

import datetime
def isoWeekToGregorian(isoYear, isoWeek, isoDayOfWeek):
    #we have to find the date of the iso year
    t0 = datetime.datetime(isoYear,1,1,0,0,0)
    t0iso = t0.isocalendar()
    if (t0iso[1] != 1):
        t0prime = t0 + datetime.timedelta(7-t0iso[2])
    else:
        t0prime = t0 - datetime.timedelta(t0iso[2])
    #we can add our weeks and days...
    return t0prime + datetime.timedelta(7*(isoWeek-1) + isoDayOfWeek)

We can create a simple test:

#TEST: we know 2004 has 53 weeks...
t0 = datetime.datetime(2004,12,26,0,0,0)
t1 = datetime.datetime(2005,1,10,0,0,0)
ndays = (t1-t0).days + 1
for i in range(ndays):
    d0 = t0 + datetime.timedelta(i)
    d0iso = d0.isocalendar()
    if (d0 != isoWeekToGregorian(d0iso[0],d0iso[1],d0iso[2])):
        print "failed: %s" % d0
    else:
        print d0, d0iso

Which correctly prints:

2004-12-26 00:00:00 (2004, 52, 7)
2004-12-27 00:00:00 (2004, 53, 1)
2004-12-28 00:00:00 (2004, 53, 2)
2004-12-29 00:00:00 (2004, 53, 3)
2004-12-30 00:00:00 (2004, 53, 4)
2004-12-31 00:00:00 (2004, 53, 5)
2005-01-01 00:00:00 (2004, 53, 6)
2005-01-02 00:00:00 (2004, 53, 7)
2005-01-03 00:00:00 (2005, 1, 1)
2005-01-04 00:00:00 (2005, 1, 2)
2005-01-05 00:00:00 (2005, 1, 3)
2005-01-06 00:00:00 (2005, 1, 4)
2005-01-07 00:00:00 (2005, 1, 5)
2005-01-08 00:00:00 (2005, 1, 6)
2005-01-09 00:00:00 (2005, 1, 7)
2005-01-10 00:00:00 (2005, 2, 1)

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