Question

我的数据格式为Thu Jun 22 09:43:06，我想推断这一年使用datetime来计算两个日期之间的时间。鉴于上述数据，有没有办法使用datetime来推断年份？

Answer 1

不，但如果您知道范围（例如2010..2017），您可以迭代多年以查看Jun 22是否属于星期四：

def find_year(start_year, end_year, month, day, week_day):
    for y in range(start_year, end_year+1):
        if datetime.datetime(y, month, day, 0, 0).weekday() == week_day:
            yield y
# weekday is 0..6 starting from Monday, so 3 stands for Thursday
print(list(find_year(2010, 2017, 6, 22, 3)))

[2017]

但是对于更长的范围，可能会有多个结果：

print(list(find_year(2000,2017, 6, 22, 3)))

[2000,2006,2017]

Answer 2

您还可以使用pd.date_range生成查找表

calendar = pd.date_range('2017-01-01', '2020-12-31')
dow = {i: d for i, d in enumerate(('Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'))}
moy = {i: d for i, d in enumerate(('Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec'), 1)}
lup = {'{} {} {:>2d}'.format(dow[d.weekday()], moy[d.month], d.day): str(d.year) for d in calendar}
date = 'Tue Jun 25'
print(lup[date])
# 2019
print(pd.Timestamp(date + ' ' + lup[date]))
# 2019-06-25 00:00:00

在ipython中对它进行基准测试，一旦生成了表，就会有一些不错的加速，但是除非您有很多要确认的日期，否则生成表的开销可能不值得。

In [28]: lup = gen_lookup('1-1-2010', '12-31-2017')

In [29]: date = 'Thu Jun 22'

In [30]: lup[date]
Out[30]: ['2017']

In [32]: list(find_year(2010, 2017, 6, 22, 3))
Out[32]: [2017]

In [33]: %timeit lup = gen_lookup('1-1-2010', '12-31-2017')
13.8 ms ± 136 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [34]: %timeit yr = lup[date]
54.1 ns ± 0.547 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

In [35]: %timeit yr = find_year(2010, 2017, 6, 22, 3)
248 ns ± 3.61 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

从python datetime开始，从星期几和日期开始

2 个答案: