如何获取两个日期之间的所有日期？

Question

我需要找出两个日期之间的所有工作日。

举个例子：

Wednesday - Friday  = Wednesday, Thursday, Friday  
        3 - 5       = 3, 4, 5

 Saturday - Tuesday = Saturday, Sunday, Monday, Tuesday
        6 - 2       = 6, 7, 1, 2

我相信一定有聪明的算法可以解决这个问题。我能想到的算法要么是用循环，要么是用if语句。

一定有一种优雅的方法来解决这个问题。我用数字1到7来表示工作日，但用0到6也可以。

我能想到的最好方法是：

def between(d1, d2):
     alldays = [0,1,2,3,4,5,6,0,1,2,3,4,5,6]    # or range(7) * 2
     offset = 8 if d1 > d2 else 1
     return alldays[d1:d2 + offset]

between(0, 4)
# [0,1,2,3,4]

between(5,2)
# [5,6,0,1,2]

Answer 1

在Stephan202的精彩回答的基础上，你可以把“圆形切片”的概念进行更广泛的应用。

>>> def circular_slice(r, s, e):
... return [r[n % len(r)] for n in range(s, e + (1 if e>s else len(r)+1))]
...
>>> circular_slice(range(0,7), 2, 4)
[2, 3, 4]
>>> circular_slice(range(0,7), 5, 1)
[5, 6, 0, 1]
>>> circular_slice('Mon Tue Wed Thu Fri Sat Sun'.split(), 5, 1)
['Sat', 'Sun', 'Mon', 'Tue']

Answer 2

这个伪代码看起来怎么样：

weekday[] = {"Mon" .. "Sun"}
for(i = wkday_start; (i % 7) != wkday_end; i = (i+1) % 7)
    printf("%s ", weekday[i]);

它的工作原理像一个循环缓冲区，wkday_start 是开始的索引（从0开始），wkday_end 是结束的索引。

希望这对你有帮助。

乔治。

Answer 3

>>> def weekdays_between(s, e):
...     return [n % 7 for n in range(s, e + (1 if e > s else 8))]
... 
>>> weekdays_between(2, 4)
[2, 3, 4]
>>> weekdays_between(5, 1)
[5, 6, 0, 1]

>>> days = 'Mon Tue Wed Thu Fri Sat Sun'.split()
>>> days_1 = {d: n for n, d in enumerate(days)}
>>> def weekdays_between(s, e): 
...     s, e = days_1[s], days_1[e]
...     return [days[n % 7] for n in range(s, e + (1 if e > s else 8))]
... 
>>> weekdays_between('Wed', 'Fri')
['Wed', 'Thu', 'Fri']
>>> weekdays_between('Sat', 'Tue')
['Sat', 'Sun', 'Mon', 'Tue']

如果你需要把时间转换成实际的天数，那就稍微复杂一些。