Question

使用Python3，我需要从任意i（i <= 6）开始依次遍历元素列表，完成整个迭代周期，从i + = 1开始重复它，依此类推。此循环不应结束。在内容上，这是从当前日期开始的星期几名称的获取。例如，如果今天是星期三，则由于迭代，我必须获得一个列表[星期三，星期四，星期五，星期六，星期日，星期一，星期二]。这似乎很简单，但我处于死胡同。我会很感谢您的提示或帮助。

Answer 1

您可以使用itertools.chain和itertools.cycle获得产生所需序列的无限迭代：

from itertools import chain, cycle

for x in chain(the_list[i:], cycle(the_list)):
    print(x)

用作：

>>> from itertools import chain, cycle
>>> the_list = range(10)
>>> i = 6
>>> for x in chain(the_list[i:], cycle(the_list)):
...     print(x)
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
[... forever ...]

等效使用cycle + islice：

>>> for x in islice(cycle(the_list), i, None):
...     print(x)
>>> for x in islice(cycle(the_list), i, 20):
...     print(x)
... 
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
[... forever ...]

Answer 2

通常可以在以下循环场景中充分利用模运算符%：

days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']

i = 2  # Wednesday

while True:
    print(days[i])
    i += 1
    i = i % len(days)  # 7 => 0

Wednesday
Thursday
Friday
Saturday
Sunday
Monday
Tuesday
Wednesday
...

Answer 3

如果我理解正确，那么您正在尝试在任何给定的日子里获得工作日您可以传递工作日名称，并以相同的轮换顺序获得列表。请看一下：-

from collections import deque
def rotate_week(week_days):
    a = deque(["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"])
    d = {}
    for i in range(len(a)):
        d[a[i]] = i
    number_to_rotate = d.get(week_days)
    a.rotate(-number_to_rotate)
    print(list(a))
rotate_week('Wednesday')

输出为：-

['Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday', 'Monday', 'Tuesday']

Answer 4

from itertools import chain, dropwhile, repeat


def endless_daynames(starting_day):
    daynames = [
        'Monday', 'Tuesday', 'Wednesday', 'Thursday', 
        'Friday', 'Saturday', 'Sunday']
    # get starting day name 
    starting_day_name = daynames[starting_day] 
    # creating an endless iterable from an origin list
    endless_list = chain.from_iterable(repeat(daynames))
    # drop first items before we meet a required one
    shifted_list = dropwhile(lambda x: x != starting_day_name, endless_list)
    # yield values from it
    yield from shifted_list

在列表上进行迭代，起点有所偏移

4 个答案: