Python：在到达结尾时循环切换列表中的元素

Question

I have a list that looks like:

a = ['01', '02', '03', '04', '05', '06', '07', '08', '09', '10']

I need to cycle through this list one element at a time but when the end of the list is reached, the cycle needs to be reversed .

For example, using itertools.cycle :

from itertools import cycle
a_cycle = cycle(a)
for _ in range(30):
    print a_cycle.next()

I get:

01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10

but what I need is:

01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 10, 09, 08, 07, 06, 05, 04, 03, 02, 01, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10

I need to cycle through a for a fixed number of times, say 200.

Answer 1

You can cycle the chain of your a and reversed a , eg:

from itertools import cycle, islice, chain

a = range(1, 11)
b = reversed(a)
c = cycle(chain(a, b))
d = list(islice(c, 100)) # `c` is infinite - hence the `islice` to stop at some point...

Which gives you:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]

Note : If a is an exhaustable iterator, you will need to make a copy of a first. But given your example, this will be fine.

Answer 2

Do you actually need to cycle through the list, as in go forward and backwards forever? Or just .reverse() the list?

print a + a[::-1]

Will do what you describe. The reversed() built-in also works, but you need to chain() it, as it returns an iterator, eg:

print list(itertools.chain(a, reversed(a)))

You can call itertools.cycle() on either result to get an infinite iterator of the list concatenated with its reverse.

Answer 3

def forwardback(lst):
    tot = len(lst)
    while 1:
        for i in xrange(tot):
            yield lst[i]
        for i in xrange(tot-1,-1,-1):
            yield lst[i]

or (using cycle 's approach, which works for all iterators)

def forwardback(lst):
    saved = []
    for elem in lst:
        yield elem
        saved.append(elem)
    while saved:
        for elem in reversed(saved):
            yield elem
        for elem in saved:
            yield elem

Answer 4

Make a copy of list a, reverse it, then append it.

a = ['01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
b = a[:]
b.reverse()
a = a + b

or based on a comment suggestion.

a = ['01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
b = a[::-1]
a = a + b