给出list = ["one", "two", "three"]，我想打印当前及其下一个元素，但顺序相反.即:
three one
two three
one two
我的脚本打印当前元素和下一个元素，但不是以相反的顺序打印:
# It prints:
# one two
# two three
for curr, nxt in zip(list, list[1:]):
print(curr, nxt)
我try 了以下几种方法:
# It prints:
# three one
for curr, nxt in zip(list[-1:], list):
print(curr, nxt)
但它只给了我一个结果.
我在这里不会 Select 使用Pythonzip作为解决方案，因为需要连接两个数组才能得到循环移位的数组(单独的Slice不能循环移位).但是，我要做的只是遍历反转列表中的所有元素，并通过它的索引获得下一个值，如下所示:
list = ["one", "two", "three"]
for i, curr in enumerate(list[::-1]): # enumerate gives you a generator of (index, value)
nxt = list[-i] # list[-i-1] would be the current value, so -i would be the next one
print(curr, nxt)
Edit:
使用list[::-1]比您通常想要的要稍微慢一些，因为它会遍历列表一次来反转它，然后另一次迭代它.更好的解决方案是:
list = ["one", "two", "three"]
for i in range(len(list)-1, -1, -1):
curr = list[i]
nxt = list[len(list) - i - 1] # list[i+1] would not work as it would be index out of range, but this way it overflows to the negative side, which python allows.
print(curr, nxt)
但是，如果您希望使用zip，则需要执行以下操作:
list = ["one", "two", "three"]
for curr, nxt in zip(list[::-1], [list[0]] + list[:0:-1]):
print(curr, nxt)
您还应该注意到，将变量命名为list并不是一个好主意，因为这样会影响到python的内置list方法，您可能应该将其命名为lst或类似的名称.