循环移位列表python

Question

我在创建一个真正的清晰度程序时遇到了麻烦。

def left():
    listL = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
    k=4
    right = listL[k::]
    left = listL[:k:]
    print(right + left)


def right():
    listL = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
    k=len(listL)-4
    right = listL[k::]
    left = listL[:k:]
    print(right + left)

我的代码根据向左或向右移动k来计算在哪里重新创建原始listL，在本例中是4。然而，我的练习问题询问...

Given a list of N numbers, write a function to shift the numbers circularly by some integer k (where k < N). The function should take the list and k as arguments and return the shifted list. 
a) Write a function that assumes the shifting is to the left. It should not print anything. 
b) Write a function that takes a third argument that specifies shifting left or right. It should not print anything. Perform whatever error-checking you consider necessary. 
c) Write a main() function that calls the above functions. It should print the lists both before and after shifting. Perform whatever error-checking you consider necessary. 

我已经满足了A部分的要求，但我不知道如何构建B部分和C部分来完整地复制问题。

解决方案示例运行：

Sample run 
>>> ================================ RESTART ================================
>>> 
original list:  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
shifted by 4, to the left: [4, 5, 6, 7, 8, 9, 0, 1, 2, 3]
shifted by 4, to the right: [6, 7, 8, 9, 0, 1, 2, 3, 4, 5]

如果有任何关于我应该如何解决b和c部分的建议，我将不胜感激！:)

Answer 1

我认为这不适用于OP，因为这听起来像是CS课程作业，但对于其他正在寻找解决方案的人来说，只需使用：

from collections import deque

d = deque([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
d.rotate(3)  # to the right
d.rotate(-3)  # back to the left

*编辑

跟进你的评论(来自deque docs)：

Deques是堆栈和队列的泛化(该名称发音为“”，是“双端队列”的缩写)。Deques支持线程安全、内存高效的附加和弹出，并且在任意方向上都具有大致相同的O(1)性能。

尽管列表对象支持类似的操作，但它们针对快速的定长操作进行了优化，并为pop(0)和insert(0，v)操作带来O(n)内存移动成本，这些操作同时更改了底层数据表示的大小和位置。

Answer 2

看看这个：

>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

循环移位4：

>>> b = a[4::] + a[:4:]
>>> b
[4, 5, 6, 7, 8, 9, 0, 1, 2, 3]

并采用两个函数的格式：

def shiftLbyn(arr, n=0):
    return arr[n::] + arr[:n:]

def shiftRbyn(arr, n=0):
    return arr[n:len(arr):] + arr[0:n:]

给他们打电话：

print shiftLbyn([1,2,3,4,5,6,7,8], 3)
print shiftRbyn([1,2,3,4,5,6,7,8], 4)

将给出输出：

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

Answer 3

首先，更改您的函数以获取参数并返回结果。例如

def left(listL, k):
    right = listL[k::]
    left = listL[:k:]
    return right + left # is this the usual meaning of left and right?

# This is how you call the function
print(left([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], 4))

现在，如果你注意到left和right有相同的最后3行。您可以这样将它们组合在一起

def shift(listL, k, direction):
    if direction == "right":
        k = len(listL) - k
    right = listL[k::]
    left = listL[:k:]
    return right + left

我猜main应该是这样的

def main(listL):
    print(listL)
    print(shift(listL, 4, "left"))
    print(shift(listL, 4, "right"))