Python制作循环列表以超出范围索引 [英] Python make a circular list to get out of range index

问题描述

我正在寻找使列表成为循环列表的方法，以便可以呼叫索引超出范围的数字.

I'm looking for ways to make a list into circular list so that I can call a number with index out of range.

例如，我目前有这个课程:

For example, I currently have this class:

class myClass(list):
  definitely __init__(self, *x):
    super().__init__(x)

用作:

>> myList = myClass(1,2,3,4,5,6,7,8,9,10)
>> print(myList)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

在创建列表后，我想要:

and after creating a list, I want to:

>> myList[2]
3
>> myList[12]
3
>> myList[302]
3
>> myList[-1]
10
>> myList[-21]
10
...

目前，无法使用索引[12]，[302]和[-21]，因为它超出了单个列表的范围，因此，如果索引超出范围，我想制作一个循环列表然后将它们变为可能.

At the moment, index [12], [302] and [-21] are not possible since it is out of range for a single list, so I want to make a circular list if the index is out of range which will then make them possible.

• myList [12] = 3，因为循环列表将是[1、2、3、4、5、6、7、8、9、10、1、2、3、4、5、6、7，8、9、10 ...].
• 该列表仅由5或10个数字组成.

推荐答案

如果您坚持为此创建整个类，请为 UserList 子类，实现 __ getitem __ 并使用取模运算符(％)，其长度为列表:

If you insist on creating an entire class for this, subclass UserList, implement __getitem__ and use the modulo operator (%) with the length of the list:

class myClass(UserList):
    def __getitem__(self, item):
        return super().__getitem__(item % len(self))

myList = myClass([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
print(myList[12])

输出

3

根据您的用法，可能需要实现其他几种 list 方法.

Depending on your usage, it's possible you'll need to implement several other list methods.

注意::请注意不要循环播放 myList .这将创建一个无限循环，因为 __ getitem __ 将始终能够提供列表中的元素.

CAVEAT: Be careful to not loop over myList. That will create an infinite loop because __getitem__ will always be able to provide an element from the list.

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