Python make a circular list to get out of range index

问题

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

For example, I currently have this class:

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

which works as:

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

and after creating a list, I want to:

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

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 because the circular list would be [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...].
The list will be only made up with 5 or 10 numbers.

回答1:

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])

outputs

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

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.

来源：https://stackoverflow.com/questions/61350386/python-make-a-circular-list-to-get-out-of-range-index

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