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- List slicing with dynamic index on [:index] 4 answers
- 列表切片与动态索引在[:索引]4答案
I try to access an array dynamically in a loop like array[n-i:-i] and it works fine as long as i != 0. In case i==0 I have array[n:0], which I would expect to output array from n to the end but it returns nothing (None i guess). How to archive the expected behaviour?
我尝试在数组[n-i:-i]这样的循环中动态访问一个数组，只要I != 0，它就可以正常工作。在i==0时，我有数组[n:0]，我期望从n到end输出数组，但它没有返回任何值(我猜没有)。如何存档预期行为?
1 个解决方案
#1
11
Use None to slice to the end; Python then'll use len(array) as the endpoint. Use or to fall back to None when -i is 0:
最后使用None切片;然后Python将使用len(数组)作为端点。当-i为0时使用或返回为None:
array[n-i:-i or None]
Numeric 0 is considered false in Python boolean contexts. The or operator short-circuits; it returns the first operand if it is a true value, otherwise it'll evaluate the second operand and return that.
在Python布尔上下文中，数字0被认为是假的。or操作符,可以终止;它返回第一个操作数，如果它是一个真值，那么它将计算第二个操作数并返回它。
#1
11
Use None to slice to the end; Python then'll use len(array) as the endpoint. Use or to fall back to None when -i is 0:
最后使用None切片;然后Python将使用len(数组)作为端点。当-i为0时使用或返回为None:
array[n-i:-i or None]
Numeric 0 is considered false in Python boolean contexts. The or operator short-circuits; it returns the first operand if it is a true value, otherwise it'll evaluate the second operand and return that.
在Python布尔上下文中，数字0被认为是假的。or操作符,可以终止;它返回第一个操作数，如果它是一个真值，那么它将计算第二个操作数并返回它。