Python 切片索引越界的问题(数组下标越界)
前言
Python语言处理字符串、数组类的问题时有一定概率需要使用切片方法,比如:Leetcode_5。
学习官方解法时发现切片的索引可以超出字符串或数组最大索引值,此时编译器不会报错。
欢迎大佬留言说明这种情况的具体原因,本文只进行一些情况的简单测试。
实例代码
a = '123' b = a[:5] print(b)
发现结果为123,编译器没有报错。而当直接使用a[5]时即报错string index out of range。下面是测试结果。
测试代码(字符串)
a = "1234567890" a1 = a[:] a2 = a[:len(a)] a3 = a[:15] a4 = a[16:16] a5 = a[:2]
运行结果:
This is the id of 'a' : 2707772994160
This is the type of 'a' : <class 'str'>
This is the value of 'a' : 1234567890This is the id of 'a1' : 2707772994160
This is the type of 'a1' : <class 'str'>
This is the value of 'a1' : 1234567890This is the id of 'a2' : 2707772994160
This is the type of 'a2' : <class 'str'>
This is the value of 'a2' : 1234567890This is the id of 'a3' : 2707772994160
This is the type of 'a3' : <class 'str'>
This is the value of 'a3' : 1234567890This is the id of 'a4' : 2707740774832
This is the type of 'a4' : <class 'str'>
This is the value of 'a4' :This is the id of 'a5' : 2707773122544
This is the type of 'a5' : <class 'str'>
This is the value of 'a5' : 12
值得注意的地方:
- 若切片后结果与原来相同,则新字符串所指向的物理地址就是原字符串的物理地址(a1、a2、a3)。
- 若切片后结果与原来不同,则新字符串指向新的物理地址(a5)。
- 若当前切片索引范围内不存在合法数值,则返回相应类型的空值(a4)。
测试代码(数组)
b = [1, 2, 3, 4, 5] b1 = b[:] b2 = b[:len(b)] b3 = b[:15] b4 = b[16:16] b5 = b[:2]
This is the id of 'b' : 2260784433096
This is the type of 'b' : <class 'list'>
This is the value of 'b' : [1, 2, 3, 4, 5]This is the id of 'b1' : 2260784432456
This is the type of 'b1' : <class 'list'>
This is the value of 'b1' : [1, 2, 3, 4, 5]This is the id of 'b2' : 2260784470920
This is the type of 'b2' : <class 'list'>
This is the value of 'b2' : [1, 2, 3, 4, 5]This is the id of 'b3' : 2260784534280
This is the type of 'b3' : <class 'list'>
This is the value of 'b3' : [1, 2, 3, 4, 5]This is the id of 'b4' : 2260784471432
This is the type of 'b4' : <class 'list'>
This is the value of 'b4' : []This is the id of 'b5' : 2260784231944
This is the type of 'b5' : <class 'list'>
This is the value of 'b5' : [1, 2]
值得注意的地方:
- 数组切片操作必定指向新的物理地址。
- 若当前切片索引范围内不存在合法数值,则返回相应类型的空值(b4)。
栏 目:Python代码
本文地址:http://www.codeinn.net/misctech/216240.html