python循环移位字符串中的字符 [英] python cyclic shifting of the characters in the string

问题描述

实现<<"的标准str类型的子类sstr和>>"方法作为字符串中字符的循环移位.尝试做的是

Subclass sstr of the standard str type that implements the "<<" and ">>" methods as a cyclic shifting of the characters in the string.What trying to do is

 >>> s1 = sstr("abcde")
 >>> s1 << 0
'abcde'
 >>> s1 >> 0
'abcde'
 >>> s1 << 2
'cdeab'
>>> s1 >> 2
'deabc'
>>> s1 >> 5
 'abcde'

# my attempt:
import string
class sstr(str):
def __new__(self, other):
    return str.__new__(self, other.upper())
def __ilshift__(self, other):
    return str.__ilshift(other)
def __rshift__(self, other):
    return str.__rshift(other)    

推荐答案

这有点像作业，所以我不打算在这里发布实际代码.但为了提供帮助，我将指出我在您的代码和算法中看到的缺陷:

This smells like homework, so I'm not going to post actual code here. But to help, I will point out flaws I see in your code and an algorithm:

我的python 2.7.2 在str 中没有报告__ilshift 或__irshift.此外，如果您试图将字符串移动一定数量的字符，那么您不应该移动您调用 other 的变量.您应该将 self 移动 other 多个字符.话虽如此，您可能最好将 other 命名为 n 或类似的名称.

My python 2.7.2 reports no __ilshift or __irshift in str. Also, if you are trying to shift a string by a certain number of characters, then you shouldn't be shifting the variable you call other. You should be shifting self by other many characters. That being said, you're probably better off naming other as n or some such.

现在，我假设您知道循环移位应该如何工作.您提供的示例很好地传达了信息.

Now, I assume you know how circular shifting is supposed to work. The examples you provide get the message across well.

作为一个简单的算法(易于阅读/理解)，试试这个(伪代码如下):

As a simple algorithm (easy to read/understand), try this (pseudo-code follows):

function __ilshift(self, n) { // self is the string to be shifted. n denotes how many characters to shift it by
    answer = copy()
    for i = 1 to n {
        answer = self[1:] //answer = everything in self excluding the first character
        answer += self[0] // append the first character of self to answer
    }
    return answer
}

上述解决方案可行.尽管如此，它的效率很低.我们知道，当一个 n 个字符的字符串被 n 移位时，移位的结果就是字符串本身.当您再仔细考虑一下时，您会意识到您最终会移动 n % lengthOfSelf.因此，for i = 1 to n 变成了 for i = 1 to n%len(self).

The above solution would work. Though, it is quite inefficient. We know that when an n-character string is shifted by n, the result of the shifting is the string itself. When you think about this a little more, you realize that you end up shifting by n % lengthOfSelf. Thus, the for i = 1 to n turns into for i = 1 to n%len(self).

不过，我们可以提高效率.要做到这一点，需要在适当的索引处拼接 self，我会让你弄清楚，因为我认为这是功课.

Still, we can make this more efficient. To do this would require splicing self at the appropriate index, which I'll let you figure out, because I think this is homework.

希望这能让你更接近！

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