[已解决]定义一个函数，与replace相同功能

xxlxxl

如何在不用replace函数的情况下达到replace的功能？

1. print(replace('abc def abc', 'def', 'abc')) # abc abc abc
   
2. print(replace('abcabcabc', 'abc', 'hi'))    # hihihi
   
3. print(replace('This is out of control!', 'out of control!', 'better!')) # This is better!

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最佳答案

月排行榜 / 总排行榜

hrp

2020-10-13 15:06:03

1. def replace(string, old, new):
   
2.     while old in string:
   
3.         ind = string.index(old)
   
4.         string = string[:ind] + new +string[ind + len(old):]
   
5.     return string
   
6. 
   
7. print(replace('abc def abc', 'def', 'abc')) # abc abc abc
   
8. print(replace('abcabcabc', 'abc', 'hi'))    # hihihi
   
9. print(replace('This is out of control!', 'out of control!', 'better!')) # This is better!

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跳转到最佳答案楼层

冬雪雪冬

1. def replace(string, old, new):
   
2.     return new.join(string.split(old))

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kogawananari

用re.sub如何（滑稽

不然得手写滑动窗口算法了

这个问题发数据结构和算法板块更好

xxlxxl

kogawananari 发表于 2020-10-13 14:27
用re.sub如何（滑稽

不然得手写滑动窗口算法了

我看了下题目，不能用re.sub，题目意思是让我用纯loop写，头秃。

xxlxxl

冬雪雪冬 发表于 2020-10-13 14:25

有没有不用join，用loop的方法啊

冬雪雪冬

xxlxxl 发表于 2020-10-13 14:45
有没有不用join，用loop的方法啊

1. def replace(string, old, new):
   
2.     lst = string.split(old)
   
3.     s = lst[0]
   
4.     for each in lst[1:]:
   
5.         s += new + each
   
6.     return s

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hrp

1. def replace(string, old, new):
   
2.     while old in string:
   
3.         ind = string.index(old)
   
4.         string = string[:ind] + new +string[ind + len(old):]
   
5.     return string
   
6. 
   
7. print(replace('abc def abc', 'def', 'abc')) # abc abc abc
   
8. print(replace('abcabcabc', 'abc', 'hi'))    # hihihi
   
9. print(replace('This is out of control!', 'out of control!', 'better!')) # This is better!

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