[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？

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

最佳答案

月排行榜 / 总排行榜

hrp

2020-10-13 15:06:03

def replace(string, old, new):
    while old in string:
        ind = string.index(old)
        string = string[:ind] + new +string[ind + len(old):]
    return string

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

跳转到最佳答案楼层

冬雪雪冬

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

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

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

hrp

def replace(string, old, new):
    while old in string:
        ind = string.index(old)
        string = string[:ind] + new +string[ind + len(old):]
    return string

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