试图设置一个字符在一个特定位置的字符到另一个字符

Question

我想尝试这样做，而不使用索引。

def SSet(s, i, c): 
#A copy of the string 's' with the character in position 'i' 
#set to character 'c' 
count = -1 
for item in s: 
    if count >= i: 
     count += 1 
    if count == i: 
     item += c 
print(s) 

print(SSet("Late", 3, "o")) 

在这个例子中，迟到应该改为Lato。

谢谢。

Answer 1

您没有累加器来保存输出，并且计数器上的逻辑关闭。以下循环遍历字符串并将字符连接到输出，除非字符索引是在给定字符的哪一点给出的索引。

def SSet(s, i, c): 
    """A copy of the string 's' with the character in position 'i' set to character 'c'""" 
    res = "" 
    count = -1 
    for item in s: 
     count += 1 
     if count == i: 
      res += c 
     else: 
      res += item 
    return res 

print(SSet("Late", 3, "o")) 

打印

Lato 

这可以罗列其去除反写更好：

def SSet(s, i, c): 
    """A copy of the string 's' with the character in position 'i' set to character 'c'""" 
    res = "" 
    for index, item in enumerate(s): 
     if index == i: 
      res += c 
     else: 
      res += item 
    return res 

它也通过附加字符的列表，然后加入进行得更快他们在最后：

def SSet(s, i, c): 
    """A copy of the string 's' with the character in position 'i' set to character 'c'""" 
    res = [] 
    for index, item in enumerate(s): 
     if index == i: 
      res.append(c) 
     else: 
      res.append(item) 
    return ''.join(res) 

它也没有提出的对，但这里是如何与切片做到这一点：

def SSet(s, i, c): 
    """A copy of the string 's' with the character in position 'i' set to character 'c'""" 
    return s[:i]+c+s[i+1:] 

Answer 2

def SSet(s, i, c): 
#A copy of the string 's' with the character in position 'i' 
#set to character 'c' 
    count = 0 
    strNew="" 
    for item in s: 
     if count == i: 
      strNew=strNew+c 
     else: 
      strNew=strNew+item 
     count=count+1 

    return strNew 

print(SSet("Late", 3, "o"))