如何替换字符串中子字符串/字符的第n次出现？ [Python 3]

Question

I was going for replacing every fifth "b" with "c" Here is my input string:

jStr = aabbbbbaa

Now here is the code

import re
m = re.search('c', jStr)
jStr1 = jStr[:m.end()]
jStr2 = jStr[:m.end()]
jStr3 = jStr[:m.end()]
jStr4 = jStr[:m.end()]
jStr5 = jStr[m.end():]
jStr6 = jStr5.replace('c', 'b', 1)
jStr == (jStr1+jStr6)

the output I keep getting is the same

aabbbbbaa

I started with?

Answer 1

This might not be the most concise way, you can find all the indices of b , take every 5th one, and then assign c . Since indices inside str are not assignable, you have to convert to list.

jStr = 'aabbbbbaa'
jStr = list(jStr)

bPos = [x for x in range(len(jStr)) if jStr[x] == 'b']

for i,x in enumerate(bPos):
   if (i+1) % 5 == 0:
      jStr[x] = 'c'

jStr = ''.join(jStr)
print(jStr)

Output:

aabbbbcaa

Answer 2

jStr = "aabbbbbaabbbbb"
count = 1
res= "" # strings are immutable so we have to create a new string.
for s in jStr:
    if count == 5 and s == "b": # if count is 5 we have our fifth "b", change to "c" and reset count
        res +=  "c"
        count = 1
    elif s == "b": # if it is a "b" but not the fifth just add b to res and increase count
        count += 1
        res += "b"
    else:           # else it is not a "b", just add to res
        res += s 
print(res)
aabbbbcaabbbbc

Finds every fifth b, counting the b's using count, when we have reached the fifth we reset the counter and go on to the next character.