python字符串按出现次数替换字母

Question

Suppose I have a string Hello, world , and I'd like to replace all instances of o by its occurrence number, ie get a sting: Hell1, w2rld ?

I have found how to reference a numbered group, \\g<1> , but it does require a group number.

Is there are way to do what I want in python?

Update: Sorry for not mentioning that I was indeed looking for a regexp solution, not just a string. I have marked the solution I liked the best, but thank for all contributions, they were cool!

Answer 1

For a regular expression solution :

import re

class Replacer:
    def __init__(self):
        self.counter = 0

    def __call__(self, mo):
        self.counter += 1
        return str(self.counter)

s = 'Hello, World!'
print(re.sub('o', Replacer(), s))

Answer 2

Split the string on the letter "o" and reassemble it by adding the index of the part in front of each part (except the first one):

string = "Hello World"

result = "".join(f"{i or ''}"+part for i,part in enumerate(string.split("o")))

output:

print(result)

# Hell1 W2rld

Answer 3

Using itertools.count

Ex:

import re
from itertools import count 

c = count(1)
s = "Hello, world"

print(re.sub(r"o", lambda x: "{}".format(next(c)), s))
#or
print(re.sub(r"o", lambda x: f"{next(c)}", s))
# --> Hell1, w2rld

Answer 4

You don't need regex for that, a simple loop will suffice:

sample = "Hello, world"

pattern = 'o'
output = ''
count = 1
for char in sample:
    if char == pattern:
        output += str(count)
        count += 1
    else:
        output += char

print(output)
>>> Hell1, w2rld