str.replace（）方法如何工作？

Question

I have this input:

'0472/91.39.17'

I want to replace my input with '1234567890' one by one like this:

'1234/56.78.90'

But my outcome is

0472/91.09.17

Here's my code

phone = '0472/91.39.17'
repl = 1234567890

for i in phone:
    if i.isdigit():
        for j in str(repl):
            x = phone.replace(i, j)
        print(x)

What is the proper way to do this?

Answer 1

You can turn repl into an iterator and use a generator expression to replace any value in phone with the next value in the repl iter if the original value is a digit. Re-combine that together with ''.join to get a string as a result, and Bob's your uncle.

repliter = iter(str(repl))

result = ''.join(next(repliter) if c.isdigit() else c for c in phone)
# the ternary expression here evaluates to:
# if c.isdigit():
#     next(repliter)
# else:
#     c

Note that this will crash with a StopIteration error if your phone number contains more than 10 digits, so consider using itertools.cycle for repliter instead.

import itertools

repliter = itertools.cycle(str(repl)) # 1 2 3 4 5 6 7 8 9 0 1 2 3 ...

Answer 2

I was thinking you could use itertools count with modulus for numbers higher than 10.

from itertools import count

c = count(0) # start number
phone = '0472/91.39.17'

''.join(str(next(c) % 10) if item.isdigit() else item for item in phone)