在Python中“定期”替换字符串中字符的最佳方法是什么？

Question

I have a string where a character ('@') needs to be replaced by characters from a list of one or more characters "in order" and "periodically". So for example I have

'ab@cde@@fghi@jk@lmno@@@p@qrs@tuvwxy@z'

and want

'ab1cde23fghi1jk2lmno312p3qrs1tuvwxy2z'

for replace_chars = ['1', '2', '3']

The problem is that in this example there are more @ in the string than I have replacers.

This is my try:

result = ''
replace_chars = ['1', '2', '3']
string = 'ab@cde@@fghi@jk@lmno@@@p@qrs@tuvwxy@z'

i = 0
for char in string:
    if char == '@':
        result += replace_chars[i]
        i += 1
    else:
        result += char

print(result)

but this only works of course if there are not more than three @ in the original string and otherwise I get IndexError .

Edit: Thanks for the answers!

Answer 1

Your code could be fixed by adding the line i = i%len(replace_chars) as the last line of your if clause. This way you will be taking the remainder from the division of i by the length of your list of replacement characters.

The shorter solution is to use a generator that periodically spits out replacement characters.

>>> from itertools import cycle
>>> s = 'ab@cde@@fghi@jk@lmno@@@p@qrs@tuvwxy@z'
>>> replace_chars = ['1', '2', '3']
>>>
>>> replacer = cycle(replace_chars)
>>> ''.join([next(replacer) if c == '@' else c for c in s])
'ab1cde23fghi1jk2lmno312p3qrs1tuvwxy2z'

For each character c in your string s , we get the next replacement character from the replacer generator if the character is an '@' , otherwise it just gives you the original character.

For an explanation why I used a list comprehension instead of a generator expression, read this .

Answer 2

Generators are fun.

def gen():
    replace_chars = ['1', '2', '3']
    while True:
        for rc in replace_chars:
            yield rc

with gen() as g:
    s = 'ab@cde@@fghi@jk@lmno@@@p@qrs@tuvwxy@z'
    s = ''.join(next(g) if c == '@' else c for c in s)

As PM 2Ring suggested, this is functionally the same as itertools.cycle . The difference is that itertools.cycle will hold an extra copy of the list in memory which may not be necessary.

itertools.cycle source:

def cycle(iterable):
    saved = []
    for element in iterable:
        yield element
        saved.append(element)
    while saved:
        for element in saved:
              yield element

Answer 3

You could also keep your index logic once you use modulo, using a list comp by using itertools.count to keep track of where you are:

from itertools import count

cn, ln = count(), len(replace_chars)

print("".join([replace_chars[next(cn) % ln] if c == "@" else c for c in string]))

ab1cde23fghi1jk2lmno312p3qrs1tuvwxy2z

Answer 4

I think it is better to not iterate character-by-character, especially for long string with lengthy parts without @ .

from itertools import cycle, chain

s = 'ab@cde@@fghi@jk@lmno@@@p@qrs@tuvwxy@z'
replace_chars = ['1', '2', '3']
result = ''.join(chain.from_iterable(zip(s.split('@'), cycle(replace_chars))))[:-1]

I don't know how to efficiently kill last char [:-1] .