Question

字典中,我有新鲜的字眼,我有问题要写一份文字,其中含有四个要素(图形、替换、替换和n)的字眼特征,并取代 occurrence。

例:

>>> replaeceit("Mississippi", "s", "l", 2)
 Mislissippi 
>>> replaeceit("Mississippi", "s", "l", 0)
 Mississippi

第2条就是这样,法典将第2条改为第1条。

坦率地说,我不知道如何执行这一公式,这是我迄今为止没有一 n的法典。

def replaceit(str,replacefrom,replaceto):
    new=""
    for letter in str:
        if letter== replacefrom:
            new=new+replaceto
        else: 
            new=new+letter
    return new

Answer 1

奥凯,现在也许我知道你正在寻求:

def replaceit(st, remove, put, pos):
outs = ""
count = 0
for letter in st:
    if letter == remove:
        count += 1
        if count == pos:
            outs += put
        else:
            outs += letter
    else:
        outs += letter
return outs

产出:

In [84]: replaceit("Mississipi", "s", "l", 2)
Out[84]:  Mislissipi

当然,你可以核实,第2和第3号论点是用第1号的透镜进行的。

Answer 2

This is second first attempt at understanding your question:

def replaceit(s, replacefrom, replaceto, n):
  new_s, count =   , 0
  for letter in s:
    if letter == replacefrom:
      count += 1
      if count == n:
        new_s += replaceto
        continue
    new_s += letter
  return new_s

这符合你的例子:

>>> replaceit("Mississippi", "s", "l", 2)
 Mislissippi 
>>> replaceit("Mississippi", "s", "l", 0)
 Mississippi

如果这不是你想要的话,请作更好的解释。

您也可以定期表达:

def replaceit(s, replacefrom, replaceto, n):
  import re
  if n <= 0:
    return s
  return re.sub( (.*?%s)%s  % (( %s.*?  % replacefrom) * (n-1), replacefrom), r 1%s  % replaceto, s)

Answer 3

每个人都总是热爱产生者的言论。

from itertools import count
def replaceit(str, replacefrom, replaceto, n):
    c = count(1)
    return   .join(replacefrom if l == replaceto and c.next() == n else l for l in str)

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