Python 拆分和追加 输入：-A89456FRERT120108A.1 预期输出：-120108AT.1

我在想下面的问题…有谁能帮我吗？如果有更简单的方法来实现这一点，我需要为字母数字字符添加T

基于。 在遇到第一个数字（本例中为120108A）后，获取“拆分[0]”的字母数字字符 将T附加到2，它将是120108AT 然后放回'split[1]'120108AT.1

---

下面是一个regex解决方案，它尝试使用您提供的相同逻辑：

import re
new_string = re.sub(r'^.*?(\d+\D*)(\..*)', r'\1T\2', orig_string)

例如：

>>> re.sub(r'^.*?(\d+\D*)(\..*)', r'\1T\2', 'A89456FRERT120108A.1')
'120108AT.1'

说明：

#regex:
    ^            # match at the start of the string
    .*?          # match any number of any character (as few as possible)
    (            # start capture group 1
      \d+          # match one or more digits
      \D*          # match any number of non-digits
    )            # end capture group 1
    (            # start capture group 2
      \..*         # match a '.', then match to the end of the string
    )            # end capture group 2

#replacement
    \1           # contents of first capture group (from digits up to the '.')
    T            # literal 'T'
    \2           # contents of second capture group ('.' to end of string)

---

下面是一个regex解决方案，它尝试使用您提供的相同逻辑：

import re
new_string = re.sub(r'^.*?(\d+\D*)(\..*)', r'\1T\2', orig_string)

例如：

>>> re.sub(r'^.*?(\d+\D*)(\..*)', r'\1T\2', 'A89456FRERT120108A.1')
'120108AT.1'

说明：

#regex:
    ^            # match at the start of the string
    .*?          # match any number of any character (as few as possible)
    (            # start capture group 1
      \d+          # match one or more digits
      \D*          # match any number of non-digits
    )            # end capture group 1
    (            # start capture group 2
      \..*         # match a '.', then match to the end of the string
    )            # end capture group 2

#replacement
    \1           # contents of first capture group (from digits up to the '.')
    T            # literal 'T'
    \2           # contents of second capture group ('.' to end of string)

---

这可能是正则表达式的情况。这可能是正则表达式的情况。