擅长:python、mysql、java
<p>除其他答案外,您还可以使用<a href="https://docs.python.org/3/library/re.html#re.sub" rel="nofollow noreferrer">^{<cd1>}</a>:</p>
<pre><code>INPUT = [
'ABCD , D.O.B: - Jun/14/1999.',
'EFGH , DOB; - Jan/10/1998,',
'IJKL , D-O-B - Jul/15/1985..',
'MNOP , (DOB)* - Dec/21/1999,',
'QRST , *DOB* - Apr/01/2000.',
'UVWX , D O B, - Feb/11/2001 '
]
pattern = r'(?i)^([a-z]+).*([a-z]{3}/\d{2}/\d{4}).*$'
OUTPUT = [re.sub(pattern, r'\1, \2', x) for x in INPUT]
# OUTPUT:
[
'ABCD, Jun/14/1999',
'EFGH, Jan/10/1998',
'IJKL, Jul/15/1985',
'MNOP, Dec/21/1999',
'QRST, Apr/01/2000',
'UVWX, Feb/11/2001'
]
</code></pre>