我是 Python 的新手。可能这可以用正则表达式来完成。我想在字符串中搜索特定的子字符串并删除字符串中前后的字符。
示例 1
Input:"This is the consignment no 1234578TP43789"
Output:"This is the consignment no TP"
示例 2
Input:"Consignment no 1234578TP43789 is on its way on vehicle no 3456MP567890"
Output:"Consignment no TP is on its way on vehicle no MP"
我有要在字符串中搜索的这些缩写词(MP、TP)的列表。
请您参考如下方法:
你可以使用 re.sub
>>> string="This is the consignment no 1234578TP43789"
>>> re.sub(r'\d+(TP|MP)\d+', r'\1', string)
'This is the consignment no TP'
>>> string="Consignment no 1234578TP43789 is on its way on vehicle no 3456MP567890"
>>> re.sub(r'\d+(TP|MP)\d+', r'\1', string)
'Consignment no TP is on its way on vehicle no MP'
它有什么作用？
\d+匹配一位或多位数字。(TP|MP)匹配TP或MP。在\1中捕获它。我们使用这个捕获的字符串来替换整个匹配的字符串。
如果任何字符可以出现在 TP/MP 之前和之后，我们可以使用 \S 来匹配除空格之外的任何内容。例如，
>>> string="Consignment no 1234578TP43789 is on its way on vehicle no 3456MP567890"
>>> re.sub(r'\S+(TP|MP)\S+', r'\1', string)
'Consignment no TP is on its way on vehicle no MP'
编辑
使用 list comprehension ，您可以遍历列表并将所有字符串替换为，
>>> list_1=["TP","MP","DCT"]
>>> list_2=["This is the consignment no 1234578TP43789","Consignment no 1234578TP43789 is on its way on vehicle no 3456MP567890"]
>>> [ re.sub(r'\d+(' + '|'.join(list_1) + ')\d+', r'\1', string) for string in list_2 ]
['This is the consignment no TP', 'Consignment no TP is on its way on vehicle no MP']
