在Python中研究特殊字符的问题

Question

我有一个文件（我只显示了一部分），我想删除一个特殊字符。

OTU1359 UniRef90_A0A095VQ09 UniRef90_A0A0C1UI80 UniRef90_A0A1M4ZSK2 UniRef90_A0A1W1CJV7 UniRef90_A0A1Z9J2X0 UniRef90_A0A1Z9THL2 UniRef90_A0A2E3B6A5 UniRef90_A0A2E5MT47 UniRef90_A0A2E5VCW9 UniRef90_A0A2E6CDK4 UniRef90_A0A2E6KTE6 UniRef90_A0A2E8AIM6 UniRef90_A0A2E8RIG1 UniRef90_A0A2E8YNS3 UniRef90_A0A2E9VEK0 UniRef90_W6RCT6

OTU0980 UniRef90_A0A084TMQ7 UniRef90_A0A090PK65 UniRef90_A0A0P1G8P0 UniRef90_A0A0P1IHL1 UniRef90_A0A286ILS7 UniRef90_A0A2A5E7H9 UniRef90_A0A2D9J217 UniRef90_H3NS47 UniRef90_H3NSN9 UniRef90_H3NSP0 UniRef90_H3NSP7 UniRef90_H3NUB2 UniRef90_H3NY28 UniRef90_H3NY47 UniRef90_UPI000C2CBC51

我想删除字符“ OTUXXXX”（它始终以OTU开头，并且始终以4个数字开头）。它可以按行显示多个OTUXXXX

我尝试过：

re.search("OTU[0-9]{4}", line)

它不起作用。有什么帮助吗？

Answer 1

您可以利用re.sub来实际执行替换或用您提供的文本替换匹配的文本。您在这里找到文档：https://docs.python.org/3/library/re.html

这里是一种可能的实现方式：

from re import compile, sub, MULTILINE

text = '''
OTU1359 UniRef90_A0A095VQ09 UniRef90_A0A0C1UI80 UniRef90_A0A1M4ZSK2 UniRef90_A0A1W1CJV7 UniRef90_A0A1Z9J2X0 UniRef90_A0A1Z9THL2 UniRef90_A0A2E3B6A5 UniRef90_A0A2E5MT47 UniRef90_A0A2E5VCW9 UniRef90_A0A2E6CDK4 UniRef90_A0A2E6KTE6 UniRef90_A0A2E8AIM6 UniRef90_A0A2E8RIG1 UniRef90_A0A2E8YNS3 UniRef90_A0A2E9VEK0 UniRef90_W6RCT6

OTU0980 UniRef90_A0A084TMQ7 UniRef90_A0A090PK65 UniRef90_A0A0P1G8P0 UniRef90_A0A0P1IHL1 UniRef90_A0A286ILS7 UniRef90_A0A2A5E7H9 UniRef90_A0A2D9J217 UniRef90_H3NS47 UniRef90_H3NSN9 UniRef90_H3NSP0 UniRef90_H3NSP7 UniRef90_H3NUB2 UniRef90_H3NY28 UniRef90_H3NY47 UniRef90_UPI000C2CBC51
'''

replacemnt = ''
regex = compile(r'OTU\d{4}', flags=MULTILINE)
cleaned = sub(regex, replacemnt, text)

Answer 2

我建议使用re.sub并找到整个单词的模式匹配项，以避免其他单词内部出现部分匹配项。

s = re.sub(r"\s*\bOTU[0-9]{4}\b", "", line).strip()

请参见regex demo。末尾的.strip()会删除在字符串的末尾/开头删除的匹配项之后剩余的多余的前导/后缀空格。

请参见regex graph：