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Python 吐出字符串并删除[n]字符后的其余字符串_Python_Regex_String - Fatal编程技术网

Python 吐出字符串并删除[n]字符后的其余字符串

Python 吐出字符串并删除[n]字符后的其余字符串,python,regex,string,Python,Regex,String,我有一个字符串srting=123d*gss6agn 如何删除=之后的所有内容并保留为空 如果要使用re或字符串模块,如何解决它请尝试以下方法: a="srting=123d*gss6agn" a.split("=")[0]+"=" 或者这个: a="srting=123d*gss6agn" a[:a.index("=")+1] 要打开文件,请执行以下操作: with open("path/to/your/file.txt","r") as f: a=f.read() # and t

我有一个字符串
srting=123d*gss6agn
如何删除
=
之后的所有内容并保留为空 如果要使用re或字符串模块,如何解决它

请尝试以下方法:

a="srting=123d*gss6agn"
a.split("=")[0]+"="
或者这个:

a="srting=123d*gss6agn"
a[:a.index("=")+1]
要打开文件,请执行以下操作:

with open("path/to/your/file.txt","r") as f:
    a=f.read() # and then use an option above

如果您想使用正则表达式。。。虽然我个人会选择另一个答案

import re 
a = "srting=123d*gss6agn"
result = re.sub(r'=.','=',a)
将等号后的所有文本替换为单个等号。

带分区()

(如果缺少“=”则没有例外。)

只是玩:

    import re
    re.sub(r"(?<==).*",lambda m: m[0][::-1],s)
    Out: 'srting=nga6ssg*d321'

(?<==) lookbehind ?<= for'='; so '=' isn't captured for replacing;
sub's 2nd arg can be a function, too, which got the match object, here 'm'.
重新导入

re.sub(r)(?
print(“srting=123d*gss6agn.split(=”[0])
?是否也要删除等号?如果要拆分的字符串i位于其他文件夹
.txt
中,该怎么办:r'=。“必须r'=*”
    import re
    re.sub(r"(?<==).*",lambda m: m[0][::-1],s)
    Out: 'srting=nga6ssg*d321'

(?<==) lookbehind ?<= for'='; so '=' isn't captured for replacing;
sub's 2nd arg can be a function, too, which got the match object, here 'm'.