enter image description here I found that StackOverflow has changed my issue, please check the image
我在使用固定的表达方式时遇到了问题。
import re
a_re = r"`(``|[^`])*`"
a_str = `abc`12`cde`45
b = re.findall(a_re,a_str)
print(b) # [ c , e ]
为什么出现这种结果?
我再次尝试其他规则。
import re
a_re = r"`(ab|[^c])*`"
a_str = `ab`1`df`
b = re.findall(a_re,a_str)
print(b) # [ f ]
更令人困惑
https://docs.python.org/3/library/re.html#re.findall”rel=“nofollow noreferer” www.un.org/Depts/DGACM/index_french.htm
其结果取决于该模式中捕获组的数量。 如果没有团体,则回到一个与整个模式相匹配的护卫清单。 如果只有一个组,则回到一个组群对应的座标清单。
捕获组的最终价值是last,在总配对中加以比对,例如,<代码>(“>][>][>][>][>][>][>]∗ 匹配(`[> >]>>。
re.findall’s API is badly designed; I recommend you use alternative functions, like re.finditer, instead.
import re
a_re = r"`((``|[^`])*)`"
a_str = `abc`12`cde`45
b = [m.group(1).replace("``", "`") for m in re.finditer(a_re, a_str)]
print(b) # [ abc , cde ]
你的回归表达是错误的,请使用另一种倒退表达方式。 你们想要什么? 我恳请你摘录 a和de,然后请你使用以下文字:
a_re = r"`([^`]*)`"
也就是说,你想要在“......”内恢复任何争执,当然你可以尝试
a_re = r"([^`]*)"
[ , abc , , 12 , , cde , , 45 , ]
或
a_re = r"`[^`]*`"
[ `abc` , `cde` ]
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