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Python';s';匹配';行输入_sre.sre_匹配输出?它能显示全场比赛吗?_Python_Regex_Match - Fatal编程技术网

Python';s';匹配';行输入_sre.sre_匹配输出?它能显示全场比赛吗?

Python';s';匹配';行输入_sre.sre_匹配输出?它能显示全场比赛吗?,python,regex,match,Python,Regex,Match,使用python 3: In [275]: blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg" In [276]: pat = re.compile("([a-z]{2,9999})") In [277]: data = re.search(pat,blah) In [278

使用python 3:

In [275]: blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg"

In [276]: pat = re.compile("([a-z]{2,9999})")

In [277]: data = re.search(pat,blah)

In [278]: data
Out[278]: <_sre.SRE_Match object; span=(0, 125), match='fffffffffffffffffffffffffffffffffffffffffffffffff>
[275]中的
:blah=“FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFG”
在[276]中:pat=re.compile(([a-z]{29999})”
在[277]中:数据=重新搜索(pat,blah)
In[278]:数据

Out[278]:您可能正在寻找.group

import re
blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg"
pat = re.compile("([a-z]{2,99})")
data = re.search(pat,blah)
if data:
    data.group(0) # returns 'fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff'
else:
    pass # do something when the string was not found!
见:


您可能正在寻找.group

import re
blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg"
pat = re.compile("([a-z]{2,99})")
data = re.search(pat,blah)
if data:
    data.group(0) # returns 'fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff'
else:
    pass # do something when the string was not found!
见:


有了这个,你就不会受到重复次数上限的限制:

import re

blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg"

r = re.compile(r'[a-z](.)\1{2,}') # no upper limit
try:
    re.search(r, blah).group()
except AttributeError:
    pass  # whatever you want to do when there is no match

有了这个,你就不会用任何重复的上限来约束自己:

import re

blah = "ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffg"

r = re.compile(r'[a-z](.)\1{2,}') # no upper limit
try:
    re.search(r, blah).group()
except AttributeError:
    pass  # whatever you want to do when there is no match

不,这是不可能的。该长度在match对象的
repr
方法的格式字符串中硬编码,不用于捕获匹配字符串的完整长度

除了编译自己的CPython构建(或任何Python风格),其精度相当于modified中
match
对象的表示。默认精度为50:

result=PyUnicode\u FromFormat(
"",
Py_类型(自身)->tp_名称,
self->mark[0],self->mark[1],group0);

正如其他人所建议的,只需使用match对象的方法访问完整的匹配字符串。

不,这是不可能的。该长度在match对象的
repr
方法的格式字符串中硬编码,不用于捕获匹配字符串的完整长度

除了编译自己的CPython构建(或任何Python风格),其精度相当于modified中
match
对象的表示。默认精度为50:

result=PyUnicode\u FromFormat(
"",
Py_类型(自身)->tp_名称,
self->mark[0],self->mark[1],group0);

正如其他人所建议的,只需使用match对象的方法访问完整的匹配字符串。

模式仅匹配99个字符,其余字符不是匹配的一部分。如果您想要更长的匹配,为什么将其限制为
{2,99}
?谢谢。我编辑了我的帖子。即使使用9999,我也看不到“g”的完整结尾。该模式仅匹配99个字符,其余字符不是匹配的一部分。如果您想要更长的匹配,为什么将其限制为
{2,99}
?谢谢。我编辑了我的帖子。即使是9999,我也看不到“g”感谢的完整结局。但是,不,我不是。谢谢。但是,不,我不是。谢谢你。那么,答案是否定的。我不关心集体方法。谢谢。那么,答案是否定的。我不关心集体方法。