Python Regex在两个索引值之间找到所有字符

发布于 2025-01-17 08:28:51 字数 476 浏览 2 评论 0 原文

寻找一种使用Python Regex提取索引之间的字符串中的所有字符的方法。我的代码如下：

import re

txt = "Hula hoops are fun."
x = re.search(r"hoops", txt)
c = x.span()
a = c[0]
b = c[1]

print(a) # prints 5
print(b) # prints 10

txt2 = "Hula loops are fun."

z = re.???(a, b, txt2) #<------ Incorrect
print(z)

我要弄清楚的是以某种方式使用 a 和 b in z =“ loops” > txt2 （ txt 的重写）。是否有Python Regex命令可以执行此操作？

原文

Looking for a way to use a Python regex to extract all the characters in a string between to indexes. My code is below:

import re

txt = "Hula hoops are fun."
x = re.search(r"hoops", txt)
c = x.span()
a = c[0]
b = c[1]

print(a) # prints 5
print(b) # prints 10

txt2 = "Hula loops are fun."

z = re.???(a, b, txt2) #<------ Incorrect
print(z)

What I am trying to figure out is to somehow use a and b to get z = "loops" in txt2 (the rewrite of txt). Is there a python regex command to do this?

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对你而言 2025-01-24 08:28:51

您可以使用 z = txt [a：b] 在 a 和 b indices之间提取所有字符。

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飞烟轻若梦 2025-01-24 08:28:51

为什么不使用切片（显而易见的方式）？

z = txt2[a:b]
print(z)  # loops

如果您确实想要使用正则表达式，则需要使用.字符a次才能到达a，因为Regex没有直接索引。然后获取下一个 b - a 字符。在您的情况下，您最终会得到 (?<=.{5}).{5} 模式。 (?<=.{5}) 部分是肯定的后向断言。

pat = rf"(?<=.{{{str(a)}}}).{{{str(b - a)}}}"
print(re.search(pat, txt2))

输出：

<re.Match object; span=(5, 10), match='loops'>

Why not using slices(the obvious way)?

z = txt2[a:b]
print(z)  # loops

If you really want to use regex, you need to consume a . character a times to reach a because Regex doesn't have indexing directly. Then get the next b - a characters. In your case you end up with (?<=.{5}).{5} pattern. (?<=.{5}) part is a positive lookbehind assertion.

pat = rf"(?<=.{{{str(a)}}}).{{{str(b - a)}}}"
print(re.search(pat, txt2))

output:

<re.Match object; span=(5, 10), match='loops'>

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归属感 2025-01-24 08:28:51

import re

txt = "Hula hoops are fun."
x = re.search(r"hoops", txt)
c = x.span()
a = c[0]
b = c[1]

print(a) # prints 5
print(b) # prints 10

txt2 = "Hula loops are fun."
txt3 = list(txt2)
xy = txt3[a:b]
z = ""
for item in xy:
    
    z = z + item

print(z)

import re

txt = "Hula hoops are fun."
x = re.search(r"hoops", txt)
c = x.span()
a = c[0]
b = c[1]

print(a) # prints 5
print(b) # prints 10

txt2 = "Hula loops are fun."
txt3 = list(txt2)
xy = txt3[a:b]
z = ""
for item in xy:
    
    z = z + item

print(z)

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