从字符串列表中，仅提取括号内的字符

Question

I have a list of strings that have variable construction but have a character sequence enclosed in square brackets. I want to extract only the sequence enclosed by the square brackets. There is only one instance of square brackets per string, which simplifies the process.

I am struggling to do so in an elegant manner, and this is clearly a simple problem with Python's large string library.

What is a simple expression to do this?

Answer 1

Check regular expression, "re"

Something like this should do the trick

import re

s = "hello_from_adele[this_is_the_string_i_am_looking_for]this_is_not_it"
match = re.search(r"\[([A-Za-z0-9_]+)\]", s)
print match.group(1)

If you provide an example, we can be more specific

Answer 2

You don't even need re to do this:

In [11]: strng = "This is some text [that has brackets] followed by more text"

In [12]: strng[strng.index("[")+1:strng.index("]")]
Out[12]: 'that has brackets'

This uses string slicing to return the characters inside the brackets. index() returns the 0-based position of its argument. Since we don't want to include the [ at the beginning, we add 1. The second argument of the slice is the stop position, but it is not included in the returned substring, so we don't need to add anything to it.

Answer 3

If you prefer not to use regex for whatever reason, it should be easy to do with string splitting since you're guaranteed to have one and only one instance of [ and ] .

s = "some[string]to check"

_, midright = s.split("[")
target, _ = midright.split("]")

or

target = s.split("[")[1].split("]")[0]  # ewww