python: parse substring in brackets from string

Question

What's a pythonic way to parse this string in the brackets:

txt = 'foo[bar]'

to get as result:

bar

What have I tried:

How I would solve it and I believe it's not very elegant:

result = txt.split('[')[1].split(']')[0]

I strongly think there is a library or method out there that has a more fault-tolerant and elegant solution for this. That's why I created this question.

Possible duplicate of [Regular expression to return all characters between two special characters](https://stackoverflow.com/questions/9889635/regular-expression-to-return-all-characters-between-two-special-characters) — Georgy, Jan 28 '19 at 09:47

score 2 · Accepted Answer · answered Jan 28 '19 at 08:38

2

Using Regex.

Ex:

import re

txt = 'foo[bar]'
print(re.findall(r"\[(.*?)\]", txt))

Output:

['bar']

answered Jan 28 '19 at 08:38

Rakesh

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DirtyBit · Answer 2 · 2019-01-28T08:40:32.597

0

One out of many ways, using slicing:

print(txt[4:].strip("[]"))

OR

import re

txt = 'foo[bar]'

m = re.search(r"\[([A-Za-z0-9_]+)\]", txt)
print(m.group(1))

OUTPUT:

bar

edited Jan 28 '19 at 08:40

answered Jan 28 '19 at 08:35

DirtyBit

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score 0 · Answer 3 · answered Jan 30 '19 at 02:35

Another take on slicing, using str.index to find the locations of the starting and ending delimiters. Once you get a Python slice, you can just it just like an index into the list, except that the slice doesn't just give a single character, but the range of characters from the start to the end.

def make_slice(s, first, second):
    floc = s.index(first)
    sloc = s.index(second, floc)
    # return a Python slice that will extract the contents between the first
    # and second delimiters
    return slice(floc+1, sloc)

txt = 'foo[bar]'

slc = make_slice(txt, '[', ']')
print(txt[slc])

Prints:

bar

python: parse substring in brackets from string

3 Answers3