Python regular expression to remove all square brackets and their contents

Question

I am trying to use this regular expression to remove all instances of square brackets (and everything in them) from strings. For example, this works when there is only one p

Answer 1

By default * (or +) matches greedily, so the pattern given in the question will match upto the last ].

>>> re.findall(r'\[[^()]*\]', "Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]")
['[a] donkey lying down among the sheep pens.[b]']

By appending ? after the repetition operator (*), you can make it match non-greedy way.

>>> import re
>>> pattern = r'\[.*?\]'
>>> s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]"""
>>> re.sub(pattern, '', s)
'Issachar is a rawboned donkey lying down among the sheep pens.'

Answer 2

Try:

import re
pattern = r'\[[^\]]*\]'
s = """Issachar is a rawboned[a] donkey lying down among the sheep pens.[b]"""
t = re.sub(pattern, '', s)
print t

Output:

Issachar is a rawboned donkey lying down among the sheep pens.

Answer 3

For Numbers inside the brackets (No Alphabets), e.g. [89], [23], [11], etc., this is the pattern to use.

import re

text = "The[TEXT] rain in[33] Spain[TEXT] falls[12] mainly in[23] the plain![45]"
pattern = "\[\d*?\]"
numBrackets = re.findall(pattern, text)

print(numBrackets)

Output:

['[33]', '[12]', '[23]', '[45]']