Python regular expression to find and replace multiple matches

Question

I have a string as follows which can have any number of spaces after the first [ or before the last ]:

my_string = " [  0.53119281  1.53762345  ]"

I have a regular expression which matches and replaces each one individually as follows:

my_regex_start = "(\[\s+)" #Find square bracket and any number of white spaces
replaced_1 = re.sub(my_regex_start, '[', my_string) --> "[0.53119281  -0.16633733  ]"

my_regex_end = "(\s+\])" #Find any number of white spaces and a square bracket
replaced_2 = re.sub(my_regex_end, ']', my_string) -->" [   0.53119281  -0.16633733]"

I have a regular expression which finds one OR the other:

my_regex_both = "(\[\s+)|(\s+\])" ##Find square bracket and any number of white spaces OR ny number of white spaces and a square bracket

How can I use this my_regex_both to replace the first one and OR the second one if any or both are found?

Djaouad · Accepted Answer · 2019-09-19 14:20:16Z

2

Instead of catching the brackets, you can replace the spaces that are preceded by [ or followed by ] with an empty string:

import re

my_string = "[  0.53119281  1.53762345  ]"

my_regex_both = r"(?<=\[)\s+|\s+(?=\])"

replaced = re.sub(my_regex_both, '', my_string)

print(replaced)

Output:

[0.53119281  1.53762345]

answered Sep 19, 2019 at 14:20

Djaouad

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Hoog · Accepted Answer · 2019-09-19 14:33:34Z

0

Another option you can use aside from MrGeek's answer would be to use a capture group to catch everything between your my_regex_start and my_regex_end like so:

import re

string1 = " [  0.53119281  1.53762345  ]"

result = re.sub(r"(\[\s+)(.*?)(\s+\])", r"[\2]", string1)
print(result)

I have just sandwiched (.*?) between your two expressions. This will lazily catch what is between which can be used with \2

OUTPUT

[0.53119281  1.53762345]

answered Sep 19, 2019 at 14:33

Hoog

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