Split concatenated functions keeping the delimiters

Question

I am trying to split strings containing python functions, so that the resulting output keeps separate functions as list elements.
s='hello()there()' should be split into ['hello()', 'there()']
To do so I use a regex lookahead to split on the closing parenthesis, but not at the end of the string.

While the lookahead seems to work, I cannot keep the ) in the resulting strings as suggested in various posts. Simply splitting with the regex discards the separator:

import re
s='hello()there()'
t=re.split(")(?!$)", s)

This results in: 'hello(', 'there()'] .

s='hello()there()'
t=re.split("())(?!$)", s)

Wrapping the separator as a group results in the ) being retained as a separate element: ['hello(', ')', 'there()']
As does this approach using the filter() function:

s='hello()there()'
u = list(filter(None, re.split("())(?!$)", s)))

resulting again in the parenthesis as a separate element: ['hello(', ')', 'there()']

How can I split such a string so that the functions remain intact in the output?

Asked By: O René

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Source

Answer 1

Use re.findall()

w+() matches one or more word characters followed by an opening and a closing parenthesis> That part matches the hello() and there()

t = re.findall(r"w+()", s)

['hello()', 'there()']

Edition:
.*? is a non-greedy match, meaning it will match as few characters as possible in the parenthesis.

s = "hello(x, ...)there(6L)now()"

t = re.findall(r"w+(.*?)", s)       
print(t)

['hello(x, ...)', 'there(6L)', 'now()']

Answered By: Jamiu S.

Answer 2

You can split on a lookbehind for () and negative lookahead for the end of the string.

t = re.split(r'(?<=())(?!$)', s)

Answered By: Unmitigated

Split concatenated functions keeping the delimiters

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