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?

Jamiu S. · Accepted Answer · 2023-02-01T09:19:37.130

5

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()']

edited Feb 01 '23 at 09:19

answered Jan 30 '23 at 17:14

Jamiu S.

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alright, that does it. can you quickly elaborate on the regex you used and how it differs from my attempts? – O René Jan 30 '23 at 17:16
1

Does the part after `|` serve any purpose here? – Fractalism Jan 30 '23 at 17:20
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`re.findall(r"\w+\(\)")` would do, too. No need to check for the end of the string here. – fsimonjetz Jan 30 '23 at 17:27
You are right @fsimonjetz – Jamiu S. Jan 30 '23 at 17:30
thank you. could this be extended to also work for other formats including arguments? such as `"hello(x, ...)there(6L)now()"` for example – O René Jan 30 '23 at 17:32
And how do you expect the output to look like? @ORené – Jamiu S. Jan 30 '23 at 17:38
I would hope for it to be `"['hello(x, ...), 'there(6L)', 'now()']"`, where the functions were again split into separate elements, on `)`, retaining the closing paranthesis – O René Jan 30 '23 at 17:40
@ORené Are keyword arguments supported? Also, are arbitrary spaces allowed between a comma and the next argument? – Fractalism Jan 30 '23 at 17:49
@Fractalism Yes exactly. I would hope for this to work for `"hello(data=x, capitalize = True)there()my(x = x)friend(x, 6L, ...)"` as well ideally – O René Jan 30 '23 at 17:52
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I found a better approach so edited the answer one more time. @ORené – Jamiu S. Jan 30 '23 at 18:08
it even works for this extended example: `re.findall(r"\w+\(.*?\)", y)` thank you so much @JamiuS., it is perfect – O René Jan 30 '23 at 18:14

score 2 · Answer 2 · answered Jan 30 '23 at 17:17

2

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

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

answered Jan 30 '23 at 17:17

Unmitigated

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Split concatenated functions keeping the delimiters

2 Answers2