converting difficult string to a list format in python

Question

I have strings from a diff file that can occur in multiple ways:

'@@ -1289,4 +1289,4 @@'

'@@ -1289,2 +1289 @@'

'@@ -1289 +1289,8 @@'

I'd like to find a nice way to cover all possibilities using regex or other tools such that I can output lists as follows for each respective case

[1289,4,1289,4]

[1289,2,1289,0]

[1289,0,1289,8]

I have tried playing around with regex queries on this website

For example (-\d+,\d+) (\+\d+,\d+). However I'm not sure how to default to zero when it can't match a value. I am quite new to regex but I'm also starting to doubt if this is the tool that I need to be resorting to.

Any elegant ways to do this? Thanks

Please, show us code sample, it'll be easier to help you. In general regexps do not have tools for default values, you need to handle that in the programming language you use. — mrzasa
– mrzasa, Commented Feb 21, 2018 at 10:13
elaborate should this 1289,4 be a single string item or 2 integer items separated by ,? — RomanPerekhrest
– RomanPerekhrest, Commented Feb 21, 2018 at 10:32

freedev · Accepted Answer · 2018-02-21 10:27:35Z

2

@@\s*([\d,+-]+)\s+([\d,+-]+)\s*@@

https://regex101.com/r/uKik9f/3

Regex explanation:

@@ matches the characters @@ literally (case sensitive)
\s* matches any whitespace character
- * Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
1st Capturing Group ([\d,+-]+) Match a single character present in the list below [\d,+-]+
- + Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
- \d matches a digit (equal to [0-9])
- ,+- matches a single character in the list ,+- (case sensitive)

\s+ matches one or more whitespace character

the 2nd Capturing Group ([\d,+-]+) and the resti is like the former part

Global pattern flags g modifier: global. All matches (don't return after first match)

edited Feb 21, 2018 at 10:27

answered Feb 21, 2018 at 10:20

freedev

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Comments

The fourth bird · Accepted Answer · 2018-03-10 18:02:43Z

1

You might use:

'@@ -(\d+),?(\d*) \+(\d+),?(\d*)? @@'

That would match

Match '@@ -
(\d+) Match one or more digits in a capturing group
,? Optional comma
(\d*) Match zero or more digits in a capturing group
\+ Match whitespace and a plus sign
(\d+) Match one or more digits in a capturing group
,? Optional comma
(\d*)? Match zero or more digits in an optional capturing group
Match @@'

This will use 4 capturing groups. If the value is missing the group will be empty. You could check for empty strings in the code and if so, add a zero.

Demo Python

answered Mar 10, 2018 at 18:02

The fourth bird

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1 Comment

user32882 Over a year ago

amazing what you did here. Makes me really want to learn more about regex

Anton Akhramovich · Accepted Answer · 2018-02-21 10:14:46Z

1

However I'm not sure how to default to zero when it can't match a value.

This is not possible with pure regex (regex can help you to find by pattern but not able to modify found result using any logic)

Also, you can try to modify your current regex to catch all cases.

(-\d+,?\d*) (\+\d+,?\d*)

answered Feb 21, 2018 at 10:14

Anton Akhramovich

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Comments

Ami Hollander · Accepted Answer · 2018-02-21 10:20:13Z

1

You can use regex as follow:

re.findall(r"[\w']+", txt)

For example:

print(re.findall(r"[\w']+", '@@ -1289,4 +1289,4 @@'))
> ['1289', '4', '1289', '4']

Notice that for your second example it won't fit cause it will result with

['1289', '2', '1289']

While you are expecting for list with length 4, in order to support so, you can right zero fill until 4, something like:

b = re.findall(r"[\w']+", '@@ -1289,2 +1289 @@')
b = b + [0] * (4 - len(b))

print(b)
> ['1289', '2', '1289', 0]

edited Feb 21, 2018 at 10:20

answered Feb 21, 2018 at 10:14

Ami Hollander

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Comments

Vikas Periyadath · Accepted Answer · 2018-02-21 10:22:28Z

1

If you want to match zero for all missing values, you could use something like this:

@@ -(\d*),?(\d*) \+(\d*),?(\d*) @@

This would return and empty string if there was no match. You'd have to replace that with zero later in your code.

Output

Your examples would then output :

['1289','4','1289','4']

['1289','2','1289','']

['1289','','1289','8']

edited Feb 21, 2018 at 10:22

Vikas Periyadath

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answered Feb 21, 2018 at 10:17

Jon

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RomanPerekhrest · Accepted Answer · 2018-02-21 10:29:12Z

1

Complete solution with re.sub() + re.findall() functions:

import re

diff_lst = ['@@ -1289,4 +1289,4 @@', '@@ -1289,2 +1289 @@', '@@ -1289 +1289,8 @@']

sub_pat, find_pat = re.compile(r'([+-]\d+)\s'), re.compile(r'[+-](\d+),(\d+)')
result = [[int(d) for t in find_pat.findall(sub_pat.sub('\\1,0', s)) for d in t]
           for s in diff_lst]

print(result)

The output:

[[1289, 4, 1289, 4], [1289, 2, 1289, 0], [1289, 0, 1289, 8]]

answered Feb 21, 2018 at 10:29

RomanPerekhrest

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Comments

Mohideen bin Mohammed · Accepted Answer · 2018-02-21 11:28:11Z

1

Here is what you want,

regex = '\-.+\+.+\d'


>>> s
'@@ -1289,4 +1289,4 @@'
>>> re.search('\-.+\+.+\d', s).group().replace(',', ' ').split(' ')
['-1289', '4', '+1289', '4']
>>>

here in re.search('\-.+\+.+\d', s).group() i am capturing number inside variable.
and then replacing , to made it equal.
finally split(' ') with empty space will give integer list

answered Feb 21, 2018 at 11:28

Mohideen bin Mohammed

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