Regex not extracting correct value from behind

Question

For the string = "4/3/09" using

df['dates'] = df['dates'].str.replace(r'([/ ]\d\d)\b', r'19\g<0>')
#or
df['dates'] = df['dates'].str.replace(r'([/ ]\d\d)$', r'19\g<0>')

I am getting 4/319/09 but I should get 4/3/1909

My data:

date_set = ['04/20/2009', '04/20/09', '4/20/09', '4/3/09',
'Mar-20-2009', 'Mar 20, 2009', 'March 20, 2009', 'Mar. 20, 2009', 
'Mar 20 2009','20 Mar 2009', '20 March 2009', '20 Mar. 2009', 
'20 March, 2009','Mar 20th, 2009', 'Mar 21st, 2009', 'Mar 22nd, 2009',
'Feb 2009', 'Sep 2009', 'Oct 2010',
'6/2008', '12/2009',
'2009', '2010']

If there is 2 digit year i need to add 1900. Ex - if year is 09, it should get replaced with 1909

I have multiple string in which dates are of form yy or yyyy. If it is yy i need to add 1900 to the year — EXODIA, Nov 07 '19 at 12:35
Ok, after you added more data, this is not your only issue here — Wiktor Stribiżew, Nov 07 '19 at 12:45

Wiktor Stribiżew · Accepted Answer · 2019-11-07T12:47:49.993

1

The ([/ ]\d\d)\b pattern matches / or space and then 2 digits up to a word boundary, and str.replace replaces the match (here, /09) with 19 + the whole match resulting in 4/3 + 19/09 => 4/319/09.

You need to use

df['dates'] = df['dates'].str.replace(r'([/ ])(\d\d)\b', r'\g<1>19\2')

See the regex demo

Here,

([/ ]) - Capturing group 1: a / or space
(\d\d) - Capturing group 2: two digits
\b - word boundary

The replacement is r'\g<1>19\2, i.e. Group 1 (here, an unambiguous backreference to Group 1 is used since the next char in the replacement pattern is a digit, see python re.sub group: number after \number) + 19 and Group 2 value (here, \2 is a regular numeric backreference is used since there is nothing following the pattern).

See re.sub Python documentation.

EDIT

After you added more data, it seems you need to only match the two digits at the end of the string.

Use

df['dates'] = df['dates'].str.replace(r'([/ ])(\d\d)$', r'\g<1>19\2')
df['dates'] = df['dates'].str.replace(r'(?<=[/ ])(?=\d\d$)', '19')

The second line removes the problem wtith backreferences since it uses lookarounds.

edited Nov 07 '19 at 12:47

answered Nov 07 '19 at 12:36

Wiktor Stribiżew

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Can you please explain about that \g or suggest a link where I should look up. I didnt understand how it is working – EXODIA Nov 07 '19 at 12:37
And this is throwing error error: invalid group reference 2 – EXODIA Nov 07 '19 at 12:38
@EXODIA Use **my code** that does **NOT** throw any errors. Do not try to alter what I posted. I added all the details and links. – Wiktor Stribiżew Nov 07 '19 at 12:41
It worked for this string but now my other stringss are filing. string - 4/20/09 output - 4/1920/1909 – EXODIA Nov 07 '19 at 12:47
it should be 4/20/1909 – EXODIA Nov 07 '19 at 12:47
@EXODIA See my EDIT – Wiktor Stribiżew Nov 07 '19 at 12:48
its working now. Is r'\g<1>19\2 and r'\g<1>19\g<2> same ? – EXODIA Nov 07 '19 at 12:52
And how is this working ? replace(r'(?<=[/ ])(?=\d\d$)', '19') – EXODIA Nov 07 '19 at 12:53
@EXODIA `r'\g<1>19\2` = `r'\g<1>19\g<2>`. `(?<=[/ ])` only checks but does not consume `/` or space immediately to the left of the current location and `(?=\d\d$)` requires two digits and end of string right after the current location. It matches an empty string between `/` or space and two digits, hence no need for backreferences. – Wiktor Stribiżew Nov 07 '19 at 12:59
Thank you for patiently understanding the doubt and explaining it. I understood and I will further go through the links to read further. – EXODIA Nov 07 '19 at 13:06
If you could share some links to understand backlinks and look around as you mentioned. It would help me. As I am unaware of these terms – EXODIA Nov 07 '19 at 13:07
1

@EXODIA [See this universal regex reference](https://www.regular-expressions.info/lookaround.html). – Wiktor Stribiżew Nov 07 '19 at 13:08

Regex not extracting correct value from behind

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