Develop Reference

regex - negative expression matching

Problem Introduction
So I've fried my brain trying to get negative look ahead/behinds to work. For the last example input, my current solution returns no match (see expected output table). I'm struggling with how to match the title part of the string when it includes a year that is not at the end of the string. To be clear, I'm only interested in matching the year if it is at the end of the string. The current regex fails on the last example, as it is matching NOT("Q" OR "\d*") in the title. However, I only want it to match NOT("Q" AND "\d{1}"). Any tips/suggestions greatly appreciated. Note using Python 3.8.
Example Input
AXP - Earnings call Q2 2021
AXP - Conference call 2021
BAC,BAC.PE,BAC.PL,BACRP,BML.PL,BML.PJ,BML.PH,BML.PG,BAC.PB,BAC.PK,BAC.PM,BAC.PN - Earnings call Q2 2021
GM - General Motors Company (GM) Presents at Deutsche Bank AutoTech Conference
AXP - American Express Company (AXP) Management Presents at Barclays 2020 Global Financial Services Conference
The period will always be of the form Q[1-4]. period and year are optional. If they do occur, they will be at the end of the string. symbol and title are always separated by - and always occur.
Expected Output
symbol
title
period
year
AXP
Earnings call
Q2
2021
AXP
Conference call
2021
BAC
Earnings call
Q2
2021
GM
General Motors Company (GM) Presents at Deutsche Bank AutoTech Conference
AXP
American Express Company (AXP) Management Presents at Barclays 2020 Global Financial Services Conference
What I've Tried
r"^(?P<symbol>[^\,]{1,8})(\,[A-Z\.]+)*\s\-\s(?P<title>[^Q\d]*)\s?(?P<period>Q\d)?\s?(?P<year>19|20\d{2})$"

You can use
^(?P<symbol>[^,]{1,8})(?:,[A-Z.]*)*\s+-\s+(?P<title>(?:(?!Q\d).)*?)\s*(?P<period>Q\d)?\s?(?P<year>(?:19|20)\d{2})?$
See the regex demo.
Note:
[^Q\d]* is wrong as it matches any zero or more chars other than Q and digit, you need to match any text up to a Q + digit, that is, a (?:(?!Q\d).)*? tempered greedy token
(?P<year>19|20\d{2}) is obligatory, but it must be optional and 19|20 are not grouped, so \d{2} is only applied to 20, (?P<year>19|20\d{2}) => (?P<year>(?:19|20)\d{2})?.
There are other small enhancements here.
Details:
^ - start of string
(?P<symbol>[^,]{1,8}) - Group "symbol": one to eight chars other than a comma
(?:,[A-Z.]*)* - zero or more repetitions of a comma and then zero or more uppercase letters/dots
\s+-\s+ - a hyphen enclosed with one or more whitespaces
(?P<title>(?:(?!Q\d).)*?) - Group "title": any char other than a line break char, zero or more but as few as possible occurrences, that does not start a Q+digit char sequence
\s* - zero or more whitespaces
(?P<period>Q\d)? - Group "period": a Q and a digit
\s? - an optional whitespace
(?P<year>(?:19|20)\d{2})? - an optional Group "year": 19 or 20 and then two digits
$ - end of string.

Python Regular expression of group to match text before amount

I am trying to write a python regular expression which captures multiple values from a few columns in dataframe. Below regular expression attempts to do the same. There are 4 parts of the string.
group 1: Date - month and day
group 2: Date - month and day
group 3: description text before amount i.e. group 4
group 4: amount - this group is optional
Some peculiar conditions for group 3 - text that
(1)the text itself might contain characters like "-" , "$". So we cannot use - & $ as the boundary of text.
(2) The text (group 3) sometimes may not be followed by amount.
(3) Empty space between group 3 and 4 is optional
Below is python function code which takes in a dataframe having 4 columns c1,c2,c3,c4 adds the columns dt, txt and amt after processing to dataframe.
def parse_values(args):
re_1='(([JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC]{3}\s{0,}[\d]{1,2})\s{0,}){2}(.*[\s]|.*[^\$]|.*[^-]){1}([-+]?\$[\d|,]+(?:\.\d+)?)?'
srch=re.search(re_1, args[0])
if srch is None:
return args
m = re.match(re_1, args[0])
args['dt']=m.group(1)
args['txt']=m.group(3)
args['amt']=m.group(4)
if m.group(4) is None:
if pd.isnull(args['c3']):
args['amt']=args.c2
else:
args['amt']=args.c3
return args
And in order to test the results I have below 6 rows which needs to return a properly formatted amt column in return.
tt=[{'c1':'OCT 7 OCT 8 HURRY CURRY THORNHILL ','c2':'$16.84'},
{'c1':'OCT 7 OCT 8 HURRY CURRY THORNHILL','c2':'$16.84'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK -$80,00,7770.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK-$2070.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK$2070.70'},
{'c1':'MAR 15 MAR 16 LOBLAWS FOODS INC - EAST YORK $80,00,7770.70'}
]
t=pd.DataFrame(tt,columns=['c1','c2','c3','c4'])
t=t.apply(parse_values,1)
t
However due to the error in my regular expression in re_1 I am not getting the amt column and txt column parsed properly as they return NaN or miss some words (as dipicted in some rows of the output image below).

How about this:
(((?:JAN|FEB|MAR|APR|MAY|JUN|JUL|AUG|SEP|OCT|NOV|DEC)\s*[\d]{1,2})\s*){2}(.*?)\s*(?=[\-$])([-+]?\$[\d|,]+(?:\.\d+)?)
As seen at regex101.com
Explanation:
First off, I've shortened the regex by changing a few minor details like using \s* instead of \s{0,}, which mean the exact same thing.
The whole [Jan|...|DEC] code was using a character class i.e. [], whcih only takes a single character from the entire set. Using non capturing groups is the correct way of selecting from different groups of multiple letters, which in your case are 'months'.
The meat of the regex: LOOKAHEADS
(?=[\-$]) tells the regex that the text before it in (.*) should match as much as it can until it finds a position followed by a dash or a dollar sign. Lookaheads don't actually match whatever they're looking for, they just tell the regex that the lookahead's arguments should be following that position.

Find date from image/text

I have dates like this and I need regex to find these types of dates
12-23-2019
29 10 2019
1:2:2018
9/04/2019
22.07.2019
here's what I did
first I removed all spaces from the text and here's what it looks like
12-23-2019291020191:02:2018
and this is my regex
re.findall(r'((\d{1,2})([.\/-])(\d{2}|\w{3,9})([.\/-])(\d{4}))',new_text)
it can find 12-23-2019 , 9/04/2019 , 22.07.2019 but cannot find 29 10 2019 and 1:02:2018

You may use
(?<!\d)\d{1,2}([.:/ -])(?:\d{1,2}|\w{3,})\1\d{4}(?!\d)
See the regex demo
Details
(?<!\d) - no digit right before
\d{1,2} - 1 or 2 digits
([.:/ -]) - a dot, colon, slash, space or hyphen (captured in Group 1)
(?:\d{1,2}|\w{3,}) - 1 or 2 digits or 3 or more word chars
\1 - same value as in Group 1
\d{4} - four digits
(?!\d) - no digit allowed right after
Python sample usage:
import re
text = 'Aaaa 12-23-2019, bddd 29 10 2019 <=== 1:2:2018'
pattern = r'(?<!\d)\d{1,2}([.:/ -])(?:\d{1,2}|\w{3,})\1\d{4}(?!\d)'
results = [x.group() for x in re.finditer(pattern, text)]
print(results) # => ['12-23-2019', '29 10 2019', '1:2:2018']

joining multiple regular expression for readability

I have following requirements in date which can be any of the following format.
mm/dd/yyyy or dd Mon YYYY
Few examples are shown below
04/20/2009 and 24 Jan 2001
To handle this I have written regular expression as below
Few text scenarios are metnioned below
txt1 = 'Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox
+ fluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical
Review of Systems Constitutional:'
txt2 = "s The patient is a 44 year old married Caucasian woman,
unemployed Decorator, living with husband and caring for two young
children, who is referred by Capitol Hill Hospital PCP, Dr. Heather
Zubia, for urgent evaluation/treatment till first visit with Dr. Toney
Winkler IN EIGHT WEEKS on 24 Jan 2001."
date = re.findall(r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}', txtData)
I am not getting 24 Jan 2001 where as if I run individually (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}' I am able to get output.
Question 1: What is bug in above expression?
Question 2: I want to combine both to make more readable as I have to parse any other formats so I used join as shown below
RE1 = '(?:\b(?<!\.)[\d{0,2}]+) (?:[/-]\d{0,}[/-]\d{2,4})'
RE2 = '(?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]* (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
regex_all = '|'.join([RE1, RE2])
regex_all = re.compile(regex_all)
date = regex_all.findall(txtData) // notice here txtData can be any one of the above string.
I am getting output as NaN in case of above for date.
Please suggest what is the mistake if I join.
Thanks for your help.

Note that it is a very bad idea to join such long patterns that also match at the same location within the string. That would cause the regex engine to backtrack too much, and possibly lead to crashes and slowdown. If there is a way to re-write the alternations so that they could only match at different locations, or even get rid of them completely, do it.
Besides, you should use grouping constructs (...) to groups sequences of patterns, and only use [...] character classes when you need to matches specific chars.
Also, your alternatives are overlapping, you may combine them easily. See the fixed regex:
\b(?<!\.)\d{1,2}(?:[/-]\d+[/-]|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{2})\b
See the regex demo.
Details
\b - a word boundary
(?<!\.) - no . immediately to the left of the current location
\d{1,2} - 1 or 2 digits
(?: - start of a non-capturing alternation group:
[/-]\d+[/-] - / or -, 1+ digits, - or /
| - or
(?:th|st|[nr]d)?\s*(?:
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*)) - th, st, nd or rd (optionally), followed with 0+ whitespaces, and then month names
\s* - 0+ whitespaces
(?:\d{4}|\d{2}) - 2 or 4 digits
\b - trailing word boundary.
Another note: if you want to match the date-like strings with two matching delimiters, you will need to capture the first one, and use a backreference to match the second one, see this regex demo. In Python, you would need a re.finditer to get those matches.
See this Python demo:
import re
rx = r"\b(?<!\.)\d{1,2}(?:([/-])\d+\1|(?:th|st|[nr]d)?\s*(?:(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z]*))\s*(?:\d{4}|\d{4})\b"
s = "Lithium 0.25 (7/11/77). LFTS wnl. Urine tox neg. Serum tox\nfluoxetine 500; otherwise neg. TSH 3.28. BUN/Cr: 16/0.83. Lipids unremarkable. B12 363, Folate >20. CBC: 4.9/36/308 Pertinent Medical\nReview of Systems Constitutional:\n\nThe patient is a 44 year old married Caucasian woman, unemployed Decorator, living with husband and caring for two young children, who is referred by Capitol Hill Hospital PCP, Dr. Heather Zubia, for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001"
print([x.group(0) for x in re.finditer(rx, s, re.I)])
# => ['7/11/77', '24 Jan 2001']

I think your approach is too complicated. I suggest using a combination of a simple regex and strptime().
import re
from datetime import datetime
date_formats = ['%m/%d/%Y', '%d %b %Y']
pattern = re.compile(r'\b(\d\d?/\d\d?/\d{4}|\d\d? \w{3} \d{4})\b')
data = "... your string ..."
for match in re.findall(pattern, data):
print("Trying to parse '%s'" % match)
for fmt in date_formats:
try:
date = datetime.strptime(match, fmt)
print(" OK:", date)
break
except:
pass
The advantage of this approach is, besides a much more manageable regex, that it won't pick dates that look plausible but do not exist, like 2/29/2000 (whereas 2/29/2004 works).

r'(?:\b(?<!\.)[\d{0,2}]+)'
'(?:[/-]\d{0,}[/-]\d{2,4}) | (?:\b(?<!\.)[\d{1,2}]+)[th|st|nd]*'
' (?:[Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec][a-z]*) \d{2,4}'
you should use raw strings (r'foo') for each string, not only the first one. This way backslashes (\) will be considered as normal character and usable by the re library.
[abc|def] matches any character between the [], while (one|two|three) matches any expression (one, two, or three)

How to Extract Date From String Python

I have a String Like
"Originally Posted on 09 May, 2016. By query 3 j...."
how can i extract date using python??
I tried this code:
dStr = "Originally Posted on 09 May, 2016. By query 3 j...."
date_st = re.findall("(\d+\ \w+,)", dStr)
printing date_st, i've got:
['09 May,']
what should i do for year??

You forget to add the year after ',', Only need to add '\d+.' is well.
re.findall("(\d+\ \w+, \d+\.)", dStr)
You will got this:
'09 May, 2016.'

You were almost there. Just add 4 digits for the year after the ,. It is better to use [a-z]+ instead of \w+ to match the month names as \w matches _ and 0-9 (along with alphabets) as well.
re.findall(r'\d+\s[a-z]+,\s\d{4}',s,re.I)