been trying to create a REGEX that will parse date and 3-letter code from a bit longer message. Here I will post examples of the messages and what I want to get:
AAA BBB 1A BY PEK14JUN18/1654 OR QQQ MF 812 XXXXX -> PEK, 14JUN18/1654
XXX/WWWW BY 05JUL 0900 BKK LT ELSE BKG WILL BE QQQQ -> BKK, 05JUL 0900
TO AZ BY 02AUG 1910 TYO OR AZ WWWW WILL BE XXX -> TYO, 02AUG 1910
BY TYO20JUL18/0355 OR CXL CA ALL QQQ -> TYO, 20JUL18/0355
BY AMS04JUL18/1954 OR CXL MF 812 L07JUL -> AMS, 04JUL18/1954
I want to be able to match 3-letter code and the date for every message. The code is always nearby the date but can be before or after the date part. Also the date part can be with or without a year.Is it possible to have one regex to use for all the above examples?
I started with this:
(\s[A-Z]{3}\d\d|\d\d[A-Z]{3}\s)
(https://regex101.com/r/LPLjgf/1) but it's not working as it should and I'm not very experienced with regex to be honest.
EDIT:
Actually I would need to use only the 3-letter codes but I need them to be connected with a date - for example in:
AAA BBB 1A BY PEK14JUN18/1654 OR QQQ MF 812 XXXXX
the AAA, BBB or QQQ shouldn't match because they arent right after / before the date as is PEK.
Same with BY TYO20JUL18/0355 OR CXL CA ALL QQQ -> only TYO should match because it's before a date and CXL shouldn't.
You may use the following pattern:
([A-Z]{3})(\d{2}[A-Z]{3}\d{2}\/\d{4})|(\d{2}[A-Z]{3} \d{4}) ([A-Z]{3})
([A-Z]{3}) Capturing group for three capital letters
(\d{2}[A-Z]{3}\d{2}\/\d{4}) Capturing group for two digits, three upper case letters, two digits, /, four digits.
| Logical OR, alternates pattern.
(\d{2}[A-Z]{3} \d{4}) Capturing group. Captures two digits, three upper case letters, whitespace and four digits.
([A-Z]{3}) Capturing group for three upper case letters.
You can try it live here.
Captured groups:
Group 1. 14-17 `PEK`
Group 2. 17-29 `14JUN18/1654`
Group 3. 83-93 `05JUL 0900`
Group 4. 94-97 `BKK`
Group 3. 151-161 `02AUG 1910`
Group 4. 162-165 `TYO`
Group 1. 211-214 `TYO`
Group 2. 214-226 `20JUL18/0355`
Group 1. 269-272 `AMS`
Group 2. 272-284 `04JUL18/1954`
Group 1. 342-345 `PEK`
Group 2. 345-357 `14JUN18/1654`
Group 1. 378-381 `TYO`
Group 2. 381-393 `20JUL18/0355`
Firstly
(\s[A-Z]{3}\d\d|\d\d[A-Z]{3}\s)
The alternation – | – means that will match \s[A-Z]{3}\d\d or \d\d[A-Z]{3}\s which is certainly not what you want. To narrow the scope of an alternation use grouping.
I would think you want to match this fairly directly:
([A-Z]{3})\d{2}[A-Z]{3}\d{2}
And that only captures the three letters in a group.
Try the following RegEx:
[A-Z]{3}(\d{2}[A-Z]{3}[\S]*)|(\d{2}[A-Z]{3}\s\d{4}\s[A-Z]{3})
It will matach 3 letters fowed by 2 numbers, followed by 3 letters OR 2 numbers followed by 3 letters, a Space, 4 numbers, a Space and 3 letters.
You can try it here
Related
I still didn't manage to find a solution for the problem that I have with my regex.
Case:
Input-
Deutsche Gesellschaftsgeschichte
Expected output:
Group 1 - Deutsche
Group 2 - Gesellschaftsgeschichte
The reason is because both word together exceed 30 chars, so they are separated into two groups.
The maximum input can be up to 90 chars.
More examples of inputs and expected outputs:
Input -
Fachlich geeignet, politisch unzuverlässig...
Output -
Fachlich geeignet, politisch
unzuverlässig...
Input -
Textbuch zur Privatrechtsgeschichte der Neuzeit
Output -
Textbuch zur
Privatrechtsgeschichte der
Neuzeit
To get what you want in up to 3 Groups, you can use this regex:
(\b.{1,30}(?=\b))(\b.{1,30}(?=\b))?(\b.{1,30}(?=\b))?
The regex starts with a Word boundary, then matches from 1 to 30 of any character, then it looks forward for a Word boundary (making sure we match and splits Whole Words).
This pattern is repeated 3 times (where the last 2 are optional).
Now you have your matches in 3 Groups, which you can access per index.
I have a list of addresses and I would like to have a regular expression that is able to capture just the name of the street without the street type, address number, or cardinal direction. There are some errors in formatting but all characters are in capital letters. So,
2038 W MAIN AVE
2038QWEW S JEFFERSON AVENUE
33 NORTH CALIFORNIA STREET
53371 SOUTH WASHINGTON
53371 S WASHINGTON AVENUE
1600 E PENNSYLVANIA AVE
WEST9 67ST ST
E171 N 23RD STREET
G171 N121ST STREET
ought to return
MAIN
JEFFERSON
CALIFORNIA
WASHINGTON
WASHINGTON
PENNSYLVANIA
67ST
23RD
121ST
So far I've got
([^ W ]|[^ E ]|[^ S ]|[^ N ])([0-9])*([A-Z]+)[^ ]
But I can't seem to only capture the first match that occurs after the street number. I feel like I need the standard greedy operators (i.e. ?, *, or +) but I can't figure out how to incorporate them.
These two links have taken me close:
Matching on every second occurence
Simple regex for street address
For the output what you want from the given (address) input, this regex will surely help: [\pL\pN]+(?=\h+[\pL\pN]+$)
This regex will match the second last word in your line where a word is "1 or more any letter or digit in any language".
For reference you could https://superuser.com/questions/1361759/matching-second-last-word-in-sentence-through-regular-expression
Logic: we are looking for the second last word (set of characters) + possible border with the symbol N
^.*?\s[N]{0,1}([-a-zA-Z0-9]+)\s*\w*$
Res:
Match 1
Full match 0-15 `2038 W MAIN AVE`
Group 1. 7-11 `MAIN`
Match 2
Full match 16-43 `2038QWEW S JEFFERSON AVENUE`
Group 1. 27-36 `JEFFERSON`
Match 3
Full match 44-70 `33 NORTH CALIFORNIA STREET`
Group 1. 53-63 `CALIFORNIA`
Match 4
Full match 71-93 `53371 SOUTH WASHINGTON`
Group 1. 83-93 `WASHINGTON`
Match 5
Full match 94-119 `53371 S WASHINGTON AVENUE`
Group 1. 102-112 `WASHINGTON`
Match 6
Full match 120-143 `1600 E PENNSYLVANIA AVE`
Group 1. 127-139 `PENNSYLVANIA`
Match 7
Full match 144-157 `WEST9 67ST ST`
Group 1. 150-154 `67ST`
Match 8
Full match 158-176 `E171 N 23RD STREET`
Group 1. 165-169 `23RD`
Match 9
Full match 177-195 `G171 N121ST STREET`
Group 1. 183-188 `121ST`
https://regex101.com/r/m2rmUQ/4
I was able to figure this out in a slightly different way
[0-9A-Z]* [0-9A-Z]*$
and then I simply split the string it created by the space. Maybe one or two steps too many but it's transparent
I have the following string:
"......(some chars) aaa bbb ###8/13/2018 ......(some chars)"
The ### in the string represent some random characters. ###'s length is unknown and it could be None (just "aaa bbb 8/13/2018").
My goal is to find the date from the string (8/13/2018) and the starting index of ###.
I currently used the following code:
m = re.search(r'\s.*?([0-9]{1,}/[0-9]{1,}/[0-9]{2,})', str)
m.groups()[0] ## The date
m.start() ## index of ###
But the regex is matching bbb ###8/13/2018 instead of ###8/13/2018
I also tried change the regex to:
r'\s(?!\s).*?[0-9]{1,}/[0-9]{1,}/[0-9]{2,}'
r'\s(?!\s)*?[0-9]{1,}/[0-9]{1,}/[0-9]{2,}'
But neither of them works.
I will be appreciated for any help or comments. Thank you.
I tend to believe you are looking for:
#*(?:\d{1,2}/){2}\d{2,4} or even \S*(?:\d{1,2}/){2}\d{2,4}
This is simply saying:
\S* start with 0 or more non-space charaters.
(?:\d{1,2}/){2} find two groups of \d{1,2}/ but do not capture them. ie not capturing: (?:..).this will match the month and date part 8/13/. \d{1,2} means atleast one digit and atmost two digits
\d{2,4} match the year .Atleast 2 digits and atmost 4 digits
Using a part of your regex, I think you mean something like this
r'\S*([0-9]+/[0-9]+/[0-9]{2,})'
https://regex101.com/r/dxF4sT/1
To find the starting index, it would be where the match was found.
Note that \S will find all consecutive non-whitespace.
You can change this to other things like [#a-zA-Z] etc..., just add it to the class.
I have a list of addresses, currently quite unclean. They take the format:
955 - 959 Fake Street
95-99 Fake Street
4-9 M4 Ln
95 - 99 Fake Street
99 Fake Street
What I would like to do is split up the street name and street number. I need a regex expression that is true for
955 - 959
95-99
4-9
95 - 99
99
I currently have this:
^[0-9][0-9]\s*+(\s*-\s*[0-9][0-9]+)
which works for the two digit addresses but does not work for the three or one digit addresses.
Thanks
I'm not sure what you're trying to do here \s*+ but you basically had the answer with the last part [0-9][0-9]+ that would find 2+ digits on the end.
Maybe try this (it's more concise). This searches for 1+ digits instead of 2+
\d+(\s*-\s*\d+)?
You can use braces {2,3} for 2-3 numbers - but also *+ isn't right.
/^(([0-9]{1,3}\s-\s)?[0-9]{1,3})\s/
I nested the braces so you only want the first result from the regex.
it breaks up like this
([0-9]{1,3}\s-\s)?
first, Is there a 1-3 digit number with a space-dash-space - OPTIONAL
then.. does it end in a 1-3 digit number followed by a space.
Starting from your regex:
^[0-9][0-9]\s*+(\s*-\s*[0-9][0-9]+)
You got an extra white space matcher in the second block:
^[0-9][0-9]\s*+(-\s*[0-9][0-9]+)
I would suggest you replace [0-9] with \d
^[\d][\d]\s*+(-\s*[\d][\d]+)
Use a + instead o 2 copies of \d meaning at least one number:
^[\d]+\s*+(-\s*[\d]+)
Make the last block optional, so it matches 99 Fake Address:
^[\d]+\s*+(-\s*[\d]+)?
If you know there's only going to be 1 white space, you could replace \s* with \s?:
^[\d]+\s?(-\s?[\d]+)?
That should match all of them :D
For your example, you can do:
/^(\d+[-\s\d]*)\s/gm
Demo
Explanation:
/^(\d+[-\s\d]*)\s/gm
^ start of line
^ at least 1 digit and as many digits as possible
^ any character of the set -, space, digit
^ zero or more
^ trailing space
^ multiline for the ^ start of line assertion
Another way could be
In [83]: s = '955 - 959 Fake Street'
In [84]: s1 = '95-99 Fake Street'
In [85]: s2 = '95 - 99 Fake Street'
In [86]: s3 = '99 Fake Street'
In [87]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s3)
In [88]: d.group()
Out[88]: '99 '
In [89]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s2)
In [90]: d.group()
Out[90]: '95 - 99'
In [91]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s1)
In [92]: d.group()
Out[92]: '95-99'
In [93]: d = re.search(r'^[0-9]+[ ]*(-[ ]*[0-9]+){0,1}', s)
In [94]: d.group()
Out[94]: '955 - 959'
the character set 0-9 cab be represented by \d like this
d = re.search(r'^[\d]+[ ]*(-[ ]*[\d]+){0,1}', s)
Here, in all the examples, we are searching at the beginning of the string, for a sequence of at least one digit followed by zero or more spaces and optionally followed by at most one sequence of only one - symbol followed by zero or more spaces and at least one or more digits.
My regex needs to parse an address which looks like this:
BLOOKKOKATU 20 A 773 00810 HELSINKI SUOMI
-------------------- ----- -------- -----
1 2 3 4*
Groups one, two and three will always exist in an address. Group 4 may not exist. I've written a regex that helps me get the first, second and third part but I would also need the fourth part. Part 4 is the country name and can either be FINLAND or SUOMI. If the fourth part didn't exist in an address the fourth group would be empty. This is my regex so far but the third group captures the country too. Any help?
(.*?)\s(\d{5})\s(.*)$
(I'm going to be using this Oracles REGEXP function)
Change the regex to:
(.*?)\s(\d{5})\s(.+?)\s?(FINLAND|SUOMI)?$
Making group three none greedy will let you match the optional space + country choices. If group 4 doesn't match I think it will be uninitialized rather than blank, that depends on language.
To match a character (or in your case group) that may or may not exist, you need to use ? after the character/subpattern/class in question. I'm answering now because RegEx is complicated and should be explained: only posting the fix without the answer isn't enough!
A question mark matches zero or one of the preceding character, class, or subpattern. Think of this as "the preceding item is optional". For example, colou?r matches both color and colour because the "u" is optional.
Above quote from http://www.autohotkey.com/docs/misc/RegEx-QuickRef.htm
Try this:
(.*?)\s(\d{5})\s(.*?)\s?([^\s]*)?$
This will match your input more tightly and each of your groups is in its own regex group:
(\w+\s\d+\s\w\s\d+)\s(\d+)\s(\w+)\s(\w*)
or if space is OK instead of "whitespace":
(\w+ \d+ \w \d+) (\d+) (\w+) (\w*)
Group 1: BLOOKKOKATU 20 A 773
Group 2: 00810
Group 3: HELSINKI
Group 4: SUOMI (optional - doesn't have to match)
(.*?)\s(\d{5})\s(\w+)\s(\w*)
An example:
SQL> with t as
2 ( select 'BLOOKKOKATU 20 A 773 00810 HELSINKI SUOMI' text from dual
3 )
4 select text
5 , regexp_replace(text,'(.*?)\s(\d{5})\s(\w+)\s(\w*)','\1**\2**\3**\4') new_text
6 from t
7 /
TEXT
-----------------------------------------
NEW_TEXT
-----------------------------------------------------------------------------------------
BLOOKKOKATU 20 A 773 00810 HELSINKI SUOMI
BLOOKKOKATU 20 A 773**00810**HELSINKI**SUOMI
1 row selected.
Regards,
Rob.