I have street address strings in different formats. I tried this old post, but did not help much. My string formats are as follows,
format 1:
string_1 = ', landlord and tenant entered into a an agreement with respect to approximately 5,569 square feet of space in the building known as "the company" located at 788 e.7th street, st. louis, missouri 55605 ( capitalized terms used herein and not otherwise defined herein shall have the respective meanings given to them in the agreement); whereas, the term of the agreement expires on may 30, 2015;'
desired output:
788 e.7th street, st. louis, missouri 55605
format 2:
string_2 = 'first floor 824 6th avenue, chicago, il where the office is located'
desired output:
824 6th avenue, chicago, il
format 3:
string_3 = 'whose address is 90 south seventh street, suite 5400, dubuque, iowa, 55402.'
desired output:
90 south seventh street, suite 5400, dubuque, iowa, 55402
So far, I tried, this for string_1,
address_match_1 = re.findall(r'((\d*)\s+(\d{1,2})(th|nd|rd).*\s([a-z]))', string_1)
I get an empty list.
For the 2nd string I tried the same and getting the empty list as follows,
address_match_2 = re.findall(r'((\d*)\s+(\d{1,2})(th|nd|rd).*\s([a-z]))', string_2)
How can I try to match using re ? They are all in different formats, how can I get suite involved in string_3? Any help would be appreciated.
Solution
This regex matches all addresses in the question:
(?i)\d+ ((?! \d+ ).)*(missouri|il|iowa)(, \d{5}| \d{5}|\b)
You would need to add all of the states and their abbreviations, as well as a better match for the zip code, which you can find if you google it. Also, this will only work for US addresses.
Here is the output for each of the given strings:
>>> m = re.findall(r"((?i)\d+ ((?! \d+ ).)*(missouri|il|iowa)(, \d{5}| \d{5}|\b))", string_1)
>>> print m
[('788 e.7th street, st. louis, missouri 55605', ' ', 'missouri', ' 55605')]
>>> m = re.findall(r"((?i)\d+ ((?! \d+ ).)*(missouri|il|iowa)(, \d{5}| \d{5}|\b))", string_2)
>>> print m
[('824 6th avenue, chicago, il', ' ', 'il', '')]
>>> m = re.findall(r"((?i)\d+ ((?! \d+ ).)*(missouri|il|iowa)(, \d{5}| \d{5}|\b))", string_3)
>>> print m
[('90 south seventh street, suite 5400, dubuque, iowa, 55402', ' ', 'iowa', ', 55402')]
>>>
The first value of each tuple has the correct address. However, this may not be exactly what you need (see Weakness below).
Detail
Assumptions:
Address starts with a number fallowed by a space
Address ends with a state, or its abbreviation, optionally followed by a 5 digit zip code
The rest of the address is in between the two parts above. This part doesn't contain any numbers surrounded by spaces (i.e. with no " \d+ ").
regex string:
r"((?i)\d+ ((?! \d+ ).)*(missouri|il|iowa)(, \d{5}| \d{5}|\b))"
r"" make string a raw string to avoid escaping special characters
(?i) to make regex case insensitive
\d+ address starts with a number followed by a space
(missouri|il|iowa)(, \d{5}| \d{5}|\b)) address ends with state optionally followed by zip code. The \b is just the 'end of word', which makes the zip code optional.
((?! \d+ ).)* any group of characters except for a number surrounded by spaces. Refer to this article for an explanation on how this works.
Weakness
Regular expressions are used to match patterns, but the addresses presented don't have much of a pattern compared with the rest of the string they may be in. Here is the pattern that I identified and that I based the solution on:
Address starts with a number fallowed by a space
Address ends with a state, or its abbreviation, optionally followed by a 5 digit zip code
The rest of the address is in between the two parts above. This part doesn't contain any numbers surrounded by spaces (i.e. with no " \d+ ").
Any address that violates these assumptions won't be matched correctly. For example:
Addresses starting with a number with letters, such as: 102A or 3B.
Addresses with numbers in between initial number and the state, such as one containing ' 7 street' instead of ' 7th street.'
Some of these weaknesses may be fixed with simple changes to the regex, but some may be more difficult to fix.
I need to extract name, street1, street2, city, state, zip
I have data in this form
JOHN m SMITH [1111 WEST OAK ROAD, SUITE 101, CITY, ST 55555]
GEORGE m JONES [222 MAIN STREET, CITY, ST 55555]
My results for JOHN should be
name="JOHN m SMITH"
street1="1111 WEST OAK ROAD"
street2="SUITE 101"
city = "CITY"
state = "ST"
zip = "55555"
This works with GEORGE's data
Regex r = new Regex(#"^(?<name>.*)\[(?<street>.*)[,]\s(?<city>.*)[,]\s(?<state>.*)\s(?<zip>\d{5})\]$");
var match = r.Match(fullNameAndAddress);
name = match.Groups["name"].Value;
street = match.Groups["street"].Value;
city = match.Groups["city"].Value;
state = match.Groups["state"].Value;
zip = match.Groups["zip"].Value;
How do I add the optional street2?
I want 1 and only 1 "street" group. I thought it should have this: (....){1}?
street2 is optional zero or 1 times. I thought it should have this (...)?
but it doesn't work with JOHN's data, both street1 & street2 are going into the street group:
^(?<name>.*)\[((?<street>.*)[,]\s){1}?((?<street2>.*)[,]\s)?(?<city>.*)[,]\s(?<state>.*)\s(?<zip>\d{5})\]$
Could you clarify what you want stored in street?
Do you want John's to look like '1111 WEST OAK ROAD, SUITE 101'?
Or do you want to stuff it into some variable you wont be using, so that street looks like '1111 WEST OAK ROAD'?
Edit: With clarification, check out this link
http://rubular.com/r/S4HaTMVFZl
What happens here I believe is that the * is greedy, grabbing as much as it can before finding the final occurence of [,]\s
Adding a ? after the .* makes it lazy, grabbing the least information possible.
The amended regex looks like this
^(?<name>.*)\[((?<street>.*?)[,]\s)((?<street2>.*)[,]\s)?(?<city>.*)[,]\s(?<state>.{2})\s(?<zip>\d{5})\]$
You'll notice I changed the Regex for state from .* to .{2}, forcing a 2-character state. Feel free to revert that if you don't want it :)
I made a couple of changes to your regex in rubular.com, and it seemed to be working on both the example strings:
^(?<name>.+)\s\[(?<street>[^,]+),\s((?<street2>[^,]+),\s+)?(?<city>[^,]+),\s(?<state>.+)\s(?<zip>\d{5})\]$
street2 = match.Groups["street2"].Value;
One trick I've learned with regex's is to use the negation of the divider (eg. [^,]* for anything but a comma) instead of .*, so it's impossible to capture multiple fields with one expression. Also, the + operator, which requires at least one match, is useful in most of the groups.
Also, the additional comma is only there if there's an street2 component of the address, which indicates that the comma should be in the same capture group as the street2 part. I added an extra capture group around the street2 capture group to account for this. You can make groups non-capturing in most languages, but it didn't seem necessary.
I am trying to write a regular expression that facilitates an address, example 21-big walk way or 21 St.Elizabeth's drive I came up with the following regular expression but I am not too keen to how to incorporate all the characters (alphanumeric, space dash, full stop, apostrophe)
"regexp=^[A-Za-z-0-99999999'
See the answer to this question on address validating with regex:
regex street address match
The problem is, street addresses vary so much in formatting that it's hard to code against them. If you are trying to validate addresses, finding if one isn't valid based on its format is mighty hard to do.
This would return the following address (253 N. Cherry St. ), anything with its same format:
\d{1,5}\s\w.\s(\b\w*\b\s){1,2}\w*\.
This allows 1-5 digits for the house number, a space, a character followed by a period (for N. or S.), 1-2 words for the street name, finished with an abbreviation (like st. or rd.).
Because regex is used to see if things meet a standard or protocol (which you define), you probably wouldn't want to allow for the addresses provided above, especially the first one with the dash, since they aren't very standard. you can modify my above code to allow for them if you wish--you could add
(-?)
to allow for a dash but not require one.
In addition, http://rubular.com/ is a quick and interactive way to learn regex. Try it out with the addresses above.
In case if you don't have a fixed format for the address as mentioned above, I would use regex expression just to eliminate the symbols which are not used in the address (like specialized sybmols - &(%#$^). Result would be:
[A-Za-z0-9'\.\-\s\,]
Just to add to Serzas' answer(since don't have enough reps. to comment).
alphabets and numbers can effectively be replaced by \w for words.
Additionally apostrophe,comma,period and hyphen doesn't necessarily need a backslash.
My requirement also involved front and back slashes so \/ and finally whitespaces with \s. The working regex for me ,as such was :
pattern: "[\w',-\\/.\s]"
Regular expression for simple address validation
^[#.0-9a-zA-Z\s,-]+$
E.g. for Address match case
#1, North Street, Chennai - 11
E.g. for Address not match case
$1, North Street, Chennai # 11
I have succesfully used ;
Dim regexString = New stringbuilder
With regexString
.Append("(?<h>^[\d]+[ ])(?<s>.+$)|") 'find the 2013 1st ambonstreet
.Append("(?<s>^.*?)(?<h>[ ][\d]+[ ])(?<e>[\D]+$)|") 'find the 1-7-4 Dual Ampstreet 130 A
.Append("(?<s>^[\D]+[ ])(?<h>[\d]+)(?<e>.*?$)|") 'find the Terheydenlaan 320 B3
.Append("(?<s>^.*?)(?<h>\d*?$)") 'find the 245e oosterkade 9
End With
Dim Address As Match = Regex.Match(DataRow("customerAddressLine1"), regexString.ToString(), RegexOptions.Multiline)
If Not String.IsNullOrEmpty(Address.Groups("s").Value) Then StreetName = Address.Groups("s").Value
If Not String.IsNullOrEmpty(Address.Groups("h").Value) Then HouseNumber = Address.Groups("h").Value
If Not String.IsNullOrEmpty(Address.Groups("e").Value) Then Extension = Address.Groups("e").Value
The regex will attempt to find a result, if there is none, it move to the next alternative. If no result is found, none of the 4 formats where present.
This one worked for me:
\d+[ ](?:[A-Za-z0-9.-]+[ ]?)+(?:Avenue|Lane|Road|Boulevard|Drive|Street|Ave|Dr|Rd|Blvd|Ln|St)\.?
The source: https://www.codeproject.com/Tips/989012/Validate-and-Find-Addresses-with-RegEx
Regex is a very bad choice for this kind of task. Try to find a web service or an address database or a product which can clean address data instead.
Related:
Address validation using Google Maps API
As a simple one line expression recommend this,
^([a-zA-z0-9/\\''(),-\s]{2,255})$
I needed
STREET # | STREET | CITY | STATE | ZIP
So I wrote the following regex
[0-9]{1,5}( [a-zA-Z.]*){1,4},?( [a-zA-Z]*){1,3},? [a-zA-Z]{2},? [0-9]{5}
This allows
1-5 Street #s
1-4 Street description words
1-3 City words
2 Char State
5 Char Zip code
I also added option , for separating street, city, state, zip
Here is the approach I have taken to finding addresses using regular expressions:
A set of patterns is useful to find many forms that we might expect from an address starting with simply a number followed by set of strings (ex. 1 Basic Road) and then getting more specific such as looking for "P.O. Box", "c/o", "attn:", etc.
Below is a simple test in python. The test will find all the addresses but not the last 4 items which are company names. This example is not comprehensive, but can be altered to suit your needs and catch examples you find in your data.
import re
strings = [
'701 FIFTH AVE',
'2157 Henderson Highway',
'Attn: Patent Docketing',
'HOLLYWOOD, FL 33022-2480',
'1940 DUKE STREET',
'111 MONUMENT CIRCLE, SUITE 3700',
'c/o Armstrong Teasdale LLP',
'1 Almaden Boulevard',
'999 Peachtree Street NE',
'P.O. BOX 2903',
'2040 MAIN STREET',
'300 North Meridian Street',
'465 Columbus Avenue',
'1441 SEAMIST DR.',
'2000 PENNSYLVANIA AVENUE, N.W.',
'465 Columbus Avenue',
'28 STATE STREET',
'P.O, Drawer 800889.',
'2200 CLARENDON BLVD.',
'840 NORTH PLANKINTON AVENUE',
'1025 Connecticut Avenue, NW',
'340 Commercial Street',
'799 Ninth Street, NW',
'11318 Lazarro Ln',
'P.O, Box 65745',
'c/o Ballard Spahr LLP',
'8210 SOUTHPARK TERRACE',
'1130 Connecticut Ave., NW, Suite 420',
'465 Columbus Avenue',
"BANNER & WITCOFF , LTD",
"CHIP LAW GROUP",
"HAMMER & ASSOCIATES, P.C.",
"MH2 TECHNOLOGY LAW GROUP, LLP",
]
patterns = [
"c\/o [\w ]{2,}",
"C\/O [\w ]{2,}",
"P.O\. [\w ]{2,}",
"P.O\, [\w ]{2,}",
"[\w\.]{2,5} BOX [\d]{2,8}",
"^[#\d]{1,7} [\w ]{2,}",
"[A-Z]{2,2} [\d]{5,5}",
"Attn: [\w]{2,}",
"ATTN: [\w]{2,}",
"Attention: [\w]{2,}",
"ATTENTION: [\w]{2,}"
]
contact_list = []
total_count = len(strings)
found_count = 0
for string in strings:
pat_no = 1
for pattern in patterns:
match = re.search(pattern, string.strip())
if match:
print("Item found: " + match.group(0) + " | Pattern no: " + str(pat_no))
found_count += 1
pat_no += 1
print("-- Total: " + str(total_count) + " Found: " + str(found_count))
UiPath Academy training video lists this RegEx for US addresses (and it works fine for me):
\b\d{1,8}(-)?[a-z]?\W[a-z|\W|\.]{1,}\W(road|drive|avenue|boulevard|circle|street|lane|waylrd\.|st\.|dr\.|ave\.|blvd\.|cir\.|In\.|rd|dr|ave|blvd|cir|ln)
I had a different use case - find any addresses in logs and scold application developers (favourite part of a devops job). I had the advantage of having the word "address" in the pattern but should work without that if you have specific field to scan
\baddress.[0-9\\\/# ,a-zA-Z]+[ ,]+[0-9\\\/#, a-zA-Z]{1,}
Look for the word "address" - skip this if not applicable
Look for first part numbers, letters, #, space - Unit Number / street number/suite number/door number
Separated by a space or comma
Look for one or more of rest of address numbers, letters, #, space
Tested against :
1 Sleepy Boulevard PO, Box 65745
Suite #100 /98,North St,Snoozepura
Ave., New Jersey,
Suite 420 1130 Connect Ave., NW,
Suite 420 19 / 21 Old Avenue,
Suite 12, Springfield, VIC 3001
Suite#100/98 North St Snoozepura
This worked for me when there were street addresses with unit/suite numbers, zip codes, only street. It also didn't match IP addresses or mac addresses. Worked with extra spaces.
This assumes users are normal people separate elements of a street address with a comma, hash sign, or space and not psychopaths who use characters like "|" or ":"!
For French address and some international address too, I use it.
[\\D+ || \\d]+\\d+[ ||,||[A-Za-z0-9.-]]+(?:[Rue|Avenue|Lane|... etcd|Ln|St]+[ ]?)+(?:[A-Za-z0-9.-](.*)]?)
I was inspired from the responses given here and came with those 2 solutions
support optional uppercase
support french also
regex structure
numbers (required)
letters, chars and spaces
at least one common address keyword (required)
as many chars you want before the line break
definitions:
accuracy
capacity of detecting addresses and not something that looks like an address which is not.
range
capacity to detect uncommon addresses.
Regex 1:
high accuracy
low range
/[0-9]+[ |[a-zà-ú.,-]* ((highway)|(autoroute)|(north)|(nord)|(south)|(sud)|(east)|(est)|(west)|(ouest)|(avenue)|(lane)|(voie)|(ruelle)|(road)|(rue)|(route)|(drive)|(boulevard)|(circle)|(cercle)|(street)|(cer\.)|(cir\.)|(blvd\.)|(hway\.)|(st\.)|(aut\.)|(ave\.)|(ln\.)|(rd\.)|(hw\.)|(dr\.)|(a\.))([ .,-]*[a-zà-ú0-9]*)*/i
regex 2:
low accuracy
high range
/[0-9]*[ |[a-zà-ú.,-]* ((highway)|(autoroute)|(north)|(nord)|(south)|(sud)|(east)|(est)|(west)|(ouest)|(avenue)|(lane)|(voie)|(ruelle)|(road)|(rue)|(route)|(drive)|(boulevard)|(circle)|(cercle)|(street)|(cer\.?)|(cir\.?)|(blvd\.?)|(hway\.?)|(st\.?)|(aut\.?)|(ave\.?)|(ln\.?)|(rd\.?)|(hw\.?)|(dr\.?)|(a\.))([ .,-]*[a-zà-ú0-9]*)*/i
This one works well for me
^(\d+) ?([A-Za-z](?= ))? (.*?) ([^ ]+?) ?((?<= )APT)? ?((?<= )\d*)?$
Source : https://community.alteryx.com/t5/Alteryx-Designer-Discussions/RegEx-Addresses-different-formats-and-headaches/td-p/360147
Here is my RegEx for address, city & postal validation rules
validation rules:
address -
1 - 40 characters length.
Letters, numbers, space and . , : ' #
city -
1 - 19 characters length
Only Alpha characters are allowed
Spaces are allowed
postalCode -
The USA zip must meet the following criteria and is required:
Minimum of 5 digits (9 digits if zip + 4 is provided)
Numeric only
A Canadian postal code is a six-character string.
in the format A1A 1A1, where A is a letter and 1 is a digit.
a space separates the third and fourth characters.
do not include the letters D, F, I, O, Q or U.
the first position does not make use of the letters W or Z.
address: ^[a-zA-Z0-9 .,#;:'-]{1,40}$
city: ^[a-zA-Z ]{1,19}$
usaPostal: ^([0-9]{5})(?:[-]?([0-9]{4}))?$
canadaPostal : ^(?!.*[DFIOQU])[A-VXY][0-9][A-Z] ?[0-9][A-Z][0-9]$
\b(\d{1,8}[a-z]?[0-9\/#- ,a-zA-Z]+[ ,]+[.0-9\/#, a-zA-Z]{1,})\n
A more dynamic approach to #micah would be the following:
(?'Address'(?'Street'[0-9][a-zA-Z\s]),?\s*(?'City'[A-Za-z\s]),?\s(?'Country'[A-Za-z])\s(?'Zipcode'[0-9]-?[0-9]))
It won't care about individual lengths of segments of code.
https://regex101.com/r/nuy7hB/1