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Insert space after the second or third capital letter python

I have a pandas dataframe containing addresses. Some are formatted correctly like 481 Rogers Rd York ON. Others have a space missing between the city quandrant and the city name, for example: 101 9 Ave SWCalgary AB or even possibly: 101 9 Ave SCalgary AB, where SW refers to south west and S to south.
I'm trying to find a regex that will add a space between second and third capital letters if they are followed by lowercase letters, or if there are only 2 capitals followed by lower case, add a space between the first and second.
So far, I've found that ([A-Z]{2,3}[a-z]) will match the situation correctly, but I can't figure out how to look back into it and sub at position 2 or 3. Ideally, I'd like to use an index to split the match at [-2:] but I can't figure out how to do this.
I found that re.findall('(?<=[A-Z][A-Z])[A-Z][a-z].+', '101 9 Ave SWCalgary AB')
will return the last part of the string and I could use a look forward regex to find the start and then join them but this seems very inefficient.
Thanks

You may use
df['Test'] = df['Test'].str.replace(r'\b([A-Z]{1,2})([A-Z][a-z])', r'\1 \2')
See this regex demo
Details
\b - a word boundary
([A-Z]{1,2}) - Capturing group 1 (later referred with \1 from the replacement pattern): one or two uppercase letters
([A-Z][a-z]) - Capturing group 2 (later referred with \2 from the replacement pattern): an uppercase letter + a lowercase one.
If you want to specifically match city quadrants, you may use a bit more specific regex:
df['Test'] = df['Test'].str.replace(r'\b([NS][EW]|[NESW])([A-Z][a-z])', r'\1 \2')
See this regex demo. Here, [NS][EW]|[NESW] matches N or S that are followed with E or W, or a single N, E, S or W.
Pandas demo:
import pandas as pd
df = pd.DataFrame({'Test':['481 Rogers Rd York ON',
'101 9 Ave SWCalgary AB',
'101 9 Ave SCalgary AB']})
>>> df['Test'].str.replace(r'\b([A-Z]{1,2})([A-Z][a-z])', r'\1 \2')
0 481 Rogers Rd York ON
1 101 9 Ave SW Calgary AB
2 101 9 Ave S Calgary AB
Name: Test, dtype: object

You can use
([A-Z]{1,2})(?=[A-Z][a-z])
to capture the first (or first and second) capital letters, and then use lookahead for a capital letter followed by a lowercase letter. Then, replace with the first group and a space:
re.sub(r'([A-Z]{1,2})(?=[A-Z][a-z])', r'\1 ', str)
https://regex101.com/r/TcB4Ph/1

regular expression to exclude 2 consecutive capital letters

I'm having difficulty using regex to solve this expression,
e.g when given below:
regex_exp(address, "OG 56432")
It should return
"OG 56432: Middle Street Pollocksville | 686"
address is an array of strings:
address = [
"622 Gordon Lane St. Louisville OH 52071",
"432 Main Long Road St. Louisville OH 43071",
"686 Middle Street Pollocksville OG 56432"
]
My solution currently looks like this (Python):
import re
def regex_exp(address, zipcode):
for i in address:
if zipcode in i:
postal_code = (re.search("[A-Z]{2}\s[0-9]{5}", x)).group(0)
# returns "OG 56432"
digits = (re.search("\d+", x)).group(0)
# returns "686"
address = (re.search("\D+", x)).group(0)
# returns "Middle Street Pollocksville OG"
print(postal_code + ":" + address + "| " + digits)
regex_exp(address, "OG 56432")
# returns OG 56432: High Street Pollocksville OG | 686
As you can see from my second paragraph, this is not the correct answer - I need the returned value to be
"OG 56432: Middle Street Pollocksville | 686"
How do I manipulate my address variable Regex search to exclude the 2 capital consecutive capital letters? I've tried things like
address = (re.search("?!\D+", x)).group(0)
to remove the two consecutive capitals based on A regular expression to exclude a word/string but I think this is a step in the wrong direction.
PS: I understand there are easier methods to solve this, but I want to use regex to improve my fundamentals

If you just want to remove the two consecutive Capital Letters which are predecessor of zip-code(a 5 digit number) then use this
import re
text = "432 Main Long PC Market Road St. Louisville OG 43071"
address = re.sub(r'([A-Z]{2}[\s]{1})(?=[\d]{5})','',text)
print(address)
# Output: 432 Main Long PC Market Road St. Louisville 43071
For removing all occurrences of two consecutive Capital Letters:
import re
text = "432 Main Long PC Market Road St. Louisville OG 43071"
address = re.sub(r'([A-Z]{2}[\s]{1})(?=[\d]{5})','',text)
print(address)
# Output: 432 Main Long Market Road St. Louisville 43071

With re.sub() and group capturing you can use:
s="686 Middle Street Pollocksville OG 56432"
re.sub(r"(\d+)(.*)\s+([A-Z]+\s+\d+)",r"\3: \2 | \1",s)
Out: 'OG 56432: Middle Street Pollocksville | 686'

How can I use a regex function to extract european streetnames, housing numbers and adjectives with pandas?

I have a dataframe with 6000 records and need to extract/split the column with streetname into: "Streetname", "Housingnumber" and "Adjectives". Unfortunately, the problem is not solved yet using regex functions because there is no structure in the notation of df["streetname"]:
**Input from df["Streetname"]**
St. edward's Lane 26
Vineyardlane3a
High Street 0-9
ParkRoad near #33
Queens Road ??
s-Georgelane9abc
Kings Road 9b
1st Park Avenue 67 near cyclelane
**Output that I would like:
df["Street"] df["housingnumber"] df["adjective"]**
St. Edward's lane 26
Vineyardlane 3 a
High Street 0-9
ParkRoad 33
Queens Road
s-Georgelane 9 abc
Kings Road 9 b
1st Park Avenue 67
I tried this:
Filter = r'(?P<S>.*)(?P<H>\s[0-9].*)'
df["Streetname"] = df["Streetname"].str.extract(Filter)
I lose a lot of data and the result is only written into one column... Hope that someone can help!

Not 100% perfect (I doubt that this will be possible without a database or machine learning algorithms) but a starting point:
^ # start of line/string
(?P<street>\w+?\D+) # [a-zA-Z0-9_]+? followed by not a number
(?P<nr>\d*[-\d]*) # a digit, followed by - and other digits, eventually
(?P<adjective>[a-zA-Z]*) # a-z
.* # consume the rest of the string
See a demo on regex101.com.
You might want to strip of #, whitespaces or ? from the end of street afterwards.

How can I extract address from raw text using NLTK in python?

I have this text
'''Hi, Mr. Sam D. Richards lives here, 44 West 22nd Street, New
York, NY 12345. Can you contact him now? If you need any help, call
me on 12345678'''
. How the address part can be extracted from the above text using NLTK? I have tried Stanford NER Tagger, which gives me only New York as Location. How to solve this?

Definitely regular expressions :)
Something like
import re
txt = ...
regexp = "[0-9]{1,3} .+, .+, [A-Z]{2} [0-9]{5}"
address = re.findall(regexp, txt)
# address = ['44 West 22nd Street, New York, NY 12345']
Explanation:
[0-9]{1,3}: 1 to 3 digits, the address number
(space): a space between the number and the street name
.+: street name, any character for any number of occurrences
,: a comma and a space before the city
.+: city, any character for any number of occurrences
,: a comma and a space before the state
[A-Z]{2}: exactly 2 uppercase chars from A to Z
[0-9]{5}: 5 digits
re.findall(expr, string) will return an array with all the occurrences found.

Pyap works best not just for this particular example but also for other addresses contained in texts.
text = ...
addresses = pyap.parse(text, country='US')

Checkout libpostal, a library dedicated to address extraction
It cannot extract address from raw text but may help in related tasks

For US address extraction from bulk text:
For US addresses in bulks of text I have pretty good luck, though not perfect with the below regex. It wont work on many of the oddity type addresses and only captures first 5 of the zip.
Explanation:
([0-9]{1,6}) - string of 1-5 digits to start off
(.{5,75}) - Any character 5-75 times. I looked at the addresses I was interested in and the vast vast majority were over 5 and under 60 characters for the address line 1, address 2 and city.
(BIG LIST OF AMERICAN STATS AND ABBERVIATIONS) - This is to match on states. Assumes state names will be Title Case.
.{1,2} - designed to accomodate many permutations of ,/s or just /s between the state and the zip
([0-9]{5}) - captures first 5 of the zip.
text = "is an individual maintaining a residence at 175 Fox Meadow, Orchard Park, NY 14127. 2. other,"
address_regex = r"([0-9]{1,5})(.{5,75})((?:Ala(?:(?:bam|sk)a)|American Samoa|Arizona|Arkansas|(?:^(?!Baja )California)|Colorado|Connecticut|Delaware|District of Columbia|Florida|Georgia|Guam|Hawaii|Idaho|Illinois|Indiana|Iowa|Kansas|Kentucky|Louisiana|Maine|Maryland|Massachusetts|Michigan|Minnesota|Miss(?:(?:issipp|our)i)|Montana|Nebraska|Nevada|New (?:Hampshire|Jersey|Mexico|York)|North (?:(?:Carolin|Dakot)a)|Ohio|Oklahoma|Oregon|Pennsylvania|Puerto Rico|Rhode Island|South (?:(?:Carolin|Dakot)a)|Tennessee|Texas|Utah|Vermont|Virgin(?:ia| Island(s?))|Washington|West Virginia|Wisconsin|Wyoming|A[KLRSZ]|C[AOT]|D[CE]|FL|G[AU]|HI|I[ADLN]|K[SY]|LA|M[ADEINOST]|N[CDEHJMVY]|O[HKR]|P[AR]|RI|S[CD]|T[NX]|UT|V[AIT]|W[AIVY])).{1,2}([0-9]{5})"
addresses = re.findall(address_regex, text)
addresses is then: [('175', ' Fox Meadow, Orchard Park, ', 'NY', '', '14127')]
You can combine these and remove spaces like so:
for address in addresses:
out_address = " ".join(address)
out_address = " ".join(out_address.split())
To then break this into a proper line 1, line 2 etc. I suggest using an address validation API like Google or Lob. These can take a string and break it into parts. There are also some python solutions for this like usaddress

FInd a US street address in text (preferably using Python regex)

Disclaimer: I read very carefully this thread:
Street Address search in a string - Python or Ruby
and many other resources.
Nothing works for me so far.
In some more details here is what I am looking for is:
The rules are relaxed and I definitely am not asking for a perfect code that covers all cases; just a few simple basic ones with assumptions that the address should be in the format:
a) Street number (1...N digits);
b) Street name : one or more words capitalized;
b-2) (optional) would be best if it could be prefixed with abbrev. "S.", "N.", "E.", "W."
c) (optional) unit/apartment/etc can be any (incl. empty) number of arbitrary characters
d) Street "type": one of ("st.", "ave.", "way");
e) City name : 1 or more Capitalized words;
f) (optional) state abbreviation (2 letters)
g) (optional) zip which is any 5 digits.
None of the above needs to be a valid thing (e.g. an existing city or zip).
I am trying expressions like these so far:
pat = re.compile(r'\d{1,4}( \w+){1,5}, (.*), ( \w+){1,5}, (AZ|CA|CO|NH), [0-9]{5}(-[0-9]{4})?', re.IGNORECASE)
>>> pat.search("123 East Virginia avenue, unit 123, San Ramondo, CA, 94444")
Don't work, and for me it's not easy to understand why. Specifically: how do I separate in my pattern a group of any words from one of specific words that should follow, like state abbrev. or street "type ("st., ave.)?
Anyhow: here is an example of what I am hoping to get:
Given
def ex_addr(text):
# does the re magic
# returns 1st address (all addresses?) or None if nothing found
for t in [
'The meeting will be held at 22 West Westin st., South Carolina, 12345 on Nov.-18',
'The meeting will be held at 22 West Westin street, SC, 12345 on Nov.-18',
'Hi there,\n How about meeting tomorr. #10am-sh in Chadds # 123 S. Vancouver ave. in Ottawa? \nThanks!!!',
'Hi there,\n How about meeting tomorr. #10am-sh in Chadds # 123 S. Vancouver avenue in Ottawa? \nThanks!!!',
'This was written in 1999 in Montreal',
"Cool cafe at 420 Funny Lane, Cupertino CA is way too cool",
"We're at a party at 12321 Mammoth Lane, Lexington MA 77777; Come have a beer!"
] print ex_addr(t)
I would like to get:
'22 West Westin st., South Carolina, 12345'
'22 West Westin street, SC, 12345'
'123 S. Vancouver ave. in Ottawa'
'123 S. Vancouver avenue in Ottawa'
None # for 'This was written in 1999 in Montreal',
"420 Funny Lane, Cupertino CA",
"12321 Mammoth Lane, Lexington MA 77777"
Could you please help?

I just ran across this in GitHub as I am having a similar problem. Appears to work and be more robust than your current solution.
https://github.com/madisonmay/CommonRegex
Looking at the code, the regex for street address accounts for many more scenarios. '\d{1,4} [\w\s]{1,20}(?:street|st|avenue|ave|road|rd|highway|hwy|square|sq|trail|trl|drive|dr|court|ct|parkway|pkwy|circle|cir|boulevard|blvd)\W?(?=\s|$)'

\d{1,4}( \w+){1,5}, (.*), ( \w+){1,5}, (AZ|CA|CO|NH), [0-9]{5}(-[0-9]{4})?
In this regex, you have one too many spaces (before ( \w+){1,5}, which already begins with one). Removing it, it matches your example.
I don't think you can assume that a "unit 123" or similar will be there, or there might be several ones (e.g. "building A, apt 3"). Note that in your initial regex, the . might match , which could lead to very long (and unwanted) matches.
You should probably accept several such groups with a limitation on the number (e.g. replace , (.*) with something like (, [^,]{1,20}){0,5}.
In any case, you will probably never get something 100% accurate that will accept any variation people might throw at them. Do lots of tests! Good luck.