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Python, pandas replace entire column with regex match of string

I'm using pandas to analyze data from 3 different sources, which are imported into dataframes and require modification to account for human error, as this data was all entered by humans and contains errors.
Specifically, I'm working with street names. Until now, I have been using .str.replace() to remove street types (st., street, blvd., ave., etc.), as shown below. This isn't working well enough, and I decided I would like to use regex to match a pattern, and transform that entire column from the original street name, to the pattern matched by regex.
df['street'] = df['street'].str.replace(r' avenue+', '', regex=True)
I've decided I would like to use regex to identify (and remove all other characters from the address column's fields): any number of integers, followed by a space, and then the first 3 number of alphabetic characters.
For example, "3762 pearl street" might become "3762 pea" if x is 3 with the following regex:
(\d+ )+\w{0,3}
How can I use panda's .str.replace to do this? I don't want to specify WHAT I want to replace with the second argument. I want to replace the original string with the pattern matched from regex.
Something that, in my mind, might work like this:
df['street'] = df['street'].str.replace(ORIGINAL STRING, r' (\d+ )+\w{0,3}, regex=True)
which might make 43 milford st. into "43 mil".
Thank you, please let me know if I'm being unclear.

you could use the extract method to overwrite the column with its own content
pat = r'(\d+\s[a-zA-Z]{3})'
df['street'] = df['street'].str.extract(pat)
Just an observation: The regex you shared (\d+ )+\w{0,3} matches the following patterns and returns some funky stuff as well
1131 1313 street
121 avenue
1 1 1 1 1 1 avenue
42
I've changed it up a bit based on what you described, but i'm not sure if that works for all your datapoints.

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.

Pandas regex replace with multiple values and spaces in the values

I have the following Pandas code where I am trying to replace the names of countries with the string <country>.
df['title_type2'] = df['title_type']
countries = open(r'countries.txt').read().splitlines() # Reads all lines into a list and removes \n.
countries = [country.replace(' ', r'\s') for country in countries]
pattern = r'\b' + '|'.join(countries) + r'\b'
df['title_type2'].str.replace(pattern, '<country>')
However I can't get countries with spaces (like South Korea) to work correctly, since they do not get replaced. The problem seems to be that my \s is turning into \\s. How can I avoid this or how can I fix the issue?

There is no need to replace any space with \s.
Your pattern should rather include:
\b - "starting" word boundary,
(?:...|...|...) a non-capturing group with country names (alternatives),
\b - "ending" word boundary,
something like:
pattern = r'\b(?:China|South Korea|Taiwan)\b'
Then you can do the replacement:
df['title_type2'].str.replace(pattern, '<country>')
I created test data as follows:
df = pd.DataFrame(['Abc Taiwan', 'Xyz China', 'Zxx South Korea', 'No country name'],
columns=['title_type'])
df['title_type2'] = df['title_type']
and got:
0 Abc <country>
1 Xyz <country>
2 Zxx <country>
3 No country name
Name: title_type2, dtype: object

Match words only if preceded by specific pattern

I have a string from a NWS bulletin:
LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
KHNX 141001 RECHNX Weather Service San Joaquin Valley
My aim is to extract a couple fields with regular expressions. In the first string I want "AAD" and from the second string I want "RECHNX". I have tried:
( )\w{3} #for the first string
and
\w{6} #for the 2nd string
But these find all 3 and 6 character strings leading up to the string I want.

Assuming the fields you want to extract are always in capital letters and preceded by 6 digits and a space, this regular expression would do the trick:
(?<=\d{6}\s)[A-Z]+
Demo: https://regex101.com/r/dsDHTs/1
Edit: if you want to match up to two alpha-numeric uppercase words preceded by 6 digits, you can use:
(?<=\d{6}\s)([A-Z0-9]+\b)\s(?:([A-Z0-9]+\b))*
Demo: https://regex101.com/r/dsDHTs/5
If you have a specific list of valid fields, you could also simply use:
(AAD|TMLB|RECHNX|RR4HNX)
https://regex101.com/r/dsDHTs/3

Since the substring you want to extract is a word that follows a number, separated by a space, you can use re.search with the following regex (given your input stored in s):
re.search(r'\b\d+ (\w+)', s).group(1)

To read first groups of word chars from each line, you can use a pattern like
(\w+) (\w+) (\w+) (\w+).
Then, from the first line read group No 4 and from the second line read group No 3.
Look at the following program. It prints four groups from each source line:
import re
txt = """LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
KHNX 141001 RECHNX Weather Service San Joaquin Valley"""
n = 0
pat = re.compile(r'(\w+) (\w+) (\w+) (\w+)')
for line in txt.splitlines():
n += 1
print(f'{n:2}: {line}')
mtch = pat.search(line)
if mtch:
gr = [ mtch.group(i) for i in range(1, 5) ]
print(f' {gr}')
The result is:
1: LTUS41 KCAR 141558 AAD TMLB Forecast for the National Parks
['LTUS41', 'KCAR', '141558', 'AAD']
2: KHNX 141001 RECHNX Weather Service San Joaquin Valley
['KHNX', '141001', 'RECHNX', 'Weather']

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