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matching content creating new column

Hello I have a dataset where I want to match my keyword with the location. The problem I am having is the location "Afghanistan" or "Kabul" or "Helmund" I have in my dataset appears in over 150 combinations including spelling mistakes, capitalization and having the city or town attached to its name. What I want to do is create a separate column that returns the value 1 if any of these characters "afg" or "Afg" or "kab" or "helm" or "are contained in the location. I am not sure if upper or lower case makes a difference.
For instance there are hundreds of location combinations like so: Jegdalak, Afghanistan, Afghanistan,Ghazni♥, Kabul/Afghanistan,
I have tried this code and it is good if it matches the phrase exactly but there is too much variation to write every exception down
keywords= ['Afghanistan','Kabul','Herat','Jalalabad','Kandahar','Mazar-i-Sharif', 'Kunduz', 'Lashkargah', 'mazar', 'afghanistan','kabul','herat','jalalabad','kandahar']
#how to make a column that shows rows with a certain keyword..
def keyword_solution(value):
strings = value.split()
if any(word in strings for word in keywords):
return 1
else:
return 0
taleban_2['keyword_solution'] = taleban_2['location'].apply(keyword_solution)
# below will return the 1 values
taleban_2[taleban_2['keyword_solution'].isin(['1'])].head(5)
Just need to replace this logic where all results will be put into column "keyword_solution" that matches either "Afg" or "afg" or "kab" or "Kab" or "kund" or "Kund"

Given the following:
Sentences from the New York Times
Remove all non-alphanumeric characters
Change everything to lowercase, thereby removing the need for different word variations
Split the sentence into a list or set. I used set because of the long sentences.
Add to the keywords list as needed
Matching words from two lists
'afgh' in ['afghanistan']: False
'afgh' in 'afghanistan': True
Therefore, the list comprehension searches for each keyword, in each word of word_list.
[True if word in y else False for y in x for word in keywords]
This allows the list of keywords to be shorter (i.e. given afgh, afghanistan is not required)
import re
import pandas as pd
keywords= ['jalalabad',
'kunduz',
'lashkargah',
'mazar',
'herat',
'mazar',
'afgh',
'kab',
'kand']
df = pd.DataFrame({'sentences': ['The Taliban have wanted the United States to pull troops out of Afghanistan Turkey has wanted the Americans out of northern Syria and North Korea has wanted them to at least stop military exercises with South Korea.',
'President Trump has now to some extent at least obliged all three — but without getting much of anything in return. The self-styled dealmaker has given up the leverage of the United States’ military presence in multiple places around the world without negotiating concessions from those cheering for American forces to leave.',
'For a president who has repeatedly promised to get America out of foreign wars, the decisions reflect a broader conviction that bringing troops home — or at least moving them out of hot spots — is more important than haggling for advantage. In his view, decades of overseas military adventurism has only cost the country enormous blood and treasure, and waiting for deals would prolong a national disaster.',
'The top American commander in Afghanistan, Gen. Austin S. Miller, said Monday that the size of the force in the country had dropped by 2,000 over the last year, down to somewhere between 13,000 and 12,000.',
'“The U.S. follows its interests everywhere, and once it doesn’t reach those interests, it leaves the area,” Khairullah Khairkhwa, a senior Taliban negotiator, said in an interview posted on the group’s website recently. “The best example of that is the abandoning of the Kurds in Syria. It’s clear the Kabul administration will face the same fate.”',
'afghan']})
# substitute non-alphanumeric characters
df['sentences'] = df['sentences'].apply(lambda x: re.sub('[\W_]+', ' ', x))
# create a new column with a list of all the words
df['word_list'] = df['sentences'].apply(lambda x: set(x.lower().split()))
# check the list against the keywords
df['location'] = df.word_list.apply(lambda x: any([True if word in y else False for y in x for word in keywords]))
# final
print(df.location)
0 True
1 False
2 False
3 True
4 True
5 True
Name: location, dtype: bool

Rearrangement of string conditionally?

I have spent the last 2 months working on a script that cleans, formats, and geocodes addresses. It is quite successful right now, however, there are some addresses that are giving me problems.
For example:
Addresses such as 7TH AVENUE 530 xxxxxxxxxxx are causing my geolocation module to fail. You can assume that the x's are other text. The other text isn't causing errors, it's purely due to the street number coming after avenue. I currently have filters in my program to truncate the address after street suffixes, such as avenue, street, etc. Due to this the program will ultimately only send 7th avenue to the cleaning module, which isn't accurate.
How can I account for instances where there are a group of numbers immediately preceding the street suffix and then move them to the front of the address. Then I can continue as I already am and truncate the string after the suffix.
You can assume that I have a list of all of the street suffixes named patterns.
Thank you. Any help is greatly appreciated.
FURTHER CLARIFICATION: I would only need to perform this rearrangement of the string if the group of numbers was 3 digits or less, because the zip code will frequently come after the address suffix, and in cases like that I wouldn't want to rearrange the string.

I am not sure if this helps, but you can start with this:
import re
address = '7TH AVENUE 530 xxxxxxxxxxx'
m = re.search('(?<=AVENUE )\d{1,3}', address)
print (m.group(0))
>>> 530
Edit based on your comment:
import re
original = '7TH AVENUE 530 xxxxxxxxxxx'
patterns = ['street', 'avenue', 'road', 'place']
regex = r'(.*)(' + '|'.join(patterns) + r')(.*)'
address = re.sub(regex, r'\1\2', original.lower()).lstrip()
new_addr = re.search(r'(?<=%s )\d{1,3} ' % address, original.lower())
resulting_address = ' '.join([new_addr.group(0).strip(),address]) if new_addr else address
address = ' '.join(map(str.capitalize, resulting_address.split()))

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

Collect last word from varied length string for filtering, auto labeling and column population?

I have been looking and reading for a few days about how to collect the last word from a varied length string. I have found lots of postings about how to collect/split the last word, but none of the content I have read addresses varied length stings.
I would like to use this function for column population, automated labeling and content filtering from inside either the field calculator or label expression interfaces.
String examples: Morgan County, Johnson Parrish, John Brown County, Rick de la Rosa City, Big Wild life Area.
Output example : County, Parish, City, Area
Nothing I have tried has worked 100%. The following code just about works, and would work if all my strings were two words long: s.split( " " )[1:][0]
I am using arcmap 10.2 / python

How about this:
# example comma separated string ending with a period
s = "Morgan County, Johnson Parrish, John Brown County, Rick de la Rosa City, Big Wild life Area."
# output list
out = []
for pair in s.replace('.', '').split(', '):
out.append(pair.split(' ')[-1])
print out
Which results in: ['County', 'Parrish', 'County', 'City', 'Area']

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.