How to show full results, rather than matched text from regex searches in python

Question

I am creating a script that searches for a file based on a keyword, my output should be the whole observation, rather than just the matched text, but I'm finding .group doesn't work on this.

import re 
import os 
 
pers_info = pd.read_csv(r".....StateWorkforceMailingList_2-7-19a.csv",encoding='utf-8')

Pers_info['State'] = Texas, Florida etc... 

 files=os.listdir(r"....\State Files")
 
Files = list of WORKFORCE_2017_ALABAMA_FILE.xlsx,...,n

matches=re.findall(pers_info.State[4], files.replace("_", " "),re.I)
print(match)

My intended output is WORKFORCE_2017_ALABAMA_FILE.xlsx Instead I get 'Alabama'

Should I try a boolean mask ?

Admin · Answer

I suppose that your Pers_info looks something like this:

Pers_info = {"state": ["Texas", "Alabama", "Florida"], "somethingelse": "stuff"}

And your files like this:

files = ["WORKFORCE_2017_ALABAMA_FILE.xlsx","WORKFORCE_2017_TEXAS_FILE.xlsx","SOMETHING.xlsx"]

(You don't need regex for this)

files = [file.lower() for file in files]
peers = [file.lower() for file in Pers_info['state']]
result = []

for x in peers:
    try:
        indx = peers.index(x)

        if any(peers[indx] in s for s in files):
            result.append(files[indx])
    except:
        break
print(result)

user_11329890 · Answer

Use

>>> import pandas as pd
>>> Pers_info = pd.DataFrame({'State':['Texas', 'Alabama', 'Florida']})
>>> Files = ['WORKFORCE_2017_ALABAMA_FILE.xlsx', 'WORKFORCE_2017_FILE.xlsx']
>>> pattern = re.compile(rf'(?>> list(filter(pattern.search, Files))
['WORKFORCE_2017_ALABAMA_FILE.xlsx']

See regex proof.

EXPLANATION

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How to show full results, rather than matched text from regex searches in python

2 Answers