python extract a file with the given regex format

Question

I was trying to extract the file which contains EOB_FILE

for example I have something like

s = "path Omega/CC/Pune/SYNTT/EOB_PROCESSED_BY_OCR/EOB_FILE/0A225618045646F2AEEFC23E74CAC253/0A225618045646F2AEEFC23E74CAC253_page1.json"

How can I get only the file name which is 0A225618045646F2AEEFC23E74CAC253_page1.json

Code I tried :

val = re.findall(r'([^.]*EOB_FILE[^.]*)', s)
val
['path Omega/CC/Pune/SYNTT/EOB_PROCESSED_BY_OCR/EOB_FILE/0A225618045646F2AEEFC23E74CAC253/0A225618045646F2AEEFC23E74CAC253_page1']

Output expected :

0A225618045646F2AEEFC23E74CAC253_page1.json

score 1 · Answer 1 · answered Mar 20 '20 at 07:10

1

import os
s = "path Omega/CC/Pune/SYNTT/EOB_PROCESSED_BY_OCR/EOB_FILE/0A225618045646F2AEEFC23E74CAC253/0A225618045646F2AEEFC23E74CAC253_page1.json"

os.path.basename(s)

os is python miscellaneous operating system interfaces. Check documentation here

answered Mar 20 '20 at 07:10

Tom Ron

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kederrac · Accepted Answer · 2020-03-20T07:27:22.177

1

you can use pathlib.Path:

from pathlib import Path

Path(s).name

output:

'0A225618045646F2AEEFC23E74CAC253_page1.json'

to check if EOB_FILE is in the path you could use:

'EOB_FILE' in Path(s).parts

or:

'EOB_FILE' in s

if 'EOB_FILE' in s:
    val = Path(s).name

edited Mar 20 '20 at 07:27

answered Mar 20 '20 at 07:22

kederrac

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python extract a file with the given regex format

2 Answers2