I am trying to extract the Name, License No., Date Of Issue and Validity from an Image I processed using Pytesseract. I am quite a lot confused with regex but still went through few documentations and codes over the web.
I got till here:
import pytesseract
import cv2
import re
import cv2
from PIL import Image
import numpy as np
import datetime
from dateutil.relativedelta import relativedelta
def driver_license(filename):
"""
This function will handle the core OCR processing of images.
"""
i = cv2.imread(filename)
newdata=pytesseract.image_to_osd(i)
angle = re.search('(?<=Rotate: )\d ', newdata).group(0)
angle = int(angle)
i = Image.open(filename)
if angle != 0:
#with Image.open("ro2.jpg") as i:
rot_angle = 360 - angle
i = i.rotate(rot_angle, expand="True")
i.save(filename)
i = cv2.imread(filename)
# Convert to gray
i = cv2.cvtColor(i, cv2.COLOR_BGR2GRAY)
# Apply dilation and erosion to remove some noise
kernel = np.ones((1, 1), np.uint8)
i = cv2.dilate(i, kernel, iterations=1)
i = cv2.erode(i, kernel, iterations=1)
txt = pytesseract.image_to_string(i)
print(txt)
text = []
data = {
'firstName': None,
'lastName': None,
'age': None,
'documentNumber': None
}
c = 0
print(txt)
#Splitting lines
lines = txt.split('\n')
for lin in lines:
c = c 1
s = lin.strip()
s = s.replace('\n','')
if s:
s = s.rstrip()
s = s.lstrip()
text.append(s)
try:
if re.match(r".*Name|.*name|.*NAME", s):
name = re.sub('[^a-zA-Z] ', ' ', s)
name = name.replace('Name', '')
name = name.replace('name', '')
name = name.replace('NAME', '')
name = name.replace(':', '')
name = name.rstrip()
name = name.lstrip()
nmlt = name.split(" ")
data['firstName'] = " ".join(nmlt[:len(nmlt)-1])
data['lastName'] = nmlt[-1]
if re.search(r"[a-zA-Z][a-zA-Z]-\d{13}", s):
data['documentNumber'] = re.search(r'[a-zA-Z][a-zA-Z]-\d{13}', s)
data['documentNumber'] = data['documentNumber'].group().replace('-', '')
if not data['firstName']:
name = lines[c]
name = re.sub('[^a-zA-Z] ', ' ', name)
name = name.rstrip()
name = name.lstrip()
nmlt = name.split(" ")
data['firstName'] = " ".join(nmlt[:len(nmlt)-1])
data['lastName'] = nmlt[-1]
if re.search(r"[a-zA-Z][a-zA-Z]\d{2} \d{11}", s):
data['documentNumber'] = re.search(r'[a-zA-Z][a-zA-Z]\d{2} \d{11}', s)
data['documentNumber'] = data['documentNumber'].group().replace(' ', '')
if not data['firstName']:
name = lines[c]
name = re.sub('[^a-zA-Z] ', ' ', name)
name = name.rstrip()
name = name.lstrip()
nmlt = name.split(" ")
data['firstName'] = " ".join(nmlt[:len(nmlt)-1])
data['lastName'] = nmlt[-1]
if re.match(r".*DOB|.*dob|.*Dob", s):
yob = re.sub('[^0-9] ', ' ', s)
yob = re.search(r'\d\d\d\d', yob)
data['age'] = datetime.datetime.now().year - int(yob.group())
except:
pass
print(data)
I need to extract the Validity and Issue Date as well. But not getting anywhere near it. Also, I have seen using regex shortens the code like a lot so is there any better optimal way for it?
My input data is a string somewhat like this:
Transport Department Government of NCT of Delhi
Licence to Drive Vehicles Throughout India
Licence No. : DL-0820100052000 (P) R
N : PARMINDER PAL SINGH GILL
: SHRI DARSHAN SINGH GILL
DOB: 10/05/1966 BG: U
Address :
104 SHARDA APPTT WEST ENCLAVE
PITAMPURA DELHI 110034
Auth to Drive Date of Issue
M.CYL. 24/02/2010
LMV-NT 24/02/2010
(Holder's Sig natu re)
Issue Date : 20/05/2016
Validity(NT) : 19/05/2021 : c
Validity(T) : NA Issuing Authority
InvCarrNo : NA NWZ-I, WAZIRPUR
Or like this:
in
Transport Department Government of NCT of Delhi
Licence to Drive Vehicles Throughout India
2
Licence No. : DL-0320170595326 () WN
Name : AZAZ AHAMADSIDDIQUIE
s/w/D : SALAHUDDIN ALI
____... DOB: 26/12/1992 BG: O
\ \ Address:
—.~J ~—; ROO NO-25 AMK BOYS HOSTEL, J.
— NAGAR, DELHI 110025
Auth to Drive Date of Issue
M.CYL. 12/12/2017
4 wt 4
Iseue Date: 12/12/2017 a
falidity(NT) < 2037
Validity(T) : NA /
Inv CarrNo : NA te sntian sana
Note: In the second example you wouldn't get the validity, will optimise the OCR for later. Any proper guide which can help me with regex which is a bit simpler would be good.
CodePudding user response：
You can use this pattern: (?<=KEY\s*:\s*)\b[^\n] and replace KEY with one of the issues of the date, License No. and others.
Also for this pattern, you need to use regex library.
Code:
import regex
text1 = """
Transport Department Government of NCT of Delhi
Licence to Drive Vehicles Throughout India
Licence No. : DL-0820100052000 (P) R
N : PARMINDER PAL SINGH GILL
: SHRI DARSHAN SINGH GILL
DOB: 10/05/1966 BG: U
Address :
104 SHARDA APPTT WEST ENCLAVE
PITAMPURA DELHI 110034
Auth to Drive Date of Issue
M.CYL. 24/02/2010
LMV-NT 24/02/2010
(Holder's Sig natu re)
Issue Date : 20/05/2016
Validity(NT) : 19/05/2021 : c
Validity(T) : NA Issuing Authority
InvCarrNo : NA NWZ-I, WAZIRPUR
"""
for key in ('Issue Date', 'Licence No\.', 'N', 'Validity\(NT\)'):
print(regex.findall(fr"(?<={key}\s*:\s*)\b[^\n] ", text1, regex.IGNORECASE))
Output:
['20/05/2016']
['DL-0820100052000 (P) R']
['PARMINDER PAL SINGH GILL']
['19/05/2021 : c']
CodePudding user response：
You can also use re with a single regex based on alternation that will capture your keys and values:
import re
text = "Transport Department Government of NCT of Delhi\nLicence to Drive Vehicles Throughout India\n\nLicence No. : DL-0820100052000 (P) R\nN : PARMINDER PAL SINGH GILL\n\n: SHRI DARSHAN SINGH GILL\n\nDOB: 10/05/1966 BG: U\nAddress :\n\n104 SHARDA APPTT WEST ENCLAVE\nPITAMPURA DELHI 110034\n\n\n\nAuth to Drive Date of Issue\nM.CYL. 24/02/2010\nLMV-NT 24/02/2010\n\n(Holder's Sig natu re)\n\nIssue Date : 20/05/2016\nValidity(NT) : 19/05/2021 : c\nValidity(T) : NA Issuing Authority\nInvCarrNo : NA NWZ-I, WAZIRPUR"
search_phrases = ['Issue Date', 'Licence No.', 'N', 'Validity(NT)']
reg = r"\b({})\s*:\W*(. )".format( "|".join(sorted(map(re.escape, search_phrases), key=len, reverse=True)) )
print(re.findall(reg, text, re.IGNORECASE))
Output of this short online Python demo:
[('Licence No.', 'DL-0820100052000 (P) R'), ('N', 'PARMINDER PAL SINGH GILL'), ('Issue Date', '20/05/2016'), ('Validity(NT)', '19/05/2021 : c')]
The regex is
\b(Validity\(NT\)|Licence\ No\.|Issue\ Date|N)\s*:\W*(. )
See its online demo.
Details:
map(re.escape, search_phrases)- escapes all special chars in your search phrases to be used as literal texts in a regex (else,.will match any chars,?won't match a?char, etc.)sorted(..., key=len, reverse=True)- sorts the search phrases by length in descending order (to get longer matches first)"|".join(...)- creates an alternation pattern,a|b|c|...r"\b({})\s*:\W*(. )".format( ... )- creates the final regex.
Regex details
\b- a word boundary (NOTE: replace with(?m)^if your matches occur at the beginning of a line)(Validity\(NT\)|Licence\ No\.|Issue\ Date|N)- Group 1: one of the search phrases\s*- zero or more whitespaces:- a colon\W*- zero or more non-word chars(. )- (capturing) Group 2: one or more chars other than line break chars, as many as possible.