CODEWITHSUNDEEP
pythonregex
Jan Yonex
Jan Yonex

Reputation: 11

How to extract duration of experience from resume using Python

I have written logic to extract dates of experiences from the resume. I have extracted experiences that have this format :

01/2017 - 04/2022

01/07/2017 - 31/07/2017

March 2017 - July 2022

Here is the code :

cur_datespan = None
    next_first_date = None
    delimeter_count = 0

    for ptoken, token in zip(tokens, tokens[1:]):
        token = str(token).lower().strip()
        ptoken = str(ptoken).lower().strip()
        tokenpair = token + " " + ptoken
        # find datespanes
        if re.search("\d+", token) != None:
            dates = search_dates(tokenpair, settings={
                                 'REQUIRE_PARTS': ['month', 'year']}) or []

        
        else:
            dates = []
        for date in dates:
            if next_first_date == None:
                next_first_date = date[1]
                delimeter_count = 0
            elif delimeter_count < 6:
                cur_datespan = (next_first_date, date[1])
                next_first_date = None
            else:
                next_first_date = date[1]
                delimeter_count = 0
        if delimeter_count > 50:
            next_first_date = None
            cur_datespan = None
        delimeter_count += len(token.split(" "))
        # find skill and add to dict with associated datespan
        if token.lower() in skills:
            skillset[cur_datespan].add(token)
        elif (ptoken + " " + token).lower() in skills:
            skillset[cur_datespan].add((ptoken + " " + token).lower())

    skilldict = {}
    for datespan, skills in skillset.items():
        for skill in skills:
            if skill not in skilldict:
                skilldict[skill] = []
            if datespan != None and datespan[1].month - datespan[0].month > 0:
                skilldict[skill].append(datespan)

    return skilldict

But I couldn't extract the experiences that have these formats for example :

March-July 2020

March 2020 - Current/Present

01/07/2017-31/07/2017 (date format "day_first")

2020-2021

From/Since 2020

From March 2020 to July 2022

Upvotes: 1

Views: 2823

Answers (2)

kathir M
kathir M

Reputation: 1

import datefinder

import re

pattern = r'(\d{1,2}\s?\d{4})|(\d{4}\s?\d{1,2})|(\d{4})'

# Find all matches in the text

matches = re.findall(pattern, text)

# Extract the matched durations

durations = [] for match in matches: for group in match: if group: durations.append(group)

print(durations)

Initialize an empty list to store extracted month and year values

extracted_dates = []

Iterate over the elements in the data list

for item in durations: matches = datefinder.find_dates(item)

# Check if any matches were found
for match in matches:
    # Extract the month and year from the match
    month = match.month
    year = match.year
    
    # Append the extracted month and year to the extracted_dates list
    extracted_dates.append((month, year))

Get unique years from the extracted_dates list

unique_years = set(year for month, year in extracted_dates)

Print the extracted month and year values for each unique year

for year in unique_years: has_month = False for month, y in extracted_dates: if y == year: has_month = True print(f"Year: {year}, Month: {month if month else '-'}") if not has_month: print(f"Year: {year}, Month: -")

output:

Year: 2017, Month: 2

Year: 2019, Month: 4

Year: 2019, Month: 4

Year: 2021, Month: 3

Year: 2021, Month: 2

Year: 2022, Month: 7

Year: 2022, Month: 6

Year: 2013, Month: 12

Upvotes: 0

visitor
visitor

Reputation: 11

You can use re.findall as follows

import re
resumes = ['01/2017 - 04/2022',
'01/07/2017 - 31/07/2017',
'March 2017 - July 2022',
'March-July 2020',
'March 2020 - Current/Present',
'01/07/2017-31/07/2017',
'2020-2021',
'From/Since 2020',
'From March 2020 to July 2022]']
pattern = r'(((Jan(uary)?|Feb(ruary)?|Mar(ch)?|Apr(il)?|May|June?|July?|Aug(ust)?|Sep(tember)?|Nov(ember)?|Dec(ember)?)|(\d{1,2}\/){0,2})[- ]?\d{4}?)'

for resume in resumes:
    res = re.findall(pattern,resume)
    if len(res) > 1:
      print('from',res[0][0],'to',res[1][0])
    else:

output

from 01/2017 to 04/2022
from 01/07/2017 to 31/07/2017
from March 2017 to July 2022
-> March-July 2020
-> March 2020 - Current/Present
from 01/07/2017 to 31/07/2017
from 2020 to -2021
-> From/Since 2020
from March 2020 to July 2022

Upvotes: 1

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