我如何从python中的字符串中提取月份和年份？

Question

输入文字：text = "Wipro Limited | Hyderabad, IN                Dec 2017 – PresentProject Analyst Infosys | Delhi, IN                Apr 2017 – Nov 2017 Software Developer HCL Technologies | Hyderabad, IN                Jun 2016 – Mar 2017 Software Engineer  "我已经为此编写了一个代码，但它显示在每个提取的单词的列表中并且无法执行任何操作。regex = re.compile('(?P<month>[a-zA-Z]+)\s+(?P<year>\d{4})\s+\–\s+(?P<month1>[a-zA-Z]+)\s+(?P<year1>\d{4})')mat = re.findall(regex, text)mat查看代码：https ://regex101.com/r/mMlgYp/1 。我希望像下面这样的输出可以预览日期并对其进行区分，然后计算总经验：此处 Present 或 Till date 应该考虑当前的月份和年份。import timePresent = time.strftime("%m-%Y")Present # output: '05-2020'#Desired outputExtracted dates: [('Dec 2017 - Present'), ('Apr 2017 - Nov 2017'), ('Jun 2016 - Mar 2017')]# and so on ...should display all the search results First experience: 1.9 years second experience: 8 monthsthird experience: 7 months# and so on ...should display all the search results Total experience: 3.4 years请帮助我，我是编程语言和 NLP、正则表达式方面的新手。

Answer 1

您可能最终希望在数据框中使用它，因为您将其标记为 pandas（请参阅Andrej 的回答），但无论哪种方式，您都可以使用内插法从字符串中解析日期：

fr"(?i)((?:{months}) *\d{{4}}) *(?:-|–) *(present|(?:{months}) *\d{{4}})"

{months}所有可能的月份名称和缩写的交替组在哪里。

import calendar

import re

from datetime import datetime

from dateutil.relativedelta import relativedelta

text = """Wipro Limited | Hyderabad, IN                Dec 2017 – Present

Project Analyst 

Infosys | Delhi, IN                Apr 2017 – Nov 2017 

Software Developer 

HCL Technologies | Hyderabad, IN                Jun 2016 – Mar 2017 

Software Engineer  

"""

def parse_date(x, fmts=("%b %Y", "%B %Y")):

    for fmt in fmts:

        try:

            return datetime.strptime(x, fmt)

        except ValueError:

            pass

months = "|".join(calendar.month_abbr[1:] + calendar.month_name[1:])

pattern = fr"(?i)((?:{months}) *\d{{4}}) *(?:-|–) *(present|(?:{months}) *\d{{4}})"

total_experience = None

for start, end in re.findall(pattern, text):

    if end.lower() == "present":

        today = datetime.today()

        end = f"{calendar.month_abbr[today.month]} {today.year}"

    duration = relativedelta(parse_date(end), parse_date(start))

    if total_experience:

        total_experience += duration

    else: 

        total_experience = duration

    print(f"{start}-{end} ({duration.years} years, {duration.months} months)")

if total_experience:

    print(f"total experience:  {total_experience.years} years, {total_experience.months} months")

else:

    print("couldn't parse text")

输出：

Dec 2017-May 2020 (2 years, 5 months)

Apr 2017-Nov 2017 (0 years, 7 months)

Jun 2016-Mar 2017 (0 years, 9 months)

total experience:  3 years, 9 months

Answer 2

import re

import numpy as np

import pandas as pd

text = '''Wipro Limited | Hyderabad, IN                Dec 2017 – Present

Project Analyst

Infosys | Delhi, IN                Apr 2017 – Nov 2017

Software Developer

HCL Technologies | Hyderabad, IN                Jun 2016 – Mar 2017

Software Engineer

'''

def pretty_format(monthts):

    return f'{monthts/12:.1f} years' if monthts > 11 else f'{monthts:.1f} months'

data = []

for employer, d1, d2 in re.findall(r'(.*?)\s*\|.*([A-Z][a-z]{2} [12]\d{3}) – (?:([A-Z][a-z]{2} [12]\d{3})|Present)', text):

    data.append({'Employer': employer, 'Begin': d1, 'End': d2 or np.nan})

df = pd.DataFrame(data)

df['Begin'] = pd.to_datetime(df['Begin'])

df['End'] = pd.to_datetime(df['End'])

df['Experience'] = ((df['End'].fillna(pd.to_datetime('now')) - df['Begin']) / np.timedelta64(1, 'M')).apply(pretty_format)

print(df)

total = np.sum(df['End'].fillna(pd.to_datetime('now')) - df['Begin']) / np.timedelta64(1, 'M')

print()

print(f'Total experience = {pretty_format(total)}')

印刷：

           Employer      Begin        End  Experience

0     Wipro Limited 2017-12-01        NaT   2.5 years

1           Infosys 2017-04-01 2017-11-01  7.0 months

2  HCL Technologies 2016-06-01 2017-03-01  9.0 months

Total experience = 3.8 years