Извлечение одинаковых групп из разных шаблонов регулярных выражений для разных форматов даты

Question

На основе фрейма данных, подобного

import pandas as pd
string_1 = 'for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001.'
string_2 = '03/25/93 Total time of visit (in minutes):'
string_3 = 'April 11, 1990 CPT Code: 90791: No medical services'
df = pd.Series([string_1,string_2,string_3])

, каждое из следующих утверждений успешно извлекает дату ровно одной строки:

print(df.str.extract(r'((?P<month>\d{1,2})[/-](?P<day>\d{1,2})[/-](?P<year>\d{2,4}))').dropna())
   0           month day year
1  03/25/93    03    25  93

print(df.str.extract(r'(?P<month>(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)[a-z\.]*)[\s\.\-\,](?P<day>\d{2})[\-\,\s]*(?P<year>\d{4})').dropna())
   month day  year
2  April  11  1990

print(df.str.extract(r'((?P<day>\d{2})\s(?P<month>(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)[a-z\.]*)[\s\.\-\,]*(?P<year>\d{4}))').dropna())
   0            day  month  year
0  24 Jan 2001  24   Jan    2001

Как можно объединить выражения для создания фрейм данных

     day   month  year
0    24    Jan    2001
1    25    03     93 
2    11    April  1990 

Где индексы должны быть исходными индексами?

Answer 1

Вы можете использовать Модуль регулярных выражений PyPi (установить с помощью pip install regex) и объединить шаблоны с ИЛИ внутри группы сброса ветви :

import regex
import pandas as pd

string_1 = 'for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001.'
string_2 = '03/25/93 Total time of visit (in minutes):'
string_3 = 'April 11, 1990 CPT Code: 90791: No medical services'
df = pd.Series([string_1,string_2,string_3])

pat1 = r'(?P<month>\d{1,2})[/-](?P<day>\d{1,2})[/-](?P<year>\d{2,4})'
pat2 = r'(?P<month>(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)[a-z.]*)[\s.,-](?P<day>\d{2})[-,\s]*(?P<year>\d{4})'
pat3 = r'(?P<day>\d{2})\s(?P<month>(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)[a-z.]*)[\s.,-]*(?P<year>\d{4})'

rx = regex.compile(r"(?|{}|{}|{})".format(pat1,pat2,pat3))

empty_val = pd.Series(["","",""], index=['month','day','year'])

def extract_regex(seq):
    m = rx.search(seq)
    if m:
        return pd.Series(list(m.groupdict().values()), index=['month','day','year'])
    else:
        return empty_val

df2 = df.apply(extract_regex)

Выход :

>>> df2
   month day  year
0    Jan  24  2001
1     03  25    93
2  April  11  1990

Answer 2

string_1 = 'for urgent evaluation/treatment till first visit with Dr. Toney Winkler IN EIGHT WEEKS on 24 Jan 2001.'
string_2 = '03/25/93 Total time of visit (in minutes):'
string_3 = 'April 11, 1990 CPT Code: 90791: No medical services'
df = pd.DataFrame([string_1,string_2,string_3])

patterns = [r'(?P<day>\d{1,2}) (?P<month>(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)) (?P<year>\d{4})',
            r'(?P<month>\d{1,2})[/-](?P<day>\d{1,2})[/-](?P<year>\d{2,4})',
            r'(?P<month>(Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sept|Oct|Nov|Dec)[a-z\.]*) (?P<day>\d{2}), (?P<year>\d{4})']


def extract_date(s):
    result = None, None, None
        for p in patterns:
        m = re.search(p, s)
        if m:
            result = m.group('year'), m.group('month'), m.group('day')
            break
    return result

df['year'], df['month'], df['day'] = zip(*df[0].apply(lambda s: extract_date(s)))