从包含大量数字的字符串中提取日期

Question

似乎有很多方法可以从字符串中提取各种格式的日期时间。但是，当字符串包含许多数字和符号时，似乎有一个问题。

下面是一个例子:

t = 'Annual Transmission Revenue Requirements and Rates Transmission Owner (Transmission Zone) Annual Transmission Revenue Requirement Network Integration Transmission Service Rate    ($/MW-Year) AE (AECO) $136,632,319 $53,775 AEP (AEP)   $1,295,660,732   $59,818.14  AP (APS) $128,000,000 $17,895 ATSI (ATSI) $659,094,666 $54,689.39 BC (BGE) $230,595,535 $35,762 ComEd, Rochelle (CE)  $702,431,433  $34,515.60 Dayton (DAY) $40,100,000 $13,295.76 Duke (DEOK) $121,250,903 $24,077 Duquesne (DLCO) $139,341,808 $51,954.44 Dominion (DOM) $1,031,382,000 $52,457.21 DPL, ODEC (DPL) $163,224,128 $42,812 East Kentucky Power Cooperative (EKPC) $83,267,903 $24,441 MAIT (METED, PENELEC) $150,858,703 $26,069.39 JCPL $135,000,000 $23,597.27  PE (PECO)  $155,439,100 $19,093 PPL, AECoop, UGI (PPL)  $435,349,329  $58,865 PEPCO, SMECO (PEPCO)   $190,876,083  $31,304.21 PS (PSEG) $1,248,819,352 $130,535.22 Rockland (RECO)  $17,724,263 $44,799 TrAILCo  $226,652,117.80  n/a  Effective June 1, 2018 '
import datefinder
m = datefinder.find_dates(t)
for match in m:
print(match)

是否有一种方法可以顺利地提取日期?如果没有更好的方法，我可以使用re来处理特定的格式。从datefinder的github来看，它似乎在一年前就被抛弃了。

Answer 1

虽然我不知道你的日期是如何格式化的，这里有一个regex解决方案，将工作与日期分隔'/'。当月和日表示为单个数字或包含前导零时，应适用于日期。

如果日期用连字符分隔，请将正则表达式的第9和第18个字符替换为连字符而不是/。(如果使用第二个print语句，替换第12和31个字符)

编辑:用更好的正则表达式添加了第二个print语句。那可能是更好的办法。

import re
mystring = r'joasidj9238nlsd93901/01/2021oijweo8939n'
print(re.findall('d{1,2}/d{1,2}/d{2,4}', mystring)) # This would probably work in most cases
print(re.findall('[0-1]{0,2}/[0-3]{0,1}d{0,1}/d{2,4}', mystring)) # This one is probably a better solution. (More protection against weirdness.)

编辑#2:这是一种拼写月份名称的方法(用全名，或3个字符的缩写)，后面跟着天，后面跟着逗号，后面跟着2或4位数字的年份。

import re
mystring = r'Jan 1, 2020'
print(re.findall(r'(?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sep(?:tember)?|Nov(?:ember)?|Dec(?:ember)?)s+d{1,2},s+d{2,4}',mystring))