Python Regex Match第一个日期，然后第二个

Question

I am working with pandas dataframes and want to create to series, Start Date and End Date from dates within the Description. I am using regex to find the occurrences of dates but can't seem to find out how to stop at the first date and then continue to find the second date.

Looking here: How to stop at first occurence of match?

yielded an answer

(?s)(\d{1,2}/\d{1,2}/\d{2,4}).*

But this didn't work for me, I still was capturing all dates instead of only the first.

Using

(\d{1,2}/\d{1,2}/\d{2,4})? 

didn't work either.

Essentially I am trying to get at

pattern_generic=re.compile('(\d{1,2}\/\d{1,2}\/\d{2,4})')   #perhaps will do start and end)
report['Start Date'] = report['Description'].apply(lambda x: re.findall(pattern_start,x))
report['End Date'] = report['Description'].apply(lambda x: re.findall(pattern_end,x))

Not sure if this is the best way to approach finding the first and second date and putting them into columns. Any help/advice is appreciated!

Edit:

Example to clarify: I have a dataframe with a column titled 'Description' with various items such as 'Purchased subscription from 1/2/13-3/4/15'. I want to capture the two dates into two columns, Start and End

 Description                                       Start Date     End Date
 'Purchased Subscription from 1/2/13-3/4/15'        1/2/13        3/4/15

Answer 1

I'd use this:

(?s)\b(\d{1,2}/\d{1,2}/\d{2,4})\b-\b(\d{1,2}/\d{1,2}/\d{2,4})\b

The start date will be in group 1 and the end date in group 2.

Answer 2

You could use the below regex,

(?s)(\d{1,2}/\d{1,2}/\d{2,4})-(\d{1,2}/\d{1,2}/\d{2,4}).*

DEMO

Assign the characters inside group index 1 to Start Date and group index 2 to End Date

>>> s = """'Purchased Subscription from 1/2/13-3/4/15'        1/2/13        3/4/15
foo 1/2/13-3/4/15'        5/2/13        6/4/15
1/2/13-3/4/15'        7/2/13        8/4/15
1/2/13-3/4/15'        9/2/13        10/4/15"""
>>> m = re.search(r'(?s)(\d{1,2}\/\d{1,2}\/\d{2,4})-(\d{1,2}\/\d{1,2}\/\d{2,4}).*', s)
>>> m.group(1)
'1/2/13'
>>> m.group(2)
'3/4/15'
>>> m = re.findall(r'(\d{1,2}\/\d{1,2}\/\d{2,4})-(\d{1,2}\/\d{1,2}\/\d{2,4}).*', s, re.DOTALL)
>>> m
[('1/2/13', '3/4/15')]

Answer 3

  .*'\s+(\d+\/\d+\/\d+)\s+(\d+\/\d+\/\d+)

Try this.Start dates will be in group1 and end dates in group2.

See Demo:

http://regex101.com/r/zN5mL9/1

Answer 4

Here is the code I used to fully solve my problem:

data['End Date'] = ''
data['Start Date']=''

pattern=re.compile('(?s)(\d{1,2}\/\d{1,2}\/\d{2,4}).*?(\d{1,2}\/\d{1,2}\/\d{2,4}).*')

first_list = []
second_list = []
for x in data['Product Description']:
  m = re.search(pattern,x)
 if m is None:
      first_list.append('')
      second_list.append('')
 else:
  first_list.append(m.group(1))
  second_list.append(m.group(2))


data['Start Date'] = Series(first_list)
data['End Date'] = Series(second_list)