I want to extract the below pattern from the dataframe:
Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009
I have written the below code to extract it:
d4=df.str.extractall(r'((?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[a-z][?:]*)((?:\d{1,2}(?:th|st|nd|rd)[,?:])\d{4})')
Unfortunately, it is not able to extract anything.
I assume that your date format would only be: MMM DDst/nd/rd/th, YYYY, thus Mar 01st, 2009 instead of Mar 1st, 2009. The following regex should work well.
\b(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) (?:[0-3][1]st|[0-2][2]nd|[0-2][3]rd|[1-3][0]th|[0-2][4-9]th), \d{4}
I saw multiple problems/doubts with your pattern, so I just rewrote it from the start as this:
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)\s+\d{1,2}(?:th|st|nd|rd),\s+\d{4}
Here is an explanation of the pattern:
(?:Jan|Feb|...|Dec) match, but do not capture, the abbreviated month name
\s+ one or more spaces
\d{1,2} day as one or two digits
(?:th|st|nd|rd) match, but do not capture, day quantifier
\s+ one or more spaces
\d{4} match a four digit year
Full code:
my_str = 'Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009'
match = re.findall(r'(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)\s+\d{1,2}(?:th|st|nd|rd),\s+\d{4}', my_str)
for item in match:
print(item)
It needs some whitespaces.
((?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec))\s+((?:\d{1,2}(?:th|st|nd|rd)[,?:])\s+\d{4})
( # (1 start)
(?:
Jan
| Feb
| Mar
| Apr
| May
| Jun
| Jul
| Aug
| Sep
| Oct
| Nov
| Dec
)
) # (1 end)
\s+
( # (2 start)
(?:
\d{1,2}
(?: th | st | nd | rd )
[,?:]
)
\s+
\d{4}
) # (2 end)
You can use re.split.
Regex: ;\s
Details:
\s Matches any whitespace characterPython code:
def Split(text):
return re.split(r';\s', text)
print(Split("Mar 20th, 2009; Mar 21st, 2009; Mar 22nd, 2009"))
Output:
['Mar 20th, 2009', 'Mar 21st, 2009', 'Mar 22nd, 2009;']
