I would like to take out only the date part of my dataframe column which has data like
$83,00010/7/2016 10/7/2016 # is the date here
$721,0002/7/2015 2/7/2015 # is the date here
$3,00012/12/2015 12/12/2015 # is the date here
I tried patterns like
^(?!,\d{3}) and (\$[,\d{3}]*\,\d{3})
but I am not able to exactly pick just the date part.
This is slightly complicated, let's start with a simple expression, then we would expand it. I'm hoping that we would have , in our money part, then we might try:
,\d{3}(\d{1,2}\/.+?\/\d{4})
or
([1-3]?\d\/.+?\/\d{4})
# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility
import re
regex = r"([1-3]?\d\/.+?\/\d{4})"
test_str = ("$83,00010/7/2016 ----> 10/7/2016 is the date here,\n"
"$721,0002/7/2015 ----> 2/7/2015 is the date here,\n"
"$3,00012/12/2015 -----> 12/12/2015 is the date here")
matches = re.finditer(regex, test_str, re.MULTILINE)
for matchNum, match in enumerate(matches, start=1):
print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))
for groupNum in range(0, len(match.groups())):
groupNum = groupNum + 1
print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))
# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
jex.im visualizes regular expressions:
The following regex will return you the date:
',\d{3}(\d{1,2}/\d{1,2}/\d{4})$'
