1import re
2import pandas as pd

1if re.search("Daisy","Daisy is a dachshund dog. Daisy is beautiful."):
2    print('match')
3else:
4    print('no match')

1re.split(" ","Daisy is a dachshund dog. Daisy is beautiful.")

1re.findall("Daisy","Daisy is a dachshund dog. Daisy is beautiful.")

1for x in re.finditer("Daisy","Daisy is a dachshund dog. Daisy is beautiful."):
2    print(x.group(0) + " at " + str(x.span()))

1re.findall("[in]","His Phone isn't 578")

1re.findall("[^in]","His Phone isn't 578")

1re.findall("[a-z]","His Phone isn't 578")

1re.findall("[A-Z]","His Phone isn't 578")

1re.findall("[0-9]","His Phone isn't 578")

1re.findall("[.]","His Phone isn't 578")

1re.findall("[.]","His Phone isn't 578.")

1re.findall("[a-z][sn]","His Phone isn't 578")

1re.findall(".","His Phone isn't 578")

1re.findall("\w","His Phone isn't 578")

1re.findall("\W","His Phone isn't 578")

1re.findall("\d","His Phone isn't 578")

1re.findall("\D","His Phone isn't 578")

1re.findall("\s","His Phone isn't 578")

1re.findall("\S","His Phone isn't 578")

1# Match first word of the string
2re.findall("^\w+","His Phone isn't 578")

1# Match last word of the string
2re.findall("\w+$","His Phone isn't 578")

1# Match all words starting with 'i' until non-space is found
2re.findall(r"\bi\S+","His Phone isn't 578")

1# Match all letters that are not at the beginning of a word 
2# starting with 'i' until non-space is found
3re.findall(r"\Bi\S+","His Phone isn't 578")

1re.findall("A-?B","AB A-B A--B A---B")

1re.findall("A-*B","AB A-B A--B A---B")

1re.findall("A-+B","AB A-B A--B A---B")

1re.findall("A-{2}B","AB A-B A--B A---B")

1re.findall("A-{2,3}B","AB A-B A--B A---B")

1# OR operator
2for x in re.finditer("(a|b)x","ax bx cx"):
3    print(x.group(0))   

1# the parentheses define groups
2for x in re.finditer("is a (\w+) (\w+)","My vehicle is a red car, her vehicle is a blue bike"):
3    print(x.group(1) + " - " + x.group(2))  

1# referencing groups in substitutions
2# note that the entire pattern is being substituted but
3# that specific components are selected in the substitution
4re.sub("is a (\w+)\s(\w+)",r"is a \2 whose colour is \1","My vehicle is a red car, her vehicle is a blue bike")

1# the first match is ignored. Only one group is returned
2for x in re.finditer("is a (?:\w+) (\w+)","My vehicle is a red car, her vehicle is a blue bike"):
3    print(x.group(1))  

1# Iteration with labels
2for x in re.finditer("(?P<name>Daisy)(?P<predicate>[\w ]*)(?P<dot>\.)","Daisy is a dachshund dog. Daisy is beautiful. She is 4."):
3    print(x.groupdict()['predicate'])

 1# Use verbose mode
 2text="""
 323423\nDaisy is black, and beautiful. Taffy is brown, and short. \nOtto is cute, and happy."
 4"""
 5pattern="""
 6(?P<name>\w*)        # Alphanumerical word
 7(\ is \ )            # followed by 'is'
 8(?P<adj1>\w*)        # followed by alphanumerical word
 9(,\ and\ )           # followed by 'and'
10(?P<adj2>\w*)        # followed by another alphanumerical word
11"""
12
13for item in re.finditer(pattern,text,re.VERBOSE):
14    print(item.groupdict())

1# Look ahead helps create user-defined, non-consuming matches 
2# 'ahead' of the expression, like '$'
3re.findall(r"\w+(?=[,!]\s?)", "One, two, three!")

1# Look behind helps create user-defined non-consuming matches 
2# 'behind' of the expression, like '^'
3re.findall("(?<=[Uu]n)\w+", "Undo it! It was unintentional!")

1# Both look ahead and look behind can be combined
2re.findall("(?<=<tag>).+(?=</tag>)", "<tag>hello world</tag>")

1valid = "c.com b.c.com a.b.c.com "
2invalid = "c c..com b..c.com a.b..com .com"
3for x in re.finditer(r"(^|(?<=\s))\w+(\.\w+)+($|(?=\s))",valid + invalid):
4    print(x.group(0))

1df = pd.DataFrame(data=["01-02-2009 First Entry",
2                        "05-12-2015 Second Entry",
3                        "30-07-2022 Third Entry"],
4                        columns=["entries"])
5df      

1pattern = r'(?P<day>\d{1,2})[/-](?P<month>\d{1,2})[/-](?P<year>\d{2,4})'
2df['entries'].str.extract(pattern)