Here's some dialogue from :fontawesome-solid-hat-wizard: Harry Potter and the Sorcerer's Stone .
quotes = """Molly Weasley: Fred, you next.
George Weasley: He's not Fred, I am!
Fred Weasley: Honestly, woman. You call yourself our mother.
Molly Weasley: Oh, I'm sorry, George.
Fred Weasley: I'm only joking, I am Fred!"""
Identify each line spoken by Fred. Then put them in a list. (Note that quotes
is a multi-line string.)
lines = [
'Fred Weasley: Honestly, woman. You call yourself our mother.' ,
'Fred Weasley: I' m only joking, I am Fred! '
]
Function
Description
Return Value
re.findall(pattern, string, flags=0)
Find all non-overlapping occurrences of pattern in string
list of strings, or list of tuples if > 1 capture group
re.finditer(pattern, string, flags=0)
Find all non-overlapping occurrences of pattern in string
iterator yielding match objects
re.search(pattern, string, flags=0)
Find first occurrence of pattern in string
match object or None
re.split(pattern, string, maxsplit=0, flags=0)
Split string by occurrences of pattern
list of strings
re.sub(pattern, repl, string, count=0, flags=0)
Replace pattern with repl
new string with the replacement(s)
Pattern
Description
[abc]
a or b or c
[^abc]
not (a or b or c )
[a-z]
a or b ... or y or z
[1-9]
1 or 2 ... or 8 or 9
\d
digits [0-9]
\D
non-digits [^0-9]
\s
whitespace [ \t\n\r\f\v]
\S
non-whitespace [^ \t\n\r\f\v]
\w
alphanumeric [a-zA-Z0-9_]
\W
non-alphanumeric [^a-zA-Z0-9_]
.
any character
x*
zero or more repetitions of x
x+
one or more repetitions of x
x?
zero or one repetitions of x
{m}
m repetitions
{m,n}
m to n repetitions
{m,n}
m to n repetitions
\\
, \.
, \*
backslash, period, asterisk
\b
word boundary
^hello
starts with hello
bye$
ends with bye
(...)
capture group
(po|go)
po or go
Solution¶
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