regular expressions

Question 1

module for regular expressions

Answer 1

re

Question 2

?

Answer 2

match 0 or 1 repetitions of the preceding RE

Question 3

make qualifiers not greedy

Answer 3

*?, +?, ??

Question 4

match from m to n repetitions of the preceding RE, attempting to match as few repetitions as possible

Answer 4

{m,n}?

Question 5

set of characters

Answer 5

[amk] [a-z] [0-9A-Fa-f] [^5]

Question 6

named group

Answer 6

(?P…)

Question 7

A backreference to a named group

Answer 7

(?P=name)

Question 8

lookahead assertion

Answer 8

(?=…)

Question 9

negative lookahead assertio

Answer 9

(?!…)

Question 10

positive lookbehind assertion

Answer 10

(?<=…)

Question 11

negative lookbehind assertion

Answer 11

(?

Question 12

yes/no pattern

Answer 12

(?(id/name)yes-pattern|no-pattern)

Question 13

Matches the contents of the group of the same number

Answer 13

\number

(.+) \1 matches ‘the the’ or ‘55 55’

Question 14

Matches the empty string, but only at the beginning or end of a wor

Answer 14

\b

r’\bfoo\b’ matches ‘foo’, ‘foo.’, ‘(foo)’, ‘bar foo baz’ but not ‘foobar’ or ‘foo3’.

Question 15

ompile a regular expression pattern into a regular expression objec

Answer 15

re.compile(pattern, flags=0)

Question 16

Scan through string looking for the first location where the regular expression pattern produces a match, and return a corresponding match object.

Answer 16

re.search(pattern, string, flags=0)
Return None if no position in the string matches the pattern; note that this is different from finding a zero-length match at some point in the string.

Question 17

Split string by the occurrences of pattern.

Answer 17

re.split(pattern, string, maxsplit=0, flags=0)

Question 18

Return all non-overlapping matches of pattern in string, as a list of strings

Answer 18

re.findall(pattern, string, flags=0)

Question 19

Return an iterator yielding match objects over all non-overlapping matches for the RE pattern in string.

Answer 19

re.finditer(pattern, string, flags=0)

Question 20

get entire found string from match

Answer 20

m.group(0), m[0]

Question 21

Return the indices of the start and end of the substring matched by group

Answer 21

Match.start([group])

Match.end([group])