Python Regular Expressions: Difference between revisions

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==Iteratively Match a Pattern against a String==
==Iteratively Match a Pattern against a String==
=The Match Object=


=Replacing Regular Expression Occurrences=
=Replacing Regular Expression Occurrences=

Revision as of 01:44, 25 March 2022

External

Internal

TODO

PROCESS: https://docs.python.org/3/howto/regex.html#regex-howto

Overview

A regular expression is specified with r"..."

Metacharacters

https://docs.python.org/3/library/re.html#regular-expression-syntax

The complete list of metacharacters:

. (dot)

It matches anything except a newline character, and there’s an alternate mode (re.DOTALL) where it will match even a newline. . Used where you want to match “any character”.

To match an actual dot, escape it:

\.

^

Complements a character set, which means match all characters not in the set. This is indicating by including ^ as the first character of the class. For example [^a] will match everything except "a". If the caret appears elsewhere in the class, it does not have a special meaning and it will represent itself.

$

*

Causes the resulting regular expression to match 0 or more repetitions of the preceding regular expression, as many repetitions as are possible.

+

?

{

}

[

Used to specify the beginning of a character class, which is a set of characters you wish to match. Characters can be listed individually:

[abc]

or as a range, by using -:

[0-9]

⚠️ Metacharacters are not active inside classes - they're stripped of their special nature. For example [abc$] will match 'a' or 'b' or 'c' or '$'.

]

Used to specify the end of a character class.

\

A backslash can be followed by various character to signal special sequences. The backslash is also used to escape all metacharacters so they can be matched in patterns. To match [, prefix it with backslash: \[.

|

(

Used to indicate the beginning of a group capture.

)

Used to indicate the end of a group capture.

Special Sequences

All the special sequences described below can be included inside a character class, and they preserve their meaning.

\d

Matches any decimal digit. It is equivalent to the class [0-9].

\D

Matches any non-digit character. It is equivalent to the class [^0-9].

\s

Matches any whitespace character. It is equivalent to the class [ \t\n\r\f\v].

\S

Matches any non-whitespace character. It is equivalent to the class [^ \t\n\r\f\v].

\w

Matches any alphanumeric character. It is equivalent to the class [a-zA-Z0-9_].

\W

Matches any non-alphanumeric character. It is equivalent to the class [^a-zA-Z0-9_].

NOT Metacharacters

The following characters are matched without any escaping:

{...}

Research this, { and } are metacharacters.

Patterns

At most one group of characters:

(...)?

Matching Modes

Before performing any match, compile the regular expression:

import re
pattern = re.compile(r'(\w+):(\w+)-(\w+)')

Once there is a compiled regular expression, in this case referred via the variable pattern, it can be used in the following modes:

  • match() determine if the regular expression matches at the beginning of the string. See Match a Pattern and Pick Up Groups below.
  • search() scan through a string looking for any location where this regular expression matches. See Scan a String below.
  • findall() find all substrings where the regular expression matches and return them as a list.
  • finditer() find all substrings where the regular expression matches and return them as an iterator.

Match a Pattern and Pick Up Groups

import re

p = re.compile(r'^(\w+):(\w+)-(\w+)$')
s = 'abc:mnp-xyz'
m = p.match(s)
if m:
    assert 'abc:mnp-xyz' == m.group(0)
    assert 'abc' == m.group(1)
    assert 'mnp' == m.group(2)
    assert 'xyz' == m.group(3)

Groups are 1-based. group(0) represents the entire expression.

Bug: when a regular expression like this one is used: '....()?()?' (two optional groups), and the last group is None, m.groups(one_based_last_group_index) throws IndexError. The solution was to retrieve the groups as a tuple before any evaluation, and use it for testing:

groups = m.groups() if m.groups(1):

 ...

... if groups[4]:

 ...

Scan a String

match = re.search(pattern, string)
if match:
    process(match)

Iteratively Match a Pattern against a String

The Match Object

Replacing Regular Expression Occurrences

https://docs.python.org/3/library/re.html#text-munging
import re

s = "this is a {{color}} car"
print(re.sub(r"{{color}}", 'blue', s))

Strip quotes:

s = "'something'"
re.sub(r"'$", '', re.sub(r"^'", '', s))

Capture groups and use them in the replacement:

s = 'this is red'
s2 = re.sub(r'^(this is).*$', '\\1 blue', s)
assert 'this is blue' == s2

To dynamically build a regular expression, use rf'...'

s = 'this is a red string'
color = 'red'
s2 = re.sub(rf'{color}', 'blue', s)
assert 'this is a blue string' == s2

Match a new line:

r'\n'

Match not a new line:

r'[^\n]'