Python Regular Expressions: Difference between revisions

From NovaOrdis Knowledge Base
Jump to navigation Jump to search
Line 90: Line 90:
</syntaxhighlight>
</syntaxhighlight>
Groups are 1-based. <code>group(0)</code> represents the entire expression.
Groups are 1-based. <code>group(0)</code> represents the entire expression.
<font color=darkkhaki>
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:
<syntaxhighlight lang='python'>
</syntaxhighlight>
groups = m.groups()
if m.groups(1):
  ...
...
if groups[4]:
  ...
</font>


=Scan a String=
=Scan a String=

Revision as of 00:29, 16 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

(...)

Used to capture groups.

^

Not a certain character, or a set of characters. [^a]

. (dot)

Stands for "any one character". To match an actual dot, escape it:

\.

NOT Metacharacters

The following characters are matched without any escaping:

{...}

Patterns

At most one group of characters:

(...)?

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]'

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)