7.15. Regex Quantifier Lazy

  • Quantifier specifies how many occurrences of preceding qualifier or character class

  • Lazy

7.15.1. SetUp

>>> import re

7.15.2. Lazy

  • Prefer shortest matches

  • Works better with text

  • Not that good results for numbers

  • Non-greedy

  • {n,m}? - minimum n repetitions, maximum m times, prefer shorter

  • {,n}? - maximum n repetitions, prefer shorter

  • {n,}? - minimum n repetitions, prefer shorter

  • {0,1}? - minimum 0 repetitions, maximum 1 repetitions (maybe)

Min/max:

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{2,4}?', TEXT)
['20', '00', '12', '00']

Nolimit:

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{2,}?', TEXT)
['20', '00', '12', '00']

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{,4}?', TEXT)
['', '', '', '', '', '', '', '', '', '', '', '', '', '1', '', '',
 '', '', '', '2', '', '0', '', '0', '', '0', '', '', '', '', '',
 '1', '', '2', '', '', '0', '', '0', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '']

7.15.3. Shorthand Lazy

  • *? - minimum 0 repetitions, no maximum, prefer shorter (alias to {0,}?)

  • +? - minimum 1 repetitions, no maximum, prefer shorter (alias to {1,}?)

  • ?? - minimum 0 repetitions, maximum 1 repetition, prefer shorter (alias to {0,1}?)

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{1,}?', TEXT)
['1', '2', '0', '0', '0', '1', '2', '0', '0']

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d+?', TEXT)
['1', '2', '0', '0', '0', '1', '2', '0', '0']

Star:

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{0,}?', TEXT)
['', '', '', '', '', '', '', '', '', '', '', '', '', '1', '', '', '',
 '', '', '2', '', '0', '', '0', '', '0', '', '', '', '', '', '1', '',
 '2', '', '', '0', '', '0', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '']

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d*?', TEXT)
['', '', '', '', '', '', '', '', '', '', '', '', '', '1', '', '', '',
 '', '', '2', '', '0', '', '0', '', '0', '', '', '', '', '', '1', '',
 '2', '', '', '0', '', '0', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '']

Question mark:

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d{0,1}?', TEXT)
['', '', '', '', '', '', '', '', '', '', '', '', '', '1', '', '', '',
 '', '', '2', '', '0', '', '0', '', '0', '', '', '', '', '', '1', '',
 '2', '', '', '0', '', '0', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '']

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d??', TEXT)
['', '', '', '', '', '', '', '', '', '', '', '', '', '1', '', '', '',
 '', '', '2', '', '0', '', '0', '', '0', '', '', '', '', '', '1', '',
 '2', '', '', '0', '', '0', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '', '',
 '', '', '', '', '', '', '', '', '']

7.15.4. Greedy vs. Lazy

  • r'\d+' - Greedy

  • r'\d+?' - Lazy

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d+', TEXT)
['1', '2000', '12', '00']

>>> TEXT = 'On Sun, Jan 1st, 2000 at 12:00 AM Alice <alice@example.com> wrote'
>>> re.findall(r'\d+?', TEXT)
['1', '2', '0', '0', '0', '1', '2', '0', '0']

>>> TEXT = 'Email from Alice. Received on Sunday.'
>>> sentence = r'[A-Z].+\.'
>>> re.findall(sentence, TEXT)
['Email from Alice. Received on Sunday.']

>>> TEXT = 'Email from Alice. Received on Sunday.'
>>> sentence = r'[A-Z].+?\.'
>>> re.findall(sentence, TEXT)
['Email from Alice.', 'Received on Sunday.']

Mind the number of sentences in each case. Without lazy quantifier it returns only one result: from first capital letter to the last possible dot. Lazy quantifier splits text into two parts. From the first capital letter to the closest dot.

7.15.5. Use Case - 1

>>> HTML = '<p>We choose to go to the Moon</p>'
>>>
>>> tag = r'<.+>'
>>> re.findall(tag, HTML)
['<p>We choose to go to the Moon</p>']
>>>
>>> tag = r'<.+?>'
>>> re.findall(tag, HTML)
['<p>', '</p>']

7.15.6. Use Case - 2

>>> import re
>>> HTML = '<h1>Header 1</h1><p>Paragraph 1</p><p>Paragraph 2</p>'

>>> re.findall(r'<p>.*</p>', HTML)
['<p>Paragraph 1</p><p>Paragraph 2</p>']

>>> re.findall(r'<p>.*?</p>', HTML)
['<p>Paragraph 1</p>', '<p>Paragraph 2</p>']

7.15.7. Use Case - 3

>>> import re
>>> HTML = '<h1>Header 1</h1><p>Paragraph 1</p><p>Paragraph 2</p>'

>>> re.findall(r'</?.*>', HTML)
['<h1>Header 1</h1><p>Paragraph 1</p><p>Paragraph 2</p>']

>>> re.findall(r'</?.*?>', HTML)
['<h1>', '</h1>', '<p>', '</p>', '<p>', '</p>']

7.15.8. Use Case - 4

>>> import re
>>> HTML = '<h1>Header 1</h1><p>Paragraph 1</p><p>Paragraph 2</p>'

>>> re.findall(r'<.+>', HTML)
['<h1>Header 1</h1><p>Paragraph 1</p><p>Paragraph 2</p>']

>>> re.findall(r'<.+?>', HTML)
['<h1>', '</h1>', '<p>', '</p>', '<p>', '</p>']

>>> re.findall(r'</?.+?>', HTML)
['<h1>', '</h1>', '<p>', '</p>', '<p>', '</p>']

>>> re.findall(r'</?(.+?)>', HTML)
['h1', 'h1', 'p', 'p', 'p', 'p']

>>> tags = re.findall(r'</?(.+?)>', HTML)
>>> sorted(set(tags))
['h1', 'p']

7.15.9. Assignments