Example of Python metacharacters usage parsing and python usage examples

Example of Python metacharacters usage parsing and python usage examples

The role of the backslash:

To process a metacharacter ^ as a common character, add a backslash

For example:

>>>import re>>>r=r'\^abc'>>>re.findall(r,'^abc ^abc ^abc')['^abc','^abc','^abc']

\ D matches any decimal number, which is equivalent to the class [0-9].
\ D matches any non-numeric character, which is equivalent to the class [^ 0-9]
\ S matches any blank character, which is equivalent to the class [\ t \ n \ r \ f \ v]
\ S matches any non-blank characters, which is equivalent to the class [^ \ t \ n \ r \ f \ v]
\ W matches any alphanumeric character, which is equivalent to the class [a-zA-Z0-9 _]
\ W matches any non-alphanumeric character, which is equivalent to the class [^ a-zA-Z0-9 _]

>>>r=r'[0-9]'>>>re.findall(r,'1234567890')['1','2','3','4','5','6','7','8','9','0']>>>r=r'\d'>>>re.findall(r,'1234567890')['1','2','3','4','5','6','7','8','9','0']

>>> R = R' ^ 010-\ d \ d'> re. findall (r, '010-87654321 ') ['010-87654321']> re. findall (r, '010-8765432 ') []> r = R' ^ 010-\ d {8}' # repeat eight times> re. findall (r, '010-12345678 ') [' ^ 010-12345678 ']

Asterisk :(*)

Match the previous character Zero or more times.

>>>r=r'ab*'>>>re.findall(r,'a')['a']>>>re.findall(r,'ab')['ab']>>>re.findall(r,'abbbbbb')['abbbbbb']

The role of the plus sign: (+)

Match once or more times.

>>>r=r'ab+'>>>re.findall(r,'a')[]>>>re.findall(r,'ab')['ab']>>>re.findall(r,'abbbb')['abbbb']

"-" In the middle of the phone number: (optional)

>>>r=r'^010-*\d{8}'>>>re.findall(r,'010-12345678')['010-12345678']>>>re.findall(r,'01012345678')['01012345678']>>>re.findall(r,'010---12345678')['010---12345678']

Question mark :(?)

Match once or zero times;

>>>r=r'^010-?\d{8}$'>>>re.findall(r,'010--12345678')[]>>>re.findall(r,'010-12345678')['010-12345678']>>>re.findall(r,'01012345678')['01012345678']

Minimum pattern matching:

Greedy pattern matching is as follows:

>>>r=r'ab+'>>>re.findall(r,'abbbbbbbbbbb')['abbbbbbbbbbb']

For non-Greedy match, use the question mark for the minimum match, as shown below:

>>>r=r'ab+?'>>>re.findall(r,'abbbbbbbbbbb')['ab']>>>r=r'ab*?'>>>re.findall(r,'abbbbbbbbbbbb')['a']

Curly braces: ({m, n })

Where m and n are decimal integers. The qualifier indicates at least m duplicates and at most n duplicates.

>>> R = r'a {1, 3} '# indicates that a repeats one to three times >>> re. findall (r, 'A') ['a']> re. findall (r, 'A') ['a']> re. findall (r, 'aaa') ['aaa']> re. findall (r, 'aaa') ['aaa', 'a']

GROUP: "(" and ")"

>>> Import re >>> email = R' \ w {3} @ \ w + (\. com | \. cn) '# define a regular expression ,(\. com | \. cn) indicates a group. The group performs ** or ** operations, or yes. com, or yes. cn> re. match (email, 'www @ owolf.com ') # match <_ sre. SRE_Match object; span = (0, 13), match = 'www @ owolf.com '>>> re. match (email, 'www @ owolf.cn ') <_ sre. SRE_Match object; span = (0, 12), match = 'www @ owolf.cn '>>> re. match (email, 'www @ owolf.org ') >>## return NULL >>> re. findall (email, 'www @ owolf.com ')['. com '] # return data in the group first when matching> re. findall (email, 'www @ owolf.cn ')['. cn '] >>>

>>> S = ''' ajhfa kasjf owolf english = chinese yes no printlafl int = 456 yes floatint = 789 yesowolf english = france yes aklfl ''' # defines the string >>> r = r'owolf english =. + yes '# define a regular expression> re. findall (r, s) # match regular ['owolf english = chinese yes ', 'owolf english = france yes'] >>> r = r'owolf english = (. +) yes '> re. findall (r, s) ['China', 'France '] # Use grouping to return data in the group first, which is often used in crawlers.

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