Character class

This chapter will discuss how to create your own custom placeholders to match limited set of characters and various metacharacters applicable inside character classes. You'll also learn about escape sequences for predefined character sets.

Custom character sets

Characters enclosed inside [] metacharacters is a character class (or set). It will result in matching any one of those characters once. It is similar to using single character alternations inside a grouping, but without the drawbacks of a capture group. In addition, character classes have their own versions of metacharacters and provide special predefined sets for common use cases. Quantifiers are applicable to character classes as well.

>>> words = ['cute', 'cat', 'cot', 'coat', 'cost', 'scuttle']

# same as: r'cot|cut' or r'c(o|u)t'
>>> [w for w in words if re.search(r'c[ou]t', w)]
['cute', 'cot', 'scuttle']

# same as: r'(a|e|o)+t'
>>> re.sub(r'[aeo]+t', 'X', 'meeting cute boat site foot')
'mXing cute bX site fX'

Range of characters

Character classes have their own metacharacters to help define the sets succinctly. Metacharacters outside of character classes like ^, $, () etc either don't have special meaning or have completely different one inside the character classes. First up, the - metacharacter that helps to define a range of characters instead of having to specify them all individually.

# all digits
>>> re.findall(r'[0-9]+', 'Sample123string42with777numbers')
['123', '42', '777']

# whole words made up of lowercase alphabets and digits only
>>> re.findall(r'\b[a-z0-9]+\b', 'coat Bin food tar12 best')
['coat', 'food', 'tar12', 'best']

# whole words made up of lowercase alphabets, but starting with 'p' to 'z'
>>> re.findall(r'\b[p-z][a-z]*\b', 'coat tin food put stoop best')
['tin', 'put', 'stoop']

# whole words made up of only 'a' to 'f' and 'p' to 't' lowercase alphabets
>>> re.findall(r'\b[a-fp-t]+\b', 'coat tin food put stoop best')
['best']

Negating character sets

Next metacharacter is ^ which has to specified as the first character of the character class. It negates the set of characters, so all characters other than those specified will be matched. As highlighted earlier, handle negative logic with care, you might end up matching more than you wanted.

# all non-digits
>>> re.findall(r'[^0-9]+', 'Sample123string42with777numbers')
['Sample', 'string', 'with', 'numbers']

# remove first two columns where : is delimiter
>>> re.sub(r'\A([^:]+:){2}', '', 'foo:123:bar:baz', count=1)
'bar:baz'

# deleting characters at end of string based on a delimiter
>>> re.sub(r'=[^=]+\Z', '', 'foo=42; baz=123', count=1)
'foo=42; baz'

>>> dates = '2020/04/25,1986/Mar/02,77/12/31'
# Note that the third character set negates comma as well
# and comma is matched optionally outside the capture groups
>>> re.findall(r'([^/]+)/([^/]+)/([^/,]+),?', dates)
[('2020', '04', '25'), ('1986', 'Mar', '02'), ('77', '12', '31')]

Sometimes, it is easier to use positive character class and negate the re.search result instead of using negated character class.

>>> words = ['tryst', 'fun', 'glyph', 'pity', 'why']

# words not containing vowel characters
>>> [w for w in words if re.search(r'\A[^aeiou]+\Z', w)]
['tryst', 'glyph', 'why']

# easier to write and maintain, note the 'not' logical operator
# but this'll match empty strings too unlike the previous solution
>>> [w for w in words if not re.search(r'[aeiou]', w)]
['tryst', 'glyph', 'why']

Matching metacharacters literally

Similar to other metacharacters, prefix \ to character class metacharacters to match them literally. Some of them can be achieved by different placement as well.

- should be first or last character or escaped using \.

>>> re.findall(r'\b[a-z-]{2,}\b', 'ab-cd gh-c 12-423')
['ab-cd', 'gh-c']

>>> re.findall(r'\b[a-z\-0-9]{2,}\b', 'ab-cd gh-c 12-423')
['ab-cd', 'gh-c', '12-423']

^ should be other than first character or escaped using \.

>>> re.findall(r'a[+^]b', 'f*(a^b) - 3*(a+b)')
['a^b', 'a+b']

>>> re.findall(r'a[\^+]b', 'f*(a^b) - 3*(a+b)')
['a^b', 'a+b']

[ can be escaped with \ or placed as last character. ] can be escaped with \ or placed as first character.

>>> re.search(r'[]a-z0-9[]+', 'words[5] = tea')[0]
'words[5]'

>>> re.search(r'[a-z\[\]0-9]+', 'words[5] = tea')[0]
'words[5]'

\ should be escaped using \.

# note that the input string is using raw-string format
>>> print(re.search(r'[a\\b]+', r'5ba\babc2')[0])
ba\bab

Escape sequence sets

Commonly used character sets have predefined escape sequences:

  • \w is similar to [a-zA-Z0-9_] for matching word characters (recall the definition for word boundaries)
  • \d is similar to [0-9] for matching digit characters
  • \s is similar to [ \t\n\r\f\v] for matching whitespace characters

These escape sequences can be used as a standalone sequence or inside a character class.

>>> re.split(r'\d+', 'Sample123string42with777numbers')
['Sample', 'string', 'with', 'numbers']

>>> re.findall(r'\d+', 'foo=5, bar=3; x=83, y=120')
['5', '3', '83', '120']

>>> ''.join(re.findall(r'\b\w', 'sea eat car rat eel tea'))
'secret'

>>> re.findall(r'[\w\s]+', 'tea sea-pit sit-lean\tbean')
['tea sea', 'pit sit', 'lean\tbean']

And negative logic strikes again. Use \W, \D, and \S respectively for their negated sets.

>>> re.sub(r'\D+', '-', 'Sample123string42with777numbers')
'-123-42-777-'

>>> re.sub(r'\W+', '', 'foo=5, bar=3; x=83, y=120')
'foo5bar3x83y120'

# this output can be achieved with normal string method too, guess which one!
>>> re.findall(r'\S+', '   1..3  \v\f  foo_baz 42\tzzz   \r\n1-2-3  ')
['1..3', 'foo_baz', '42', 'zzz', '1-2-3']

warning As mentioned before, the examples and description assume input made up of ASCII characters only, unless otherwise specified. These sets would behave differently depending on flags used. See Unicode chapter for more details.

Numeric ranges

Character classes can also be used to construct numeric ranges. However, it is easy to miss corner cases and some ranges are complicated to design.

# numbers between 10 to 29
>>> re.findall(r'\b[12]\d\b', '23 154 12 26 98234')
['23', '12', '26']

# numbers >= 100
>>> re.findall(r'\b\d{3,}\b', '23 154 12 26 98234')
['154', '98234']

# numbers >= 100 if there are leading zeros
>>> re.findall(r'\b0*[1-9]\d{2,}\b', '0501 035 154 12 26 98234')
['0501', '154', '98234']

If numeric range is difficult to construct, better to convert the matched portion to appropriate numeric format first.

# numbers < 350
>>> m_iter = re.finditer(r'\d+', '45 349 651 593 4 204')
>>> [m[0] for m in m_iter if int(m[0]) < 350]
['45', '349', '4', '204']

# note that return value is string and s[0] is used to get matched portion
>>> def num_range(s):
...     return '1' if 200 <= int(s[0]) <= 650 else '0'
... 

# numbers between 200 and 650
# note that only function name is supplied, () is not used
# re.Match object is automatically passed as argument
>>> re.sub(r'\d+', num_range, '45 349 651 593 4 204')
'0 1 0 1 0 1'

info See regular-expressions: matching numeric ranges for more examples.

Cheatsheet and Summary

NoteDescription
[ae;o]match any of these characters once
quantifiers are applicable to character classes too
[3-7]range of characters from 3 to 7
[^=b2]match other than = or b or 2
[a-z-]- should be first/last or escaped using \ to match literally
[+^]^ shouldn't be first character or escaped using \
[a\[\]][ can be escaped with \ or placed as last character
[a\[\]]] can be escaped with \ or placed as first character
[a\\b]\ should be escaped using \
\wsimilar to [a-zA-Z0-9_] for matching word characters
\dsimilar to [0-9] for matching digit characters
\ssimilar to [ \t\n\r\f\v] for matching whitespace characters
\W, \D, and \S for their opposites respectively

This chapter focused on how to create custom placeholders for limited set of characters. Grouping and character classes can be considered as two levels of abstractions. On the one hand, you can have character sets inside [] and on the other, you can have multiple alternations grouped inside () including character classes. As anchoring and quantifiers can be applied to both these abstractions, you can begin to see how regular expressions is considered a mini-programming language. In coming chapters, you'll even see how to negate groupings similar to negated character class in certain scenarios.

Exercises

a) For the list items, filter all elements starting with hand and ending with s or y or le.

>>> items = ['-handy', 'hand', 'handy', 'unhand', 'hands', 'handle']

##### add your solution here
['handy', 'hands', 'handle']

b) Replace all whole words reed or read or red with X.

>>> ip = 'redo red credible :read: rod reed'

##### add your solution here
'redo X credible :X: rod X'

c) For the list words, filter all elements containing e or i followed by l or n. Note that the order mentioned should be followed.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'unicorn', 'eel']

d) For the list words, filter all elements containing e or i and l or n in any order.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'unicorn', 'newer', 'eel']

e) Extract all hex character sequences, with 0x optional prefix. Match the characters case insensitively, and the sequences shouldn't be surrounded by other word characters.

>>> str1 = '128A foo 0xfe32 34 0xbar'
>>> str2 = '0XDEADBEEF place 0x0ff1ce bad'

>>> hex_seq = re.compile()        ##### add your solution here

##### add your solution here for str1
['128A', '0xfe32', '34']

##### add your solution here for str2
['0XDEADBEEF', '0x0ff1ce', 'bad']

f) Delete from ( to the next occurrence of ) unless they contain parentheses characters in between.

>>> str1 = 'def factorial()'
>>> str2 = 'a/b(division) + c%d(#modulo) - (e+(j/k-3)*4)'
>>> str3 = 'Hi there(greeting). Nice day(a(b)'

>>> remove_parentheses = re.compile()      ##### add your solution here

>>> remove_parentheses.sub('', str1)
'def factorial'
>>> remove_parentheses.sub('', str2)
'a/b + c%d - (e+*4)'
>>> remove_parentheses.sub('', str3)
'Hi there. Nice day(a'

g) For the list words, filter all elements not starting with e or p or u.

>>> words = ['surrender', 'unicorn', 'newer', 'door', 'empty', 'eel', 'pest']

##### add your solution here
['surrender', 'newer', 'door']

h) For the list words, filter all elements not containing u or w or ee or -.

>>> words = ['p-t', 'you', 'tea', 'heel', 'owe', 'new', 'reed', 'ear']

##### add your solution here
['tea', 'ear']

i) The given input strings contain fields separated by , and fields can be empty too. Replace last three fields with WHTSZ323.

>>> row1 = '(2),kite,12,,D,C,,'
>>> row2 = 'hi,bye,sun,moon'

>>> pat = re.compile()      ##### add your solution here

>>> pat.sub()       ##### add your solution here for row1
'(2),kite,12,,D,WHTSZ323'
>>> pat.sub()       ##### add your solution here for row2
'hi,WHTSZ323'

j) Split the given strings based on consecutive sequence of digit or whitespace characters.

>>> str1 = 'lion \t Ink32onion Nice'
>>> str2 = '**1\f2\n3star\t7 77\r**'

>>> pat = re.compile()       ##### add your solution here

>>> pat.split(str1)
['lion', 'Ink', 'onion', 'Nice']
>>> pat.split(str2)
['**', 'star', '**']

k) Delete all occurrences of the sequence <characters> where characters is one or more non > characters and cannot be empty.

>>> ip = 'a<apple> 1<> b<bye> 2<> c<cat>'

##### add your solution here
'a 1<> b 2<> c'

l) \b[a-z](on|no)[a-z]\b is same as \b[a-z][on]{2}[a-z]\b. True or False? Sample input lines shown below might help to understand the differences, if any.

>>> print('known\nmood\nknow\npony\ninns')
known
mood
know
pony
inns

m) For the given list, filter all elements containing any number sequence greater than 624.

>>> items = ['hi0000432abcd', 'car00625', '42_624 0512', '3.14 96 2 foo1234baz']

##### add your solution here
['car00625', '3.14 96 2 foo1234baz']

n) Count the maximum depth of nested braces for the given strings. Unbalanced or wrongly ordered braces should return -1. Note that this will require a mix of regular expressions and Python code.

>>> def max_nested_braces(ip):
##### add your solution here

>>> max_nested_braces('a*b')
0
>>> max_nested_braces('}a+b{')
-1
>>> max_nested_braces('a*b+{}')
1
>>> max_nested_braces('{{a+2}*{b+c}+e}')
2
>>> max_nested_braces('{{a+2}*{b+{c*d}}+e}')
3
>>> max_nested_braces('{{a+2}*{\n{b+{c*d}}+e*d}}')
4
>>> max_nested_braces('a*{b+c*{e*3.14}}}')
-1

o) By default, str.split method will split on whitespace and remove empty strings from the result. Which re module function would you use to replicate this functionality?

>>> ip = ' \t\r  so  pole\t\t\t\n\nlit in to \r\n\v\f  '

>>> ip.split()
['so', 'pole', 'lit', 'in', 'to']
##### add your solution here
['so', 'pole', 'lit', 'in', 'to']

p) Convert the given input string to two different lists as shown below.

>>> ip = 'price_42 roast^\t\n^-ice==cat\neast'

##### add your solution here
['price_42', 'roast', 'ice', 'cat', 'east']

##### add your solution here
['price_42', ' ', 'roast', '^\t\n^-', 'ice', '==', 'cat', '\n', 'east']

q) Filter all elements whose first non-whitespace character is not a # character. Any element made up of only whitespace characters should be ignored as well.

>>> items = ['    #comment', '\t\napple #42', '#oops', 'sure', 'no#1', '\t\r\f']

##### add your solution here
['\t\napple #42', 'sure', 'no#1']