Remove duplicate chars using regex?
Let's say I want to remove all duplicate chars (of a particular char) in a string using regular expressions. This is simple -
import re
re.sub("a*", "a", "aaaa") # gives 'a'
What if I want to replace all duplicate chars (i.e. a,z) with that respectiv开发者_Go百科e char? How do I do this?
import re
re.sub('[a-z]*', <what_to_put_here>, 'aabb') # should give 'ab'
re.sub('[a-z]*', <what_to_put_here>, 'abbccddeeffgg') # should give 'abcdefg'
NOTE: I know this remove duplicate approach can be better tackled with a hashtable or some O(n^2) algo, but I want to explore this using regexes
>>> import re
>>> re.sub(r'([a-z])\1+', r'\1', 'ffffffbbbbbbbqqq')
'fbq'
The ()
around the [a-z]
specify a capture group, and then the \1
(a backreference) in both the pattern and the replacement refer to the contents of the first capture group.
Thus, the regex reads "find a letter, followed by one or more occurrences of that same letter" and then entire found portion is replaced with a single occurrence of the found letter.
On side note...
Your example code for just a
is actually buggy:
>>> re.sub('a*', 'a', 'aaabbbccc')
'abababacacaca'
You really would want to use 'a+'
for your regex instead of 'a*'
, since the *
operator matches "0 or more" occurrences, and thus will match empty strings in between two non-a
characters, whereas the +
operator matches "1 or more".
In case you are also interested in removing duplicates of non-contiguous occurrences you have to wrap things in a loop, e.g. like this
s="ababacbdefefbcdefde"
while re.search(r'([a-z])(.*)\1', s):
s= re.sub(r'([a-z])(.*)\1', r'\1\2', s)
print s # prints 'abcdef'
A solution including all category:
re.sub(r'(.)\1+', r'\1', 'aaaaabbbbbb[[[[[')
gives:
'ab['
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