Say you have a string like this: "(hello) (yes) (yo diddly)".
You want a list like this: ["hello", "yes", "yo diddly"]
How would you do this with Python?
import re
pattern = re.compile(r'\(([^)]*)\)')
The pattern matches the parentheses in your string (\(...\)) and these need to be escaped.
Then it defines a subgroup ((...)) - these parentheses are part of the regex-syntax.
The subgroup matches all characters except a right parenthesis ([^)]*)
s = "(hello) (yes) (yo diddly)"
pattern.findall(s)
gives
['hello', 'yes', 'yo diddly']
UPDATE:
It is probably better to use [^)]+ instead of [^)]*. The latter would also match an empty string.
Using the non-greedy modifiers, as DSM suggested, makes the pattern probably better to read: pattern = re.compile(r'\((.+?)\)')
I would do it like this:
"(hello) (yes) (yo diddly)"[1:-1].split(") (")
First, we cut off the first and last characters (since they should be removed anyway). Next, we split the resulting string using ") (" as the delimiter, giving the desired list.
This will give you words from any string :
>>> s="(hello) (yes) (yo diddly)"
>>> import re
>>> words = re.findall(r'\((.*?\))',s)
>>> words
['hello', 'yes', 'yo diddly']
as D.S.M said.
? in the regex to make it non-greedy.
Related
I want to split strings only by suffixes. For example, I would like to be able to split dord word to [dor,wor].
I though that \wd would search for words that end with d. However this does not produce the expected results
import re
re.split(r'\wd',"dord word")
['do', ' wo', '']
How can I split by suffixes?
x='dord word'
import re
print re.split(r"d\b",x)
or
print [i for i in re.split(r"d\b",x) if i] #if you dont want null strings.
Try this.
As a better way you can use re.findall and use r'\b(\w+)d\b' as your regex to find the rest of word before d:
>>> re.findall(r'\b(\w+)d\b',s)
['dor', 'wor']
Since \w also captures digits and underscore, I would define a word consisting of just letters with a [a-zA-Z] character class:
print [x.group(1) for x in re.finditer(r"\b([a-zA-Z]+)d\b","dord word")]
See demo
If you're wondering why your original approach didn't work,
re.split(r'\wd',"dord word")
It finds all instances of a letter/number/underscore before a "d" and splits on what it finds. So it did this:
do[rd] wo[rd]
and split on the strings in brackets, removing them.
Also note that this could split in the middle of words, so:
re.split(r'\wd', "said tendentious")
would split the second word in two.
Hey guys I'm trying to find all words with a specific character in the middle of the word. The word cannot begin or end with the specified character.
lets use 'x' for example. My current regex looks like this:
r'\b(?!x)\w+x(?<!x)\b'
the \w+x is not returning any results. Anyone have an idea why?
Try this:
>>> z = 'hello welxtra xcra crax extra'
>>> re.findall(r'[^x ]\w*x\w*[^x ]', z)
['welxtra', 'extra']
You can use something like this:
import re
print re.match(r'[^]+x+[^]','provxa')
print re.match(r'[^]+x+[^]','xprova')
Output:
<_sre.SRE_Match object at 0x10eaa0bf8>
None
where [^] is any char. So it will basically match an 'x' only is it between something else. You can change [^] with [a-z] to specify lowercase letters instead any char.
\b(?!x)\w+x\w+(?<!x)\b
^^
You missed the \w+after x.See demo.
https://regex101.com/r/nS2lT4/34
I am trying to replace this string to become this
import re
s = "haha..hehe.hoho"
s = re.sub('[..+]+',' ', s)
my output i get haha hehe hoho
desired output
haha hehe.hoho
What am i doing wrong?
Test on sites like regexpal: http://regexpal.com/
It's easier to get the output and check if the regex is right.
You should change your regex to something like: '\.\.' if you want to remove only double dots.
If you want to remove when there's at least 2 dots you can use '\.{2,}'.
Every character you put inside a [] will be checked against your expression
And the dot character has a special meaning on a regex, to avoid this meaning you should prefix it with a escape character: \
You can read more about regular expressions metacharacters here: https://www.hscripts.com/tutorials/regular-expression/metacharacter-list.php
[a-z] A range of characters. Matches any character in the specified
range.
. Matches any single character except "n".
\ Specifies the next character as either a special character, a literal, a back reference, or an octal escape.
Your new code:
import re
s = "haha..hehe.hoho"
#pattern = '\.\.' #If you want to remove when there's 2 dots
pattern = '\.{2,}' #If you want to remove when there's at least 2 dots
s = re.sub(pattern, ' ', s)
Unless you are constrained to use regex, then I find the replace() function much simpler:
s = "haha..hehe.hoho"
print s.replace('..',' ')
gives your desired output:
haha hehe.hoho
Change:
re.sub('[..+]+',' ', s)
to:
re.sub('\.\.+',' ', s)
[..+]+ , this meaning in regex is that use the any in the list at least one time. So it matches the .. as well as . in your input. Make the changes as below:
s = re.sub('\.\.+',' ', s)
[] is a character class and will match on anything in it (meaning any 1 .).
I'm guessing you used it because a simple . wouldn't work, because it's a meta character meaning any character. You can simply escape it to mean a literal dot with a \. As such:
s = re.sub('\.\.',' ', s)
Here is what your regex means:
So, you allow for 1 or more literal periods or plus symbols, which is not the case.
You do not have to repeat the same symbol when looking for it, you can use quantifiers, like {2}, which means "exactly 2 occurrences".
You can use split and join, see sample working program:
import re
s = "haha..hehe.hoho"
s = " ".join(re.split(r'\.{2}', s))
print s
Output:
haha hehe.hoho
Or you can use the sub with the regex, too:
s = re.sub(r'\.{2}', ' ', "haha..hehe.hoho")
In case you have cases with more than 2 periods, you should use \.{2,} regex.
What would be the easiest way to replace a substring within a string when I don't know the exact substring I am looking for and only know the delimiting strings? For example, if I have the following:
mystr = 'wordone wordtwo "In quotes"."Another word"'
I basically want to delete the first quoted words (including the quotes) and the period (.) following so the resulting string is:
'wordone wordtwo "Another word"'
Basically I want to delete the first quoted words and the quotes and the following period.
You are looking for regular expressions here, using the re module:
import re
quoted_plus_fullstop = re.compile(r'"[^"]+"\.')
result = quoted_plus_fullstop.sub('', mystr)
The pattern matches a literal quote, followed by 1 or more characters that are not quotes, followed by another quote and a full stop.
Demo:
>>> import re
>>> mystr = 'wordone wordtwo "In quotes"."Another word"'
>>> quoted_plus_fullstop = re.compile(r'"[^"]+"\.')
>>> quoted_plus_fullstop.sub('', mystr)
'wordone wordtwo "Another word"'
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 respective 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['