I want my code to only return the special characters [".", "*", "=", ","]
I want to remove all digits/alphabetical characters ("\W") and all white spaces ("\S")
import re
original_string = "John is happy. He owns 3*4=12, apples"
new_string = re.findall("\W\S",original_string)
print(new_string)
But instead I get this as my output:
[' i', ' h', ' H', ' o', ' 3', '*4', '=1', ' a']
I have absolutely no idea why this happens. Hence I have two questions:
1) Is it possible to achieve my goal using regular expressions
2) What is actually going on with my code?
You were close, but you need to specify these escape sequences inside a character class.
re.findall(r'[^\w\s]', original_string)
# ['.', '*', '=', ',']
Note that the caret ^ indicates negation (i.e., don't match these characters).
Alternatively, instead of removing what you don't need, why not extract what you do?
re.findall(r'[.*=,]', original_string)
# ['.', '*', '=', ',']
Here, we can also add our desired special chars in a [], swipe everything else, and then collect only those chars:
([\s\S].*?)([.*=,])?
Python Test
# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility
import re
regex = r"([\s\S].*?)([.*=,])?"
test_str = "John is happy. He owns 3*4=12, apples"
subst = "\\2"
# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 0, re.MULTILINE)
if result:
print (result)
# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.
JavaScript Demo
const regex = /([\s\S].*?)([.*=,])?/gm;
const str = `John is happy. He owns 3*4=12, apples`;
const subst = `$2`;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
RegEx
If this wasn't our desired expression, we can modify/change it in regex101.com.
RegEx Circuit
We can also visualize expressions in jex.im:
Demo
The regular expression \W\S matches a sequence of two characters; one non-word, and one non-space. If you want to combine them, that's [^\w\s] which matches one character which does not belong to either the word or the whitespace group.
However, there are many characters which are not one of the ones you enumerate which match this expression. If you want to remove characters which are not in your set, the character class containing exactly all those characters is simply [^.*=,]
Perhaps it's worth noting that inside [...] you don't need to (and in fact should not) backslash-escape e.g. the literal dot. By default, a character class cannot match a newline character, though there is an option re.DOTALL to change this.
If you are trying to extract and parse numerical expressions, regex can be a useful part of the lexical analysis, but you really want a proper parser.
Related
I want to get the content between single quotes, but only if it contains a certain word (i.e 'sample_2'). It additionally should not match ones with white space.
Input example: (The following should match and return only: ../sample_2/file and sample_2/file)
['asdf', '../sample_2/file', 'sample_2/file', 'example with space', sample_2, sample]
Right now I just have that matched the first 3 items in the list:
'(.\S*?)'
I can't seem to find the right regex that would return those containing the word 'sample_2'
If you want specific words/characters you need to have them in the regular expression and not use the '\S'. The \S is the equivalent to [^\r\n\t\f\v ] or "any non-whitespace character".
import re
teststr = "['asdf', '../sample_2/file', 'sample_2/file', 'sample_2 with spaces','example with space', sample_2, sample]"
matches = re.findall(r"'([^\s']*sample_2[^\s]*?)',", teststr)
# ['../sample_2/file', 'sample_2/file']
Based on your wording, you suggest the desired word can change. In that case, I would recommend using re.compile() to dynamically create a string which then defines the regular expression.
import re
word = 'sample_2'
teststr = "['asdf', '../sample_2/file', 'sample_2/file', ' sample_2 with spaces','example with space', sample_2, sample]"
regex = re.compile("'([^'\\s]*"+word+"[^\\s]*?)',")
matches = regex.findall(teststr)
# ['../sample_2/file', 'sample_2/file']
Also if you haven't heard of this tool yet, check out regex101.com. I always build my regular expressions here to make sure I get them correct. It gives you the references, explanation of what is happening and even lets you test it right there in the browser.
Explanation of regex
regex = r"'([^\s']*sample_2[^\s]*?)',"
Find first apostrophe, start group capture. Capture anything except a whitespace character or the corresponding ending apostrophe. It must see the letters "sample_2" before accepting any non-whitespace character. Stop group capture when you see the closing apostrophe and a comma.
Note: In python, a string " or ' prepositioned with the character 'r' means the text is compiled as a regular expression. Strings with the character 'r' also do not require double-escape '\' characters.
My goal is to find a group in a string using regex and replace it with a space.
The group I am looking to find is a group of symbols only when they fall between strings. When I use re.findall() it works exactly as expected
word = 'This##Is # A # Test#'
print(word)
re.findall(r"[a-zA-Z\s]*([\$\#\%\!\s]*)[a-zA-Z]",word)
>>> ['##', '# ', '# ', '']
But when I use re.sub(), instead of replacing the group, it replaces the entire regex.
x = re.sub(r"[a-zA-Z\s]*([\$\#\%\!\s]*)[a-zA-Z]",r' ',word)
print(x)
>>> ' #'
How can I use regular expressions to replace ONLY the group? The outcome I expect is:
'This Is A Test#'
First, there's no need to escape every "magic" character within a character class, [$#%!\s]* is equally fine and much more readable.
Second, matching (i.e. retrieving) is different from replacing and you could use backreferences to achieve your goal.
Third, if you only want to have # at the end, you could help yourself with a much easier expression:
(?:[\s#](?!\Z))+
Which would then need to be replaced by a space, see a demo on regex101.com.
In Python this could be:
import re
string = "This##Is # A # Test#"
rx = re.compile(r'(?:[\s#](?!\Z))+')
new_string = rx.sub(' ', string)
print(new_string)
# This Is A Test#
You can group the portions of the pattern you want to retain and use backreferences in your replacement string instead:
x = re.sub(r"([a-zA-Z\s]*)[\$\#\%\!\s]*([a-zA-Z])", r'\1 \2', word)
The problem is that your regex matches the wrong thing entirely.
x = re.sub(r'\b[$#%!\s]+\b', ' ', word)
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.
I'm fighting too long on this regex now.
The split should use blank as separator
but maintain the remaining ones in a blank sequence to the next token
'123 45 678 123.0'
=>
'123', '45', ' 678', ' 123.0'
My numbers are floats as well and the group count is unknown.
What about using a lookbehind assertion?:
>>> import re
>>> regex = re.compile(r'(?<=[^\s])\s')
>>> regex.split('this is a string')
['this', ' is', 'a', ' string']
regex breakdown:
(?<=...) #lookbehind. Only match if the `...` matches before hand
[^\s] #Anything that isn't whitespace
\s #single whitespace character
In english, this translates to "match a single whitespace character if it isn't preceded by a whitespace character."
Or you can use a negative lookbehind assertion:
regex = re.compile(r'(?<!\s)\s')
which might be slightly nicer (as suggested in the comments), and should be relatively easy to figure out how it works since it is very similar to the above.
I'm trying to match a specific pattern using the re module in python.
I wish to match a full sentence (More correctly I would say that they are alphanumeric string sequences separated by spaces and/or punctuation)
Eg.
"This is a regular sentence."
"this is also valid"
"so is This ONE"
I'm tried out of various combinations of regular expressions but I am unable to grasp the working of the patterns properly, with each expression giving me a different yet inexplicable result (I do admit I am a beginner, but still).
I'm tried:
"((\w+)(\s?))*"
To the best of my knowledge this should match one or more alpha alphanumerics greedily followed by either one or no white-space character and then it should match this entire pattern greedily. This is not what it seems to do, so clearly I am wrong but I would like to know why. (I expected this to return the entire sentence as the result)
The result I get for the first sample string mentioned above is [('sentence', 'sentence', ''), ('', '', ''), ('', '', ''), ('', '', '')].
"(\w+ ?)*"
I'm not even sure how this one should work. The official documentation(python help('re')) says that the ,+,? Match x or x (greedy) repetitions of the preceding RE.
In such a case is simply space the preceding RE for '?' or is '\w+ ' the preceding RE? And what will be the RE for the '' operator? The output I get with this is ['sentence'].
Others such as "(\w+\s?)+)" ; "((\w*)(\s??)) etc. which are basically variation of the same idea that the sentence is a set of alpha numerics followed by a single/finite number of white spaces and this pattern is repeated over and over.
Can someone tell me where I go wrong and why, and why the above expressions do not work the way I was expecting them to?
P.S I eventually got "[ \w]+" to work for me but With this I cannot limit the number of white-space characters in continuation.
Your reasoning about the regex is correct, your problem is coming from using capturing groups with *. Here's an alternative:
>>> s="This is a regular sentence."
>>> import re
>>> re.findall(r'\w+\s?', s)
['This ', 'is ', 'a ', 'regular ', 'sentence']
In this case it might make more sense for you to use \b in order to match word boundries.
>>> re.findall(r'\w+\b', s)
['This', 'is', 'a', 'regular', 'sentence']
Alternatively you can match the entire sentence via re.match and use re.group(0) to get the whole match:
>>> r = r"((\w+)(\s?))*"
>>> s = "This is a regular sentence."
>>> import re
>>> m = re.match(r, s)
>>> m.group(0)
'This is a regular sentence'
Here's an awesome Regular Expression tutorial website:
http://regexone.com/
Here's a Regular Expression that will match the examples given:
([a-zA-Z0-9,\. ]+)
Why do you want to limit the number of white space character in continuation? Because a sentence can have any number of words (sequences of alphanumeric characters) and spaces in a row, but rather a sentence is the area of text that ends with a punctuation mark or rather something that is not in the above sequence including white space.
([a-zA-Z0-9\s])*
The above regex will match a sentence wherein it is a series or spaces in series zero or more times. You can refine it to be the following though:
([a-zA-Z0-9])([a-zA-Z0-9\s])*
Which simply states that the above sequence must be prefaced with a alphanumeric character.
Hope this is what you were looking for.
Maybe this will help:
import re
source = """
This is a regular sentence.
this is also valid
so is This ONE
how about this one followed by this one
"""
re_sentence = re.compile(r'[^ \n.].*?(\.|\n| +)')
def main():
i = 0
for s in re_sentence.finditer(source):
print "%d:%s" % (i, s.group(0))
i += 1
if __name__ == '__main__':
main()
I am using alternation in the expression (\.|\n| +) to describe the end-of-sentence condition. Note the use of two spaces in the third alternation. The second space has the '+' meta-character so that two or more spaces in a row will be an end-of-sentence.