This sounds very simple, I know, but for some reason I can't get all the results I need
Word in this case is any char but white-space that is separetaed with white-space
for example in the following string: "Hello there stackoverflow."
the result should be: ['Hello','there','stackoverflow.']
My code:
import re
word_pattern = "^\S*\s|\s\S*\s|\s\S*$"
result = re.findall(word_pattern,text)
print result
but after using this pattern on a string like I've shown it only puts the first and the last words in the list and not the words separeted with two spaces
What is the problem with this pattern?
Use the \b boundary test instead:
r'\b\S+\b'
Result:
>>> import re
>>> re.findall(r'\b\S+\b', 'Hello there StackOverflow.')
['Hello', 'there', 'StackOverflow']
or not use a regular expression at all and just use .split(); the latter would include the punctiation in a sentence (the regex above did not match the . in the sentence).
to find all words in a string best use split
>>> "Hello there stackoverflow.".split()
['Hello', 'there', 'stackoverflow.']
but if you must use regular expressions, then you should change your regex to something simpler and faster: r'\b\S+\b'.
r turns the string to a 'raw' string. meaning it will not escape your characters.
\b means a boundary, which is a space, newline, or punctuation.
\S you should know, is any non-whitespace character.
+ means one or more of the previous.
so together it means find all visible sets of characters (words/numbers).
How about simply using -
>>> s = "Hello there stackoverflow."
>>> s.split()
['Hello', 'there', 'stackoverflow.']
The other answers are good. Depending on what you want (eg. include/exclude punctuation or other non-word characters) an alternative could be to use a regex to split by one or more whitespace characters:
re.split(r'\s+', 'Hello there StackOverflow.')
['Hello', 'There', 'StackOverflow.']
Related
i want to split a string by all spaces and punctuation except for the apostrophe sign. Preferably a single quote should still be used as a delimiter except for when it is an apostrophe. I also want to keep the delimeters.
example string
words = """hello my name is 'joe.' what's your's"""
Here is my re pattern thus far splitted = re.split(r"[^'-\w]",words.lower())
I tried throwing the single quote after the ^ character but it is not working.
My desired output is this. splitted = [hello,my,name,is,joe,.,what's,your's]
It might be simpler to simply process your list after splitting without accounting for them at first:
>>> words = """hello my name is 'joe.' what's your's"""
>>> split_words = re.split(r"[ ,.!?]", words.lower()) # add punctuation you want to split on
>>> split_words
['hello', 'my', 'name', 'is', "'joe.'", "what's", "your's"]
>>> [word.strip("'") for word in split_words]
['hello', 'my', 'name', 'is', 'joe.', "what's", "your's"]
One option is to make use of lookarounds to split at the desired positions, and use a capture group what you want to keep in the split.
After the split, you can remove the empty entries from the resulting list.
\s+|(?<=\s)'|'(?=\s)|(?<=\w)([,.!?])
The pattern matches
\s+ Match 1 or more whitespace chars
| Or
(?<=\s)' Match ' preceded by a whitespace char
| Or
'(?=\s) Match ' when followed by a whitespace char
| Or
(?<=\w)([,.!?]) Capture one of , . ! ? in group 1, when preceded by a word character
See a regex demo and a Python demo.
Example
import re
pattern = r"\s+|(?<=\s)'|'(?=\s)|(?<=\w)([,.!?])"
words = """hello my name is 'joe.' what's your's"""
result = [s for s in re.split(pattern, words) if s]
print(result)
Output
['hello', 'my', 'name', 'is', 'joe', '.', "what's", "your's"]
I love regex golf!
words = """hello my name is 'joe.' what's your's"""
splitted = re.findall(r"\b(?:\w'\w|\w)+\b", words)
The part in the parenthesis is a group that matches either an apostrophe surrounded by letters or a single letter.
EDIT:
This is more flexible:
re.findall(r"\b(?:(?<=\w)'(?=\w)|\w)+\b", words)
It's getting a bit unreadable at this point though, in practice you should probably use Woodford's answer.
I got a long string and i need to find words which contain the character 'd' and afterwards the character 'e'.
l=[" xkn59438","yhdck2","eihd39d9","chdsye847","hedle3455","xjhd53e","45da","de37dp"]
b=' '.join(l)
runs1=re.findall(r"\b\w?d.*e\w?\b",b)
print(runs1)
\b is the boundary of the word, which follows with any char (\w?) and etc.
I get an empty list.
You can massively simplify your solution by applying a regex based search on each string individually.
>>> p = re.compile('d.*e')
>>> list(filter(p.search, l))
Or,
>>> [x for x in l if p.search(x)]
['chdsye847', 'hedle3455', 'xjhd53e', 'de37dp']
Why didn't re.findall work? You were searching one large string, and your greedy match in the middle was searching across strings. The fix would've been
>>> re.findall(r"\b\S*d\S*e\S*", ' '.join(l))
['chdsye847', 'hedle3455', 'xjhd53e', 'de37dp']
Using \S to match anything that is not a space.
You can filter the result :
import re
l=[" xkn59438","yhdck2","eihd39d9","chdsye847","hedle3455","xjhd53e","45da","de37dp"]
pattern = r'd.*?e'
print(list(filter(lambda x:re.search(pattern,x),l)))
output:
['chdsye847', 'hedle3455', 'xjhd53e', 'de37dp']
Something like this maybe
\b\w*d\w*e\w*
Note that you can probably remove the word boundary here because
the first \w guarantees a word boundary before.
The same \w*d\w*e\w*
For example, if I have a string:
"I really..like something like....that"
I want to get only:
"I something"
Any suggestion?
If you want to do it with regex; you can to use below regex to remove them:
r"[^\.\s]+\.{2,}[^\.\s]+"g
[ Regex Demo ]
Regex explanation:
[^\.\s]+ at least one of any character instead of '.' and a white space
\.{2,} at least two or more '.'
[^\.\s]+ at least one of any character instead of '.' and a white space
or this regex:
r"\s+[^\.\s]+\.{2,}[^\.\s]+"g
^^^ for including spaces before those combination
[ Regex Demo ]
If you want to use a regex explicitly you could use the following.
import re
string = "I really..like something like....that"
with_dots = re.findall(r'\w+[.]+\w+', string)
split = string.split()
without_dots = [word for word in split if word not in with_dots]
The solution provided by rawing also works in this case.
' '.join(word for word in text.split() if '..' not in word)
You may very well use boundaries in combination with lookarounds:
\b(?<!\.)(\w+)\b(?!\.)
See a demo on regex101.com.
Broken apart, this says:
\b # a word boundary
(?<!\.) # followed by a negative lookbehind making sure there's no '.' behind
\w+ # 1+ word characters
\b # another word boundary
(?!\.) # a negative lookahead making sure there's no '.' ahead
As a whole Python snippet:
import re
string = "I really..like something like....that"
rx = re.compile(r'\b(?<!\.)(\w+)\b(?!\.)')
print(rx.findall(string))
# ['I', 'something']
I'm trying to do this:
import re
sentence = "How are you?"
print(re.split(r'\b', sentence))
The result being
[u'How are you?']
I want something like [u'How', u'are', u'you', u'?']. How can this be achieved?
Unfortunately, Python cannot split by empty strings.
To get around this, you would need to use findall instead of split.
Actually \b just means word boundary.
It is equivalent to (?<=\w)(?=\W)|(?<=\W)(?=\w).
That means, the following code would work:
import re
sentence = "How are you?"
print(re.findall(r'\w+|\W+', sentence))
import re
split = re.findall(r"[\w']+|[.,!?;]", "How are you?")
print(split)
Output:
['How', 'are', 'you', '?']
Ideone Demo
Regex101 Demo
Regex Explanation:
"[\w']+|[.,!?;]"
1st Alternative: [\w']+
[\w']+ match a single character present in the list below
Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
\w match any word character [a-zA-Z0-9_]
' the literal character '
2nd Alternative: [.,!?;]
[.,!?;] match a single character present in the list below
.,!?; a single character in the list .,!?; literally
Here is my approach to split on word boundaries:
re.split(r"\b\W\b", "How are you?") # Reprocess list to split on special characters.
# Result: ['How', 'are', 'you?']
and using findall on word boundaries
re.findall(r"\b\w+\b", "How are you?")
# Result: ['How', 'are', 'you']
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.