Looks very obvious but couldn't find anything similar. I want to split some text and want the pattern of the split condition to be part of the first split part.
some_text = "Hi there. It's a nice weather. Have a great day."
pattern = re.compile(r'\.')
splitted_text = pattern.split(some_text)
returns:
['Hi there', " It's a nice weather", ' Have a great day', '']
What I want is that it returns:
['Hi there.', " It's a nice weather.", ' Have a great day.']
btw: I am only interested in the re solution and not some nltk library what is doing it with other methods.
It would be simpler and more efficient to use re.findall instead of splitting in this case:
re.findall(r'[^.]*\.', some_text)
This returns:
['Hi there.', " It's a nice weather.", ' Have a great day.']
You can use capture groups with re.split:
>>> re.split(r'([^.]+\.)', some_text)
['', 'Hi there.', '', " It's a nice weather.", '', ' Have a great day.', '']
If you want to also strip the leading spaces from the second two sentences, you can have \s* outside the capture group:
>>> re.split(r'([^.]+\.)\s*', some_text)
['', 'Hi there.', '', "It's a nice weather.", '', 'Have a great day.', '']
Or, (with Python 3.7+ or with the regex module) use a zero width lookbehind that will split immediately after a .:
>>> re.split(r'(?<=\.)', some_text)
['Hi there.', " It's a nice weather.", ' Have a great day.', '']
That will split the same even if there is no space after the ..
And you can filter the '' fields to remove the blank results from splitting:
>>> [field for field in re.split(r'([^.]+\.)', some_text) if field]
['Hi there.', " It's a nice weather.", ' Have a great day.']
You can split on the whitespace with a lookbehind to account for the period. Additionally, to account for the possibility of no whitespace, a lookahead can be used:
import re
some_text = "Hi there. It's a nice weather. Have a great day.It is a beautify day."
result = re.split('(?<=\.)\s|\.(?=[A-Z])', some_text)
Output:
['Hi there.', "It's a nice weather.", 'Have a great day', 'It is a beautify day.']
re explanation:
(?<=\.) => position lookbehind, a . must be matched for the next sequence to be matched.
\s => matches whitespace ().
| => Conditional that will attempt to match either the expression to its left or its right, depending on what side matches first.
\. => matches a period
(?=[A-Z]) matches the latter period if the next character is a capital letter.
If each sentence always ends with a ., it would be simpler and more efficient to use the str.split method instead of using any regular expression at all:
[s + '.' for s in some_text.split('.') if s]
This returns:
['Hi there.', " It's a nice weather.", ' Have a great day.']
Related
I have this following string:
string = 'This is one sentence ${w_{1},..,w_{i}}$. This is another sentence. '
Now, I want to split it into two sentence.
However, when I do:
string.split('.')
I get:
['This is one sentence ${w_{1},',
'',
',w_{i}}$',
' This is another sentence',
' ']
Anyone has an idea of how to improve it, in order to not detect the "." within the $ $ ?
Also, how would you go about this:
string2 = 'This is one sentence ${w_{1},..,w_{i}}$! This is another sentence. Is this a sentence? Maybe ! '
EDIT 1:
The desired outputs would be:
For string 1:
['This is one sentence ${w_{1},..,w_{i}}$','This is another sentence']
For string 2:
['This is one sentence ${w_{1},..,w_{i}}$','This is another sentence', 'Is this a sentence', 'Maybe ! ']
For the more general case, you could use re.split like so:
import re
mystr = 'This is one sentence ${w_{1},..,w_{i}}$. This is another sentence. '
re.split("[.!?]\s{1,}", mystr)
# ['This is one sentence ${w_{1},..,w_{i}}$', 'This is another sentence', '']
str2 = 'This is one sentence ${w_{1},..,w_{i}}$! This is another sentence. Is this a sentence? Maybe ! '
re.split("[.!?]\s{1,}", str2)
['This is one sentence ${w_{1},..,w_{i}}$', 'This is another sentence', 'Is this a sentence', 'Maybe ', '']
Where the chars in the brackets are what you pick as your punctuation, and you add at least one space at the end \s{1,} to ignore the other .'s, which have no spacing. This will also handle your exclamation point case
Here's a (somewhat hacky) way to get the punctuation back
punct = re.findall("[.!?]\s{1,}", str2)
['! ', '. ', '? ', '! ']
sent = [x+y for x,y in zip(re.split("[.!?]\s{1,}", str2), punct)]
sent
['This is one sentence ${w_{1},..,w_{i}}$! ', 'This is another sentence. ', 'Is this a sentence? ', 'Maybe ! ']
You can use re.findall with an alternation pattern. To ensure that the sentence starts and ends with a non-whitespace, use a positive lookahead pattern at the start and a positive lookbehind pattern at the end:
re.findall(r'((?=[^.!?\s])(?:$.*?\$|[^.!?])*(?<=[^.!?\s]))\s*[.!?]', string)
This returns, for the first string:
['This is one sentence ${w_{1},..,w_{i}}$', 'This is another sentence']
and for the second string:
['This is one sentence ${w_{1},', ',w_{i}}$', 'This is another sentence', 'Is this a sentence', 'Maybe']
Use '. ' (with a space after the .) because that only exists when a sentence ends, not mid-sentence.
string = 'This is one sentence ${w_{1},..,w_{i}}$. This is another sentence. '
string.split('. ')
this returns:
['This is one sentence ${w_{1},..,w_{i}}$', 'This is another sentence', '']
I want to separate a string on commas and/or periods in nltk. I've tried with sent_tokenize() but it separates only on periods.
I've also tried this code
from nltk.tokenize.punkt import PunktSentenceTokenizer, PunktLanguageVars
ex_sent = "This is an example showing sentence filtration.This is how it is done, in case of Python I want to learn more. So, that i can have some experience over it, by it I mean python."
class CommaPoint(PunktLanguageVars):
sent_end_chars = ('.','?','!',',')
tokenizer = PunktSentenceTokenizer(lang_vars = CommaPoint())
n_w=tokenizer.tokenize(ex_sent)
print n_w
The output for the code above is
['This is an example showing sentence filtration.This is how it is done,' 'in case of Python I want to learn more.' 'So,' 'that i can have some experience over it,' 'by it I mean python.\n']
When I try to give '.' without any space it is taking it as a word
I want the output as
['This is an example showing sentence filtration.' 'This is how it is done,' 'in case of Python I want to learn more.' 'So,' 'that i can have some experience over it,' 'by it I mean python.']
How about something simpler with re:
>>> import re
>>> sent = "This is an example showing sentence filtration.This is how it is done, in case of Python I want to learn more. So, that i can have some experience over it, by it I mean python."
>>> re.split(r'[.,]', sent)
['This is an example showing sentence filtration', 'This is how it is done', ' in case of Python I want to learn more', ' So', ' that i can have some experience over it', ' by it I mean python', '']
To keep the delimiter, you can use group:
>>> re.split(r'([.,])', sent)
['This is an example showing sentence filtration', '.', 'This is how it is done', ',', ' in case of Python I want to learn more', '.', ' So', ',', ' that i can have some experience over it', ',', ' by it I mean python', '.', '']
in this case you maybe can replace all commas with dots in the string and then tokenize it:
from nltk.tokenize import sent_tokenize
ex_sent = "This is an example showing sentence filtration.This is how it is done, in case of Python I want to learn more. So, that i can have some experience over it, by it I mean python."
ex_sent = ex_sent.replace(",", ".")
n_w = sent_tokenize(texto2, 'english')
print(n_w)
I've been attempting to strip out the \n plus the whitespace before and after the words from a string, but it is not working for some reason.
This is what I tried:
.strip(my_string)
and
re.sub('\n', '', my string)
I have tried using .strip and re in order to get it working, but it simply returns the same string.
Example input:
\\n The people who steal our cards already know all of this...\\n
\\n , \\n I\'m sure every fraud minded person in America is taking notes.\\n
\\n
Expected output would be:
The people who steal our cards already know all of this..., I\'m sure every fraud minded person in America is taking notes.
You're probably looking for something like this:
re.sub(r'\s+', r' ', x)
A usage example follows:
In [10]: x
Out[10]: 'hello \n world \n blue'
In [11]: re.sub(r'\s+', r' ', x)
Out[11]: 'hello world blue'
If you'd also like to grab the sequence of characters r'\n', then let's grab them as well:
re.sub(r'(\s|\\n)+', r' ', x)
And the output:
In [14]: x
Out[14]: 'hello \\n world \n \\n blue'
In [15]: re.sub(r'(\s|\\n)+', r' ', x)
Out[15]: 'hello world blue'
How can a string be split on a separator expression while leaving that separator on the preceding string?
>>> text = "This is an example. Is it made up of more than once sentence? Yes, it is."
>>> re.split("[\.\?!] ", text)
['This is an example', 'Is it made up of more than one sentence', 'Yes, it is.']
I would like the result to be.
['This is an example.', 'Is it made up of more than one sentence?', 'Yes, it is.']
So far I have only tried a lookahead assertion but this fails to split at all.
>>> re.split("(?<=[\.\?!]) ", text)
['This is an example.', 'Is it made up of more than once sentence?', 'Yes, it is.']
The crucial thing is the use of a look-behind assertion with ?<=.
import re
text = "This is an example.A particular case.Made up of more "\
"than once sentence?Yes, it is.But no blank !!!That's"\
" a problem ????Yes.I think so! :)"
for x in re.split("(?<=[\.\?!]) ", text):
print repr(x)
print '\n'
for x in re.findall("[^.?!]*[.?!]|[^.?!]+(?=\Z)",text):
print repr(x)
result
"This is an example.A particular case.Made up of more than once sentence?Yes, it is.But no blank !!!That'sa problem ????Yes.I think so!"
':)'
'This is an example.'
'A particular case.'
'Made up of more than once sentence?'
'Yes, it is.'
'But no blank !'
'!'
'!'
"That's a problem ?"
'?'
'?'
'?'
'Yes.'
'I think so!'
' :)'
.
EDIT
Also
import re
text = "! This is an example.A particular case.Made up of more "\
"than once sentence?Yes, it is.But no blank !!!That's"\
" a problem ????Yes.I think so! :)"
res = re.split('([.?!])',text)
print [ ''.join(res[i:i+2]) for i in xrange(0,len(res),2) ]
gives
['!', ' This is an example.', 'A particular case.', 'Made up of more than once sentence?', 'Yes, it is.', 'But no blank !', '!', '!', "That's a problem ?", '?', '?', '?', 'Yes.', 'I think so!', ' :)']
Here is what I have so far
text = "Hello world. It is a nice day today. Don't you think so?"
re.findall('\w{3,}\s{1,}\w{3,}',text)
#['Hello world', 'nice day', 'you think']
The desired output would be ['Hello world', 'nice day', 'day today', 'today Don't', 'Don't you', 'you think']
Can this be done with a simple regex pattern?
import itertools as it
import re
three_pat=re.compile(r'\w{3}')
text = "Hello world. It is a nice day today. Don't you think so?"
for key,group in it.groupby(text.split(),lambda x: bool(three_pat.match(x))):
if key:
group=list(group)
for i in range(0,len(group)-1):
print(' '.join(group[i:i+2]))
# Hello world.
# nice day
# day today.
# today. Don't
# Don't you
# you think
It not clear to me what you want done with all punctuation. On the one hand, it looks like you want periods to be removed, but single quotation marks to be kept. It would be easy to implement the removal of periods, but before I do, would you clarify what you want to happen to all punctuation?
map(lambda x: x[0] + x[1], re.findall('(\w{3,}(?=(\s{1,}\w{3,})))',text))
May be you can rewrite the lambda for shorter (like just '+')
And BTW ' is not part of \w or \s
Something like this with additional checks for list boundaries should do:
>>> text = "Hello world. It is a nice day today. Don't you think so?"
>>> k = text.split()
>>> k
['Hello', 'world.', 'It', 'is', 'a', 'nice', 'day', 'today.', "Don't", 'you', 'think', 'so?']
>>> z = [x for x in k if len(x) > 2]
>>> z
['Hello', 'world.', 'nice', 'day', 'today.', "Don't", 'you', 'think', 'so?']
>>> [z[n]+ " " + z[n+1] for n in range(0, len(z)-1, 2)]
['Hello world.', 'nice day', "today. Don't", 'you think']
>>>
There are two problems with your approach:
Neither \w nor \s matches punctuation.
When you match a string with a regular expression using findall, that part of the string is consumed. Searching for the next match commences immediately after the end of the previous match. Because of this a word can't be included in two separate matches.
To solve the first issue you need to decide what you mean by a word. Regular expressions aren't good for this sort of parsing. You might want to look at a natural language parsing library instead.
But assuming that you can come up with a regular expression that works for your needs, to fix the second problem you can use a lookahead assertion to check the second word. This won't return the entire match as you want but you can at least find the first word in each word pair using this method.
re.findall('\w{3,}(?=\s{1,}\w{3,})',text)
^^^ ^
lookahead assertion