I wrote a tokenize function that basically reads a string representation and splits it into list of words.
My code:
def tokenize(document):
x = document.lower()
return re.findall(r'\w+', x)
My output:
tokenize("Hi there. What's going on? first-class")
['hi', 'there', 'what', 's', 'going', 'on', 'first', 'class']
Desired Output:
['hi', 'there', "what's", 'going', 'on', 'first-class']
Basically I want the apostrophed words and hypen words to remain as a single word in list along with double quotes. How can i change my function to get the desired output.
\w+ matches one or more word characters; this does not include apostrophes or hyphens.
You need to use a character set here to tell Python exactly what you want to match:
>>> import re
>>> def tokenize(document):
... return re.findall("[A-Za-z'-]+", document)
...
>>> tokenize("Hi there. What's going on? first-class")
['hi', 'there', "what's", 'going', 'on', 'first-class']
>>>
You'll notice too that I removed the x = document.lower() line. This is no longer necessary since we can match uppercase characters by simply adding A-Z to the character set.
Related
I have a function made to tokenize the words given; however, the problem ensures in the fact that it does separate 'isn't' and 'o'brian' and similar words. This is the code:
from typing import List
import re
def tokenize(text: str) -> List[str]:
return re.sub(r'(\W+)', r' \1 ', text.lower()).split()
When I put in something like
"He said 'Isn't O'Brian the best?'"
it'll come as
['he', 'said', "'", "isn't", "o'brian", 'the', 'best', '?', "'"],
not
['he', 'said', "'", 'isn', "'", 't', 'o', "'", 'brian', 'the', 'best', "?'"]
I'm really lost because I've tried to do more than re.sub to separate the words and tried to split them, but it seemingly is not working.
You need a better definition of "word", for example:
word_re = r"\b[A-Za-z]+(?:'[A-Za-z]+)?\b"
that is, a word boundary, then some letters, then optionally an apostrophe, followed by letters. Once you've got it, use findall to extract words:
words = [w.lower() for w in re.findall(word_re, text)]
if the apostrophes arent important to the meaning of the word, ie "obrien" and "o'brien" are the same for your purposes you could do some basic text pre-processing to remove all the apostrophes using text.replace("'","") see here
if they do matter, you could replace them with Unicode (the unicode for ' is U+0027) and translate them out of Unicode after you've tokenized them, or just text.replace("'","U+0027") before you tokenize, and text.replace("U+0027","'") after
hope that helps, goodluck!
Question: please debug logic to reflect expected output
import re
text = "Hello there."
word_list = []
for word in text.split():
tmp = re.split(r'(\W+)', word)
word_list.extend(tmp)
print(word_list)
OUTPUT is :
['Hello', 'there', '.', '']
Problem: needs to be expected without space
Expected: ['Hello', 'there', '.']
First of all the actual output you shared is not right, it is ['Hello', ' ', 'there', '.', ''] because-
The \W, Matches anything other than a letter, digit or underscore. Equivalent to [^a-zA-Z0-9_] so it is splitting your string by space(\s) and literal dot(.) character
So if you want to get the expected output you need to do some further processing like the below-
With Earlier Code:
import re
s = "Hello there."
l = list(filter(str.strip,re.split(r"(\W+)", s)))
print(l)
With Edited code:
import re
text = "Hello there."
word_list = []
for word in text.split():
tmp = re.split(r'(\W+)', word)
word_list.extend(tmp)
print(list(filter(None,word_list)))
Output:
['Hello', 'there', '.']
Working Code: https://rextester.com/KWJN38243
assuming word is "Hello there.", the results make sense. See the split function documentation: Split string by the occurrences of pattern. If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.
You have put capturing parenthesis in the pattern, so you are splitting the string on non-word characters, and also return the characters used for splitting.
Here is the string:
Hello there.
Here is how it is split:
Hello|there|
that means you have three values: hello there and an empty string '' in the last place.
And the values you split on are a space and a period
So the output should be the three values and the two characters that we split on:
hello - space - there - period - empty string
which is exactly what I get.
import re
s = "Hello there."
t = re.split(r"(\W+)", s)
print(t)
output:
['Hello', ' ', 'there', '.', '']
Further Explanation
From your question is may be that you think because the string ends with a non-word character that there would be nothing "after" it, but this is not how splitting works. If you think back to CSV files (which have been around forever, and consider a CSV file like this:
date,product,qty,price
20220821,P1,10,20.00
20220821,P2,10,
The above represents a csv file with four fields, but in line two the last field (which definitely exists) is missing. And it would be parsed as an empty string if we split on the comma.
I'm trying to write a regex that takes in a string of words containing quoted substrings like "green lizards" like to sit "in the sun", tokenizes it into words and quoted substrings (using either single or double quotes) separated by spaces, and then returns a list [['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']] where the list items are either single words or nested lists of words where a quoted substrings was encountered.
I am new to regex, and was able to find a solution that captures the quoted parts: re.findall('"([^"]*)"', '"green lizards" like to sit "in the sun"') ... which returns: ['green lizards', 'in the sun']
But this doesn't capture the individual words, and also doesn't tokenize them (returning a single string instead of list of words, which requires me to split() them each separately.
How would I make a regex that correctly returns the type of list I'm wanting? Also, I'm open to better methods/tools than regex for parsing these sorts of strings if anyone has suggestions.
Thanks!
With re.findall() function and built-in str methods:
import re
s = '"green lizards" like to sit "in the sun"'
result = [i.replace('"', "").split() if i.startswith('"') else i
for i in re.findall(r'"[^"]+"|\S+', s)]
print(result)
The output:
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']]
Another approach (supporting both single and double quotes):
import re
sentence = """"green lizards" like to sit "in the sun" and 'single quotes' remain alone"""
rx = re.compile(r"""(['"])(.*?)\1|\S+""")
tokens = [m.group(2).split()
if m.group(2) else m.group(0)
for m in rx.finditer(sentence)]
print(tokens)
Yielding
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun'], 'and', ['single', 'quotes'], 'remain', 'alone']
The idea here is:
(['"]) # capture a single or a double quote
(.*?) # 0+ characters lazily
\1 # up to the same type of quote previously captured
| # ...or...
\S+ # not a whitespace
In the list comprehension we check which condition was met.
You can use re.split and then a final str.split:
import re
s = '"green lizards" like to sit "in the sun"'
new_s = [[i[1:-1].split()] if i.startswith('"') else i.split() for i in re.split('(?<=")\s|\s(?=")', s)]
last_result = [i for b in new_s for i in b]
Output:
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']]
I have this code
import re
str1 = "These should be counted as a single-word, b**m !?"
match_pattern = re.findall(r'\w{1,15}', str1)
print(match_pattern)
I want the output to be:
['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']
The output should exclude non-words such as the "!?" what are the other validation should I use to match and achieve the desired output?
I would use word boundaries (\b) filled with 1 or more non-space:
match_pattern = re.findall(r'\b\S+\b', str1)
result:
['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']
!? is skipped thanks to word boundary magic, which don't consider that as a word at all either.
Probably you want something like [^\s.!?] instead of \w but what exactly you want is not evident from a single example. [^...] matches a single character which is not one of those between the brackets and \s matches whitespace characters (space, tab, newline, etc).
You can also achieve a similar result not using RegEx:
string = "These should be counted as a single-word, b**m !?"
replacements = ['.',',','?','!']
for replacement in replacements:
if replacement in string:
string = string.replace(replacement, "");
print string.split()
>>> ['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']
I have a string composed of numbers and letters: string = 'this1234is5678it', and I would like the string.split() output to give me a list like ['this', '1234', 'is', '5678', 'it'], splitting at where numbers and letter meet. Is there an easy way to do this?
You can use Regex for this.
import re
s = 'this1234is5678it'
re.split('(\d+)',s)
Running example http://ideone.com/JsSScE
Outputs ['this', '1234', 'is', '5678', 'it']
Update
Steve Rumbalski mentioned in the comment the importance of the parenthesis in the regex. He quotes from the documentation:
If capturing parentheses are used in pattern, then the text of all
groups in the pattern are also returned as part of the resulting
list." Without the parenthesis the result would be ['this', 'is',
'it'].