I\'m building a backend and trying to crunch the following problem.
2000
characters on average)
I faced an almost identical problem with my own chat page system. I wanted to be able to add a link to a number of keywords (with slight variations) that were present in the text. I only had around 200 phrases
though to check.
I decided to try using a standard regular expression for the problem to see how fast it would be. The main bottleneck was in constructing the regular expression. I decided to pre-compile this and found the match time was very fast for shorter texts.
The following approach takes a list of phrases
, where each contains phrase
and link
keys. It first constructs a reverse lookup dictionary:
{'phrase to match' : 'link_url', 'another phrase' : 'link_url2'}
Next it compiles a regular expression in the following form, this allows for matches which contain different amounts of white space between words:
(phrase\s+to\s+match|another\s+phrase)
Then for each piece of text (e.g. 2000 words each), it uses finditer()
to get each match. The match
object gives you .span()
giving the start and end location of the matching text and group(1)
gives the matched text. As the text can possibly have extra whitespace, re_whitespace
is first applied to remove it and bring it back to the form stored in the reverse
dictionary. With this, it is possible to automatically look up the required link
:
import re
texts = ['this is a phrase to match', 'another phrase this is']
phrases = [{'phrase': 'phrase to match', 'link': 'link_url'}, {'phrase': 'this is', 'link': 'link_url2'}]
reverse = {d['phrase']:d['link'] for d in sorted(phrases, key=lambda x: x['phrase'])}
re_whitespace = re.compile(r'\s+')
re_phrases = re.compile('({})'.format('|'.join(d['phrase'].replace(' ', r'\s+') for d in phrases)))
for text in texts:
matches = [(match.span(), reverse[re_whitespace.sub(' ', match.group(1))]) for match in re_phrases.finditer(text)]
print(matches)
Which would display the matches for the two texts as:
[((0, 7), 'link_url2'), ((10, 30), 'link_url')]
[((15, 23), 'link_url2')]
To test how this scales, I have tested it by importing a list of English words from nltk
and automatically creating 80,000
two to six word phrases along with unique links. I then timed it on two suitably long texts:
import re
import random
from nltk.corpus import words
import time
english = words.words()
def random_phrase(l=2, h=6):
return ' '.join(random.sample(english, random.randint(l, h)))
texts = ['this is a phrase to match', 'another phrase this is']
# Make texts ~2000 characters
texts = ['{} {}'.format(t, random_phrase(200, 200)) for t in texts]
phrases = [{'phrase': 'phrase to match', 'link': 'link_url'}, {'phrase': 'this is', 'link': 'link_url2'}]
#Simulate 80k phrases
for x in range(80000):
phrases.append({'phrase': random_phrase(), 'link': 'link{}'.format(x)})
construct_time = time.time()
reverse = {d['phrase']:d['link'] for d in phrases}
re_whitespace = re.compile(r'\s+')
re_phrases = re.compile('({})'.format('|'.join(d['phrase'].replace(' ', r'\s+') for d in sorted(phrases, key=lambda x: len(x['phrase'])))))
print('Time to construct:', time.time() - construct_time)
print()
for text in texts:
start_time = time.time()
print('{} characters - "{}..."'.format(len(text), text[:60]))
matches = [(match.span(), reverse[re_whitespace.sub(' ', match.group(1))]) for match in re_phrases.finditer(text)]
print(matches)
print('Time taken:', time.time() - start_time)
print()
This takes ~17 seconds to construct the regular expression and reverse lookup (which is only needed once). It then takes about 6 seconds per text. For very short text it takes ~0.06 seconds per text.
Time to construct: 16.812477111816406
2092 characters - "this is a phrase to match totaquine externize intoxatio..."
[((0, 7), 'link_url2'), ((10, 30), 'link_url')]
Time taken: 6.000027656555176
2189 characters - "another phrase this is political procoracoidal playstead as..."
[((15, 23), 'link_url2')]
Time taken: 6.190425715255737
This will at least give you an idea to compare against.