How can I get words after and before a specific token?

Question

I currently work on a project which is simply creating basic corpus databases and tokenizes texts. But it seems I am stuck in a matter. Assume that we have those things:

import os, re

texts = []

for i in os.listdir(somedir): # Somedir contains text files which contain very large plain texts.
    with open(i, 'r') as f:
        texts.append(f.read())

Now I want to find the word before and after a token.

myToken = 'blue'
found = []
for i in texts:
    fnd = re.findall('[a-zA-Z0-9]+ %s [a-zA-Z0-9]+|\. %s [a-zA-Z0-9]+|[a-zA-Z0-9]+ %s\.' %(myToken, myToken, myToken), i, re.IGNORECASE|re.UNICODE)
    found.extend(fnd)

print myToken
for i in found:
    print '\t\t%s' %(i)

I thought there would be three possibilities: The token might start sentence, the token might end sentence or the token might appear somewhere in the sentence, so I used the regex rule above. When I run, I come across those things:

blue
    My blue car # What I exactly want.
    he blue jac # That's not what I want. That must be "the blue jacket."
    eir blue phone # Wrong! > their
    a blue ali # Wrong! > alien
    . Blue is # Okay.
    is blue. # Okay.
    ...

I also tried \\b\\w\\b or \\b\\W\\b things, but unfortunately those did not return any results instead of returning wrong results. I tried:

'\b\w\b%s\b[a-zA-Z0-9]+|\.\b%s\b\w\b|\b\w\b%s\.'
'\b\W\b%s\b[a-zA-Z0-9]+|\.\b%s\b\W\b|\b\W\b%s\.'

I hope question is not too blur.

Answer 1

Let's say token is test.

        (?=^test\s+.*|.*?\s+test\s+.*?|.*?\s+test$).*

You can use lookahead.It will not eat up anything and at the same time validate as well.

http://regex101.com/r/wK1nZ1/2

Answer 2

I think what you want is:

1. (Optionally) a word and a space;
2. (Always) 'blue' ;
3. (Optionally) a space and a word.

Therefore one appropriate regex would be:

r'(?i)((?:\w+\s)?blue(?:\s\w+)?)'

For example:

>>> import re
>>> text = """My blue car
the blue jacket
their blue phone
a blue alien
End sentence. Blue is
is blue."""
>>> re.findall(r'(?i)((?:\w+\s)?{0}(?:\s\w+)?)'.format('blue'), text)
['My blue car', 'the blue jacket', 'their blue phone', 'a blue alien', 'Blue is', 'is blue']

See demo and token-by-token explanation here .

Answer 3

Regex can be sometimes slow (if not implemented correctly) and moreover accepted answer did not work for me in several cases.

So I went for the brute force solution (not saying it is the best one), where keyword can be composed of several words:

@staticmethod
def find_neighbours(word, sentence):
    prepost_map = []

    if word not in sentence:
        return prepost_map

    split_sentence = sentence.split(word)
    for i in range(0, len(split_sentence) - 1):
        prefix = ""
        postfix = ""

        prefix_list = split_sentence[i].split()
        postfix_list = split_sentence[i + 1].split()

        if len(prefix_list) > 0:
            prefix = prefix_list[-1]

        if len(postfix_list) > 0:
            postfix = postfix_list[0]

        prepost_map.append([prefix, word, postfix])

    return prepost_map

Empty string before or after the keyword indicates that keyword was the first or the last word in the sentence, respectively.