Finding Multiple Substrings In A String Without Iterating Over It Multiple Times

I need to find if items from a list appear in a string, and then add the items to a different list. This code works: data =[] line = 'akhgvfalfhda.dhgfa.lidhfalihflaih**Thing1**aou

Solution 1:

One way I could think of to improve is:

• Get all unique lengths of the words in _legal
• Build a dictionary of words from line of those particular lengths using a sliding window technique. The complexity should be O( len(line)*num_of_unique_lengths ), this should be better than brute force.
• Now look for each thing in the dictionary in O(1).

Code:

line = 'thing1 thing2 456 xxualt542l lthin. dfjladjfj lauthina '
_legal = ['thing1', 'thing2', 'thing3', 'thing4', 't5', '5', 'fj la']
ul = {len(i) for i in _legal}
s=set()
for l in ul:
    s = s.union({line[i:i+l] for i inrange(len(line)-l)})
print(s.intersection(set(_legal)))

Output:

{'thing1', 'fj la', 'thing2', 't5', '5'}

Solution 2:

One approach is to build a very simple regex pattern, and use re.findall() to find/extract any matched words in the string.

import re

line = 'akhgvfalfhda.dhgfa.lidhfalihflaih**Thing1**aoufgyafkugafkjhafkjhflahfklh**Thing2**dlfkhalfhafli...'
_legal = ['thing1', 'thing2', 'thing3', 'thing4']

exp = re.compile(r'|'.join(_legal), re.IGNORECASE)
exp.findall(line)

>>> ['Thing1', 'Thing2']

Solution 3:

The following should work quite efficiently. It implements the suggestion from @SomeDude given above

lens=set([len(i) for i in _legal])
d={}
for k in lens:
    d[k]=[line[i:i+k] for i inrange(len(line)-k)]
s=set(sum(d.values(), []))
result=list(s.intersection(set(_legal)))

for the following data (i changed "line" a bit as it returned an empty list due to uppecase in Thing1 and Thing2)

line = 'akhgvfalfhda.dhgfa.lidhfalihflaih**thing1**aoufgyafkugafkjhafkjhflahfklh**thing2**dlfkhalfhafli...'_legal = ['thing1', 'thing2', 'thing3', 'thing4',...]

Output is:

print(result)

['thing2', 'thing1']

Explanation: We saved all possible lengths of the words in substrings. For these lengths, we created all possible substrings in text (set s). Finally we found common items in s and substrings, which is the answer to the question.