Fastest way to check if a string contains a string from a list

For this I'd suggest firstly tokenize the string with RegexpTokenizer to remove all special characters and then use sets to find the intersection:

from nltk.tokenize import RegexpTokenizer
test_string = "Hello! This is a test. I love to eat apples."

tokenizer = RegexpTokenizer(r'\w+')
test_set = set(tokenizer.tokenize(test_string))
# {'Hello', 'I', 'This', 'a', 'apples', 'eat', 'is', 'love', 'test', 'to'}

Having tokenized the string and constructed a set find the set.intersection:

set(['apples', 'oranges', 'bananas']) & test_set
# {'apples'}

Yes. you can decrease your iterations like this :

print(any(fruit in frozenset(test_string.replace('.',' ').lower().split()) for fruit in fruits))

Sets are probably your best bet for speed when using the in operator.

For building a set containing only words, we need to:

1) remove the punctuation from the string;

2) split the string in whitespaces.

For removing punctuation, this answer probably has the fastest solution (using str.makestrans and string.punctuation).

Here's an example using your test case:

import string

test_string = "Hello! This is a test. I love to eat apples."
test_string_no_punctuation = test_string.translate(str.maketrans('', '', string.punctuation))
word_set = set(test_string_no_punctuation.split())

fruits = ['apples', 'oranges', 'bananas'] 

for fruit in fruits:
    if fruit in word_set:
        print(fruit+" contains in the string")

You might want to wrap the verbose operations of removing punctuations + splitting the string into a function:

def word_set(input_string):
    return set(input_string.translate(str.maketrans('', '', string.punctuation)).split())

the text is usually bigger than the list of words you are searching for.

for fruit in fruits:
    if fruit in test_string:
        print(fruit+" contains in the string")

this is really inefficient because you are actually looping over the whole text for each fruit in the fruits list, it may not be a problem for short sentences but if you were searching long texts this process would take so much longer.

a better way is to iterate through the text one time and catch all words that are in the fruits list along the way.

If you're only interested if a word is present:

>>> words = set(test_string.replace('.',' ').lower().split())
>>> any(fruit in words for fruit in fruits)
True

You could of course loop over each fruit to check which ones can be found in the fruitcake. So you'd change if fruit in test_string to if fruit in words in your loop example.

You could do something like this:

import re

fruits = ['apples', 'oranges', 'bananas']
test_string = "Hello! This is a test. I love to eat apples."

basket = set(fruits)
words = re.compile('\w+')

for match in words.finditer(test_string):
    fruit = match.group()
    if fruit in basket:
        print(fruit + " contains in the string")

Output

apples contains in the string