Check if substring is in a list of strings?

Question

I have found some answers to this question before, but they seem to be obsolete for the current Python versions (or at least they don’t work for me).

I want to check if a substring is contained in a list of strings. I only need the boolean result.

I found this solution:

word_to_check = 'or'
wordlist = ['yellow','orange','red']

result = any(word_to_check in word for word in worldlist)

From this code I would expect to get a True value. If the word was “der”, then the output should be False.

However, the result is a generator function, and I can’t find a way to get the True value.

Any idea?

Asked By: Álvaro

||

Source

Answer 1

You can import any from __builtin__ in case it was replaced by some other any:

>>> from  __builtin__ import any as b_any
>>> lst = ['yellow', 'orange', 'red']
>>> word = "or"
>>> b_any(word in x for x in lst)
True

Note that in Python 3 __builtin__ has been renamed to builtins.

Answered By: Ashwini Chaudhary

Answer 2

You could use next instead:

colors = ['yellow', 'orange', 'red'] 
search = "or"

result = next((True for color in colors if search in color), False)

print(result) # True

To show the string that contains the substring:

colors = ['yellow', 'orange', 'red'] 
search = "or"

result = [color for color in colors if search in color]  

print(result) # Orange

Answered By: stderr

Answer 3

Posted code

The OP’s posted code using any() is correct and should work. The spelling of "worldlist" needs to be fixed though.

Alternate approach with str.join()

That said, there is a simple and fast solution to be had by using the substring search on a single combined string:

>>> wordlist = ['yellow','orange','red']
>>> combined = 't'.join(wordlist)

>>> 'or' in combined
True
>>> 'der' in combined
False

For short wordlists, this is several times faster than the approach using any.

And if the combined string can be precomputed before the search, the in-operator search will always beat the any approach even for large wordlists.

Alternate approach with sets

The O(n) search speed can be reduced to O(1) if a substring set is precomputed in advance and if we don’t mind using more memory.

Precomputed step:

from itertools import combinations

def substrings(word):
    for i, j in combinations(range(len(word) + 1), 2):
        yield word[i : j]

wordlist = ['yellow','orange','red']
word_set = set().union(*map(substrings, wordlist))

Fast O(1) search step:

>>> 'or' in word_set
True
>>> 'der' in word_set
False

Answered By: Raymond Hettinger

Answer 4

Also if someone wants to check if any of the values of a dictionary exists as a substring in a list of strings, can use this:

list_a = [
    'Copy of snap-009ecf9feb43d902b from us-west-2',
    'Copy of snap-0fe999422014504b6 from us-west-2',
    'Copy of snap-0fe999422014cscx504b6 from us-west-2',
    'Copy of snap-0fe999422sdad014504b6 from us-west-2'
]
dict_b = {
    '/dev/xvda': 'snap-0fe999422014504b6',
    '/dev/xvdsdsa': 'snap-sdvcsdvsdvs'
}

for b1 in dict_b.itervalues():
    result = next( ("found" for a1 in a if b1 in a1), "not found")
    print result

It prints

not found
found

Answered By: Kostas Demiris

Check if substring is in a list of strings?

Question:

Answers:

Posted code

Alternate approach with str.join()

Alternate approach with sets