CODEWITHSUNDEEP
pythonstringlistsubstring
Matt Fisher
Matt Fisher

Reputation: 303

Python - Remove any element from a list of strings that is a substring of another element

So starting with a list of strings, as below

string_list = ['rest', 'resting', 'look', 'looked', 'it', 'spit']

I want to remove any element from the list that is a substring of another element, giving the result for instance...

string_list = ['resting', 'looked', 'spit']

I have some code that acheives this but it's embarrassingly ugly and probably needlessly complex. Is there a simple way to do this in Python?

Upvotes: 17

Views: 18810

Answers (8)

Carl Bellingan
Carl Bellingan

Reputation: 1480

Here's an un-optimal way, only use if the lists are small:

for str1 in string_list:
    for str2 in string_list:
        if str1 in str2 and str1 != str2:
            string_list.remove(str1)

Upvotes: 0

aiwabdn
aiwabdn

Reputation: 2230

Here's another way to do it. Assuming you have a sorted list to start with and you don't have to do the sieving inplace, we can just choose the longest strings in one pass:

string_list = sorted(string_list)
sieved = []
for i in range(len(string_list) - 1):
    if string_list[i] not in string_list[i+1]:
        sieved.append(string_list[i])

Upvotes: -1

Liyan Chang
Liyan Chang

Reputation: 8051

First building block: substring.

You can use in to check:

>>> 'rest' in 'resting'
True
>>> 'sing' in 'resting'
False

Next, we're going to choose the naive method of creating a new list. We'll add items one by one into the new list, checking if they are a substring or not.

def substringSieve(string_list):
    out = []
    for s in string_list:
        if not any([s in r for r in string_list if s != r]):
            out.append(s)
    return out

You can speed it up by sorting to reduce the number of comparisons (after all, a longer string can never be a substring of a shorter/equal length string):

def substringSieve(string_list):
    string_list.sort(key=lambda s: len(s), reverse=True)
    out = []
    for s in string_list:
        if not any([s in o for o in out]):
            out.append(s)
    return out

Upvotes: 14

Aaron Hall
Aaron Hall

Reputation: 394775

Here's a possible solution:

string_list = ['rest', 'resting', 'look', 'looked', 'it', 'spit']
def string_set(string_list):
    return set(i for i in string_list 
               if not any(i in s for s in string_list if i != s))

print(string_set(string_list))

prints out:

set(['looked', 'resting', 'spit'])

Note I create a set (using a generator expression) to remove possibly duplicated words as it appears that order does not matter.

Upvotes: 3

Shan Valleru
Shan Valleru

Reputation: 3121

Here's is the efficient way of doing it (relative to the above solutions ;) ) as this approach reduces the number of comparisons between the list elements a lot. If I have a huge list, I'd definitely go with this and of course you can morph this solution into a lambda function to make it look small:

string_list = ['rest', 'resting', 'look', 'looked', 'it', 'spit']
for item in string_list: 
  for item1 in string_list:
    if item in item1 and item!= item1:
      string_list.remove(item)

print string_list

Output:

>>>['resting', 'looked', 'spit']

Hope it helps !

Upvotes: 0

placeybordeaux
placeybordeaux

Reputation: 2216

Another one liner:

[string for string in string_list if len(filter(lambda x: string in x,string_list)) == 1]

should be fairly readable, just not that pythonic.

Upvotes: 2

Sunny Nanda
Sunny Nanda

Reputation: 2382

Here is a one-liner that does what you want:

filter(lambda x: [x for i in string_list if x in i and x != i] == [], string_list)

Example:

>>> string_list = ['rest', 'resting', 'look', 'looked', 'it', 'spit']
>>> filter(lambda x: [x for i in string_list if x in i and x != i] == [], string_list)
['resting', 'looked', 'spit']

Upvotes: 0

Michael0x2a
Michael0x2a

Reputation: 63978

Here's one method:

def find_unique(original):
    output = []

    for a in original:
        for b in original:
            if a == b:
                continue     # So we don't compare a string against itself
            elif a in b:
                break
        else:
            output.append(a) # Executed only if "break" is never hit

    return output

if __name__ == '__main__':
    original = ['rest', 'resting', 'look', 'looked', 'it', 'split']
    print find_unique(original)

It exploits the fact that we can easily check if one string is a substring of another by using the in operator. It essentially goes through each string, checks to see if it's a substring of another, and appends itself to an output list if it isn't.

This prints out ['resting', 'looked', 'split']

Upvotes: 0

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