Check if multiple strings exist in another string


How can I check if any of the strings in an array exists in another string?

For example:

a = ['a', 'b', 'c']
s = "a123"
if a in s:
    print("some of the strings found in s")
    print("no strings found in s")

How can I replace the if a in s: line to get the appropriate result?

Asked By: jahmax



You can use any:

a_string = "A string is more than its parts!"
matches = ["more", "wholesome", "milk"]

if any([x in a_string for x in matches]):

Similarly to check if all the strings from the list are found, use all instead of any.

Answered By: Mark Byers

You need to iterate on the elements of a.

a = ['a', 'b', 'c']
str = "a123"
found_a_string = False
for item in a:    
    if item in str:
        found_a_string = True

if found_a_string:
    print "found a match"
    print "no match found"
Answered By: Seamus Campbell
a = ['a', 'b', 'c']
str =  "a123"

a_match = [True for match in a if match in str]

if True in a_match:
  print "some of the strings found in str"
  print "no strings found in str"
Answered By: mluebke

You should be careful if the strings in a or str gets longer. The straightforward solutions take O(S*(A^2)), where S is the length of str and A is the sum of the lenghts of all strings in a. For a faster solution, look at Aho-Corasick algorithm for string matching, which runs in linear time O(S+A).

Answered By: jbernadas

Just to add some diversity with regex:

import re

if any(re.findall(r'a|b|c', str, re.IGNORECASE)):
    print 'possible matches thanks to regex'
    print 'no matches'

or if your list is too long – any(re.findall(r'|'.join(a), str, re.IGNORECASE))

Answered By: Shankar ARUL

any() is by far the best approach if all you want is True or False, but if you want to know specifically which string/strings match, you can use a couple things.

If you want the first match (with False as a default):

match = next((x for x in a if x in str), False)

If you want to get all matches (including duplicates):

matches = [x for x in a if x in str]

If you want to get all non-duplicate matches (disregarding order):

matches = {x for x in a if x in str}

If you want to get all non-duplicate matches in the right order:

matches = []
for x in a:
    if x in str and x not in matches:
Answered By: zondo

It depends on the context
suppose if you want to check single literal like(any single word a,e,w,..etc) in is enough

original_word ="hackerearcth"
for 'h' in original_word:

if you want to check any of the character among the original_word:
make use of

if any(your_required in yourinput for your_required in original_word ):

if you want all the input you want in that original_word,make use of all

original_word = ['h', 'a', 'c', 'k', 'e', 'r', 'e', 'a', 'r', 't', 'h']
yourinput = str(input()).lower()
if all(requested_word in yourinput for requested_word in original_word):
Answered By: Trinadh Koya

jbernadas already mentioned the Aho-Corasick-Algorithm in order to reduce complexity.

Here is one way to use it in Python:

  1. Download from here

  2. Put it in the same directory as your main Python file and name it

  3. Try the alrorithm with the following code:

    from aho_corasick import aho_corasick #(string, keywords)
    print(aho_corasick(string, ["keyword1", "keyword2"]))

Note that the search is case-sensitive

Answered By: Domi W
flog = open('test.txt', 'r')
flogLines = flog.readlines()
strlist = ['SUCCESS', 'Done','SUCCESSFUL']
res = False
for line in flogLines:
     for fstr in strlist:
         if line.find(fstr) != -1:
            res = True

if res:
    print('res true')
    print('res false')

output example image

Answered By: LeftSpace

I would use this kind of function for speed:

def check_string(string, substring_list):
    for substring in substring_list:
        if substring in string:
            return True
    return False
Answered By: Ivan Mikhailov
data = "firstName and favoriteFood"
mandatory_fields = ['firstName', 'lastName', 'age']

# for each
for field in mandatory_fields:
    if field not in data:
        print("Error, missing req field {0}".format(field));

# still fine, multiple if statements
if ('firstName' not in data or 
    'lastName' not in data or
    'age' not in data):
    print("Error, missing a req field");

# not very readable, list comprehension
missing_fields = [x for x in mandatory_fields if x not in data]
if (len(missing_fields)>0):
    print("Error, missing fields {0}".format(", ".join(missing_fields)));
Answered By: Robert I

Just some more info on how to get all list elements availlable in String

a = ['a', 'b', 'c']
str = "a123" 
list(filter(lambda x:  x in str, a))
Answered By: Nilesh Birari

A surprisingly fast approach is to use set:

a = ['a', 'b', 'c']
str = "a123"
if set(a) & set(str):
    print("some of the strings found in str")
    print("no strings found in str")

This works if a does not contain any multiple-character values (in which case use any as listed above). If so, it’s simpler to specify a as a string: a = 'abc'.

Answered By: Berislav Lopac

Yet another solution with set. using set.intersection. For a one-liner.

subset = {"some" ,"words"} 
text = "some words to be searched here"
if len(subset & set(text.split())) == len(subset):
   print("All values present in text")

if subset & set(text.split()):
   print("Atleast one values present in text")
Answered By: sjd

The regex module recommended in python docs, supports this

words = {'he', 'or', 'low'}
p = regex.compile(r"L<name>", name=words)
m = p.findall('helloworld')


['he', 'low', 'or']

Some details on implementation: link

Answered By: balki

A compact way to find multiple strings in another list of strings is to use set.intersection. This executes much faster than list comprehension in large sets or lists.

>>> astring = ['abc','def','ghi','jkl','mno']
>>> bstring = ['def', 'jkl']
>>> a_set = set(astring)  # convert list to set
>>> b_set = set(bstring)
>>> matches = a_set.intersection(b_set)
>>> matches
{'def', 'jkl'}
>>> list(matches) # if you want a list instead of a set
['def', 'jkl']
Answered By: Jerald Cogswell

If you want exact matches of words then consider word tokenizing the target string. I use the recommended word_tokenize from nltk:

from nltk.tokenize import word_tokenize

Here is the tokenized string from the accepted answer:

a_string = "A string is more than its parts!"
tokens = word_tokenize(a_string)
Out[46]: ['A', 'string', 'is', 'more', 'than', 'its', 'parts', '!']

The accepted answer gets modified as follows:

matches_1 = ["more", "wholesome", "milk"]
[x in tokens for x in matches_1]
Out[42]: [True, False, False]

As in the accepted answer, the word "more" is still matched. If "mo" becomes a match string, however, the accepted answer still finds a match. That is a behavior I did not want.

matches_2 = ["mo", "wholesome", "milk"]
[x in a_string for x in matches_1]
Out[43]: [True, False, False]

Using word tokenization, "mo" is no longer matched:

[x in tokens for x in matches_2]
Out[44]: [False, False, False]

That is the additional behavior that I wanted. This answer also responds to the duplicate question here.

Answered By: Spirit of the Void
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