Fastest way to check if a string can be created with a list of characters in python

Question

I need to check if a string can be created with a list of characters and return True or False.

I am using different solutions with list.count or collections.Counter.

I am also using this solution which I dont need to read through the list of characters:

def check(message, characters):
    try:
        [list(characters).remove(m) for m in message]
        return True
    except:
        return False

Is there a fastest way? for a very very very big list of characters. Counter and list count seems slower. Dont know if there is a fast pythonic way to do this.

Example:

message = "hello"
characters = "hheellooasdadsfgfdgfdhgfdlkgkfd"

check(message, characters) # this should return True or False
# characters can be a veeeeery long string

Duplicates matter so for example characters = "hheloo" would not work for message = "hello"

Answer 1

You could use collections.Counter() . Just build two counters and use the subtract() method to check if there're any negative counts:

>>> c1 = Counter(characters)
>>> c2 = Counter(message)
>>> c1.subtract(c2)
>>> all(v >= 0 for v in c1.values())
False

This should work in linear time.

Answer 2

This is not feasible in linear time, as the length of both strings matter and they need to be iterated for each character. Without having checked its actual implementation, I assume remove() is logarithmic.

def check(msg, chars):
    c = list(chars)  # Creates a copy
    try:
        for m in msg:
            c.remove(m)
    except ValueError:
        return False
    return True

if __name__ == '__main__':
    print(check('hello', 'ehlo'))
    print(check('hello', 'ehlol'))
    print(check('hello', 'ehloijin2oinscubnosinal'))

Answer 3

Here is another solution compared to eugene's solution and jbndlr's solution.

def test1(input_word, alphabet):
    alp_set = set(list(alphabet))
    in_set = set(list(input_word))
    return in_set.issubset(alp_set)

def test2(input_word, alphabet):
    c1 = collections.Counter(alphabet)
    c2 = collections.Counter(input_word)
    c1.subtract(c2)
    return all(v >= 0 for v in c1.values())

def check(msg, chars):
    c = list(chars)  # Creates a copy
    try:
        for m in msg:
            c.remove(m)
    except ValueError:
        return False
    return True

input_word = "hello"
alphabet = "hheellooasdadsfgfdgfdhgfdlkgkfd"


start_time = time.time()
for i in range(10000):
    test1(input_word,alphabet)
print("--- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
for i in range(10000):
    test2(input_word,alphabet)
print("--- %s seconds ---" % (time.time() - start_time))

start_time = time.time()
   for i in range(10000):
       check(input_word,alphabet)
   print("--- %s seconds ---" % (time.time() - start_time))

>> --- 0.03100299835205078 seconds ---
>> --- 0.24402451515197754 seconds ---
>> --- 0.022002220153808594 seconds ---

⇒ jbndlr's solution is the fastest - for this test case.

Another testcase:

input_word = "hellohellohellohellohellohellohellohellohellohellohellohellohello"
alphabet =   

""

>> --- 0.21964788436889648 seconds ---
>> --- 0.518169641494751 seconds ---
>> --- 1.3148927688598633 seconds ---

⇒ test1 is fastest

Answer 4

There is maybe a faster way of doing this, apparently due to the cost of creating the all() generator ( Why is Python's 'all' function so slow? ) perhaps a for loop is faster, Expanding on @eugene y's answer:

from collections import Counter
import time

message = "hello"
characters = "hheeooasdadsfgfdgfdhgfdlkgkfd"

def check1(message,characters):
    c1 = Counter(characters)
    c2 = Counter(message)
    c1.subtract(c2)
    return all(v > -1 for v in c1.values())

def check2(message,characters):
    c1 = Counter(characters)
    c2 = Counter(message)
    c1.subtract(c2)
    for v in c1.values():
        if v < 0:
            return False
    return True

st = time.time()
for i in range(350000):
    check1(message,characters)
end = time.time()
print ("all(): "+str(end-st))

st = time.time()
for i in range(350000):
    check2(message,characters)
end = time.time()
print ("for loop: "+str(end-st))

results:

all(): 5.201688051223755
for loop: 4.864434719085693