Python String Search Regardless Of Character Sequence

Question

I want to create an application that checks if word that the user typed in contains word/words from a separate text file (e.g. input = \'teeth\', separate file contains the word

Answer 1

I know it's unlikely, but if performance really mattered for very large inputs, you could avoid needing to create the second Counter and just iterate directly over the characters of the substring, allowing the possibility of terminating early if you run out of a given character.

In [26]: def contains2(string, substring):
    ...:     c = Counter(string)
    ...:     for char in substring:
    ...:         if c[char] > 0:
    ...:             c[char] -= 1
    ...:         else:
    ...:             return False
    ...:     return True
    ...: 

In [27]: contains2("teeth", "eeh")
Out[27]: True

In [28]: contains2("teeth", "ehe")
Out[28]: True

In [29]: contains2("teth", "ehe")
Out[29]: False

In [30]: contains2("teth", "eeh")
Out[30]: False

In [31]: def contains(string, substring):
    ...:     c1 = collections.Counter(string)
    ...:     c2 = collections.Counter(substring)
    ...:     return not(c2-c1)
    ...: 

In [32]: contains("teth", "ehe")
Out[32]: False

In [33]: contains("teeth", "ehe")
Out[33]: True

In [34]: contains("teeth", "eeh")
Out[34]: True

In [35]: %timeit contains("teeth", "eeh")
19.6 µs ± 94.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [36]: %timeit contains2("teeth", "eeh")
9.59 µs ± 29.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [37]: %timeit contains("friday is a good day", "ss a")
22.9 µs ± 121 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [38]: %timeit contains2("friday is a good day", "ss a")
9.52 µs ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 2

I would create a collections.Counter object from both strings, counting the characters, then substract the dicts, test if resulting dict is empty (which means string contains substring with cardinality respected)

import collections

def contains(substring, string):
    c1 = collections.Counter(string)
    c2 = collections.Counter(substring)
    return not(c2-c1)

print(contains("eeh","teeth"))
print(contains("eeh","teth"))

result:

True
False

Note that your example isn't representative as

>>> "eet" in "teeth"
True

that's why I changed it.