6
\$\begingroup\$

I have a piece of code that checks if two strings are anagrams. I would like this to be as pythonic as possible.

def are_two_string_anagram(string1, string2):
    if len(string1) != len(string2):
        return False

    def char_count(string):
        char_count = {}
        for s in string:
            char_count[s] = char_count.get(s, 0) + 1
        return char_count

    chr_count = char_count(string1)
    chr1_count = char_count(string2)
    return chr_count == chr1_count

string1 = "mary"
string2 = "yram"
print(are_two_string_anagram(string1, string2))

string1 = "albert"
string2 = "abelrt"
print(are_two_string_anagram(string1, string2))

string1 = "something"
string2 = "anotherus"
print(are_two_string_anagram(string1, string2))
\$\endgroup\$
1

1 Answer 1

5
\$\begingroup\$

Most python users would just use sorted(string1) == sorted(string2). That's \$O(n log n)\$, which is not that much worse than your \$O(n)\$ solution.

But the code you've actually written is just a reimplementation of:

collections.Counter(string1) == collections.Counter(string2)

That said, your short-circuit if the lengths mismatch is a good idea no matter which implementation you use.

There are some style issues that jump out with your implementation:

  • char_count doesn't have a pressing need to be a nested function, I would put it at top level.
  • Don't have a local variable named char_count inside the same-named function.
  • Don't cross numbers like chr_count = char_count(string1);chr1_count = char_count(string2). If you need numbered variables, use the same number for all transformed forms also.

For your demonstration code, I would have written:

def test_two_string_anagrams():
    assert are_two_string_anagram("mary", "yram")
    assert are_two_string_anagram("albert", "abelrt")
    assert are_two_string_anagram("something", "anotherus")

Or possibly:

def test_two_string_anagrams():
    for s1, s2 in [
        ("mary", "yram"),
        ("albert", "abelrt"),
        ("something", "anotherus"),

    ]:
        assert are_two_string_anagram(s1, s2)

That way it can be run directly with py.test (and possibly other harnesses, I know the standard unittest requires wrapping them in a dummy class, but I haven't used any others).

\$\endgroup\$
3
  • \$\begingroup\$ What if I cannot use collections.Counter? is my implementation correct? I was asked for a job interview that I can only use basic functions. \$\endgroup\$
    – toy
    Commented Jul 15, 2015 at 3:10
  • 1
    \$\begingroup\$ @toy Yeah, you've written the same algorithm as collections.Counter uses. \$\endgroup\$
    – o11c
    Commented Jul 15, 2015 at 3:57
  • \$\begingroup\$ anotherus should return false. \$\endgroup\$ Commented Jul 15, 2015 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.