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Easy Anagram Dictionaries Practice

I do not use dictionaries very often. Friday, I was without internet all day, so I took the opportunity to play with dir() and help() to discover some dictionary properties. My short-lived obsession with Draw Something on the iPhone has gotten me interested in anagrams (kicked the habit by reading programming books). I believe using dictionaries is the fastest and most accurate way to determine if two words are anagrams of each other, at least without importing any other modules.

A dictionary is an unordered set of key: value pairs. Keys must be an immutable type. Values can be anything. Being unordered causes some interesting properties for working with dictionaries, different from any other python data structure. Instead of being indexed by numbers, dictionaries are indexed by keys. Because they are indexed by keys, each key is unique within it's dictionary. If two dictionaries with the same keys are added to each other, the values of the same data type combine. This is convenient for our anagram activity. But first, some dictionary review.

>>> sample_dict = {}        # creates an empty dictionary
>>> type( sample_dict )
<type 'dict'>
>>> sample_dict['Name'] = 'Jessie'        # creating a key:value pair
>>> sample_dict['Age'] = 23        # another key:value pair
>>> sample_dict
{'Age': 23, 'Name': 'Jessie'}
>>> sample_dict2 = {'Name': 'Jessie', 'Age': 23}        # another way to create dict
>>> type (sample_dict2)
<type 'dict'>
>>> sample_dict2
{'Age': 23, 'Name': 'Jessie'}
>>> sample_dict + sample_dict2        # cannot add dictionaries, only values
Traceback (most recent call last):
  File "<console>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'dict' and 'dict'
>>> sample_dict['Age'] + sample_dict2['Age']        # adds values
46
>>> sample_dict.keys()
['Age', 'Name']
>>> sample_dict.values()
[23, 'Jessie']
>>> type( sample_dict.values() )        # keys and values are returned as lists
<type 'list'>
>>> sample_dict.get('Age')        # gets the value at a specific key
23
>>> type( sample_dict.get('Age'))         # value maintains data type in dictionary
<type 'int'>
>>> sample_dict.has_key('Age')        # D.has_key() returns boolean
True
>>> sample_dict3 = {'Children': 'Graham'}
>>> sample_dict3.update(sample_dict)        # update keys and values
>>> sample_dict
{'Age': 23, 'Name': 'Jessie'}
>>> sample_dict3        # Children field is added as a key:value pair
{'Age': 23, 'Children': 'Graham', 'Name': 'Jessie'}
>>> {'Age': 23, 'Name': 'Jessie'} == {'Name': 'Jessie', 'Age': 23}        # different order is equal
True

How do we know if two words are anagrams? Consider the anagrams odor and door. We could say that they are reshuffled strings. Each word uses the same letters, but in a different order: 2 o's, 1 r, and 1 d. My simple program creates empty dictionaries for the two words being compared, stores the letters as keys, and adds to the value for each occurrence of the same letter, then checks that the dictionaries are equivalent. I have not included exception handling and I made the design decision to count white space as part of the anagram such that 'abc def' is not an anagram of 'fdeabc,' but is an anagram of 'abc fed.'

def get_dict(word, count):
    for i in word.lower():
        if count.has_key(i):
            count[i] += 1
        else:
            count[i] = 1
    return count

def main():
    word1 = raw_input("What is the first word? \n")
    word2 = raw_input("What is the second? \n")
    count1 = {}
    count2 = {}
    count1 = get_dict(word1, count1)
    count2 = get_dict(word2, count2)
    if count1 == count2:
        print("Yes, those are anagrams!\n")
    else:
        print("No, you've failed \n")

if __name__ == "__main__":
    main()

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