CODEWITHSUNDEEP
pythonpython-3.xfunctiondictionary
Ryan Murphy
Ryan Murphy

Reputation: 39

Comparing to dictionaries in Python

I have a function that takes a string that counts the individual characters and puts the character along with the amount of time it shows up into a string. E.g:

isanagram("surfer")

returns

w1 = {'a': 0, 'b': 0, 'c': 0, 'd': 0, 'e': 1, 'f': 1, 'g': 0, 'h': 0, 'i': 0, 'j': 0, 'k': 0, 'l': 0, 'm': 0, 'n': 0, 'o': 0, 'p': 0, 'q': 0, 'r': 2, 's': 1, 't': 0, 'u': 1, 'v': 0, 'w': 0, 'x': 0, 'y': 0, 'z': 0}

Although when I compare this function with two different parameters the print statement outputs True instead of False, which it clearly should be. Could anyone take a look. Here is my code:

alphabetlist = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
    alphabetdict = {}
    newalphadict = {}    
              
    def isanagram(aword):
        for i in alphabetlist:
            count = word.count(i) 
            alphabetdict[i] = count
            
        return alphabetdict
    
     print(isanagram("computer") == isanagram("science")) #outputting True. Should be outputting False.

Upvotes: 0

Views: 53

Answers (2)

anon01
anon01

Reputation: 11171

There are a few issues with your code. Here is a better version:

from collections import Counter

def is_anagram(word1, word2):    

    c1 = Counter(word1)
    c2 = Counter(word2)
    return c1 == c2

is_anagram("computer", "science") # False

This solves the following issues with your current code:

  • newalphadict, alphabetdict reside outside your function, so letters get added to them every time isanagram is called (a bug)
  • is_anagram compares two words. It should probably take two words as arguments (and nothing else).
  • aword != word
  • you don't have to pre-populate the dictionary. You could do this with a bit of try/accept logic and a dict, or use a defaultdict, but Counter already exists so lets use it.

Upvotes: 1

Makkoveev
Makkoveev

Reputation: 59

alphabetlist = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
        
def isanagram(word=''):
    alphabetdict = {}
    for i in alphabetlist:
        count = word.count(i) 
        alphabetdict[i] = count
    
    return alphabetdict

print(isanagram("computer") == isanagram("science")) #-> False

Upvotes: 1

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