boolean loopd keys strings comparator

Question

i am doing a python project and i wanna compare a string to a list of keys and values - if all the keys were to match up a certain number of times with a word string, it should return True for example - def compareTo(word, hand):

   kal = False 
   for d in wordlists: 
       if d in word: 
          kal = True 
   return kal

it always returns false how can i make it return true?!? please help!!!

so if

word = "hammer"

and

hand =  {'a': 1, 'h': 1, 'r': 1, 'm': 2, 'e': 1}

values represent the frequency of each letter if I insert parameters which are true, how can I make it return true and not false...

comapreTo("hammer",{'a': 1, 'h': 1, 'r': 1, 'm': 2, 'e': 1})

should return True and not False and comapreTo("hammers",{'a': 1, 'h': 1, 'r': 1, 'd': 2, 'e': 1}) should return false and not true!

nneonneo · Accepted Answer · 2013-03-09 20:10:51Z

1

You can simply use Counter to count letter frequencies:

from collections import Counter

def compareTo(word, hand):
    return Counter(word) == hand

For example, Counter("hammer") is Counter({'m': 2, 'a': 1, 'h': 1, 'r': 1, 'e': 1}); since Counter is dictionary-like, this will compare equal to {'a': 1, 'h': 1, 'r': 1, 'm': 2, 'e': 1}.

Counter is built in to Python 2.7 and Python 3.x. If you need to use it in Python 2.6 or earlier, you can get it from here: http://code.activestate.com/recipes/576611/.

edited Mar 9, 2013 at 20:10

answered Mar 9, 2013 at 20:03

nneonneo

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1 Comment

kkonrad Over a year ago

Nice. I would add info that Counter is in Python since 2.7. In some cases this might be a deal breaker ...

kkonrad · Accepted Answer · 2013-03-09 21:36:07Z

0

This should work:

def compareTo(word, hand):
    d = hand.copy()
    for c in word:
        d[c] = d.get(c, 0) - 1
        if d[c] < 0:
            return False
    return True

word = "hammer"
hand =  {'a': 1, 'h': 1, 'r': 1, 'm': 2, 'e': 1}

print compareTo(word, hand)  # => True    
print compareTo("hammers", hand)  # => False
print compareTo("plum", {'f': 1, 'h': 1, 'k': 1, 'm': 1, 'l': 1, 'p': 1, 'u': 1, 'y': 1})  # => True

edited Mar 9, 2013 at 21:36

answered Mar 9, 2013 at 20:01

kkonrad

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9 Comments

user2152315 Over a year ago

how come it doesn't work for this case - >>> plum = "plum" >>> pluma = {'f': 1, 'h': 1, 'k': 1, 'm': 1, 'l': 1, 'p': 1, 'u': 1, 'y': 1} >>> print compareTo(plum, pluma) False

kkonrad Over a year ago

in what case it doesn't work? I'm getting outputs that are in comments ... If you want to make it work if hand is superset of minimum matching one than last return should be just return True

kkonrad Over a year ago

@user2152315 to make it work for your case you need to apply change from comment above

user2152315 Over a year ago

in the case "plum" and {'f': 1, 'h': 1, 'k': 1, 'm': 1, 'l': 1, 'p': 1, 'u': 1, 'y': 1} it returns False when it should return True

kkonrad Over a year ago

@user2152315 I would recommend changing problem description to strengthen that numbers in those hand dict are lower bounds - not equal values

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