CODEWITHSUNDEEP
pythonlistfunction
R. Mercy
R. Mercy

Reputation: 509

I need to assume for an empty sequence in this code for python

The function prints the keyword length, then prints a sorted list of all the dictionary keywords of the required length, which have the highest frequency, followed by the frequency. For example, the following code:

 word_frequencies =  {"fish":9,  "parrot":8,  "frog":9,  "cat":9,  "stork":1,  "dog":4, "bat":9,  "rat":4}

 print_most_frequent(word_frequencies,3)
 print_most_frequent(word_frequencies,4)
 print_most_frequent(word_frequencies,5)
 print_most_frequent(word_frequencies,6)
 print_most_frequent(word_frequencies, 7)

prints outs:

 3 letter keywords: ['bat', 'cat'] 9
 4 letter keywords: ['fish', 'frog'] 9
 5 letter keywords: ['stork'] 1
 6 letter keywords: ['parrot'] 8
 7 letter keywords: [] 0

So far I have this code:

def print_most_frequent(words_dict, word_len):
    right_length = {}
    for k, v in words_dict.items():
        if len(k) == word_len:
             right_length[k] = v
    max_freq = max(right_length.values())
    max_words = {}
    for k, v in right_length.items():
         if v == max_freq:
              max_words[k] = v
    print (str(word_len) + " letter keywords: " + str(sorted(max_words.keys())) + " " + str(max_freq))

and this gives me:

 3 letter keywords: ['bat', 'cat'] 9
 4 letter keywords: ['fish', 'frog'] 9
 5 letter keywords: ['stork'] 1
 6 letter keywords: ['parrot'] 8

but not the empty sequence. How do I make an addition to my code so that it counts for the empty sequence.

 7 letter keywords: [] 0

Upvotes: 1

Views: 39

Answers (1)

TigerhawkT3
TigerhawkT3

Reputation: 49330

Simply add a fallback condition to the max() call so that it never tries to find the largest of an empty iterable:

max_freq = max(right_length.values() or [0])

If right_length.values() is empty, [0] (which is not empty) will be used instead, and max() doesn't produce an error for that.

Upvotes: 1

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