打印字典中的单词列表

Question

(Sorry for the long question)

print_most_common() function which is passed two parameters, a dictionary containing words and their corresponding frequencies eg

{"fish":9,  "parrot":8,  "frog":9,  "cat":9,  "stork":1,  "dog":4, "bat":9,  "rat":3}

and, an integer, the required number of characters. The function gets a list of all the words of the required number of characters which are keys of the dictionary and have the highest frequency for words of that length. The function first prints the string made up of the word length (the second parameter), followed by " letter keywords: ", then prints a list of all the words of the required length (keys from the dictionary) which have the highest frequency followed by the frequency value. The list of words must be sorted alphabetically.

eg

word_frequencies = {"fish":9, "parrot":8, "frog":9, "cat":9,
                                           "stork":1, "dog":4, "bat":9, "rat":3}
print_most_common(word_frequencies, 3)
print_most_common(word_frequencies, 4)
print_most_common(word_frequencies, 5)

Will print:

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

How would I define the print_most_common(words_dict, word_len) function?

Answer 1

how about this.

freq_dict = {k: v for k, v in word_frequencies.items() if len(k) == word_len}

for example:

>> freq_dict = {k: v for k, v in word_frequencies.items() if len(k) == 3}
>> print(freq_dict)
>> {'bat': 9, 'cat': 9, 'dog': 4, 'rat': 3}

Answer 2

This should work for a Python 2 implementation at least, updating to 3 should not be difficult.

Asav provides a way to get dictionary with a words of word_len and their corresponding frequency. You can then retrieve the max value from the frequencies and consequently retrieve the list of words with that frequency.

def print_most_common(words_dict, word_len):
    wl_dict = {k: v for k, v in words_dict.items() if len(k) == word_len}
    max_value = wl_dict[max(wl_dict, key=wl_dict.get)]
    res_list = [key for key,val in wl_dict.items() if val == max_value]
    print '%d letter keywords %s %d' % (word_len, res_list, max_value)

Do let me know if you want further decomposition or explanation.

Answer 3

Here is a possible solution:

1. Get all words of the required length.

    filtered_words = {k: v for k, v in words_dict.items() if len(k) == word_len} 

2. Get the maximum count for that length.

    max_count = max(filtered_words.values()) 

3. Filter the words with that count.

    [k for k, v in filtered_words.items() if v == max_count] 

Full Code

def print_most_common(words_dict, word_len):
  filtered_words = {k: v for k, v in words_dict.items() if len(k) == word_len}
  max_count = max(filtered_words.values())
  print word_len, 'letter keywords:', [k for k, v in filtered_words.items() if v == max_count], max_count