Question

该函数打印关键字长度，然后打印所需长度的所有字典关键字的排序列表，其中频率最高，频率最高。例如，以下代码：

for (int priority : pMap.keySet()) {
     System.out.print(priority);   //prints the priority
     System.out.print(pMap.get(priority));  // prints the product for the priority
     System.out.print(priceMap.get(priority));  //prints the price 
     System.out.println();  // changes the line
}

打印输出：

 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)

到目前为止，我有这段代码：

 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

这给了我：

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))

但不是空序列。如何添加我的代码，以便计算空序列。

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

Answer 1

只需在max()调用中添加一个回退条件，以便它永远不会尝试找到空的可迭代中的最大值：

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

如果right_length.values()为空，则会使用[0]（非空），而max()不会产生错误。

我需要假设这个代码中的空序列为python

1 个答案: