字典问题 Python

Question

问题：

定义print_most_common()函数，它传递两个参数，一个包含单词及其对应频率的字典，例如，

{"and":15,  "talon":7,  "frog":1,  "cat":15,  "tests":1,  "dog":2, "bat":14,  "rat":15}

和一个整数，字符数。

该函数获取所需字符数的所有单词的列表，这些单词是字典的键并且对于该长度的单词具有最高频率。

该函数首先打印由单词长度（第二个参数）组成的字符串，后跟“字母单词：”，然后打印字典中所有具有最高频率的所需长度单词的列表由频率值。例如以下代码：

word_frequencies =  {"and":15, "talon":7, "frog":1, "cat":15, "tests":1,  "dog":2, "bat":14, "rat":15}

print_most_common(word_frequencies, 3)

print_most_common(word_frequencies, 4)

print_most_common(word_frequencies, 5)

print_most_common(word_frequencies, 6)

打印以下三行：

3 letter words: ['and', 'cat', 'rat'] 15

4 letter words: ['frog'] 1

5 letter words: ['talon'] 7

6 letter words: [] 0

我尝试了这个问题，并提出了以下代码（我意识到这不是最好的方法，但我们还没有详细学习 python 编程）：

def print_most_common(words_dict, word_len): 
    words_list = list(words_dict.keys())
    frequency = 0
    max_word_list = []
    for word in words_list:
        if word_len == len(word) and frequency <= words_dict[word]:
            frequency = words_dict[word]
    for word, value in words_dict.items():
        if value == frequency:
            max_word_list += [word]
    print (word_len, "letter words:", max_word_list, frequency)

我遇到的问题是代码在频率 3、5、6、0 下工作正常。但是，对于频率 4，它也打印出一个 5 个字母的单词：

预期结果：

3 letter words: ['and', 'cat', 'rat'] 15
4 letter words: ['frog'] 1
5 letter words: ['talon'] 7
6 letter words: [] 0

实际结果：

3 letter words: ['and', 'rat', 'cat'] 15
4 letter words: ['tests', 'frog'] 1
5 letter words: ['talon'] 7
6 letter words: [] 0

预期结果：

3 letter words: ['and', 'cat', 'pig', 'rat'] 15
4 letter words: ['bird', 'frog'] 7
5 letter words: ['front', 'tests'] 1

实际结果：

3 letter words: ['and', 'pig', 'rat', 'cat'] 15
4 letter words: ['talons', 'frog', 'bird'] 7
5 letter words: ['tests', 'front'] 1

我不知道我的代码哪里出错了。

Answer 1

计算出最大频率后，然后选择具有该频率的所有个单词，无论长度如何：

for word, value in words_dict.items():
    if value == frequency:
        max_word_list += [word]

您需要再次测试长度：

for word, value in words_dict.items():
    if len(word) == word_len and value == frequency:
        max_word_list += [word]

您可以一次性收集给定频率和字长的所有单词，但是：

def print_most_common(words_dict, word_len):
    results = []
    max_frequency = 0
    for word, freq in words_dict.items():
        if len(word) != word_len or freq < max_frequency:
            # wrong length or not frequent enough
            continue
        if freq == max_frequency:
            results.append(word)
        else:
            # right word length and *higher* frequency
            max_frequency = freq
            results = [word]
    print (word_len, "letter words:", results, max_frequency)

Answer 2

您需要检查字典循环中的单词长度，即and len(word) == word_len，频率可能匹配但单词可以是任何长度，因为您不检查：

for word, value in words_dict.items():
      # make sure the word_len is also equal
      if value == frequency and len(word) == word_len:
          max_word_list.append(word)

你也可以每次都简单地追加单词，不需要将单词包裹在列表中并扩展，for循环也可以做成推导式。

max_word_list = [word  for word, value in words_dict.items() if value == frequency and len(word) == word_len]

你也可以只取最大值：

def print_most_common(words_dict, word_len):
    mx = max(v for k, v in words_dict.items() if len(k) == word_len)
    max_word_list = [word for word, value in words_dict.items() if value == mx and len(word) == word_len]
    print(word_len, "letter words:", max_word_list, mx)