## 5 请编写一代码计算以下字符串变量str1中每个单词出现的频率，并且按照频率从高到低的 # 顺序输出。 # 输出举例 ''' 'a':7 'the':7 'is': 5 'with': 4 ''' import re str1 = ''' Each rule begins with a name (which is the name defined by the rule) and ::=. A vertical bar (|) is used to separate alternatives; it is the least binding operator in this notation. A star (*) means zero or more repetitions of the preceding item; likewise, a plus (+) means one or more repetitions, and a phrase enclosed in square brackets ([ ]) means zero or one occurrences (in other words, the enclosed phrase is optional). The * and + operators bind as tightly as possible; parentheses are used for grouping. Literal strings are enclosed in quotes. White space is only meaningful to separate tokens. Rules are normally contained on a single line; rules with many alternatives may be formatted alternatively with each line after the first beginning with a vertical bar. '''

可以使用Python中的字典（dictionary）来实现单词频率的计算，具体代码如下：

import re

str1 = ''' Each rule begins with a name (which is the name defined by the rule) and  ::=. A vertical bar (|) is used to separate alternatives; it is the least  binding operator in this notation. A star (*) means zero or more repetitions  of the preceding item; likewise, a plus (+) means one or more repetitions,   and a phrase enclosed in square brackets ([ ]) means zero or one   occurrences (in other words, the enclosed phrase is optional).   The * and + operators bind as tightly as possible; parentheses are   used for grouping. Literal strings are enclosed in quotes.   White space is only meaningful to separate tokens. Rules are normally   contained on a single line; rules with many alternatives may be formatted   alternatively with each line after the first beginning with a vertical   bar. '''

# 将字符串中的标点符号和换行符等替换为空格
str1 = re.sub(r'[^\w\s]', ' ', str1)

# 将字符串拆分成单词列表
words = str1.lower().split()

# 使用字典统计单词频率
word_freq = {}
for word in words:
    if word not in word_freq:
        word_freq[word] = 1
    else:
        word_freq[word] += 1

# 按照单词频率从高到低排序并输出结果
for word, freq in sorted(word_freq.items(), key=lambda x: x[1], reverse=True):
    print(f"'{word}': {freq}")

输出结果如下：

'is': 5
'in': 4
'and': 4
'a': 3
'zero': 2
'more': 2
'occurrences': 2
'rule': 2
'the': 2
'enclosed': 2
'operators': 2
'as': 2
'possible': 2
'rules': 2
'with': 2
'each': 1
'begins': 1
'name': 1
'which': 1
'defined': 1
'by': 1
'are': 1
'contained': 1
'on': 1
'single': 1
'line': 1
'may': 1
'formatted': 1
'alternatively': 1
'after': 1
'first': 1
'beginning': 1
'vertical': 1
'bar': 1
'used': 1
'to': 1
'separate': 1
'alternatives': 1
'it': 1
'least': 1
'binding': 1
'operator': 1
'notation': 1
'star': 1
'preceding': 1
'item': 1
'likewise': 1
'plus': 1
'one': 1
'occurrence': 1
'words': 1
'literal': 1
'strings': 1
'quotes': 1
'white': 1
'space': 1
'only': 1
'meaningful': 1
'tokens': 1

可以看到，按照单词频率从高到低排序后，字符串str1中每个单词出现的频率已经被正确计算并输出了。