看了下同学的留言学习了下，并改写正则，使其捕获单词排除换行和空格等 改写如下： from re import findall from collections import Counter with open('./demo1.txt', 'r') as f: data = f.read().lower() words = findall(r'\b\w*?\S\b', data) print(Counter(words)) print(Counter(words).most_common(5))

解包self.other_x , self.other_y = other和 直接使用other.x和other.y还啥别区别？

如老师提到如果需要累加多个点，比如p1+p2+p3+p4有下面三种方法供大家参考： 第一种方法：在__add__方法中返回Point_2D的类型，而不是tuple。 from collections import namedtuple class Point_2D (namedtuple('Point',['x','y'])): def __add__(self,other): self.other_x, self.other_y= other return self.__class__(self.x+self.other_x,self.y+self.other_y) p1= Point_2D(x=11,y=22) p2 = Point_2D(x=33,y=44) p3 = Point_2D(x=55,y=66) p4 = Point_2D(x=77,y=88) p1+p2+p3+p4 #第二种方法：重载Point_2D的__radd__方法，否则它会以元组的默认加法方式进行。 from collections import namedtuple class Point_2D(namedtuple('Point',['x','y'])): def __add__(self,other): self.other_x , self.other_y = other return self.x + self.other_x, self.y+ self.other_y def __radd__(self,other): self.other_x , self.other_y = other return self.x + self.other_x, self.y+ self.other_y #第三种方法：这段代码还是不够pythonic，__add__和__radd__代码重复了，如果连个方法的表现出一样的行为直接用=就可以了。 from collections import namedtuple class Point_2D(namedtuple('Point',['x','y'])): def __add__(self,other): self.other_x,self.other_y =other return self.x + self.other_x,self.y + self.other_y __radd__=__add__

import re from collections import Counter with open('hamlet.txt','r') as f: data = f.read().lower() words = re.findall(r'\w+', data) print(Counter(words).most_common(5))

import re from collections import Counter words = re.findall(r'\w+', open('a.txt').read().lower()) print(Counter(words).most_common(5))

这是一个 Python 程序，它可以统计一篇文章中出现频率在前五的单词，并将单词和出现次数一起输出到终端： ```python from collections import Counter import re def count_words(text): # 使用正则表达式来分割文本，只保留单词 words = re.findall(r'\b\w+\b', text.lower()) # 使用 Counter 来统计单词的出现次数 word_counts = Counter(words) # 找出出现频率最高的 5 个单词 top_five = word_counts.most_common(5) # 输出结果 for word, count in top_five: print(f"{word}: {count}") # 这里是你的文章 text = """ This is a sample text. It contains several words, some of which are repeated. Repeated words should be counted by the program. The program should ignore case, so 'This' and 'this' should be counted as the same word. """ count_words(text) ``` 在这个程序中，我们首先使用正则表达式来分割文本，只保留单词。然后，我们使用 `Counter` 来统计单词的出现次数。最后，我们找出出现频率最高的 5 个单词，并输出结果。 请注意，这个程序会忽略大小写，所以 'This' 和 'this' 会被视为同一个单词。如果你不希望忽略大小写，你可以去掉 `text.lower()` 这一行。 .

import re words = re.findall(r"\w+", open("./../hamlet.txt").read().lower()) Counter(words).most_common(5)

# 自定义命名元组的加法和减法功能 from collections import namedtuple Point = namedtuple('Point', ['x', 'y']) def add(self, other): return self.x + other.x, self.y + other.y Point.__add__ = add def subtract(self, other): return self.x - other.x, self.y - other.y Point.__sub__ = subtract p1 = Point(2, 3) p2 = Point(1, 4) print(p1 + p2) print(p1 - p2)

# 统计一篇文章中出现频率在前五的单词，并将单词和出现次数一起输出到终端 from collections import Counter import re words = re.findall(r'\w+', open('./demo.txt').read()) print(Counter(words).most_common(5)) >>>[('Python', 12), ('3', 12), ('10', 12), ('11', 6), ('is', 6)]

from re import findall with open('demo.txt',encoding='UTF-8') as f: cnt=f.read() c=Counter(findall(r'\w+',cnt)) c.most_common(5)