如何按受欢迎程度对网站排序？Python脚本实现求助

你现在的脚本已经能计算各网站的访问量，要按访问量（受欢迎程度）排序，可以用Python的sorted()函数处理存储访问量的字典Walk_Number。以下是修改后的完整脚本（注释已翻译成中文）：

import random

# 创建超链接字典
Hypertext = {}
# 创建存储访问次数的字典
Walk_Number = {}
# 总访问次数变量
Total_Walk = 0
# 网站列表
Websites = ["A","B","C","D","E","F"]
# 超链接配置
# 字典的键是网站名称，值是该网站指向的其他网站列表
Hypertext["A"] = ["B","C","E"]
Hypertext["B"] = ["F"]
Hypertext["C"] = ["A","E"]
Hypertext["D"] = ["B","C"]
Hypertext["E"] = ["A","B","C","D","F"]
Hypertext["F"] = ["E"]
print(Hypertext)

# 初始化各网站访问次数为0.0
for site in Websites:
    Walk_Number[site] = 0.0

i = 0
while i < 1000:
    x = random.choice(Websites)
    while random.random() < 0.85:
        Walk_Number[x] = Walk_Number[x] + 1
        Total_Walk = Total_Walk + 1     
        x = random.choice(Hypertext[x])
    i = i + 1

print("原始访问量数据：")
print(Walk_Number)
print(f"总访问次数：{Total_Walk}")

# 按访问量从高到低排序网站
sorted_sites = sorted(Walk_Number.items(), key=lambda item: item[1], reverse=True)
print("\n按访问量排序的网站（从高到低）：")
for site, count in sorted_sites:
    print(f"{site}: {count}")

关键说明： #

• Walk_Number.items()会返回字典的所有键值对（比如('A', 120.0)这样的元组）
• key=lambda item: item[1]指定排序依据是每个元组的第二个元素（也就是访问量数值）
• reverse=True让结果按降序排列（如果要升序去掉这个参数即可）

内容的提问来源于stack exchange，提问作者Yamighawara Kazeru