为什么我的这段代码content没有值：from lxml import etree import requests urls=[] for i in range(0,5,1): i=i*20 url = 'https://movie.douban.com/review/best/?start={}'.format(i) urls.append(url) import requests from lxml import etree herders = { "Content-Type":"text/html; charset=utf-8", "uesr-agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/114.0.0.0 Safari/537.36 Edg/114.0.1823.67" } detail_urls = [] for url in urls: # 发送请求 response= requests.get(url,headers=herders) # 编码转码 content = response.content.decode('utf8') # 解析html字符串 html =etree.HTML(content) # 利用xpath提取每个电影影评的url detail_url = html.xpath('//h2/a/@href') detail_urls.append(detail_url) print(detail_urls) break

到sqrt(n)。在每次迭代中，我们检查 `$n` 是否可以整除当前的迭代变量 `$i`。如果可以整除，则将 `$i` 加入到数组中，并将 `$n` 更新为 `$n` 除以 `$i` 的结果。

最后，如果 `$n` 大于2，则说明 `$n` 本身也是一个质因子，我们将其加入到数组中。

最后，我们使用 `echo` 语句输出找到的质因子。

如果你运行这段代码，并将 `$n` 设置为36，它将输出：

正整数 36 的质因子为: 2 2 3 3

希望这能解决你的问题！如果你还有其他疑问，请随时提问。

相关问题

import requests from lxml import etree urls = ['https://www.xxddxs.com/book/{}.html'.format(i) for i in range(1, 20)] path = r'D:\py\venv\Scripts\xiaoshuo' def get_text(url: object) -> object: r = requests.get(url) r.encoding = 'utf-8' selector = etree.HTML(r.text) # Get the book title from the page title = selector.xpath('//div[@class="bookname"]/h1/text()') # If no title found, print an error message and return early if len(title) == 0: print(f"No title found for {url}") return # Get the text of the chapter text = selector.xpath('//div[@id="content"]/text()') # Print the text to standard output print(text) # Write the text to a file in the specified directory with open(path + "\\" + title[0], 'w', encoding='utf-8') as f: for i in text: f.write(i) if __name__ == '__main__': for url in urls: get_text(url) 帮我修改并可以爬取到标题

Sure! 以下是修改后的代码：

import requests
from lxml import etree

urls = ['https://www.xxddxs.com/book/{}.html'.format(i) for i in range(1, 20)]
path = r'D:\py\venv\Scripts\xiaoshuo'

def get_text(url):
    r = requests.get(url)
    r.encoding = 'utf-8'
    selector = etree.HTML(r.text)

    # Get the book title from the page
    title = selector.xpath('//div[@class="bookname"]/h1/text()')

    # If no title found, print an error message and return early
    if len(title) == 0:
        print(f"No title found for {url}")
        return

    # Get the text of the chapter
    text = selector.xpath('//div[@id="content"]/text()')

    # Print the text to standard output
    print(text)

    # Write the text to a file in the specified directory
    with open(path + "\\" + title[0].strip() + ".txt", 'w', encoding='utf-8') as f:
        for i in text:
            f.write(i)

if __name__ == '__main__':
    for url in urls:
        get_text(url)

在原有代码的基础上，我在 `open()` 方法中添加了 `.strip()`，以去除标题中的空白字符。此外，我还为文件添加了 `.txt` 扩展名，以方便后续的操作。

import requests from lxml import etree #发起请求 headers = {'user-agent': 'Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/113.0.0.0 Mobile Safari/537.36 Edg/113.0.1774.57'} def get_fist_teat(list): try: return list[0].strip() except: return " " urls = ['https://movie.douban.com/top250?start={}&filter='.format(str(i*25)) for i in range(10)] cont = 1 for url in urls: res = requests.get(url=url,headers=headers) html = etree.HTML(res.text) lis = html.xpath('//*[@id="content"]/div/div[1]/ol/li') #解析数据 for li in lis: title = get_fist_teat(li.xpath('./div/div[2]/div[1]/a/span[1]/text()')) scr = get_fist_teat(li.xpath('./div/div[2]/div[1]/a/@href')) #获取合副属性+@href dictor = get_fist_teat(li.xpath('./div/div[2]/div[2]/p[1]/text()')) scort = get_fist_teat(li.xpath('./div/div[2]/div[2]/div/span[2]/text()')) commen = get_fist_teat(li.xpath('./div/div[2]/div[2]/div/span[4]/text()')) print(cont,title,scr,dictor,scort,commen) cont += 1如何实现可视化，比如加上柱状图

要将数据可视化为柱状图，可以使用 Python 中的 Matplotlib 库。以下是修改后的代码示例，可以将每部电影的评分作为柱状图进行可视化：

import requests 
from lxml import etree 
import matplotlib.pyplot as plt

# 发起请求  
headers = {'user-agent': 'Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/113.0.0.0 Mobile Safari/537.36 Edg/113.0.1774.57'} 

def get_fist_teat(list):     
    try:         
        return list[0].strip()     
    except:         
        return " " 

urls = ['https://movie.douban.com/top250?start={}&filter='.format(str(i*25)) for i in range(10)] 
scores = []  # 保存电影评分
cont = 1 

for url in urls:     
    res = requests.get(url=url,headers=headers)     
    html = etree.HTML(res.text)     
    lis = html.xpath('//*[@id="content"]/div/div[1]/ol/li') 
    
    # 解析数据     
    for li in lis:         
        title = get_fist_teat(li.xpath('./div/div[2]/div[1]/a/span[1]/text()'))         
        scr = get_fist_teat(li.xpath('./div/div[2]/div[1]/a/@href'))                                    #获取合副属性+@href         
        dictor = get_fist_teat(li.xpath('./div/div[2]/div[2]/p[1]/text()'))         
        scort = get_fist_teat(li.xpath('./div/div[2]/div[2]/div/span[2]/text()'))         
        commen = get_fist_teat(li.xpath('./div/div[2]/div[2]/div/span[4]/text()'))         
        
        # 保存评分数据
        scores.append(float(scort))
        
        print(cont,title,scr,dictor,scort,commen)         
        cont += 1

# 绘制柱状图
plt.bar(range(len(scores)), scores)
plt.title('Top 250 Movies on Douban')
plt.xlabel('Movie Rank')
plt.ylabel('Movie Score')
plt.show()

在上面的代码中，我们首先创建了一个空列表 scores，用于保存每部电影的评分。在解析每部电影的数据时，我们将电影的评分转换为浮点数，并将其添加到 scores 列表中。最后，我们使用 Matplotlib 库的 `bar` 函数将 scores 列表中的评分数据绘制成柱状图，并添加了一些必要的标签和标题。最后使用 `show` 函数显示图形。