BeautifulSoup元素输出到列表

Question

我有一个使用BeautifulSoup的输出。

1. 我需要将'type''bs4.element.Tag'的输出转换为列表并将列表导出到名为COLUMN_A的DataFrame列中

2. 我希望我的输出停在第14个元素（最后三个h2没用）

我的代码：

import requests
from bs4 import BeautifulSoup


url = 'https://www.planetware.com/tourist-attractions-/oslo-n-osl-oslo.htm'
url_get = requests.get(url)
soup = BeautifulSoup(url_get.content, 'html.parser')
attraction_place=soup.find_all('h2', class_="sitename")    

for attraction in attraction_place:
    print(attraction.text)
    type(attraction)

输出：

1  Vigeland Sculpture Park
2  Akershus Fortress
3  Viking Ship Museum
4  The National Museum
5  Munch Museum
6  Royal Palace
7  The Museum of Cultural History
8  Fram Museum
9  Holmenkollen Ski Jump and Museum
10  Oslo Cathedral
11  City Hall (Rådhuset)
12  Aker Brygge
13  Natural History Museum & Botanical Gardens
14  Oslo Opera House and Annual Music Festivals
Where to Stay in Oslo for Sightseeing
Tips and Tours: How to Make the Most of Your Visit to Oslo
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我希望列表如下：

attraction=[Vigeland Sculpture Park, Akershus Fortress, ......]

非常感谢你提前。

Answer 1

new = []
count = 1
for attraction in attraction_place:
    while count < 15:
        text = attraction.text
        new.append(text)
        count += 1

Answer 2

一个很简单的方法是采用照片的alt属性。 这样可以获得干净的文本输出，只需要14而不需要切片/索引。

from bs4 import BeautifulSoup
import requests

r = requests.get('https://www.planetware.com/tourist-attractions-/oslo-n-osl-oslo.htm')
soup = bs(r.content, 'lxml')
attractions = [item['alt'] for item in soup.select('.photo [alt]')]
print(attractions)

Answer 3

你可以使用切片。

for attraction in attraction_place[:14]:
    print(attraction.text)
    type(attraction)