Python：将解析的信息返回到列表？

我的代码：

from urllib2 import urlopen
from bs4 import BeautifulSoup

url = "https://realpython.com/practice/profiles.html"

html_page = urlopen(url)
html_text = html_page.read()

soup = BeautifulSoup(html_text)

links = soup.find_all('a', href = True)

files = []

def page_names():
    for a in links:
        files.append(a['href'])
        return files


page_names()

print files[:]

base = "https://realpython.com/practice/"

print base + files[:]

我试图解析出三个网页文件名并将它们附加到“文件”列表中，然后以某种方式将它们附加或添加到基本url的末尾，以便进行简单打印

我曾尝试将“base”作为一个单独的项目列表，以便添加，但我对Python还是比较陌生，我认为我把for语句搞砸了

目前我得到：

print files[:]
TypeError: 'type' object has no attribute '__getitem__'

---

最后定义了

list[：]

，这是完全错误的，因为

list

是用于创建实际列表的内置关键字

from urllib2 import urlopen
from bs4 import BeautifulSoup

url = "https://realpython.com/practice/profiles.html"

html_page = urlopen(url)
html_text = html_page.read()

soup = BeautifulSoup(html_text)

links = soup.find_all('a', href = True)

files = []

def page_names():
    for a in links:
        files.append(a['href'])


page_names()


base = "https://realpython.com/practice/"
for i in files:
    print base + i

输出：

https://realpython.com/practice/aphrodite.html
https://realpython.com/practice/poseidon.html
https://realpython.com/practice/dionysus.html

您不需要创建中间列表来存储链接或文件，只需使用列表即可

from urllib2 import urlopen
from bs4 import BeautifulSoup
url = "https://realpython.com/practice/profiles.html"
html_page = urlopen(url)
html_text = html_page.read()
soup = BeautifulSoup(html_text)
files = [i['href'] for i in soup.find_all('a', href = True)]
base = "https://realpython.com/practice/"
for i in files:
    print base + i

---

links=soup.find_all（'a'，href=True）

仍然是相同的错误，但谢谢。您尚未定义

列表，因此它仍然是内置类型。你是说
打印链接
（打印时不需要浅拷贝）？