Python 如何从列表中逐项提取并对其执行新功能

如何在posts\u list=[]中的每个项目上执行新功能 假设我有这个函数

def hello():
    print('Hello post from posts_list')

我想给posts\u list=[]

titles = driver.find_elements_by_xpath("//*[@class='item__ad--title']")
posts_list = []
def xxx():
    for title in titles:
        x = title.get_attribute('href')
        posts_list.append(x)

        for p in posts_list:
            print(f"this is {p}")
    print(len(posts_list))

    xxx()

---

这就是你要找的吗

def xxx():
    for title in titles:
        x = title.get_attribute('href')
        posts_list.append(x)

    # this should come out of the titles loop, 
    # since you were currently adding data, 
    # and then you can traverse through it 
    for p in posts_list:
       # Calling your help(), assuming it does accepts p from your list
       help(p)
    print(len(posts_list))

# Indentation matters, please look into the Python Indentation, 
# this is where you will be calling your method outside your def xxx()
xxx()

您的帮助功能

# def help should accept an argument p, which you will use to print the data
def help(p):
  print(f"this is {p}")

我希望这就是你要找的。请仔细阅读本教程，以复习Python的基础知识，因为在编写代码时这确实很重要

---

感谢并愉快地学习：）

原始帖子中的缩进严重扭曲

您试图解决的问题最好通过列表理解来解决：

posts_list = [title.get_attribute('href') for title in titles]
[help(p) for p in posts_list]

如果您对帖子本身不感兴趣，而只对应用

help

的结果感兴趣，则可以将这两个循环组合起来：

[help(title.get_attribute('href')) for title in titles]

---

你想在

posts\u list

的循环中调用

hello（）

，你想把

p

传递给

hello（）

对吗？是的，这就是我想要的答案@Ihab Hamed，让我知道你的情况