为什么我的函数只循环列表中的三个元素

发布于 2024-11-27 13:13:38 字数 841 浏览 1 评论 0原文

我试图循环遍历列表中的所有元素（其中 8 个），但我的函数只为我提供其中 4 个。这是我作为个人项目制作的应用程序。

import urllib


def get_followers():
    count = 0
    link = ['http://twitter.com/statuses/user_timeline.xml?screen_name=beratmahmuzlu',     'http://twitter.com/statuses/user_timeline.xml?screen_name=lidiazuin', 'http://twitter.com/statuses/user_timeline.xml?screen_name=kelewele_boham', 'http://twitter.com/statuses/user_timeline.xml?screen_name=AwangHafizam', 'http://twitter.com/statuses/user_timeline.xml?screen_name=BRAYANLVN', 'http://twitter.com/statuses/user_timeline.xml?screen_name=zezol_pl', 'http://twitter.com/statuses/user_timeline.xml?screen_name=brysonwong', 'http://twitter.com/statuses/user_timeline.xml?screen_name=racsozara']
    while count < len(link):
        print link[count]
        link.pop()
        count = count + 1

原文

I am trying to loop through all the elements (8 of them) in my list, but my function is only providing me with 4 of them. This is for an application I am making as a personal project.

import urllib


def get_followers():
    count = 0
    link = ['http://twitter.com/statuses/user_timeline.xml?screen_name=beratmahmuzlu',     'http://twitter.com/statuses/user_timeline.xml?screen_name=lidiazuin', 'http://twitter.com/statuses/user_timeline.xml?screen_name=kelewele_boham', 'http://twitter.com/statuses/user_timeline.xml?screen_name=AwangHafizam', 'http://twitter.com/statuses/user_timeline.xml?screen_name=BRAYANLVN', 'http://twitter.com/statuses/user_timeline.xml?screen_name=zezol_pl', 'http://twitter.com/statuses/user_timeline.xml?screen_name=brysonwong', 'http://twitter.com/statuses/user_timeline.xml?screen_name=racsozara']
    while count < len(link):
        print link[count]
        link.pop()
        count = count + 1

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从来不烧饼 2024-12-04 13:13:38

您正在弹出列表并根据列表的计数进行循环。

尝试使用 for 循环：

for lnk in link:
    print lnk

You are popping the list and basing your loop off of the count of the list.

try a for loop instead:

for lnk in link:
    print lnk

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一影成城 2024-12-04 13:13:38

link.pop() 删除一个元素，而 len(link) 在每次迭代时为您提供列表的新长度，因此仅循环列表的一半。

def get_followers():
    count = 0
    link = [0, 1, 2, 3, 4, 5, 6, 7, 8]
    l = len(link)
    while count < l:
        print link.pop()
        count = count + 1

这是正确的实现，尽管有许多更简洁的方法可以在 python 中迭代列表，但这是最简单的方法之一：

def get_followers():
    link = [0, 1, 2, 3, 4, 5, 6, 7, 8]
    for l in link:
        print l

link.pop() removes an element and len(link) gives you the new length of the list at each iteration, looping thus thru only the half of your list.

def get_followers():
    count = 0
    link = [0, 1, 2, 3, 4, 5, 6, 7, 8]
    l = len(link)
    while count < l:
        print link.pop()
        count = count + 1

This is the correct implementation, although there are many more cleaner way to iterate over a list in python, this is one of the simplest:

def get_followers():
    link = [0, 1, 2, 3, 4, 5, 6, 7, 8]
    for l in link:
        print l

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彻夜缠绵 2024-12-04 13:13:38

我正在尝试循环遍历列表中的所有元素（其中 8 个）

然后执行此操作。不要设置计数器变量，重复使用它来索引列表、从列表中删除元素并递增计数器。只需循环遍历元素即可。

如果我让你打碎一打鸡蛋，你不需要在上面写数字，也不需要考虑你已经打碎了多少个鸡蛋。你就这么做吧。

所以就这么做吧。

for link in links:
    print link

I am trying to loop through all the elements (8 of them) in my list

Then do that. Don't set up a counter variable, repeatedly use it to index into the list, remove elements from the list and increment the counter. Just loop through the elements.

If I asked you to crack a dozen eggs, you would not need to write numbers on them, or think about how many eggs you'd already cracked. You'd just do it.

So just do it.