Python循环:列表索引超出范围 [英] Python Loop: List Index Out of Range

问题描述

提供以下列表

a = [0, 1, 2, 3]

我想创建一个新列表b，该列表由将a的当前值和下一个值相加的元素组成.该元素将比a少 1 个元素.

I'd like to create a new list b, which consists of elements for which the current and next value of a are summed. It will contain 1 less element than a.

赞:

b = [1, 3, 5]

(从0 + 1、1 + 2和2 + 3开始)

(from 0+1, 1+2, and 2+3)

这是我尝试过的:

b = []
for i in a:
   b.append(a[i + 1] - a[i])
b

问题是我不断收到此错误:

The trouble is I keep getting this error:

IndexError: list index out of range

我很确定会发生这种情况，因为当我得到(3)的最后一个元素时，我无法将其添加到任何东西中，因为这样做超出了它的值范围(之后没有任何值3添加).因此，我需要告诉代码停止在2处，同时仍要引用3进行计算.

I'm pretty sure it occurs because by the time I get the the last element of a (3), I can't add it to anything because doing so goes outside of the value of it (there is no value after 3 to add). So I need to tell the code to stop at 2 while still referring to 3 for the calculation.

推荐答案

1. 在for循环中，您正在遍历列表a的元素.但是在循环的主体中，当您实际需要索引时，您将使用这些项目为该列表建立索引.
   想象一下，如果列表a包含5个项目，则其中将包含100，并且for循环将到达该项目.实际上，您将尝试检索列表a的第100个元素，该元素显然不存在.这将给您一个IndexError.

1. In your for loop, you're iterating through the elements of a list a. But in the body of the loop, you're using those items to index that list, when you actually want indexes.
   Imagine if the list a would contain 5 items, a number 100 would be among them and the for loop would reach it. You will essentially attempt to retrieve the 100th element of the list a, which obviously is not there. This will give you an IndexError.

我们可以通过遍历一系列索引来解决此问题:

We can fix this issue by iterating over a range of indexes instead:

for i in range(len(a))

，然后访问a的项目，例如:a[i].这不会出现任何错误.

and access the a's items like that: a[i]. This won't give any errors.

1. 在循环的主体中，不仅要索引a[i]，而且要索引a[i+1].这也是潜在错误的地方.如果您的列表包含5个项目，并且像我在第1点中所示的那样对它进行迭代，则会得到IndexError.为什么?因为range(5)本质上是0 1 2 3 4，所以当循环达到4时，您将尝试获取a[5]项.由于Python中的索引从0开始且您的列表包含5个项目，因此最后一个项目的索引为4，因此获取a[5]意味着获取不存在的第六个元素.

1. In the loop's body, you're indexing not only a[i], but also a[i+1]. This is also a place for a potential error. If your list contains 5 items and you're iterating over it like I've shown in the point 1, you'll get an IndexError. Why? Because range(5) is essentially 0 1 2 3 4, so when the loop reaches 4, you will attempt to get the a[5] item. Since indexing in Python starts with 0 and your list contains 5 items, the last item would have an index 4, so getting the a[5] would mean getting the sixth element which does not exist.

要解决此问题，应从len(a)中减去1以获取范围序列0 1 2 3.由于您使用的是索引i+1，因此您仍然会获得最后一个元素，但是这样做可以避免错误.

To fix that, you should subtract 1 from len(a) in order to get a range sequence 0 1 2 3. Since you're using an index i+1, you'll still get the last element, but this way you will avoid the error.

1. 有许多不同的方法可以完成您要在此处完成的工作.其中一些非常优雅，更"pythonic"，例如列表理解:

b = [a[i] + a[i+1] for i in range(len(a) - 1)]

这只能完成一行工作.

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