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

问题描述

给定以下列表

a = [0, 1, 2, 3]

我想创建一个新列表 b，其中包含对 a 的当前值和下一个值求和的元素.它将包含比 a 少 1 的元素.

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.

像这样:

b = [1, 3, 5]

(从 0+1、1+2 和 2+3)

(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

问题是我一直收到这个错误:

The trouble is I keep getting this error:

IndexError: list index out of range

我很确定它会发生，因为当我获得 a (3) 的最后一个元素时，我无法将其添加到任何内容中，因为这样做超出了它的值(之后没有任何值)3 添加).所以我需要告诉代码在 2 处停止，同时仍然参考 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. 在您的 for 循环中，您将遍历列表 a 的元素.但是在循环体中，当您真正需要索引时，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 将包含 5 个项目，其中有一个数字 100，并且 for 循环将到达它.您实际上将尝试检索列表 a 的第 100 个元素，这显然不存在.这会给你一个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.

我们可以通过迭代一系列索引来解决这个问题:

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

for i in range(len(a))

并像这样访问a的项目:a[i].这不会出错.

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

1. 在循环体中，您不仅索引 a[i]，还索引 a[i+1].这也是潜在错误的地方.如果您的列表包含 5 个项目并且您像我在第 1 点中显示的那样对其进行迭代，您将得到一个 IndexError.为什么?因为range(5)本质上是0 1 2 3 4，所以当循环到4时，你会尝试得到a[5] 项目.由于 Python 中的索引从 0 开始并且您的列表包含 5 个项目，因此最后一个项目的索引为 4，因此获取 a[5] 将意味着获取不存在的第六个元素.莉>

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.

要解决这个问题，您应该从 len(a) 中减去 1 以获得范围序列 0 1 2 3.由于您使用的是索引 i+1，您仍然会得到最后一个元素，但这样您就可以避免错误.

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在此处尝试执行的操作.其中一些非常优雅且更pythonic"，例如列表推导式:

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

这仅在一行中完成工作.

This does the job in only one line.

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