Python 循环：列表索引超出范围

Python Loop: List Index Out of Range

提问by Dance Party2

Given the following list

鉴于以下列表

a = [0, 1, 2, 3]

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

我想创建一个新的 list b，它由当前和下一个值a相加的元素组成。它将包含比 少1 个元素a。

Like this:

像这样：

b = [1, 3, 5]

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

（从 0+1、1+2 和 2+3）

Here's what I've tried:

这是我尝试过的：

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

The trouble is I keep getting this error:

问题是我不断收到此错误：

IndexError: list index out of range

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.

我很确定它会发生，因为当我得到 a (3) 的最后一个元素时，我无法将它添加到任何东西中，因为这样做超出了它的值（在 3 之后没有要添加的值） ）。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。

回答by illright

1. In your forloop, 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 awould 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.

1. 在您的for循环中，您正在遍历 list 的元素a。但是在循环体中，当您真正需要索引时，您正在使用这些项目来索引该列表。
   想象一下，如果列表a将包含 5 个项目，那么其中就有一个数字 100，并且 for 循环会到达它。您实际上将尝试检索 list 的第 100 个元素a，这显然不存在。这会给你一个IndexError.

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

我们可以通过迭代一系列索引来解决这个问题：

for i in range(len(a))

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

并a像这样访问的项目：a[i]。这不会给出任何错误。

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.

2. 在循环体中，您不仅要索引a[i]，还要索引a[i+1]. 这也是一个潜在错误的地方。如果您的列表包含 5 个项目并且您正在迭代它，就像我在第 1 点中显示的那样，您将得到一个IndexError. 为什么？因为range(5)本质上是0 1 2 3 4，所以当循环达到 4 时，您将尝试获取该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 thea[5]将意味着获取不存在的第六个元素。

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.

要解决这个问题，您应该从中减去 1len(a)以获得范围序列0 1 2 3。由于您使用的是 index 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。

1. There are many different ways to accomplish what you're trying to do here. Some of them are quite elegant and more "pythonic", like list comprehensions:

3. 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)]

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

This does the job in only one line.

这仅在一行中完成工作。

回答by K. Menyah

Try reducing the range of the for loop to range(len(a)-1):

尝试将 for 循环的范围缩小到range(len(a)-1)：

a = [0,1,2,3]
b = []

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

print(b)

This can also be written as a list comprehension:

这也可以写成列表理解：

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

回答by Everyone_Else

When you call for i in a:, you are getting the actual elements, not the indexes. When we reach the last element, that is 3, b.append(a[i+1]-a[i])looks for a[4], doesn't find one and then fails. Instead, try iterating over the indexes while stopping just short of the last one, like

当您调用 时for i in a:，您获得的是实际元素，而不是索引。当我们到达最后一个元素，那就是3，b.append(a[i+1]-a[i])验看a[4]，没有找到一个，然后失败。相反，尝试迭代索引，同时在最后一个索引之前停止，例如

for i in range(0, len(a)-1): Do something

for i in range(0, len(a)-1): Do something

Your current code won't work yet for the do something part though ;)

不过，您当前的代码还不能用于做某事部分；)

回答by miradulo

You are accessing the list elements and then using them to attempt to index your list. This is not a good idea. You already have an answer showing how you could use indexing to get your sum list, but another option would be to zipthe list with a slice of itself such that you can sum the pairs.

您正在访问列表元素，然后使用它们尝试为您的列表建立索引。这不是一个好主意。您已经有了一个答案，显示了如何使用索引来获取总和列表，但另一种选择是zip使用自身切片的列表，以便您可以对这些对进行求和。

b = [i + j for i, j in zip(a, a[1:])]