问题：为什么此迭代列表增长代码使IndexError：列表分配索引超出范围？
请考虑以下代码：
i = [1, 2, 3, 5, 8, 13]
j = []
k = 0
for l in i:
j[k] = l
k += 1
print j
输出（Win 7 32位上的Python 2.6.6）为：
> Traceback (most recent call last):
> j[k] = l IndexError: list assignment index out of range
我想这很简单，我不明白。有人可以清理吗？
Please consider the following code:
i = [1, 2, 3, 5, 8, 13]
j = []
k = 0
for l in i:
j[k] = l
k += 1
print j
The output (Python 2.6.6 on Win 7 32-bit) is:
> Traceback (most recent call last):
> j[k] = l IndexError: list assignment index out of range
I guess it’s something simple I don’t understand. Can someone clear it up?
回答 0
j是一个空列表，但您正在尝试[0]在第一次迭代中写入元素，但该迭代尚不存在。
请尝试以下操作，以将新元素添加到列表的末尾：
for l in i:
j.append(l)
当然，如果您只想复制一个现有列表，则您永远都不会这样做。您只需：
j = list(i)
另外，如果您想像其他语言的数组一样使用Python列表，则可以预先创建一个列表，并将其元素设置为空值（None在下面的示例中），然后在特定位置覆盖值：
i = [1, 2, 3, 5, 8, 13]
j = [None] * len(i)
#j == [None, None, None, None, None, None]
k = 0
for l in i:
j[k] = l
k += 1
要实现的事情是，list对象将不允许您将值分配给不存在的索引。
j is an empty list, but you’re attempting to write to element [0] in the first iteration, which doesn’t exist yet.
Try the following instead, to add a new element to the end of the list:
for l in i:
j.append(l)
Of course, you’d never do this in practice if all you wanted to do was to copy an existing list. You’d just do:
j = list(i)
Alternatively, if you wanted to use the Python list like an array in other languages, then you could pre-create a list with its elements set to a null value (None in the example below), and later, overwrite the values in specific positions:
i = [1, 2, 3, 5, 8, 13]
j = [None] * len(i)
#j == [None, None, None, None, None, None]
k = 0
for l in i:
j[k] = l
k += 1
The thing to realise is that a list object will not allow you to assign a value to an index that doesn’t exist.
回答 1
您的另一个选择是初始化j：
j = [None] * len(i)
Your other option is to initialize j:
j = [None] * len(i)
回答 2
做j.append(l)而不是j[k] = l避免k。
Do j.append(l) instead of j[k] = l and avoid k at all.
回答 3
您还可以使用列表理解：
j = [l for l in i]
或使用以下语句进行复制：
j = i[:]
You could also use a list comprehension:
j = [l for l in i]
or make a copy of it using the statement:
j = i[:]
回答 4
j.append(l)
还要避免使用小写的“ L”，因为它们很容易与1混淆。
j.append(l)
Also avoid using lower-case “L’s” because it is easy for them to be confused with 1’s
回答 5
我认为您正在寻找Python方法插入项：
在位置i处插入元素x。list.insert（i，x）
array = [1,2,3,4,5]
array.insert(1,20)
print(array)
# prints [1,2,20,3,4,5]
I think the Python method insert is what you’re looking for:
Inserts element x at position i.
list.insert(i,x)
array = [1,2,3,4,5]
array.insert(1,20)
print(array)
# prints [1,2,20,3,4,5]
回答 6
您可以为j使用字典（类似于关联数组）
i = [1, 2, 3, 5, 8, 13]
j = {} #initiate as dictionary
k = 0
for l in i:
j[k] = l
k += 1
print j
将打印：
{0: 1, 1: 2, 2: 3, 3: 5, 4: 8, 5: 13}
You could use a dictionary (similar to an associative array) for j
i = [1, 2, 3, 5, 8, 13]
j = {} #initiate as dictionary
k = 0
for l in i:
j[k] = l
k += 1
print j
will print :
{0: 1, 1: 2, 2: 3, 3: 5, 4: 8, 5: 13}
回答 7
另一种方式：
j=i[0]
for k in range(1,len(i)):
j = numpy.vstack([j,i[k]])
在这种情况下j将是一个numpy数组
One more way:
j=i[0]
for k in range(1,len(i)):
j = numpy.vstack([j,i[k]])
In this case j will be a numpy array
回答 8
也许你需要extend（）
i=[1,3,5,7]
j=[]
j.extend(i)
Maybe you need extend()
i=[1,3,5,7]
j=[]
j.extend(i)
