在python中创建具有特定大小的空列表 [英] Create an empty list in python with certain size

问题描述

我想创建一个可以容纳10个元素的空列表(或最好的方法).

I want to create an empty list (or whatever is the best way) that can hold 10 elements.

之后，我想在该列表中分配值，例如应该显示0到9:

After that I want to assign values in that list, for example this is supposed to display 0 to 9:

s1 = list();
for i in range(0,9):
   s1[i] = i

print  s1

但是当我运行这段代码时，它会生成一个错误，或者在另一种情况下，它只会显示[](空).

But when I run this code, it generates an error or in another case it just displays [] (empty).

有人可以解释为什么吗?

Can someone explain why?

推荐答案

除非已使用至少i+1个元素初始化了列表，否则您无法分配给类似lst[i] = something的列表.您需要使用append将元素添加到列表的末尾. lst.append(something).

You cannot assign to a list like lst[i] = something, unless the list already is initialized with at least i+1 elements. You need to use append to add elements to the end of the list. lst.append(something).

(如果您使用字典，则可以使用分配符号).

(You could use the assignment notation if you were using a dictionary).

创建一个空列表:

>>> l = [None] * 10
>>> l
[None, None, None, None, None, None, None, None, None, None]

为上述列表中的现有元素分配值:

Assigning a value to an existing element of the above list:

>>> l[1] = 5
>>> l
[None, 5, None, None, None, None, None, None, None, None]

请记住，由于l[15] = 5之类的列表只有10个元素，因此仍然会失败.

Keep in mind that something like l[15] = 5 would still fail, as our list has only 10 elements.

range(x)从[0，1，2，... x-1]创建一个列表

range(x) creates a list from [0, 1, 2, ... x-1]

# 2.X only. Use list(range(10)) in 3.X.
>>> l = range(10)
>>> l
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

使用函数创建列表:

>>> def display():
...     s1 = []
...     for i in range(9): # This is just to tell you how to create a list.
...         s1.append(i)
...     return s1
... 
>>> print display()
[0, 1, 2, 3, 4, 5, 6, 7, 8]

列表理解(使用正方形，因为对于范围您不需要执行所有这些操作，您只需返回range(0,9)即可):

List comprehension (Using the squares because for range you don't need to do all this, you can just return range(0,9) ):

>>> def display():
...     return [x**2 for x in range(9)]
... 
>>> print display()
[0, 1, 4, 9, 16, 25, 36, 49, 64]

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