>>> n = 5                     #length of list
>>> list = [None] * n         #populate list, length n with n entries "None"
>>> print(list)
[None, None, None, None, None]

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[None, None, None, None, 1]

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[None, None, None, 1, 1]

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[None, None, 1, 1, 1]

或者一开始就什么都没有:

>>> n = 5                     #length of list
>>> list = []                 # create list
>>> print(list)
[]

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[1]

在append的第4次迭代中:

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[1,1,1,1]

5及所有后续:

>>> list.append(1)            #append 1 to right side of list
>>> list = list[-n:]          #redefine list as the last n elements of list
>>> print(list)
[1,1,1,1,1]

我发现很容易做的一件事是我设置一个数组 例如，我喜欢的大小的空字符串

代码:

import numpy as np

x= np.zeros(5,str)
print x

输出:

['' '' '' '' '']

希望这对你有帮助:)

这样做:

>>> d = [ [ None for y in range( 2 ) ] for x in range( 2 ) ]
>>> d
[[None, None], [None, None]]
>>> d[0][0] = 1
>>> d
[[1, None], [None, None]]

其他的解决方案会导致这样的问题:

>>> d = [ [ None ] * 2 ] * 2
>>> d
[[None, None], [None, None]]
>>> d[0][0] = 1
>>> d
[[1, None], [1, None]]

一个简单的解决方案是x = [None]*length，但请注意，它将所有列表元素初始化为None。如果大小是固定的，你也可以执行x=[None,None,None,None]。但严格来说，无论采用哪种方式，都不会得到未定义的元素，因为Python中不存在这种麻烦。

你可以使用:

>>> lst = [None] * 5
>>> lst
[None, None, None, None, None]

最好的办法是使用numpy库。

from numpy import ndarray

a = ndarray((5,),int)