Initialising an array of fixed size in Python [duplicate]

Question

You can use:

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

Why don't these questions get answered with the obvious answer?

a = numpy.empty(n, dtype=object)

This creates an array of length n that can store objects. It can't be resized or appended to. In particular, it doesn't waste space by padding its length. This is the Python equivalent of Java's

Object[] a = new Object[n];

If you're really interested in performance and space and know that your array will only store certain numeric types then you can change the dtype argument to some other value like int. Then numpy will pack these elements directly into the array rather than making the array reference int objects.

Do this:

>>> 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]]

The other solutions will lead to this kind of problem:

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

The best bet is to use the numpy library.

from numpy import ndarray

a = ndarray((5,),int)

>>> import numpy
>>> x = numpy.zeros((3,4))
>>> x
array([[ 0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.]])
>>> y = numpy.zeros(5)   
>>> y
array([ 0.,  0.,  0.,  0.,  0.])

x is a 2-d array, and y is a 1-d array. They are both initialized with zeros.

An easy solution is x = [None]*length, but note that it initializes all list elements to None. If the size is really fixed, you can do x=[None,None,None,None,None] as well. But strictly speaking, you won't get undefined elements either way because this plague doesn't exist in Python.

>>> 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]

or with really nothing in the list to begin with:

>>> 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]

on the 4th iteration of append:

>>> 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 and all subsequent:

>>> 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]