Python 2.7 creating a multidimensional list

Question

In Python I want an intuitive way to create a 3 dimensional list.

I want an (n by n) list. So for n = 4 it should be:

x = [[[],[],[],[]],[[],[],[],[]],[[],[],[],[]],[[],[],[],[]]]

I've tried using:

y = [n*[n*[]]]    
y = [[[]]* n for i in range(n)]

Which both appear to be creating copies of a reference. I've also tried naive application of the list builder with little success:

y = [[[]* n for i in range(n)]* n for i in range(n)]
y = [[[]* n for i in range(1)]* n for i in range(n)]

I've also tried building up the array iteratively using loops, with no success. I also tried this:

y = []
for i in range(0,n):
    y.append([[]*n for i in range(n)])

Is there an easier or more intuitive way of doing this?

Solution 1

I think your list comprehension versions were very close to working. You don't need to do any list multiplication (which doesn't work with empty lists anyway). Here's a working version:

>>> y = [[[] for i in range(n)] for i in range(n)]
>>> print y
[[[], [], [], []], [[], [], [], []], [[], [], [], []], [[], [], [], []]]

Solution 2

looks like the most easiest way is as follows:

def create_empty_array_of_shape(shape):
    if shape: return [create_empty_array_of_shape(shape[1:]) for i in xrange(shape[0])]

it's work for me

Solution 3

A very simple and elegant way is:

a = [([0] * 5) for i in range(5)]
a
[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]

Solution 4

In Python I made a little factory method to create a list of variable dimensions and variable sizes on each of those dimensions:

def create_n_dimensional_matrix(self, n):
  dimensions = len(n)
  if (dimensions == 1): 
    return [0 for i in range(n[0])]

  if (dimensions == 2): 
    return [[0 for i in range(n[0])] for j in range(n[1])]

  if (dimensions == 3): 
    return [[[0 for i in range(n[0])] for j in range(n[1])] for k in range(n[2])]

  if (dimensions == 4): 
    return [[[[0 for i in range(n[0])] for j in range(n[1])] for k in range(n[2])] for l in range(n[3])]

run it like this:

print(str(k.create_n_dimensional_matrix([2,3])))
print(str(k.create_n_dimensional_matrix([3,2])))
print(str(k.create_n_dimensional_matrix([1,2,3])))
print(str(k.create_n_dimensional_matrix([3,2,1])))
print(str(k.create_n_dimensional_matrix([2,3,4,5])))
print(str(k.create_n_dimensional_matrix([5,4,3,2])))    

Which prints:

1. The two dimensional lists (2x3), (3x2)
2. The three dimensional lists (1x2x3),(3x2x1)

3. The four dimensional lists (2x3x4x5),(5x4x3x2)

   [[0, 0], [0, 0], [0, 0]]
   
   [[0, 0, 0], [0, 0, 0]]
   
   [[[0], [0]], [[0], [0]], [[0], [0]]]
   
   [[[0, 0, 0], [0, 0, 0]]]
   
   [[[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]], [[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]], [[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]], [[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]], [[[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]], [[0, 0], [0, 0], [0, 0]]]]
   
   [[[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]], [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]], [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]], [[[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]], [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]], [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]]]]
   

Solution 5

i found this:

Matrix = [[0 for x in xrange(5)] for x in xrange(5)]

You can now add items to the list:

Matrix[0][0] = 1
Matrix[4][0] = 5

print Matrix[0][0] # prints 1
print Matrix[4][0] # prints 5

from here: How to define two-dimensional array in python

View more solutions

Comments

[Blckknght]

That was mine, from before your edits. I've removed it now, since your answer is at least sensible now. I think it's still technically wrong, since the questioner specifically wanted an n by n by 0 three dimensional structure and you're only making an n by n two dimensional one.

[user1505695]

fair enough. but this answer can be easily extended to 3D, no? or maybe i misinterpreted the question.

[pterodragon]

I'm not sure if SO should handle cross-links smarter :)

[DavidW]

Possibly. I've deleted my comment anyway since it no longer applies... :)

[orange]

See user2114402's answer. That's a generic way post more than 2 years before your's. "I am amazed that no one" reads previously posted answers ;-).

[pterodragon]

whoops...missed that :D. I originally posted the answer for the other question and I copied that here. Upvoted user2114402's for the short recursive answer.

[Vladimir]

I am interested in an even more generic answer to that question. Imagine we have N dimensions and we need to both get and set an element. Getting an element will be a relatively simple recursion, while setting an element is not that easy (correct me if I am wrong). "l[1][1][1]" won't work simply because we don't know the number of dimensions in advance.

[pterodragon]

I don't quite get exactly what you want. You assumed that the list is an N (known) dimensional list and said you can't set the element in the list because N is unknown?