Python: initialize multi-dimensional list

Question

I want to initialize a multidimensional list. Basically, I want a 10x10 grid - a list of 10 lists each containing 10 items.

Each list value should be initialized to

Answer 1

I found that to get what you mean you need to youse

import copy

def n_dim_list(dims, init_val):
    if not dims:
        return []
    lst = [init_val for i in range(dims[-1])]
    for d in dims[::-1][1::]:
        lst = [copy.deepcopy(lst) for i in range(d)]
return lst

Where dims is a list of length of the number of the dimentions you want and the content is the size of the requiered nd-list in every dimention. Not the most elegant but clear and does the job.

Answer 2

You might actually need an array instead of some lists. Almost every time I see this "presized nested list" pattern, something is not quite right.

Answer 3

You can do it quite efficiently with a list comprehension:

a = [[0] * number_cols for i in range(number_rows)]

Answer 4

Yet another method, but using the OP's rejected method.

import numpy as np
myList = [[0]*10]*10
myList = np.array(myList)
l=myList.tolist()
myList=l

The output and testing below:

>>> l
[[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]]
>>> l[0][0]=100
>>> l
[[100, 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]]

The output is unlike the expected clone of l[0].

Although this is not time efficient. It takes nearly 7 seconds for a 1000X1000 list, where as list comprehensions took only 0.0052158 seconds for the same.

Answer 5

Two common and short way to do this:

First:

[[0] * n] * m

Second:

[[0 for column in range(n)] for row in range(m)]

Answer 6

Just thought I'd add an answer because the question asked for the general n-dimensional case and I don't think that was answered yet. You can do this recursively for any number of dimensions with the following example:

n_dims = [3, 4, 5]

empty_list = 0
for n in n_dims:
    empty_list = [empty_list] * n

>>>empty_list
>>>[[[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]]]