I also don't recommend this, but you could use a numpy.chararray for this:

import numpy as np
arr = np.chararray((100, 12, 31, 24, 60, 60), itemsize=100)

arr[52, 7, 12, 12, 44, 54] = 'year 1950+52, 7th month, 12th day, 12th hour, 44th minute, 54th second'

I'm not exactly sure what your desired structure is, but the string I inserted into the array should explain the structure I proposed, and you can change it however you need. Note that itemsize limits how many characters you can put in at any index.

Again, with a caveat that this is not necessarily the most efficient thing in the world to do, but if you wish to store lists of ints and/or floats in that array (as per your comment), one way to do it would be to convert that list to strings, and then when retrieving it, re-transform back to a list:

data_to_insert = [1,2,3,4.5]
# store as string
arr[52, 7, 12, 12, 44, 54] = ','.join(map(str, data_to_insert))
# retrieve
arr[52, 7, 12, 12, 44, 54].decode('utf-8').split(',')

This should be pretty fast

You need to use [] instead of 0 as:

n = 5
empty_list = [[[ [] for _ in range(n)] for _ in range(n)] for _ in range(n)]

Value hold by empty_list will be:

[[[[], [], [], [], []], 
  [[], [], [], [], []], 
  [[], [], [], [], []], 
  [[], [], [], [], []], 
  [[], [], [], [], []]], 
# Repeated 4 more times 

For example:

>>> empty_list[0][0][0]  # Access element with nest level 3 in list
[]

def createList(l: list):
    new_l = [[] for i in l]
    return new_l

my_list = [[1, 2, 3, 4], [5, 6], [7, 8, 9],[1],[]]
empty_2dimen_list = createList(my_list)
print(empty_2dimen_list)
>>>  [[], [], [], [], []]

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
[[[], [], [], []], [[], [], [], []], [[], [], [], []], [[], [], [], []]]

example code:

lst = [[]]
for x in a:
    if x != '\n':
        lst[-1].append(x)
    else:
        lst.append([])
print(lst)

output:

[[1, 2, 3, 4, 5], [6, 7, 8, 9, 0], [3, 45, 6, 7, 2]]

I'd suggest using np.full_like to choose the fill-value directly...

x = np.full_like((3, 1), None, dtype=object)

... of course the dtype you chose kind of defines what you mean by "empty"

You get the IndexError because you can't assign to an index in the list beyond the current length of the list. Since grid2 is initialized to an empty list, any attempt to index it will raise this error.

One correct way to write your nested list comprehension using for loops would be to construct the inner list first for each row, then append this to grid2:

grid2 = []
for i in range(rows):
    inner = []
    for j in range(cols):
        inner.append(0)
    grid2.append(inner)

This has nothing to do with your specific use case, you make the beginner's mistake of assigning to the lowest non-existing index instead of using append. Assignment to an index (so list[n] = something) is only possible when there's already an item at that index.

myArray = [[1],[2],[3],[],[5]]

for item in myArray:
    try:
        print(item[0])
    except IndexError:
        item.append(0)
        print(item[0])

Note that you should always except: the specific exception(s) you're expecting, don't use catch-all's (or if you do, print the actual exception). As when you let except: catch everything, then handle those using some silent operation, it will then also handle bugs silently causing frustration why the code works without error but returns incorrect results.