I would like to create a 3D array in Python (2.7) to use like this:
distance[i][j][k] And the sizes of the array should be the size of a variable I have. (nnn)
I tried using:
distance = [[[]*n]*n] but that didn't seem to work.
I can only use the default libraries, and the method of multiplying (i.e.,[[0]*n]*n) wont work because they are linked to the same pointer and I need all of the values to be individual
10 Answers
You should use a list comprehension:
>>> import pprint >>> n = 3 >>> distance = [[[0 for k in xrange(n)] for j in xrange(n)] for i in xrange(n)] >>> pprint.pprint(distance) [[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]]] >>> distance[0][1] [0, 0, 0] >>> distance[0][1][2] 0 You could have produced a data structure with a statement that looked like the one you tried, but it would have had side effects since the inner lists are copy-by-reference:
>>> distance=[[[0]*n]*n]*n >>> pprint.pprint(distance) [[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]]] >>> distance[0][0][0] = 1 >>> pprint.pprint(distance) [[[1, 0, 0], [1, 0, 0], [1, 0, 0]], [[1, 0, 0], [1, 0, 0], [1, 0, 0]], [[1, 0, 0], [1, 0, 0], [1, 0, 0]]] numpy.arrays are designed just for this case:
numpy.zeros((i,j,k)) will give you an array of dimensions ijk, filled with zeroes.
depending what you need it for, numpy may be the right library for your needs.
4The right way would be
[[[0 for _ in range(n)] for _ in range(n)] for _ in range(n)] (What you're trying to do should be written like (for NxNxN)
[[[0]*n]*n]*n but that is not correct, see @Adaman comment why).
1d3 = [[[0 for col in range(4)]for row in range(4)] for x in range(6)] d3[1][2][1] = 144 d3[4][3][0] = 3.12 for x in range(len(d3)): print d3[x] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 144, 0, 0], [0, 0, 0, 0]] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [3.12, 0, 0, 0]] [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]] """ Create 3D array for given dimensions - (x, y, z) @author: Naimish Agarwal """ def three_d_array(value, *dim): """ Create 3D-array :param dim: a tuple of dimensions - (x, y, z) :param value: value with which 3D-array is to be filled :return: 3D-array """ return [[[value for _ in xrange(dim[2])] for _ in xrange(dim[1])] for _ in xrange(dim[0])] if __name__ == "__main__": array = three_d_array(False, *(2, 3, 1)) x = len(array) y = len(array[0]) z = len(array[0][0]) print x, y, z array[0][0][0] = True array[1][1][0] = True print array Prefer to use numpy.ndarray for multi-dimensional arrays.
You can also use a nested for loop like shown below
n = 3 arr = [] for x in range(n): arr.append([]) for y in range(n): arr[x].append([]) for z in range(n): arr[x][y].append(0) print(arr) There are many ways to address your problem.
- First one as accepted answer by @robert. Here is the generalised solution for it:
def multi_dimensional_list(value, *args): #args dimensions as many you like. EG: [*args = 4,3,2 => x=4, y=3, z=2] #value can only be of immutable type. So, don't pass a list here. Acceptable value = 0, -1, 'X', etc. if len(args) > 1: return [ multi_dimensional_list(value, *args[1:]) for col in range(args[0])] elif len(args) == 1: #base case of recursion return [ value for col in range(args[0])] else: #edge case when no values of dimensions is specified. return None Eg:
>>> multi_dimensional_list(-1, 3, 4) #2D list [[-1, -1, -1, -1], [-1, -1, -1, -1], [-1, -1, -1, -1]] >>> multi_dimensional_list(-1, 4, 3, 2) #3D list [[[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]], [[-1, -1], [-1, -1], [-1, -1]]] >>> multi_dimensional_list(-1, 2, 3, 2, 2 ) #4D list [[[[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]]], [[[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]], [[-1, -1], [-1, -1]]]] P.S If you are keen to do validation for correct values for args i.e. only natural numbers, then you can write a wrapper function before calling this function.
- Secondly, any multidimensional dimensional array can be written as single dimension array. This means you don't need a multidimensional array. Here are the function for indexes conversion:
def convert_single_to_multi(value, max_dim): dim_count = len(max_dim) values = [0]*dim_count for i in range(dim_count-1, -1, -1): #reverse iteration values[i] = value%max_dim[i] value /= max_dim[i] return values def convert_multi_to_single(values, max_dim): dim_count = len(max_dim) value = 0 length_of_dimension = 1 for i in range(dim_count-1, -1, -1): #reverse iteration value += values[i]*length_of_dimension length_of_dimension *= max_dim[i] return value Since, these functions are inverse of each other, here is the output:
>>> convert_single_to_multi(convert_multi_to_single([1,4,6,7],[23,45,32,14]),[23,45,32,14]) [1, 4, 6, 7] >>> convert_multi_to_single(convert_single_to_multi(21343,[23,45,32,14]),[23,45,32,14]) 21343 - If you are concerned about performance issues then you can use some libraries like pandas, numpy, etc.
n1=np.arange(90).reshape((3,3,-1)) print(n1) print(n1.shape) def n_arr(n, default=0, size=1): if n is 0: return default return [n_arr(n-1, default, size) for _ in range(size)] arr = n_arr(3, 42, 3) assert arr[2][2][2], 42 If you insist on everything initializing as empty, you need an extra set of brackets on the inside ([[]] instead of [], since this is "a list containing 1 empty list to be duplicated" as opposed to "a list containing nothing to duplicate"):
distance=[[[[]]*n]*n]*n 
