I have a numpy array D of dimensions 4x4
I want a new numpy array based on an user defined value v
If v=2, the new numpy array should be [D D].
If v=3, the new numpy array should be [D D D]
How do i initialise such a numpy array as numpy.zeros(v) dont allow me to place arrays as elements?
If I understand correctly, you want to take a 2D array and tile it v times in the first dimension? You can use np.repeat:
# a 2D array
D = np.arange(4).reshape(2, 2)
print D
# [[0 1]
# [2 3]]
# tile it 3 times in the first dimension
x = np.repeat(D[None, :], 3, axis=0)
print x.shape
# (3, 2, 2)
print x
# [[[0 1]
# [2 3]]
# [[0 1]
# [2 3]]
# [[0 1]
# [2 3]]]
If you wanted the output to be kept two-dimensional, i.e. (6, 2), you could omit the [None, :] indexing (see this page for more info on numpy's broadcasting rules).
print np.repeat(D, 3, axis=0)
# [[0 1]
# [0 1]
# [0 1]
# [2 3]
# [2 3]
# [2 3]]
Another alternative is np.tile, which behaves slightly differently in that it will always tile over the last dimension:
print np.tile(D, 3)
# [[0, 1, 0, 1, 0, 1],
# [2, 3, 2, 3, 2, 3]])
You can do that as follows:
import numpy as np
v = 3
x = np.array([np.zeros((4,4)) for _ in range(v)])
>>> print 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. 0. 0. 0.]
[ 0. 0. 0. 0.]]
[[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]]]
Here you go, see if this works for you.
import numpy as np
v = raw_input('Enter: ')
To intialize the numpy array of arrays from user input (obviously can be whatever shape you're wanting here):
b = np.zeros(shape=(int(v),int(v)))
I know this isn't initializing a numpy array but since you mentioned wanting an array of [D D] if v was 2 for example, just thought I'd throw this in as another option as well.
new_array = []
for x in range(0, int(v)):
new_array.append(D)
np.concatenate, np.vstack, np.hstack etc. Either way, it's more efficient to use np.tile or np.repeat to get there in one call.
[D D]is supposed to mean.