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 [DD]
. If v=3
, the new numpy array should be [DDD]
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, ie (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 [DD] 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)
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