Python: Concatenate (or clone) a numpy array N times

Question

I want to create an MxN numpy array by cloning a Mx1 ndarray N times. Is there an efficient pythonic way to do that instead of looping?

Btw the following way doesn’t work for me (X is my Mx1 array) :

   numpy.concatenate((X, numpy.tile(X,N)))

since it created a [M*N,1] array instead of [M,N]

Asked By: stelios

||

Source

Answer 1

Have you tried this:

n = 5
X = numpy.array([1,2,3,4])
Y = numpy.array([X for _ in xrange(n)])
print Y
Y[0][1] = 10
print Y

prints:

[[1 2 3 4]
 [1 2 3 4]
 [1 2 3 4]
 [1 2 3 4]
 [1 2 3 4]]

[[ 1 10  3  4]
 [ 1  2  3  4]
 [ 1  2  3  4]
 [ 1  2  3  4]
 [ 1  2  3  4]]

Answered By: Samy Arous

Answer 2

You are close, you want to use np.tile, but like this:

a = np.array([0,1,2])
np.tile(a,(3,1))

Result:

array([[0, 1, 2],
   [0, 1, 2],
   [0, 1, 2]])

If you call np.tile(a,3) you will get concatenate behavior like you were seeing

array([0, 1, 2, 0, 1, 2, 0, 1, 2])

http://docs.scipy.org/doc/numpy/reference/generated/numpy.tile.html

Answered By: Cory Kramer

Answer 3

You could use vstack:

numpy.vstack([X]*N)

e.g.

>>> import numpy as np
>>> X = np.array([1,2,3,4])
>>> N = 7
>>> np.vstack([X]*N)
array([[1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4],
       [1, 2, 3, 4]])

Answered By: atomh33ls

Answer 4

An alternative to np.vstack is np.array used this way (also mentioned by @bluenote10 in a comment):

x = np.arange([-3,4]) # array([-3, -2, -1,  0,  1,  2,  3])
N = 3 # number of time you want the array repeated
X0 = np.array([x] * N)

gives:

array([[-3, -2, -1,  0,  1,  2,  3],
       [-3, -2, -1,  0,  1,  2,  3],
       [-3, -2, -1,  0,  1,  2,  3]])

You can also use meshgrid this way (granted it’s longer to write, and kind of pulling hairs but you get yet another possibility and you may learn something new along the way):

X1,_ = np.meshgrid(a,np.empty([N]))

>>> X1 shows:

array([[-3, -2, -1,  0,  1,  2,  3],
       [-3, -2, -1,  0,  1,  2,  3],
       [-3, -2, -1,  0,  1,  2,  3]])

Checking that all these are equivalent:

meshgrid and np.array approach

X0 == X1

result:

array([[ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True]])

np.array and np.vstack approach

X0 == np.vstack([x] * 3)

result:

array([[ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True]])

np.array and np.tile approach

X0 == np.tile(x,(N,1))

result:

array([[ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True,  True,  True]])

Answered By: calocedrus

Python: Concatenate (or clone) a numpy array N times

Question:

Answers: