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]

The (num)Pythonic way is not to do this but to use broadcasting instead of tile and repmat and the like. — YXD
– YXD, Commented Mar 25, 2014 at 14:44
You might not need to expand it. If, for example, it is added or multiplied with a [M,N] or [1,N] matrix, the result will be [M,N]. numpy broadcasts it for you. In fact you could use that to expand the array: X + np.zeros(N). — hpaulj
– hpaulj, Commented Mar 25, 2014 at 19:17

Cory Kramer · Accepted Answer · 2014-03-25 12:22:40Z

163

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 Mar 25, 2014 at 12:22

Cory Kramer

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Lee · Accepted Answer · 2023-05-14 20:06:36Z

36

You could use vstack:

numpy.vstack([X]*N)

or array (credit to bluenote10 below):

numpy.array([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]])

edited May 14, 2023 at 20:06

answered Mar 25, 2014 at 13:05

Lee

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3 Comments

Sidharth N. Kashyap Over a year ago

vstack works when we need a multidimensional array to get repeated, for example: a=[[1,2,3,4][5,6,7,8]] becomes [[1,2,3,4][5,6,7,8][1,2,3,4][5,6,7,8]] with np.vstack([a]*2). The other approaches get you [[1,2,3,4][1,2,3,4][5,6,7,8][5,6,7,8]]

bluenote10 Over a year ago

Is vstack even needed? Why not just np.array([X] * N)?

Evidlo Over a year ago

This doesn't work for N=1. You should just use np.array instead of np.vstack

Samy Arous · Accepted Answer · 2014-03-25 12:22:14Z

4

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 Mar 25, 2014 at 12:22

Samy Arous

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Comments

calocedrus · Accepted Answer · 2021-07-23 07:15:58Z

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]])

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