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

69,734

Solution 1

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

Solution 2

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]])

Solution 3

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]]

Solution 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]])
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stelios
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stelios

Updated on July 05, 2022

Comments

  • stelios
    stelios over 1 year

    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]

  • Sidharth N. Kashyap
    Sidharth N. Kashyap over 3 years
    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
    bluenote10 over 2 years
    Is vstack even needed? Why not just np.array([X] * N)?