Python: Concatenate (or clone) a numpy array N times
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]])
stelios
Updated on July 05, 2022Comments
-
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 over 3 yearsvstack 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 2 yearsIs
vstack
even needed? Why not justnp.array([X] * N)
?