Python Numpy repeating an arange array

Question

so say I do this

x = np.arange(0, 3)

which gives

array([0, 1, 2])

but what can I do like

x = np.arange(0, 3)*repeat(N=3)times

to get

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

Answer 1

EDIT: Refer to hpaulj's answer. It is frankly better.

The simplest way is to convert back into a list and use:

list(np.arange(0,3))*3

Which gives:

>> [0, 1, 2, 0, 1, 2, 0, 1, 2]

Or if you want it as a numpy array:

np.array(list(np.arange(0,3))*3)

Which gives:

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

Answer 2

how about this one?

arr = np.arange(3)
res = np.hstack((arr, ) * 3)

Output

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

Not much overhead I would say.

Answer 3

I've seen several recent questions about resize. It isn't used often, but here's one case where it does just what you want:

In [66]: np.resize(np.arange(3),3*3)
Out[66]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])

There are many other ways of doing this.

In [67]: np.tile(np.arange(3),3)
Out[67]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])
In [68]: (np.arange(3)+np.zeros((3,1),int)).ravel()
Out[68]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])

np.repeat doesn't repeat in the way we want

In [70]: np.repeat(np.arange(3),3)
Out[70]: array([0, 0, 0, 1, 1, 1, 2, 2, 2])

but even that can be reworked (this is a bit advanced):

In [73]: np.repeat(np.arange(3),3).reshape(3,3,order='F').ravel()
Out[73]: array([0, 1, 2, 0, 1, 2, 0, 1, 2])