Repeating numpy arrays

Question

I have a 3D np arrays like this :

 x= array([[[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]]])

I would like to repeat my array n times ( eg 3 times) like this :

array([[[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]],

    [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]]

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]]])

I have tried like this :

xx=np.vstack([x]*3)
print xx.reshape(6,4,3)


array([[[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]],

   [[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]],

   [[  1.,   2.,   3.],
    [  4.,   5.,   6.],
    [  7.,   8.,   9.],
    [ 10.,  11.,  12.]],

   [[ 13.,  14.,  15.],
    [ 16.,  17.,  18.],
    [ 19.,  20.,  21.],
    [ 22.,  23.,  24.]]])

How can I get in the order what I want, there should be the easy way to do this. Thanks in advance for your suggestions.

Addy · Accepted Answer · 2016-12-31 04:18:49Z

4

After a bit of trial and error I have found a way to do it:

np.tile(x.reshape(2,12), [1,3]).reshape(6,4,3)

answered Dec 31, 2016 at 4:18

Addy

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Comments

akuiper · Accepted Answer · 2016-12-31 16:14:19Z

2

You can use np.repeat with axis = 0:

np.repeat(x, [3, 3], axis = 0)  # or more generally np.repeat(x, [n] * len(x), axis = 0)
                                # here n is the repeat times
Out[514]:
array([[[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]]])

Another option would be to index it as:

x[[0,0,0,1,1,1]]

Or programmatically:

x[[i for i in range(len(x)) for j in range(3)]]
Out[518]:
array([[[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[  1.,   2.,   3.],
        [  4.,   5.,   6.],
        [  7.,   8.,   9.],
        [ 10.,  11.,  12.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]],

       [[ 13.,  14.,  15.],
        [ 16.,  17.,  18.],
        [ 19.,  20.,  21.],
        [ 22.,  23.,  24.]]])

edited Dec 31, 2016 at 16:14

answered Dec 31, 2016 at 4:18

akuiper

216k33 gold badges362 silver badges379 bronze badges

2 Comments

hpaulj Over a year ago

Starting with a (2,4,3) you shouldn't need the reshape after repeat.

akuiper Over a year ago

@hpaulj Thanks for the comment. It works with [3,3] along axis = 0 without reshape.

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