Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

generating batch of clones from image numpy

Tags:

python

numpy

I have a numpy array (an image) called a with this size:

[3,128,192]

now i want create a numpy array that contains n copies of a which will have this dimension:

[n,3,128,192]

exist a numpy function that can help me with this problem without using loop instructions?

like image 482
Alberto Merciai Avatar asked Apr 12 '17 14:04

Alberto Merciai


2 Answers

Simply use np.stack

# say you need 10 copies of a 3D array `a`
In [267]: n = 10

In [266]: np.stack([a]*n)

Alternatively, you should use np.concatenate if you're really concerned about the performance.

In [285]: np.concatenate([a[np.newaxis, :, :]]*n)

Example:

In [268]: a
Out[268]: 
array([[[ 0,  1,  2,  3],
        [ 4,  5,  6,  7],
        [ 8,  9, 10, 11],
        [12, 13, 14, 15]],

       [[16, 17, 18, 19],
        [20, 21, 22, 23],
        [24, 25, 26, 27],
        [28, 29, 30, 31]],

       [[32, 33, 34, 35],
        [36, 37, 38, 39],
        [40, 41, 42, 43],
        [44, 45, 46, 47]]])

In [271]: a.shape
Out[271]: (3, 4, 4)

In [269]: n = 10

In [270]: np.stack([a]*n).shape
Out[270]: (10, 3, 4, 4)

In [285]: np.concatenate([a[np.newaxis, :, :]]*n).shape
Out[285]: (10, 3, 4, 4)

Performance:

# ~ 4x faster than using `np.stack`
In [292]: %timeit np.concatenate([a[np.newaxis, :, :]]*n)
100000 loops, best of 3: 10.7 µs per loop

In [293]: %timeit np.stack([a]*n)
10000 loops, best of 3: 41.1 µs per loop
like image 109
kmario23 Avatar answered Oct 08 '22 08:10

kmario23


Approach #1 : One way would be with np.repeat to repeat along the first axis after extending the input array to have one more dimension, a singleton one with None/np.newaxis -

np.repeat(a[None],n,axis=0)

Sample runs -

1) 2D case :

In [209]: a
Out[209]: 
array([[7, 8, 0, 1],
       [5, 0, 1, 0],
       [4, 3, 0, 1]])

In [210]: np.repeat(a[None],2,axis=0)
Out[210]: 
array([[[7, 8, 0, 1],
        [5, 0, 1, 0],
        [4, 3, 0, 1]],

       [[7, 8, 0, 1],
        [5, 0, 1, 0],
        [4, 3, 0, 1]]])

2) 3D case :

In [214]: a
Out[214]: 
array([[[7, 2, 4, 2],
        [6, 7, 7, 6],
        [6, 8, 2, 1]],

       [[1, 5, 8, 5],
        [8, 0, 6, 4],
        [1, 2, 8, 8]]])

In [215]: np.repeat(a[None],2,axis=0)
Out[215]: 
array([[[[7, 2, 4, 2],
         [6, 7, 7, 6],
         [6, 8, 2, 1]],

        [[1, 5, 8, 5],
         [8, 0, 6, 4],
         [1, 2, 8, 8]]],


       [[[7, 2, 4, 2],
         [6, 7, 7, 6],
         [6, 8, 2, 1]],

        [[1, 5, 8, 5],
         [8, 0, 6, 4],
         [1, 2, 8, 8]]]])

Approach #2 : If you don't mind a read-only version of the input array, we could use np.broadcast_to -

np.broadcast_to(a, (n,) + a.shape)

Approach #3 : If you don't mind a view into the input array, here's one with NumPy strides -

def strided_repeat_newaxis0(a, n):
    s0,s1,s2 = a.strides
    shp = (n,) + a.shape
    return np.lib.index_tricks.as_strided(a, shape=shp, strides=(0,s0,s1,s2))

Runtime test

In [290]: a = np.random.randint(0,9,(3,128,192))

In [291]: %timeit np.repeat(a[None],100,axis=0)
100 loops, best of 3: 6.15 ms per loop

In [292]: %timeit strided_repeat_newaxis0(a, 100)
100000 loops, best of 3: 4.69 µs per loop

In [293]: %timeit np.broadcast_to(a, (n,) + a.shape)
100000 loops, best of 3: 3.03 µs per loop
like image 40
Divakar Avatar answered Oct 08 '22 10:10

Divakar