How to pad a 2d array with variable length per row on the left and right to form a bigger 2d array

Question

I have a 2d array like

small = np.arange(9).reshape((3, 3))

array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

I want to apply a padding, each row to pad variable 0s on the left. And to make sure the result 2d array is of shape 3 x 8. (padding 0s on the right)

offset = np.array([1, 3, 2])

So that the result looks like

array([[ 0.,  0.,  1.,  2.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  3.,  4.,  5.,  0.,  0.],
       [ 0.,  0.,  6.,  7.,  8.,  0.,  0.,  0.]])

What's the best way to achieve that?

thanks to @Divakar solution. I ran some benchmarks on the following methods.

def f1(small, offset, ncols):
    nrows, num_small_cols = small.shape
    big = np.zeros((nrows, ncols))
    inner = np.empty_like(small, dtype=np.int64)
    for i in range(num_small_cols):
        inner[:, i] = offset + i
    big[np.arange(nrows)[:, None], inner] = small
    return big

def f2(small, offset, ncols):
    n = small.shape[1]
    r = np.arange(ncols)
    offset2 = offset[:,None]
    # This took a lot of time
    mask = (offset2 <= r) & (offset2 + n > r)
    out = np.zeros_like(mask, dtype=np.float64)
    out[mask] = small.ravel()
    return out

def f3(small, offset, ncols):
    n = small.shape[1]
    m = ncols - n
    small_pad = np.zeros((len(small), n + 2*m))
    small_pad[:,m:m+n] = small    
    w = view_as_windows(small_pad, (1,ncols))[:,:,0]
    return w[np.arange(len(offset)), ncols-offset-n]

n = 10000
offset = np.repeat(np.array([1, 3, 2]), n)
small = np.random.rand(n * 3, 5)

%timeit f1(small, offset, 9)
# 1.32 ms

%timeit f2(small, offset, 9)
# 2.24 ms

%timeit f3(small, offset, 9)
# 1.3 ms

As it turns out using an array for indexing is faster for the strided one. Edited my post with it. So, could you please re-run those tests? — Divakar
– Divakar, Commented Apr 26, 2018 at 16:14
@Divakar Oh, I didn't know np.arange is so much better than range in this case. — colinfang
– colinfang, Commented Apr 26, 2018 at 17:27

Divakar · Accepted Answer · 2018-04-26 16:13:18Z

Approach #1

We can use broadcasting to create a mask for assigning into those positions and then assign into a zeros-intialized array -

def pad_offsetpos(small, ncols):
    n = small.shape[1]
    r = np.arange(ncols)
    mask = (offset[:,None] <= r) & (offset[:,None]+n > r)
    out = np.zeros(mask.shape)
    out[mask] = small.ravel()
    return out

Sample run -

In [327]: small
Out[327]: 
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

In [328]: offset
Out[328]: array([1, 3, 2])

In [329]: pad_offsetpos(small, ncols=8)
Out[329]: 
array([[0., 0., 1., 2., 0., 0., 0., 0.],
       [0., 0., 0., 3., 4., 5., 0., 0.],
       [0., 0., 6., 7., 8., 0., 0., 0.]])

Approach #2

We can also leverage np.lib.stride_tricks.as_strided based scikit-image's view_as_windows for efficient patch extraction after padding the input array with enough zeros on either sides -

from skimage.util.shape import view_as_windows

def pad_offsetpos_strided(small, ncols):
    n = small.shape[1]
    m = ncols - n
    small_pad = np.zeros((len(small), n + 2*m))
    small_pad[:,m:m+n] = small    
    w = view_as_windows(small_pad, (1,ncols))[:,:,0]
    return w[np.arange(len(offset)), ncols-offset-n]

p.deman · Accepted Answer · 2018-04-26 12:10:57Z

0

PaddedMat=numpy.zeros(shape=[3,8],dtype="float")

and then loop to fill it.

PaddedMat[i,offset[i]:offset[i]+3]=small[i,:]

etc...

edited Apr 26, 2018 at 12:10

answered Apr 26, 2018 at 12:01

p.deman

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