python how to pad numpy array with zeros

Question

I want to know how I can pad a 2D numpy array with zeros using python 2.6.6 with numpy version 1.5.0. But these are my limitations. Therefore I cannot use np.pad. For example, I want to pad a with zeros such that its shape matches b. The reason why I want to do this is so I can do:

b-a

such that

>>> a
array([[ 1.,  1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.,  1.],
       [ 1.,  1.,  1.,  1.,  1.]])
>>> b
array([[ 3.,  3.,  3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.,  3.,  3.],
       [ 3.,  3.,  3.,  3.,  3.,  3.]])
>>> c
array([[1, 1, 1, 1, 1, 0],
       [1, 1, 1, 1, 1, 0],
       [1, 1, 1, 1, 1, 0],
       [0, 0, 0, 0, 0, 0]])

The only way I can think of doing this is appending, however this seems pretty ugly. is there a cleaner solution possibly using b.shape?

Edit, Thank you to MSeiferts answer. I had to clean it up a bit, and this is what I got:

def pad(array, reference_shape, offsets):
    """
    array: Array to be padded
    reference_shape: tuple of size of ndarray to create
    offsets: list of offsets (number of elements must be equal to the dimension of the array)
    will throw a ValueError if offsets is too big and the reference_shape cannot handle the offsets
    """

    # Create an array of zeros with the reference shape
    result = np.zeros(reference_shape)
    # Create a list of slices from offset to offset + shape in each dimension
    insertHere = [slice(offsets[dim], offsets[dim] + array.shape[dim]) for dim in range(array.ndim)]
    # Insert the array in the result at the specified offsets
    result[insertHere] = array
    return result

MSeifert · Answer 1 · 2019-02-07T13:21:04.030

NumPy 1.7.0 (when numpy.pad was added) is pretty old now (it was released in 2013) so even though the question asked for a way without using that function I thought it could be useful to know how that could be achieved using numpy.pad.

It's actually pretty simple:

>>> import numpy as np
>>> a = np.array([[ 1.,  1.,  1.,  1.,  1.],
...               [ 1.,  1.,  1.,  1.,  1.],
...               [ 1.,  1.,  1.,  1.,  1.]])
>>> np.pad(a, [(0, 1), (0, 1)], mode='constant')
array([[ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.]])

In this case I used that 0 is the default value for mode='constant'. But it could also be specified by passing it in explicitly:

>>> np.pad(a, [(0, 1), (0, 1)], mode='constant', constant_values=0)
array([[ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.]])

Just in case the second argument ([(0, 1), (0, 1)]) seems confusing: Each list item (in this case tuple) corresponds to a dimension and item therein represents the padding before (first element) and after (second element). So:

[(0, 1), (0, 1)]
         ^^^^^^------ padding for second dimension
 ^^^^^^-------------- padding for first dimension

  ^------------------ no padding at the beginning of the first axis
     ^--------------- pad with one "value" at the end of the first axis.

In this case the padding for the first and second axis are identical, so one could also just pass in the 2-tuple:

>>> np.pad(a, (0, 1), mode='constant')
array([[ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.]])

In case the padding before and after is identical one could even omit the tuple (not applicable in this case though):

>>> np.pad(a, 1, mode='constant')
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.]])

Or if the padding before and after is identical but different for the axis, you could also omit the second argument in the inner tuples:

>>> np.pad(a, [(1, ), (2, )], mode='constant')
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

However I tend to prefer to always use the explicit one, because it's just to easy to make mistakes (when NumPys expectations differ from your intentions):

>>> np.pad(a, [1, 2], mode='constant')
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

Here NumPy thinks you wanted to pad all axis with 1 element before and 2 elements after each axis! Even if you intended it to pad with 1 element in axis 1 and 2 elements for axis 2.

I used lists of tuples for the padding, note that this is just "my convention", you could also use lists of lists or tuples of tuples, or even tuples of arrays. NumPy just checks the length of the argument (or if it doesn't have a length) and the length of each item (or if it has a length)!

That's really well explained. Far better than the original documentation. Thanks. — Innat, Nov 11 '18 at 22:40
`mode='constant'` is the sensible default, so padding with zeros can be achieved without the need for any optional keyword, leading to slightly more readable code. — divenex, Jan 10 '20 at 14:50
how can I add padding only to the third dimension of a 3D numpy array? — Ramsha Siddiqui, Sep 14 '20 at 06:36
@RamshaSiddiqui you can use 0s for the dimensions that should not be padded. — MSeifert, Sep 18 '20 at 20:37

MSeifert · Accepted Answer · 2016-03-04T00:21:24.540

Very simple, you create an array containing zeros using the reference shape:

result = np.zeros(b.shape)
# actually you can also use result = np.zeros_like(b) 
# but that also copies the dtype not only the shape

and then insert the array where you need it:

result[:a.shape[0],:a.shape[1]] = a

and voila you have padded it:

print(result)
array([[ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.]])

You can also make it a bit more general if you define where your upper left element should be inserted

result = np.zeros_like(b)
x_offset = 1  # 0 would be what you wanted
y_offset = 1  # 0 in your case
result[x_offset:a.shape[0]+x_offset,y_offset:a.shape[1]+y_offset] = a
result

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

but then be careful that you don't have offsets bigger than allowed. For x_offset = 2 for example this will fail.

If you have an arbitary number of dimensions you can define a list of slices to insert the original array. I've found it interesting to play around a bit and created a padding function that can pad (with offset) an arbitary shaped array as long as the array and reference have the same number of dimensions and the offsets are not too big.

def pad(array, reference, offsets):
    """
    array: Array to be padded
    reference: Reference array with the desired shape
    offsets: list of offsets (number of elements must be equal to the dimension of the array)
    """
    # Create an array of zeros with the reference shape
    result = np.zeros(reference.shape)
    # Create a list of slices from offset to offset + shape in each dimension
    insertHere = [slice(offset[dim], offset[dim] + array.shape[dim]) for dim in range(a.ndim)]
    # Insert the array in the result at the specified offsets
    result[insertHere] = a
    return result

And some test cases:

import numpy as np

# 1 Dimension
a = np.ones(2)
b = np.ones(5)
offset = [3]
pad(a, b, offset)

# 3 Dimensions

a = np.ones((3,3,3))
b = np.ones((5,4,3))
offset = [1,0,0]
pad(a, b, offset)

Just to summarize case I needed: if inserting at origin, arbitrary dimensions: `padded = np.zeros(b.shape)` `padded[tuple(slice(0,n) for n in a.shape)] = a` — shaneb, Jun 24 '19 at 16:17

Juan Leni · Answer 3 · 2017-05-05T06:05:21.117

10

I understand that your main problem is that you need to calculate d=b-a but your arrays have different sizes. There is no need for an intermediate padded c

You can solve this without padding:

import numpy as np

a = np.array([[ 1.,  1.,  1.,  1.,  1.],
              [ 1.,  1.,  1.,  1.,  1.],
              [ 1.,  1.,  1.,  1.,  1.]])

b = np.array([[ 3.,  3.,  3.,  3.,  3.,  3.],
              [ 3.,  3.,  3.,  3.,  3.,  3.],
              [ 3.,  3.,  3.,  3.,  3.,  3.],
              [ 3.,  3.,  3.,  3.,  3.,  3.]])

d = b.copy()
d[:a.shape[0],:a.shape[1]] -=  a

print d

Output:

[[ 2.  2.  2.  2.  2.  3.]
 [ 2.  2.  2.  2.  2.  3.]
 [ 2.  2.  2.  2.  2.  3.]
 [ 3.  3.  3.  3.  3.  3.]]

edited May 05 '17 at 06:05

answered Mar 02 '16 at 15:43

Juan Leni

6,982
5
55
87

True, for his specific case, he does not necessarily need to pad but that's one of the very few arithmetic operations where padding and your approach are equivalent. Nevertheless nice answer! – MSeifert Mar 02 '16 at 16:22
1

Not only that. This could be also more memory efficient than zero-padding. – norok2 Aug 31 '18 at 12:10

score 2 · Answer 4 · answered May 22 '22 at 16:53

TL;DR

def pad_n_cols_left_of_2d_matrix(arr, n):
    """Adds n columns of zeros to left of 2D numpy array matrix.

    :param arr: A two dimensional numpy array that is padded.
    :param n: the number of columns that are added to the left of the matrix.
    """
    padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
    padded_array[:, n:] = arr
    return padded_array


def pad_n_cols_right_of_2d_matrix(arr, n):
    """Adds n columns of zeros to right of 2D numpy array matrix.

    :param arr: A two dimensional numpy array that is padded.
    :param n: the number of columns that are added to the right of the matrix.
    """
    padded_array = np.zeros((arr.shape[0], arr.shape[1] + n))
    padded_array[:, : arr.shape[1]] = arr
    return padded_array


def pad_n_rows_above_2d_matrix(arr, n):
    """Adds n rows of zeros above 2D numpy array matrix.

    :param arr: A two dimensional numpy array that is padded.
    :param n: the number of rows that are added above the matrix.
    """
    padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
    padded_array[n:, :] = arr
    return padded_array


def pad_n_rows_below_2d_matrix(arr, n):
    """Adds n rows of zeros below 2D numpy array matrix.

    :param arr: A two dimensional numpy array that is padded.
    :param n: the number of rows that are added below the matrix.
    """
    padded_array = np.zeros((arr.shape[0] + n, arr.shape[1]))
    padded_array[: arr.shape[0], :] = arr
    return padded_array

Output

Original array:
[[0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.4 0.6 0.8 1.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.4 0.6 0.8 1.  0.  0.  0.  0.  0.  0. ]]
Pad left:
[[0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.4 0.6 0.8 1.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.5 0.8 1.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.2 0.4 0.6 0.8 1.  0.  0.  0.  0.  0.  0. ]]
Pad right:
[[0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.5 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
 [0.  0.  0.  0.3 0.7 1.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0.  0. ]
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Motivation

I came here searching for how to pad an array and found a lot of text, even though my objective, like the question was simple: pad a 2D array with n rows or columns. I think the explanations are better because they help build understanding. However, just to allow some people to save some time if they like, here is a copy-pastable function.

Code Description

Each function adds n rows or columns to the incoming array. The code assumes the incoming array is a 2D numpy array. The direction of the padding is according to the function definition/names. For a more detailed explanation I would like to refer the reader to the answer by MSeifert.

score 1 · Answer 5 · edited May 15 '18 at 20:50

1

In case you need to add a fence of 1s to an array:

>>> mat = np.zeros((4,4), np.int32)
>>> mat
array([[0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0],
       [0, 0, 0, 0]])
>>> mat[0,:] = mat[:,0] = mat[:,-1] =  mat[-1,:] = 1
>>> mat
array([[1, 1, 1, 1],
       [1, 0, 0, 1],
       [1, 0, 0, 1],
       [1, 1, 1, 1]])

edited May 15 '18 at 20:50

MSeifert

145,886
38
333
352

answered Nov 15 '17 at 09:12

Qasim Wani · Answer 6 · 2020-09-25T00:40:12.770

I know I'm a bit late to this, but in case you wanted to perform relative padding (aka edge padding), here's how you can implement it. Note that the very first instance of assignment results in zero-padding, so you can use this for both zero-padding and relative padding (this is where you copy the edge values of the original array into the padded array).

def replicate_padding(arr):
    """Perform replicate padding on a numpy array."""
    new_pad_shape = tuple(np.array(arr.shape) + 2) # 2 indicates the width + height to change, a (512, 512) image --> (514, 514) padded image.
    padded_array = np.zeros(new_pad_shape) #create an array of zeros with new dimensions
    
    # perform replication
    padded_array[1:-1,1:-1] = arr        # result will be zero-pad
    padded_array[0,1:-1] = arr[0]        # perform edge pad for top row
    padded_array[-1, 1:-1] = arr[-1]     # edge pad for bottom row
    padded_array.T[0, 1:-1] = arr.T[0]   # edge pad for first column
    padded_array.T[-1, 1:-1] = arr.T[-1] # edge pad for last column
    
    #at this point, all values except for the 4 corners should have been replicated
    padded_array[0][0] = arr[0][0]     # top left corner
    padded_array[-1][0] = arr[-1][0]   # bottom left corner
    padded_array[0][-1] = arr[0][-1]   # top right corner 
    padded_array[-1][-1] = arr[-1][-1] # bottom right corner

    return padded_array

Complexity Analysis:

The optimal solution for this is numpy's pad method. After averaging for 5 runs, np.pad with relative padding is only 8% better than the function defined above. This shows that this is fairly an optimal method for relative and zero-padding padding.


#My method, replicate_padding
start = time.time()
padded = replicate_padding(input_image)
end = time.time()
delta0 = end - start

#np.pad with edge padding
start = time.time()
padded = np.pad(input_image, 1, mode='edge')
end = time.time()
delta = end - start


print(delta0) # np Output: 0.0008790493011474609 
print(delta)  # My Output: 0.0008130073547363281
print(100*((delta0-delta)/delta)) # Percent difference: 8.12316715542522%

score 1 · Answer 7 · answered Jan 09 '21 at 04:50

Tensorflow also implemented functions for resizing/padding images tf.image.pad tf.pad.

padded_image = tf.image.pad_to_bounding_box(image, top_padding, left_padding, target_height, target_width)

padded_image = tf.pad(image, paddings, "CONSTANT")

These functions work just like other input-pipeline features of tensorflow and will work much better for machine learning applications.

python how to pad numpy array with zeros

7 Answers7

TL;DR

Output

Motivation

Code Description

Complexity Analysis:

Linked

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