How to create an array with values along specified axis?

Question

numpy.full() is a great function which allows us to generate an array of specific shape and values. For example,

>>>np.full((2,2),[1,2])
array([[1,2],
       [1,2]])

However, it does not have a built-in option to apply values along a specific axis. So, the following code would not work:

>>>np.full((2,2),[1,2],axis=0)
array([[1,1],
       [2,2]])

Hence, I am wondering how I can create a 10x48x271x397 multidimensional array with values [1,2,3,4,5,6,7,8,9,10] inserted along axis=0? In other words, an array with [1,2,3,4,5,6,7,8,9,10] repeated along the first dimensional axis. Is there a way to do this using numpy.full() or an alternative method?

#Does not work, no axis argument in np.full()
values=[1,2,3,4,5,6,7,8,9,10]
np.full((10, 48, 271, 397), values, axis=0)

np.full supports broadcasting: np.full((2,2),[[1],[2]]) for your first example. So np.full((10, 48, 271, 397), np.array(values)[:,None,None,None]). — Michael Szczesny
– Michael Szczesny, Commented Jun 15, 2022 at 18:55

paime · Accepted Answer · 2022-06-16 05:27:54Z

2

Edit: adding ideas from Michael Szczesny

import numpy as np

shape = (10, 48, 271, 397)
root = np.arange(shape[0])

You can use np.full or np.broadcast_to (only get a view at creation time):

arr1 = np.broadcast_to(root, shape[::-1]).T
arr2 = np.full(shape[::-1], fill_value=root).T

%timeit np.broadcast_to(root, shape[::-1]).T
%timeit np.full(shape[::-1], fill_value=root).T

# 3.56 µs ± 18.2 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)
# 75.6 ms ± 243 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

And instead of getting the shape backwards and the array backwards again, you can use singleton dimension, but it seems less generalizable:

root = root[:, None, None, None]

arr3 = np.broadcast_to(root, shape)
arr4 = np.full(shape, fill_value=root)

root = np.arange(shape[0])
%timeit root_ = root[:, None, None, None]; np.broadcast_to(root_, shape)
%timeit root_ = root[:, None, None, None]; np.full(shape, fill_value=root_)

# 3.61 µs ± 6.36 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)
# 57.5 ms ± 114 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Checks that everything is equal and actually what we want:

assert arr1.shape == shape

for i in range(shape[0]):
    sub = arr1[i]
    assert np.all(sub == i)

assert np.all(arr1 == arr2)
assert np.all(arr1 == arr3)
assert np.all(arr1 == arr4)

edited Jun 16, 2022 at 5:27

answered Jun 15, 2022 at 17:50

paime

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1 Comment

Michael Szczesny Over a year ago

This is 6120x faster than np.full on my machine because it's a read-only view of root. But if you only need a view, it is a very nice solution. On the other hand, 5164176 pointers referencing the same element seems to have a rather limited field of application.

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