Using numpy.vectorize() to rotate all elements of a NumPy array

Question

I am in the beginning phases of learning NumPy. I have a Numpy array of 3x3 matrices. I would like to create a new array where each of those matrices is rotated 90 degrees. I've studied this answer but I still can't figure out what I am doing wrong.

import numpy as np

# 3x3
m = np.array([[1,2,3], [4,5,6], [7,8,9]])

# array of 3x3
a = np.array([m,m,m,m])

# rotate a single matrix counter-clockwise
def rotate90(x):
    return np.rot90(x)

# function that can be called on all elements of an np.array
# Note: I've tried different values for otypes= without success
f = np.vectorize(rotate90)

result = f(a)
# ValueError: Axes=(0, 1) out of range for array of ndim=0.
# The error occurs in NumPy's rot90() function.

Note: I realize I could do the following but I'd like to understand the vectorized option.

t = np.array([ np.rot90(x, k=-1) for x in a])

Your array, `a`, is 3d. Check its shape (4,3,3) and dtype, jnt. So its elements are numbers, not the `m` 2d arrays. `vectorize` feeds those numbers, not 3x3 arrays to your function. It is not a tool for iterating on just one dimension of an array. — hpaulj, Jun 25 '17 at 13:04
@hpaulj Thank you. Is there no way to set this up so that my array is just an "array of 3x3 matrices"? (i.e. so that vectorize would treat each element as a separate entity) — RobertJoseph, Jun 25 '17 at 13:09
Look again at the `vectorize` docs. It doesn't promise you the sun moon and stars.- speed. — hpaulj, Jun 25 '17 at 14:52
But even `map` doesn't promise speed. It's just an alternative to a list comprehension. Anyways, `np.vectorize` is probably misnamed, and often misused. But the speed disclaimer is quite explicit. As for fairness - comments don't count one way or other to your reputation. — hpaulj, Jun 25 '17 at 19:33
On rereading the `vectorize` docs I found a way of using it with your problem - a new `signature` parameter. — hpaulj, Jun 26 '17 at 02:56

Willem Van Onsem · Accepted Answer · 2017-06-25T13:05:56.497

No need to do the rotations individually: numpy has a builtin numpy.rot90(m, k=1, axes=(0, 1)) function. By default the matrix is thus rotate over the first and second dimension.

If you want to rotate one level deeper, you simply have to set the axes over which rotation happens, one level deeper (and optionally swap them if you want to rotate in a different direction). Or as the documentation specifies:

axes: (2,) array_like

The array is rotated in the plane defined by the axes. Axes must be different.

So we rotate over the y and z plane (if we label the dimensions x, y and z) and thus we either specify (2,1) or (1,2).

All you have to do is set the axes correctly, when you want to rotate to the right/left:

np.rot90(a,axes=(2,1)) # right
np.rot90(a,axes=(1,2)) # left

This will rotate all matrices, like:

>>> np.rot90(a,axes=(2,1))
array([[[7, 4, 1],
        [8, 5, 2],
        [9, 6, 3]],

       [[7, 4, 1],
        [8, 5, 2],
        [9, 6, 3]],

       [[7, 4, 1],
        [8, 5, 2],
        [9, 6, 3]],

       [[7, 4, 1],
        [8, 5, 2],
        [9, 6, 3]]])

Or if you want to rotate to the left:

>>> np.rot90(a,axes=(1,2))
array([[[3, 6, 9],
        [2, 5, 8],
        [1, 4, 7]],

       [[3, 6, 9],
        [2, 5, 8],
        [1, 4, 7]],

       [[3, 6, 9],
        [2, 5, 8],
        [1, 4, 7]],

       [[3, 6, 9],
        [2, 5, 8],
        [1, 4, 7]]])

Note that you can only specify the axes from numpy 1.12 and (probably) future versions.

hpaulj · Answer 2 · 2017-06-26T08:16:43.913

Normally np.vectorize is used to apply a scalar (Python, non-numpy) function to all elements of an array, or set of arrays. There's a note that's often overlooked:

The vectorize function is provided primarily for convenience, not for performance. The implementation is essentially a for loop.

In [278]: m = np.array([[1,2,3],[4,5,6]])
In [279]: np.vectorize(lambda x:2*x)(m)
Out[279]: 
array([[ 2,  4,  6],
       [ 8, 10, 12]])

This multiplies each element of m by 2, taking care of the looping paper-work for us.

Better yet, when given several arrays, it broadcasts (a generalization of 'outer product').

In [280]: np.vectorize(lambda x,y:2*x+y)(np.arange(3), np.arange(2)[:,None])
Out[280]: 
array([[0, 2, 4],
       [1, 3, 5]])

This feeds (x,y) scalar tuples to the lambda for all combinations of a (3,) array broadcasted against a (2,1) array, resulting in a (2,3) array. It can be viewed as a broadcasted extension of map.

The problem with np.vectorize(np.rot90) is that rot90 takes a 2d array, but vectorize will feed it scalars.

However I see in the docs that for v1.12 they've added a signature parameter. This is the first time I used it.

Your problem - apply np.rot90 to 2d elements of a 3d array:

In [266]: m = np.array([[1,2,3],[4,5,6]])
In [267]: a = np.stack([m,m])
In [268]: a
Out[268]: 
array([[[1, 2, 3],
        [4, 5, 6]],

       [[1, 2, 3],
        [4, 5, 6]]])

While you could describe this a as an array of 2d arrays, it's better to think of it as a 3d array of integers. That's how the np.vectorize(myfun)(a) sees it, giving myfun each number.

Applied to a 2d m:

In [269]: np.rot90(m)
Out[269]: 
array([[3, 6],
       [2, 5],
       [1, 4]])

With the Python work horse, the list comprehension:

In [270]: [np.rot90(i) for i in a]
Out[270]: 
[array([[3, 6],
        [2, 5],
        [1, 4]]), array([[3, 6],
        [2, 5],
        [1, 4]])]

The result is a list, but we could wrap that in np.array.

Python map does the same thing.

In [271]: list(map(np.rot90, a))
Out[271]: 
[array([[3, 6],
        [2, 5],
        [1, 4]]), array([[3, 6],
        [2, 5],
        [1, 4]])]

The comprehension and map both iterate on the 1st dimension of a, action on the resulting 2d element.

vectorize with signature:

In [272]: f = np.vectorize(np.rot90, signature='(n,m)->(k,l)')
In [273]: f(a)
Out[273]: 
array([[[3, 6],
        [2, 5],
        [1, 4]],

       [[3, 6],
        [2, 5],
        [1, 4]]])

The signature tells it to pass a 2d array and expect back a 2d array. (I should explore how signature plays with the otypes parameter.)

Some quick time comparisons:

In [287]: timeit np.array([np.rot90(i) for i in a])
10000 loops, best of 3: 40 µs per loop
In [288]: timeit np.array(list(map(np.rot90, a)))
10000 loops, best of 3: 41.1 µs per loop
In [289]: timeit np.vectorize(np.rot90, signature='(n,m)->(k,l)')(a)
1000 loops, best of 3: 234 µs per loop
In [290]: %%timeit f=np.vectorize(np.rot90, signature='(n,m)->(k,l)')
     ...: f(a)
     ...: 
1000 loops, best of 3: 196 µs per loop

So for a small array, the Python list methods are faster, by quite a bit. Sometimes, numpy approaches do better with larger arrays, though I doubt in this case.

rot90 with the axes parameter is even better, and will do well with larger arrays:

In [292]: timeit np.rot90(a,axes=(1,2))
100000 loops, best of 3: 15.7 µs per loop

Looking at the np.rot90 code, I see that it is just doing np.flip (reverse) and np.transpose, in various combinations depending on the k. In effect for this case it is doing:

In [295]: a.transpose(0,2,1)[:,::-1,:]
Out[295]: 
array([[[3, 6],
        [2, 5],
        [1, 4]],

       [[3, 6],
        [2, 5],
        [1, 4]]])

(this is even faster than rot90.)

I suspect vectorize with the signature is doing something like:

In [301]: b = np.zeros(2,dtype=object)
In [302]: b[...] = [m,m]
In [303]: f = np.frompyfunc(np.rot90, 1,1)
In [304]: f(b)
Out[304]: 
array([array([[3, 6],
       [2, 5],
       [1, 4]]),
       array([[3, 6],
       [2, 5],
       [1, 4]])], dtype=object)

np.stack(f(b)) will convert the object array into a 3d array like the other code.

frompyfunc is the underlying function for vectorize, and returns an array of objects. Here I create an array like your a except it is 1d, containing multiple m arrays. It is an array of arrays, as opposed to a 3d array.

Using numpy.vectorize() to rotate all elements of a NumPy array

2 Answers2

Linked