I would like to create a function that has input: x.shape==(2,2), and outputs y.shape==(2,2,3).
For example:
#np.vectorize
def foo(x):
#This function doesn't work like I want
return x,x,x
a = np.array([[1,2],[3,4]])
print(foo(a))
#desired output
[[[1 1 1]
[2 2 2]]
[[3 3 3]
[4 4 4]]]
#actual output
(array([[1, 2],
[3, 4]]), array([[1, 2],
[3, 4]]), array([[1, 2],
[3, 4]]))
Or maybe:
#np.vectorize
def bar(x):
#This function doesn't work like I want
return np.array([x,2*x,5])
a = np.array([[1,2],[3,4]])
print(bar(a))
#desired output
[[[1 2 5]
[2 4 5]]
[[3 6 5]
[4 8 5]]]
Note that foo is just an example. I want a way to map over a numpy array (which is what vectorize is supposed to do), but have that map take a 0d object and shove a 1d object in its place. It also seems to me that the dimensions here are arbitrary, as one might wish to take a function that takes a 1d object and returns a 3d object, vectorize it, call it on a 5d object, and get back a 7d object.... However, my specific use case only requires vectorizing a 0d to 1d function, and mapping it appropriately over a 2d array.
It would help, in your question, to show both the actual result and your desired result. As written that isn't very clear.
In [79]: foo(np.array([[1,2],[3,4]]))
Out[79]:
(array([[1, 2],
[3, 4]]), array([[1, 2],
[3, 4]]), array([[1, 2],
[3, 4]]))
As indicated in the vectorize docs, this has returned a tuple of arrays, corresponding to the tuple of values that your function returned.
Your bar returns an array, where as vectorize expected it to return a scalar (or single value):
In [82]: bar(np.array([[1,2],[3,4]]))
ValueError: setting an array element with a sequence.
vectorize takes an otypes parameter that sometimes helps. For example if I say that bar (without the wrapper) returns an object, I get:
In [84]: f=np.vectorize(bar, otypes=[object])
In [85]: f(np.array([[1,2],[3,4]]))
Out[85]:
array([[array([1, 2, 5]), array([2, 4, 5])],
[array([3, 6, 5]), array([4, 8, 5])]], dtype=object)
A (2,2) array of (3,) arrays. The (2,2) shape matches the shape of the input.
vectorize has a relatively new parameter, signature
In [90]: f=np.vectorize(bar, signature='()->(n)')
In [91]: f(np.array([[1,2],[3,4]]))
Out[91]:
array([[[1, 2, 5],
[2, 4, 5]],
[[3, 6, 5],
[4, 8, 5]]])
In [92]: _.shape
Out[92]: (2, 2, 3)
I haven't used this much, so am still getting a feel for how it works. When I've tested it, it is slower than the original scalar version of vectorize. Neither offers any speed advantage of explicit loops. However vectorize does help when 'broadcasting', allowing you to use a variety of input shapes. That's even more useful when your function takes several inputs, not just one as in this case.
In [94]: f(np.array([1,2]))
Out[94]:
array([[1, 2, 5],
[2, 4, 5]])
In [95]: f(np.array(3))
Out[95]: array([3, 6, 5])
For best speed, you want to use existing numpy whole-array functions where possible. For example your foo case can be done with:
In [97]: np.repeat(a[:,:,None],3, axis=2)
Out[97]:
array([[[1, 1, 1],
[2, 2, 2]],
[[3, 3, 3],
[4, 4, 4]]])
np.stack([a]*3, axis=2) also works.
And your bar desired result:
In [100]: np.stack([a, 2*a, np.full(a.shape, 5)], axis=2)
Out[100]:
array([[[1, 2, 5],
[2, 4, 5]],
[[3, 6, 5],
[4, 8, 5]]])
2*a takes advantage of the whole-array multiplication. That's true 'numpy-onic' thinking.
Just repeating the value into another dimension is quite simple:
import numpy as np
x = a = np.array([[1,2],[3,4]])
y = np.repeat(x[:,:,np.newaxis], 3, axis=2)
print y.shape
print y
(2L, 2L, 3L)
[[[1 1 1]
[2 2 2]]
[[3 3 3]
[4 4 4]]]
This seems to work for the "f R0 -> R1 mapped over a nd array giving a (n+1)d one"
def foo(x):
return np.concatenate((x,x))
np.apply_along_axis(foo,2,x.reshape(list(x.shape)+[1]))
doesn't generalize all that well, though
Related
I want to add one numpy array two another so it will look like this:
a = [3, 4]
b = [[6, 5], [2, 1]]
output:
[[3, 4], [[6, 5], [2, 1]]]
It should look like the output above and not like [[3,4],[6,5],[2,1]].
How do I do that with numpy arrays?
Work with pure python lists, and not numpy arrays.
It doesn't make sense to have a numpy array holding two list objects. There's literally no gain in doing that.
If you directly instantiate an array like so:
np.array([[3, 4], [[6, 5], [2, 1]]])
You get
array([[3, 4],
[list([6, 5]), list([2, 1])]], dtype=object)
which is an array with dtype=object. Most of numpy's power is lost in this case. For more information on examples and why, take a look at this thread.
If you work with pure python lists, then you can easily achieve what you want:
>>> a + b
[[3, 4], [[6, 5], [2, 1]]]
Numpy as built-in stack commands that are (in my opinion) slightly easier to use:
>>> a = np.array([3, 4])
>>> b = np.array([[6, 5], [2, 1]])
>>> np.row_stack([b, a])
array([[3, 4],
[6, 5],
[2, 1]])
There's also a column stack.
Ref: https://numpy.org/doc/stable/reference/generated/numpy.ma.row_stack.html
You can't stack arrays of different shapes in one array the way you want (or you have to fill in gaps with NaNs or zeroes), so if you want to iterate over them, consider using list.
a = np.array([3, 4])
b = np.array([[6, 5], [2, 1]])
c = [a, b]
for arr in c:
...
If you still want a numpy array, you can try this:
>>> a = np.array([3, 4])
>>> b = np.array([[6, 5], [2, 1]])
>>> a.resize(b.shape,refcheck=False)
>>> c = np.array([a, b])
array([[[3, 4],
[0, 0]],
[[6, 5],
[2, 1]]])
I am trying to find a realization of accessing elements of numpy arrays corresponding to a feature of Matlab.
Suppose given a (2,2,2) Matlab matrix m in the form
m(:,:,1) = [1,2;3,4]
m(:,:,2) = [5,6;7,8]
Even though this is a 3-d array, Matlab allows accessing its column in the fashion like
m(:,1) = [1;3]
m(:,2) = [2;4]
m(:,3) = [5;7]
m(:,4) = [6;8]
I am curious to know that if numpy supports such indexing so that given the following array
m = array([[[1, 2],
[3, 4]],
[[5, 6],
[7, 8]]])
One can also access columns in the fashion as Matlab listed above.
My answer to this question is as following, suppose given the array listed as in the question
m = array([[[1, 2],
[3, 4]],
[[5, 6],
[7, 8]]])
One can create a list, I call it m_list in the form such that
m_list = [m[i][:,j] for i in range(m.shape[0]) for j in range(m.shape[-1])]
This will output m_list in the form such that
m_list = [array([1, 3]), array([2, 4]), array([7, 9]), array([ 8, 10])]
Now we can access elements of m_list exactly as the fashion as Matlab as listed in the question.
In [41]: m = np.arange(1,9).reshape(2,2,2)
In [42]: m
Out[42]:
array([[[1, 2],
[3, 4]],
[[5, 6],
[7, 8]]])
Indexing the equivalent blocks:
In [47]: m[0,:,0]
Out[47]: array([1, 3])
In [48]: m[0,:,1]
Out[48]: array([2, 4])
In [49]: m[1,:,0]
Out[49]: array([5, 7])
In [50]: m[1,:,1]
Out[50]: array([6, 8])
We can reshape, to "flatten" one pair of dimensions:
In [84]: m = np.arange(1,9).reshape(2,2,2)
In [85]: m.reshape(2,4)
Out[85]:
array([[1, 2, 3, 4],
[5, 6, 7, 8]])
In [87]: m.reshape(2,4)[:,2]
Out[87]: array([3, 7])
and throw in a transpose:
In [90]: m.transpose(1,0,2).reshape(2,4)
Out[90]:
array([[1, 2, 5, 6],
[3, 4, 7, 8]])
MATLAB originally was strictly 2d. Then sometime around v3.9 (2000) they allowed for more, but in a kludgy way. They added a way to index the trailing dimension as though it was multidimensional. In another recent SO I noticed that when reshaping to (2,2,1,1) the result remained (2,2). Trailing size 1 dimensions are squeeze out.
I suspect the m(:,3) is a consequence of that as well.
Testing a 4d MATLAB
>> m=reshape(1:36,2,3,3,2);
>> m(:,:,1)
ans =
1 3 5
2 4 6
>> reshape(m,2,3,6)(:,:,1)
ans =
1 3 5
2 4 6
>> m(:,17)
ans =
33
34
>> reshape(m,2,18)(:,17)
ans =
33
34
I'd like to copy a numpy 2D array into a third dimension. For example, given the 2D numpy array:
import numpy as np
arr = np.array([[1, 2], [1, 2]])
# arr.shape = (2, 2)
convert it into a 3D matrix with N such copies in a new dimension. Acting on arr with N=3, the output should be:
new_arr = np.array([[[1, 2], [1,2]],
[[1, 2], [1, 2]],
[[1, 2], [1, 2]]])
# new_arr.shape = (3, 2, 2)
Probably the cleanest way is to use np.repeat:
a = np.array([[1, 2], [1, 2]])
print(a.shape)
# (2, 2)
# indexing with np.newaxis inserts a new 3rd dimension, which we then repeat the
# array along, (you can achieve the same effect by indexing with None, see below)
b = np.repeat(a[:, :, np.newaxis], 3, axis=2)
print(b.shape)
# (2, 2, 3)
print(b[:, :, 0])
# [[1 2]
# [1 2]]
print(b[:, :, 1])
# [[1 2]
# [1 2]]
print(b[:, :, 2])
# [[1 2]
# [1 2]]
Having said that, you can often avoid repeating your arrays altogether by using broadcasting. For example, let's say I wanted to add a (3,) vector:
c = np.array([1, 2, 3])
to a. I could copy the contents of a 3 times in the third dimension, then copy the contents of c twice in both the first and second dimensions, so that both of my arrays were (2, 2, 3), then compute their sum. However, it's much simpler and quicker to do this:
d = a[..., None] + c[None, None, :]
Here, a[..., None] has shape (2, 2, 1) and c[None, None, :] has shape (1, 1, 3)*. When I compute the sum, the result gets 'broadcast' out along the dimensions of size 1, giving me a result of shape (2, 2, 3):
print(d.shape)
# (2, 2, 3)
print(d[..., 0]) # a + c[0]
# [[2 3]
# [2 3]]
print(d[..., 1]) # a + c[1]
# [[3 4]
# [3 4]]
print(d[..., 2]) # a + c[2]
# [[4 5]
# [4 5]]
Broadcasting is a very powerful technique because it avoids the additional overhead involved in creating repeated copies of your input arrays in memory.
* Although I included them for clarity, the None indices into c aren't actually necessary - you could also do a[..., None] + c, i.e. broadcast a (2, 2, 1) array against a (3,) array. This is because if one of the arrays has fewer dimensions than the other then only the trailing dimensions of the two arrays need to be compatible. To give a more complicated example:
a = np.ones((6, 1, 4, 3, 1)) # 6 x 1 x 4 x 3 x 1
b = np.ones((5, 1, 3, 2)) # 5 x 1 x 3 x 2
result = a + b # 6 x 5 x 4 x 3 x 2
Another way is to use numpy.dstack. Supposing that you want to repeat the matrix a num_repeats times:
import numpy as np
b = np.dstack([a]*num_repeats)
The trick is to wrap the matrix a into a list of a single element, then using the * operator to duplicate the elements in this list num_repeats times.
For example, if:
a = np.array([[1, 2], [1, 2]])
num_repeats = 5
This repeats the array of [1 2; 1 2] 5 times in the third dimension. To verify (in IPython):
In [110]: import numpy as np
In [111]: num_repeats = 5
In [112]: a = np.array([[1, 2], [1, 2]])
In [113]: b = np.dstack([a]*num_repeats)
In [114]: b[:,:,0]
Out[114]:
array([[1, 2],
[1, 2]])
In [115]: b[:,:,1]
Out[115]:
array([[1, 2],
[1, 2]])
In [116]: b[:,:,2]
Out[116]:
array([[1, 2],
[1, 2]])
In [117]: b[:,:,3]
Out[117]:
array([[1, 2],
[1, 2]])
In [118]: b[:,:,4]
Out[118]:
array([[1, 2],
[1, 2]])
In [119]: b.shape
Out[119]: (2, 2, 5)
At the end we can see that the shape of the matrix is 2 x 2, with 5 slices in the third dimension.
Use a view and get free runtime! Extend generic n-dim arrays to n+1-dim
Introduced in NumPy 1.10.0, we can leverage numpy.broadcast_to to simply generate a 3D view into the 2D input array. The benefit would be no extra memory overhead and virtually free runtime. This would be essential in cases where the arrays are big and we are okay to work with views. Also, this would work with generic n-dim cases.
I would use the word stack in place of copy, as readers might confuse it with the copying of arrays that creates memory copies.
Stack along first axis
If we want to stack input arr along the first axis, the solution with np.broadcast_to to create 3D view would be -
np.broadcast_to(arr,(3,)+arr.shape) # N = 3 here
Stack along third/last axis
To stack input arr along the third axis, the solution to create 3D view would be -
np.broadcast_to(arr[...,None],arr.shape+(3,))
If we actually need a memory copy, we can always append .copy() there. Hence, the solutions would be -
np.broadcast_to(arr,(3,)+arr.shape).copy()
np.broadcast_to(arr[...,None],arr.shape+(3,)).copy()
Here's how the stacking works for the two cases, shown with their shape information for a sample case -
# Create a sample input array of shape (4,5)
In [55]: arr = np.random.rand(4,5)
# Stack along first axis
In [56]: np.broadcast_to(arr,(3,)+arr.shape).shape
Out[56]: (3, 4, 5)
# Stack along third axis
In [57]: np.broadcast_to(arr[...,None],arr.shape+(3,)).shape
Out[57]: (4, 5, 3)
Same solution(s) would work to extend a n-dim input to n+1-dim view output along the first and last axes. Let's explore some higher dim cases -
3D input case :
In [58]: arr = np.random.rand(4,5,6)
# Stack along first axis
In [59]: np.broadcast_to(arr,(3,)+arr.shape).shape
Out[59]: (3, 4, 5, 6)
# Stack along last axis
In [60]: np.broadcast_to(arr[...,None],arr.shape+(3,)).shape
Out[60]: (4, 5, 6, 3)
4D input case :
In [61]: arr = np.random.rand(4,5,6,7)
# Stack along first axis
In [62]: np.broadcast_to(arr,(3,)+arr.shape).shape
Out[62]: (3, 4, 5, 6, 7)
# Stack along last axis
In [63]: np.broadcast_to(arr[...,None],arr.shape+(3,)).shape
Out[63]: (4, 5, 6, 7, 3)
and so on.
Timings
Let's use a large sample 2D case and get the timings and verify output being a view.
# Sample input array
In [19]: arr = np.random.rand(1000,1000)
Let's prove that the proposed solution is a view indeed. We will use stacking along first axis (results would be very similar for stacking along the third axis) -
In [22]: np.shares_memory(arr, np.broadcast_to(arr,(3,)+arr.shape))
Out[22]: True
Let's get the timings to show that it's virtually free -
In [20]: %timeit np.broadcast_to(arr,(3,)+arr.shape)
100000 loops, best of 3: 3.56 µs per loop
In [21]: %timeit np.broadcast_to(arr,(3000,)+arr.shape)
100000 loops, best of 3: 3.51 µs per loop
Being a view, increasing N from 3 to 3000 changed nothing on timings and both are negligible on timing units. Hence, efficient both on memory and performance!
This can now also be achived using np.tile as follows:
import numpy as np
a = np.array([[1,2],[1,2]])
b = np.tile(a,(3, 1,1))
b.shape
(3,2,2)
b
array([[[1, 2],
[1, 2]],
[[1, 2],
[1, 2]],
[[1, 2],
[1, 2]]])
A=np.array([[1,2],[3,4]])
B=np.asarray([A]*N)
Edit #Mr.F, to preserve dimension order:
B=B.T
Here's a broadcasting example that does exactly what was requested.
a = np.array([[1, 2], [1, 2]])
a=a[:,:,None]
b=np.array([1]*5)[None,None,:]
Then b*a is the desired result and (b*a)[:,:,0] produces array([[1, 2],[1, 2]]), which is the original a, as does (b*a)[:,:,1], etc.
Summarizing the solutions above:
a = np.arange(9).reshape(3,-1)
b = np.repeat(a[:, :, np.newaxis], 5, axis=2)
c = np.dstack([a]*5)
d = np.tile(a, [5,1,1])
e = np.array([a]*5)
f = np.repeat(a[np.newaxis, :, :], 5, axis=0) # np.repeat again
print('b='+ str(b.shape), b[:,:,-1].tolist())
print('c='+ str(c.shape),c[:,:,-1].tolist())
print('d='+ str(d.shape),d[-1,:,:].tolist())
print('e='+ str(e.shape),e[-1,:,:].tolist())
print('f='+ str(f.shape),f[-1,:,:].tolist())
b=(3, 3, 5) [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
c=(3, 3, 5) [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
d=(5, 3, 3) [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
e=(5, 3, 3) [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
f=(5, 3, 3) [[0, 1, 2], [3, 4, 5], [6, 7, 8]]
Good luck
I wish to vstack a numpy.array (like building a list) but, I cannot initialize the numpy.array with the correct shape to use numpy.append(numpy.empty/zero/like_empty, etc. did not do the trick... anyway. Finally, I figure the two pieces of code below. Is there someyhing more pythonic? I am using python 3.6.9
import numpy as np
a=[]
n=4
for i in range(n):
'''
some calculation resultinng for example in an numpy.array([[i,i+1,i+2])
'''
a.append(np.array([i,i+1,i+2]))
a=np.array(a).reshape(3,n)
print(a)
or because I prefer to mantain it as a numpy array inside the loop:
import numpy as np
a=np.array([])
n=4
for i in range(n):
'''
some calculation resultinng for example in an numpy.array [i,i+1,i+2]
'''
if a.size == 0:
a=np.array([i,i+1,i+2])
else:
a=np.vstack((a,np.array([i,i+1,i+2])))
print(a)
both output:
[[0 1 2]
[1 2 3]
[2 3 4]
[3 4 5]]
Your first use, with list append:
In [146]: alist=[]
In [147]: for i in range(4):
...: alist.append(np.arange(i,i+3))
...:
In [148]: alist
Out[148]: [array([0, 1, 2]), array([1, 2, 3]), array([2, 3, 4]), array([3, 4, 5])]
and make the array:
In [149]: np.array(alist)
Out[149]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5]])
or since vstack is happy with a list of arrays:
In [150]: np.vstack(alist)
Out[150]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5]])
You could use vstack in the loop:
In [151]: arr = np.zeros((0,3),int)
In [152]: for i in range(4):
...: arr = np.vstack((arr, np.arange(i,i+3)))
...:
In [153]: arr
Out[153]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4],
[3, 4, 5]])
This has two problems:
it is slower; list append operates in-place simply adding a pointer to the list. vstack makes whole new array each time, with a full copy!
it is harder to initialize, as you found out. You actually have to understand array shapes, and what concatenate does when it combines 2 or more arrays. Here I started with a (0,3) array.
np.array([np.arange(i, i+3) for i in range(n)])
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])
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.
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.