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I am studying image-processing using NumPy and facing a problem with filtering with convolution.
I would like to convolve a gray-scale image. (convolve a 2d Array with a smaller 2d Array)
Does anyone have an idea to refine my method?
I know that SciPy supports convolve2d but I want to make a convolve2d only by using NumPy.
What I have done
First, I made a 2d array the submatrices.
a = np.arange(25).reshape(5,5) # original matrix
submatrices = np.array([
[a[:-2,:-2], a[:-2,1:-1], a[:-2,2:]],
[a[1:-1,:-2], a[1:-1,1:-1], a[1:-1,2:]],
[a[2:,:-2], a[2:,1:-1], a[2:,2:]]])
the submatrices seems complicated but what I am doing is shown in the following drawing.
Next, I multiplied each submatrices with a filter.
conv_filter = np.array([[0,-1,0],[-1,4,-1],[0,-1,0]])
multiplied_subs = np.einsum('ij,ijkl->ijkl',conv_filter,submatrices)
and summed them.
np.sum(np.sum(multiplied_subs, axis = -3), axis = -3)
#array([[ 6, 7, 8],
# [11, 12, 13],
# [16, 17, 18]])
Thus this procedure can be called my convolve2d.
def my_convolve2d(a, conv_filter):
submatrices = np.array([
[a[:-2,:-2], a[:-2,1:-1], a[:-2,2:]],
[a[1:-1,:-2], a[1:-1,1:-1], a[1:-1,2:]],
[a[2:,:-2], a[2:,1:-1], a[2:,2:]]])
multiplied_subs = np.einsum('ij,ijkl->ijkl',conv_filter,submatrices)
return np.sum(np.sum(multiplied_subs, axis = -3), axis = -3)
However, I find this my_convolve2d troublesome for 3 reasons.
Generation of the submatrices is too awkward that is difficult to read and can only be used when the filter is 3*3
The size of the variant submatrices seems to be too big, since it is approximately 9 folds bigger than the original matrix.
The summing seems a little non intuitive. Simply said, ugly.
Thank you for reading this far.
Kind of update. I wrote a conv3d for myself. I will leave this as a public domain.
def convolve3d(img, kernel):
# calc the size of the array of submatrices
sub_shape = tuple(np.subtract(img.shape, kernel.shape) + 1)
# alias for the function
strd = np.lib.stride_tricks.as_strided
# make an array of submatrices
submatrices = strd(img,kernel.shape + sub_shape,img.strides * 2)
# sum the submatrices and kernel
convolved_matrix = np.einsum('hij,hijklm->klm', kernel, submatrices)
return convolved_matrix
You could generate the subarrays using as_strided:
import numpy as np
a = np.array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19],
[20, 21, 22, 23, 24]])
sub_shape = (3,3)
view_shape = tuple(np.subtract(a.shape, sub_shape) + 1) + sub_shape
strides = a.strides + a.strides
sub_matrices = np.lib.stride_tricks.as_strided(a,view_shape,strides)
To get rid of your second "ugly" sum, alter your einsum so that the output array only has j and k. This implies your second summation.
conv_filter = np.array([[0,-1,0],[-1,5,-1],[0,-1,0]])
m = np.einsum('ij,ijkl->kl',conv_filter,sub_matrices)
# [[ 6 7 8]
# [11 12 13]
# [16 17 18]]
Cleaned up using as_strided and #Crispin 's einsum trick from above. Enforces the filter size into the expanded shape. Should even allow non-square inputs if the indices are compatible.
def conv2d(a, f):
s = f.shape + tuple(np.subtract(a.shape, f.shape) + 1)
strd = numpy.lib.stride_tricks.as_strided
subM = strd(a, shape = s, strides = a.strides * 2)
return np.einsum('ij,ijkl->kl', f, subM)
You can also use fft (one of the faster methods to perform convolutions)
from numpy.fft import fft2, ifft2
import numpy as np
def fft_convolve2d(x,y):
""" 2D convolution, using FFT"""
fr = fft2(x)
fr2 = fft2(np.flipud(np.fliplr(y)))
m,n = fr.shape
cc = np.real(ifft2(fr*fr2))
cc = np.roll(cc, -m/2+1,axis=0)
cc = np.roll(cc, -n/2+1,axis=1)
return cc
https://gist.github.com/thearn/5424195
you must pad the filter to be the same size as image ( place it in the middle of a zeros_like mat.)
cheers,
Dan
https://laurentperrinet.github.io/sciblog/posts/2017-09-20-the-fastest-2d-convolution-in-the-world.html
Check out all the convolution methods and their respective performances here.
Also, I found the below code snippet to be simpler.
from numpy.fft import fft2, ifft2
def np_fftconvolve(A, B):
return np.real(ifft2(fft2(A)*fft2(B, s=A.shape)))
I have a matrix (2d numpy ndarray, to be precise):
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
And I want to roll each row of A independently, according to roll values in another array:
r = np.array([2, 0, -1])
That is, I want to do this:
print np.array([np.roll(row, x) for row,x in zip(A, r)])
[[0 0 4]
[1 2 3]
[0 5 0]]
Is there a way to do this efficiently? Perhaps using fancy indexing tricks?
Sure you can do it using advanced indexing, whether it is the fastest way probably depends on your array size (if your rows are large it may not be):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:, np.newaxis]
result = A[rows, column_indices]
numpy.lib.stride_tricks.as_strided stricks (abbrev pun intended) again!
Speaking of fancy indexing tricks, there's the infamous - np.lib.stride_tricks.as_strided. The idea/trick would be to get a sliced portion starting from the first column until the second last one and concatenate at the end. This ensures that we can stride in the forward direction as needed to leverage np.lib.stride_tricks.as_strided and thus avoid the need of actually rolling back. That's the whole idea!
Now, in terms of actual implementation we would use scikit-image's view_as_windows to elegantly use np.lib.stride_tricks.as_strided under the hoods. Thus, the final implementation would be -
from skimage.util.shape import view_as_windows as viewW
def strided_indexing_roll(a, r):
# Concatenate with sliced to cover all rolls
a_ext = np.concatenate((a,a[:,:-1]),axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = a.shape[1]
return viewW(a_ext,(1,n))[np.arange(len(r)), (n-r)%n,0]
Here's a sample run -
In [327]: A = np.array([[4, 0, 0],
...: [1, 2, 3],
...: [0, 0, 5]])
In [328]: r = np.array([2, 0, -1])
In [329]: strided_indexing_roll(A, r)
Out[329]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])
Benchmarking
# #seberg's solution
def advindexing_roll(A, r):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]
return A[rows, column_indices]
Let's do some benchmarking on an array with large number of rows and columns -
In [324]: np.random.seed(0)
...: a = np.random.rand(10000,1000)
...: r = np.random.randint(-1000,1000,(10000))
# #seberg's solution
In [325]: %timeit advindexing_roll(a, r)
10 loops, best of 3: 71.3 ms per loop
# Solution from this post
In [326]: %timeit strided_indexing_roll(a, r)
10 loops, best of 3: 44 ms per loop
In case you want more general solution (dealing with any shape and with any axis), I modified #seberg's solution:
def indep_roll(arr, shifts, axis=1):
"""Apply an independent roll for each dimensions of a single axis.
Parameters
----------
arr : np.ndarray
Array of any shape.
shifts : np.ndarray
How many shifting to use for each dimension. Shape: `(arr.shape[axis],)`.
axis : int
Axis along which elements are shifted.
"""
arr = np.swapaxes(arr,axis,-1)
all_idcs = np.ogrid[[slice(0,n) for n in arr.shape]]
# Convert to a positive shift
shifts[shifts < 0] += arr.shape[-1]
all_idcs[-1] = all_idcs[-1] - shifts[:, np.newaxis]
result = arr[tuple(all_idcs)]
arr = np.swapaxes(result,-1,axis)
return arr
I implement a pure numpy.lib.stride_tricks.as_strided solution as follows
from numpy.lib.stride_tricks import as_strided
def custom_roll(arr, r_tup):
m = np.asarray(r_tup)
arr_roll = arr[:, [*range(arr.shape[1]),*range(arr.shape[1]-1)]].copy() #need `copy`
strd_0, strd_1 = arr_roll.strides
n = arr.shape[1]
result = as_strided(arr_roll, (*arr.shape, n), (strd_0 ,strd_1, strd_1))
return result[np.arange(arr.shape[0]), (n-m)%n]
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
r = np.array([2, 0, -1])
out = custom_roll(A, r)
Out[789]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])
By using a fast fourrier transform we can apply a transformation in the frequency domain and then use the inverse fast fourrier transform to obtain the row shift.
So this is a pure numpy solution that take only one line:
import numpy as np
from numpy.fft import fft, ifft
# The row shift function using the fast fourrier transform
# rshift(A,r) where A is a 2D array, r the row shift vector
def rshift(A,r):
return np.real(ifft(fft(A,axis=1)*np.exp(2*1j*np.pi/A.shape[1]*r[:,None]*np.r_[0:A.shape[1]][None,:]),axis=1).round())
This will apply a left shift, but we can simply negate the exponential exponant to turn the function into a right shift function:
ifft(fft(...)*np.exp(-2*1j...)
It can be used like that:
# Example:
A = np.array([[1,2,3,4],
[1,2,3,4],
[1,2,3,4]])
r = np.array([1,-1,3])
print(rshift(A,r))
Building on divakar's excellent answer, you can apply this logic to 3D array easily (which was the problematic that brought me here in the first place). Here's an example - basically flatten your data, roll it & reshape it after::
def applyroll_30(cube, threshold=25, offset=500):
flattened_cube = cube.copy().reshape(cube.shape[0]*cube.shape[1], cube.shape[2])
roll_matrix = calc_roll_matrix_flattened(flattened_cube, threshold, offset)
rolled_cube = strided_indexing_roll(flattened_cube, roll_matrix, cube_shape=cube.shape)
rolled_cube = triggered_cube.reshape(cube.shape[0], cube.shape[1], cube.shape[2])
return rolled_cube
def calc_roll_matrix_flattened(cube_flattened, threshold, offset):
""" Calculates the number of position along time axis we need to shift
elements in order to trig the data.
We return a 1D numpy array of shape (X*Y, time) elements
"""
# armax(...) finds the position in the cube (3d) where we are above threshold
roll_matrix = np.argmax(cube_flattened > threshold, axis=1) + offset
# ensure we don't have index out of bound
roll_matrix[roll_matrix>cube_flattened.shape[1]] = cube_flattened.shape[1]
return roll_matrix
def strided_indexing_roll(cube_flattened, roll_matrix_flattened, cube_shape):
# Concatenate with sliced to cover all rolls
# otherwise we shift in the wrong direction for my application
roll_matrix_flattened = -1 * roll_matrix_flattened
a_ext = np.concatenate((cube_flattened, cube_flattened[:, :-1]), axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = cube_flattened.shape[1]
result = viewW(a_ext,(1,n))[np.arange(len(roll_matrix_flattened)), (n - roll_matrix_flattened) % n, 0]
result = result.reshape(cube_shape)
return result
Divakar's answer doesn't do justice to how much more efficient this is on large cube of data. I've timed it on a 400x400x2000 data formatted as int8. An equivalent for-loop does ~5.5seconds, Seberg's answer ~3.0seconds and strided_indexing.... ~0.5second.
I would like to know if there is an efficient method to get sub-arrays from a larger numpy array.
What I have is an application of np.where. I iterate 'manually' over x and y as offsets and apply where with a kernel to each rectangle extracted from the larger array with proper dimensions.
But is there a more direct approach in numpy's collection of methods?
import numpy as np
example = np.arange(20).reshape((5, 4))
# e.g. a cross kernel
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
np.where(a_kernel, example[1:4, 1:4], 0)
# returns
# array([[ 0, 6, 0],
# [ 9, 10, 11],
# [ 0, 14, 0]])
def arrays_from_kernel(a, a_kernel):
width, height = a_kernel.shape
y_max, x_max = a.shape
return [np.where(a_kernel, a[y:(y + height), x:(x + width)], 0)
for y in range(y_max - height + 1)
for x in range(x_max - width + 1)]
sub_arrays = arrays_from_kernel(example, a_kernel)
This returns the arrays I need for further processing.
# [array([[0, 1, 0],
# [4, 5, 6],
# [0, 9, 0]]),
# array([[ 0, 2, 0],
# [ 5, 6, 7],
# [ 0, 10, 0]]),
# ...
# array([[ 0, 9, 0],
# [12, 13, 14],
# [ 0, 17, 0]]),
# array([[ 0, 10, 0],
# [13, 14, 15],
# [ 0, 18, 0]])]
The context: similar to 2D convolution I would like to apply a custom function on each of the subarrays (e.g. product of squared numbers).
At the moment, you're manually advancing a sliding window over the data - stride tricks to the rescue! (And no, I didn't just make that up - there's actually a submodule called stride_tricks in numpy!) Instead of manually building windows into the data, and calling np.where() on them, if you had the windows in an array, you could call np.where() just once. Stride tricks allow you to create such an array without even having to copy the data.
Let me explain. Normal slices in numpy create views into the original data instead of copies. This is done by referring to the original data, but changing the strides used to access the data (ie. how much to jump between two elements or two rows, and so on). Stride tricks allow you to modify those strides more freely than just slicing and reshaping does, so you can eg. iterate over the same data more than once, which is useful here.
Let me demonstrate:
import numpy as np
example = np.arange(20).reshape((5, 4))
a_kernel = np.array([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
def sliding_window(data, win_shape, **kwargs):
assert data.ndim == len(win_shape)
shape = tuple(dn - wn + 1 for dn, wn in zip(data.shape, win_shape)) + win_shape
strides = data.strides * 2
return np.lib.stride_tricks.as_strided(data, shape=shape, strides=strides, **kwargs)
def arrays_from_kernel(a, a_kernel):
windows = sliding_window(a, a_kernel.shape)
return np.where(a_kernel, windows, 0)
sub_arrays = arrays_from_kernel(example, a_kernel)
The scipy.ndimage module offers a number of filters -- one of which might meet your needs. If none of those filters do what you want, you could use ndimage.generic_filter
to call a custom function on each subarray. ndimage.generic_filter is not as fast as the other ndimage filters, however.
For example,
import numpy as np
example = np.arange(20).reshape((5, 4))
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
# def arrays_from_kernel(a, a_kernel):
# width, height = a_kernel.shape
# y_max, x_max = a.shape
# return [np.where(a_kernel, a[y:(y + height), x:(x + width)], 0)
# for y in range(y_max - height + 1)
# for x in range(x_max - width + 1)]
# sub_arrays = arrays_from_kernel(example, a_kernel)
# for arr in sub_arrays:
# print(arr)
# print('-'*80)
import scipy.ndimage as ndimage
def func(x):
# reject subarrays that extend beyond the border of the `example` array
if not np.isnan(x).any():
y = np.zeros_like(a_kernel, dtype=example.dtype)
np.put(y, np.flatnonzero(a_kernel), x)
print(y)
# Instead or returning 0, you can perform your desired computation on the subarray here.
# Note that you may not need the 2D array y; often, you only need the values in the 1D array x
return 0
result = ndimage.generic_filter(example, func, footprint=a_kernel, mode='constant', cval=np.nan)
For the particular problem of computing the product of squares for each subarray, you
could convert the product into a sum by taking advantage of the fact that A * B = exp(log(A)+log(B)). This would allow you to express the computation as a normal convolution. Now using ndimage.convolve can improve performance a lot. The amount of the improvement depends on the size of example:
import numpy as np
import scipy.ndimage as ndimage
import perfplot
a_kernel = np.asarray([[0, 1, 0], [1, 1, 1], [0, 1, 0]])
def orig(example, a_kernel=a_kernel):
def arrays_from_kernel(a, a_kernel):
width, height = a_kernel.shape
y_max, x_max = a.shape
return [
np.where(a_kernel, a[y : (y + height), x : (x + width)], 1)
for y in range(y_max - height + 1)
for x in range(x_max - width + 1)
]
return [np.prod(x) ** 2 for x in arrays_from_kernel(example, a_kernel)]
def alt(example, a_kernel=a_kernel):
logged = np.log(example)
result = ndimage.convolve(logged, a_kernel, mode="constant", cval=0)[1:-1, 1:-1]
return (np.exp(result) ** 2).ravel()
def make_example(N):
return np.random.random(size=(N, N))
def check(A, B):
return np.allclose(A, B)
perfplot.show(
setup=make_example,
kernels=[orig, alt],
n_range=[2 ** k for k in range(2, 11)],
logx=True,
logy=True,
xlabel="len(example)",
equality_check=check,
)
For the Code below, I am wondering how to make a circular kernel instead of a rectangular one. I am currently looking at something circular, and I want to find the BGR average values for it. By adjusting my kernel, my data will be more accurate.
for center in c_1:
b = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 0]
g = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 1]
r = img2[center[0]-4: center[0]+5, center[1]-4: center[1]+5, 2]
From: https://docs.opencv.org/3.0-beta/doc/py_tutorials/py_imgproc/py_morphological_ops/py_morphological_ops.html
We manually created a structuring elements in the previous examples with help of Numpy. It is rectangular shape. But in some cases, you may need elliptical/circular shaped kernels. So for this purpose, OpenCV has a function, cv2.getStructuringElement(). You just pass the shape and size of the kernel, you get the desired kernel.
# Elliptical Kernel
>>> cv2.getStructuringElement(cv2.MORPH_ELLIPSE,(5,5))
array([[0, 0, 1, 0, 0],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[0, 0, 1, 0, 0]], dtype=uint8)
Get the circle region when given the center, you could try the following function:
def circleAverage(center, r = 4):
"""
"""
for i in range(center[0]-r, center[0]+r):
for j in range(center[1]-r, center[1] + r):
if (center[0] - i) ** 2 + (center[1] - j) ** 2 <= r**2:
// do your computation here.
Hope this helps you.
Came here to find how to make a circular (symmetric) kernel. Ended up with my own implementation.
import numpy as np
def get_circular_kernel(diameter):
mid = (diameter - 1) / 2
distances = np.indices((diameter, diameter)) - np.array([mid, mid])[:, None, None]
kernel = ((np.linalg.norm(distances, axis=0) - mid) <= 0).astype(int)
return kernel
Note that for low diameters, behavior is perhaps unexpected. Variable mid when used for the second time can for example be replaced by diameter / 2.
I've implemented it in a following way:
r = 16
kernel = np.fromfunction(lambda x, y: ((x-r)**2 + (y-r)**2 <= r**2)*1, (2*r+1, 2*r+1), dtype=int).astype(np.uint8)
Extra type conversion is needed to avoid overflow
I already found two solutions for the strides moving windows which can compute mean, max, min, variance, etc. Now, I look to add a count of unique value function by axis. By axis, I mean compute all 2D arrays in single pass.
len(numpy.unique(array)) can make it but a lot of iterations will be needed to compute all arrays. I may work with image as big as 2000 x 2000, so iterations are not a good option. It's all about performance and memory effectiveness.
Here is the two solutions for the strides moving windows:
First is directly taken from Erik Rigtorp's at http://www.mail-archive.com/numpy-discussion#scipy.org/msg29450.html
import numpy as np
def rolling_window_lastaxis(a, window):
if window < 1:
raise ValueError, "`window` must be at least 1."
if window > a.shape[-1]:
raise ValueError, "`window` is too long."
shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
strides = a.strides + (a.strides[-1],)
return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
def rolling_window(a, window):
if not hasattr(window, '__iter__'):
return rolling_window_lastaxis(a, window)
for i, win in enumerate(window):
if win > 1:
a = a.swapaxes(i, -1)
a = rolling_window_lastaxis(a, win)
a = a.swapaxes(-2, i)
return a
filtsize = (3, 3)
a = np.zeros((10,10), dtype=np.float)
a[5:7,5] = 1
b = rolling_window(a, filtsize)
blurred = b.mean(axis=-1).mean(axis=-1)
Second is from Alex Rogozhnikov at http://gozhnikov.github.io/2015/09/30/NumpyTipsAndTricks2.html.
def compute_window_mean_and_var_strided(image, window_w, window_h):
w, h = image.shape
strided_image = np.lib.stride_tricks.as_strided(image,
shape=[w - window_w + 1, h - window_h + 1, window_w, window_h],
strides=image.strides + image.strides)
# important: trying to reshape image will create complete 4-dimensional compy
means = strided_image.mean(axis=(2,3))
mean_squares = (strided_image ** 2).mean(axis=(2, 3))
maximums = strided_image.max(axis=(2,3))
variations = mean_squares - means ** 2
return means, maximums, variations
image = np.random.random([500, 500])
compute_window_mean_and_var_strided(image, 20, 20)
Is there a way to add/implement a count of unique value function in one or both solutions?
Clarification: Basically, I need a Unique Value filter for a 2D array, just like numpy.ndarray.mean.
Thanks you
Alex
Here's one approach with scikit-image's view_as_windows for efficient sliding window extraction.
Steps involved :
Get sliding windows.
Reshape into 2D array. Note that this would make a copy and thus we would lose the efficiency of views, but keep it vectorized.
Sort along the axis of merged block axes.
Get the differentiation along that axes and count the number of different elements, which when added with 1 would be the count of unique values in each of those sliding windows and hence the final expected result.
The implementation would be like so -
from skimage.util import view_as_windows as viewW
def sliding_uniq_count(a, BSZ):
out_shp = np.asarray(a.shape) - BSZ + 1
a_slid4D = viewW(a,BSZ)
a_slid2D = np.sort(a_slid4D.reshape(-1,np.prod(BSZ)),axis=1)
return ((a_slid2D[:,1:] != a_slid2D[:,:-1]).sum(1)+1).reshape(out_shp)
Sample run -
In [233]: a = np.random.randint(0,10,(6,7))
In [234]: a
Out[234]:
array([[6, 0, 5, 7, 0, 8, 5],
[3, 0, 7, 1, 5, 4, 8],
[5, 0, 5, 1, 7, 2, 3],
[5, 1, 3, 3, 7, 4, 9],
[9, 0, 7, 4, 9, 1, 1],
[7, 0, 4, 1, 6, 3, 4]])
In [235]: sliding_uniq_count(a, [3,3])
Out[235]:
array([[5, 4, 4, 7, 7],
[5, 5, 4, 6, 7],
[6, 6, 6, 6, 6],
[7, 5, 6, 6, 6]])
Hybrid approach
To make it work with very large arrays, to accommodate everything into memory, we might have to keep one loop that would iterate along each row of the input data, like so -
def sliding_uniq_count_oneloop(a, BSZ):
S = np.prod(BSZ)
out_shp = np.asarray(a.shape) - BSZ + 1
a_slid4D = viewW(a,BSZ)
out = np.empty(out_shp,dtype=int)
for i in range(a_slid4D.shape[0]):
a_slid2D_i = np.sort(a_slid4D[i].reshape(-1,S),-1)
out[i] = (a_slid2D_i[:,1:] != a_slid2D_i[:,:-1]).sum(-1)+1
return out
Hybrid approach - Version II
Another version of hybrid one, with the explicit usage of np.lib.stride_tricks.as_strided -
def sliding_uniq_count_oneloop(a, BSZ):
S = np.prod(BSZ)
out_shp = np.asarray(a.shape) - BSZ + 1
strd = np.lib.stride_tricks.as_strided
m,n = a.strides
N = out_shp[1]
out = np.empty(out_shp,dtype=int)
for i in range(out_shp[0]):
a_slid3D = strd(a[i], shape=((N,) + tuple(BSZ)), strides=(n,m,n))
a_slid2D_i = np.sort(a_slid3D.reshape(-1,S),-1)
out[i] = (a_slid2D_i[:,1:] != a_slid2D_i[:,:-1]).sum(-1)+1
return out
np.mean operates on a given axis without making any copies. Looking at just the shape of the as_strided array it looks much bigger than the original array. But because each 'window' is a view, it doesn't take up any additional space. Reduction operators like mean work fine with that kind of view.
But note that your second example warns about reshape. That creates a copy; it replicates the values in all of those windows.
unique starts with
ar = np.asanyarray(ar).flatten()
so right off the bat is is making a reshapened copy. It's a copy, and 1d. Then it sorts elements, looks for duplicates etc.
There are ways of finding unique rows, but they require converting rows into large structured array elements. In effect turning a 2d array into a 1d that unique can work with.