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I have two matrices. The first has the following structure:
[[1, 0, a],
[0, 1, b],
[1, 0, c],
[0, 1, d]]
where 1, 0, a, b, c, and d are scalars. The matrix is 4 by 3
The second is just a 2 by 3 matrix:
[[r1],
[r2]]
where r1 and r2 are the first and second rows respectively, each having 3 elements.
I would like the output to be:
[[r1, 0, a*r1],
[0, r1, b*r1],
[r2, 0, c*r2],
[0, r2, d*r2]]
which would be a 4 by 9 matrix.
This is similar to the Kronecker product, except separately for each row of the second matrix. Of course this could be done with cumbersome loops which I want to avoid.
How can I do this concisely?
You can do exactly what you said in the last line: do a separate Kronecker product for each row of the second column and then concatenate the results.
Let's assume that the two matrices are called x (4 by 3) and y (2 by 3). The first thing to do is to split x in two parts because only half matrix participates in each part of the product.
x = x.reshape(2, 2, 3)
Then you can calculate the two products separately:
z0 = np.kron(x[0], y[0])
z1 = np.kron(x[1], y[1])
Finally, concatenate the two results along the first axis:
z = np.concatenate([z0, z1], axis=0)
Or if, like me, you enjoy big ugly one-liners you can do:
z = np.concatenate([np.kron(xr, yr) for xr, yr in zip(x.reshape(2, 2, 3), y)], axis=0)
In the general case you mentioned in the comments, it would become:
z = np.concatenate([np.kron(xr, yr) for xr, yr in zip(x.reshape(int(n / 2), 2, 3), y)], axis=0)
This gives equal results to the explicit loop, which can be numba.jit compiled I believe:
def solve_explicit(x, y):
# sanity checks
assert x.shape[0] == 2*y.shape[0]
assert x.shape[1] == y.shape[1]
n = x.shape[0]
z = np.zeros((n, 9))
for i in range(n):
for j in range(3):
for k in range(3):
z[i, k + 3 * j] = x[i, j] * y[int(i / 2), k]
return z
Using broadcasting, with x.shape (n, 3), and y.shape (n//2, 3):
out = (x.reshape(-1, 2, 3, 1) * y.reshape(-1, 1, 1, 3)).reshape(-1, 9)
I personally would use np.einsum in this situation because I think it's easier to understand than broadcasting.
import numpy as np
(a, b, c, d) = np.random.rand(4)
x = np.array([[1, 0, a], [0, 1, b], [1, 0, c], [0, 1, d]])
y = np.random.rand(2, 3)
z = np.einsum("ij,ik->ijk", x.reshape(-1, 6), y).reshape(-1, 9)
# timeit magic commands.
# %timeit -n 50000 np.einsum("ij,ik->ijk", x.reshape(-1, 6), y).reshape(-1, 9)
# %timeit -n 50000 (x.reshape(-1, 2, 3, 1) * y.reshape(-1, 1, 1, 3)).reshape(-1, 9)
Some good references on Einstein summation in NumPy: [2, 3, 4].
I have two numpy arrays which look like this:
x = [v1, v2, v3, ..., vm]
y = [w1, w2, w3, ..., wn]
where vi, wj are numpy arrays of length 3.
I want to perform a pairwise summation of v's and w's and get a final array
z = [v1+w1, v1+w2,...,v1+wn,v2+w1, ..., vi+wj, ..., vm+wn]
A simple way of obtaining z is as follows:
z = np.zeros ((m*n, 3))
for i in range(m):
for j in range(n):
z[n*i+j] = x[i] + y[j]
This computation is not feasible is m, n are very large.
I know scipy.spatial has methods to enumerate pairwise distances using distance_matrix in a vectorized fashion.
I want to ask if there is a vectorized version of performing such pairwise additions for numpy arrays?
You can take advantage of broadcasting, creating a 2D array, then you can easily get z[i,j] = x[i] + y[j]
x = np.reshape(x, (-1, 1)) # shape (N, 1)
y = np.reshape(y, (-1, 1)) # shape (N, 1)
z = x + y.T # shape (N, N)
If you want to have z as a 1D array you can do z.reshape(-1).
If x is mx3 matrix, y is a nx3
x.shape # (m,3)
y.shape # (n,3)
x1 = x.reshape(m,1,3)
y1 = y.reshape(1,n,3)
z = x1 + y1 # shape (m,n,3)
z1 = z.reshape(-1,3) # (m*n, 3)
equivalently
z = x[:,None]+y
test:
In [263]: x=np.arange(12).reshape(4,3); y=np.arange(6).reshape(2,3)
In [264]: z = x[:,None]+y
In [265]: z.shape
Out[265]: (4, 2, 3)
In [266]: z
Out[266]:
array([[[ 0, 2, 4],
[ 3, 5, 7]],
[[ 3, 5, 7],
[ 6, 8, 10]],
[[ 6, 8, 10],
[ 9, 11, 13]],
[[ 9, 11, 13],
[12, 14, 16]]])
Can I index NumPy N-D array with fallback to default values for out-of-bounds indexes? Example code below for some imaginary np.get_with_default(a, indexes, default):
import numpy as np
print(np.get_with_default(
np.array([[1,2,3],[4,5,6]]), # N-D array
[(np.array([0, 0, 1, 1, 2, 2]), np.array([1, 2, 2, 3, 3, 5]))], # N-tuple of indexes along each axis
13, # Default for out-of-bounds fallback
))
should print
[2 3 6 13 13 13]
I'm looking for some built-in function for this. If such not exists then at least some short and efficient implementation to do that.
I arrived at this question because I was looking for exactly the same. I came up with the following function, which does what you ask for 2 dimension. It could likely be generalised to N dimensions.
def get_with_defaults(a, xx, yy, nodata):
# get values from a, clipping the index values to valid ranges
res = a[np.clip(yy, 0, a.shape[0] - 1), np.clip(xx, 0, a.shape[1] - 1)]
# compute a mask for both x and y, where all invalid index values are set to true
myy = np.ma.masked_outside(yy, 0, a.shape[0] - 1).mask
mxx = np.ma.masked_outside(xx, 0, a.shape[1] - 1).mask
# replace all values in res with NODATA, where either the x or y index are invalid
np.choose(myy + mxx, [res, nodata], out=res)
return res
xx and yy are the index array, a is indexed by (y,x).
This gives:
>>> a=np.zeros((3,2),dtype=int)
>>> get_with_defaults(a, (-1, 1000, 0, 1, 2), (0, -1, 0, 1, 2), -1)
array([-1, -1, 0, 0, -1])
As an alternative, the following implementation achieves the same and is more concise:
def get_with_default(a, xx, yy, nodata):
# get values from a, clipping the index values to valid ranges
res = a[np.clip(yy, 0, a.shape[0] - 1), np.clip(xx, 0, a.shape[1] - 1)]
# replace all values in res with NODATA (gets broadcasted to the result array), where
# either the x or y index are invalid
res[(yy < 0) | (yy >= a.shape[0]) | (xx < 0) | (xx >= a.shape[1])] = nodata
return res
I don't know if there is anything in NumPy to do that directly, but you can always implement it yourself. This is not particularly smart or efficient, as it requires multiple advanced indexing operations, but does what you need:
import numpy as np
def get_with_default(a, indices, default=0):
# Ensure inputs are arrays
a = np.asarray(a)
indices = tuple(np.broadcast_arrays(*indices))
if len(indices) <= 0 or len(indices) > a.ndim:
raise ValueError('invalid number of indices.')
# Make mask of indices out of bounds
mask = np.zeros(indices[0].shape, np.bool)
for ind, s in zip(indices, a.shape):
mask |= (ind < 0) | (ind >= s)
# Only do masking if necessary
n_mask = np.count_nonzero(mask)
# Shortcut for the case where all is masked
if n_mask == mask.size:
return np.full_like(a, default)
if n_mask > 0:
# Ensure index arrays are contiguous so masking works right
indices = tuple(map(np.ascontiguousarray, indices))
for ind in indices:
# Replace masked indices with zeros
ind[mask] = 0
# Get values
res = a[indices]
if n_mask > 0:
# Replace values of masked indices with default value
res[mask] = default
return res
# Test
print(get_with_default(
np.array([[1,2,3],[4,5,6]]),
(np.array([0, 0, 1, 1, 2, 2]), np.array([1, 2, 2, 3, 3, 5])),
13
))
# [ 2 3 6 13 13 13]
I also needed a solution to this, but I wanted a solution that worked in N dimensions. I made Markus' solution work for N-dimensions, including selecting from an array with more dimensions than the coordinates point to.
def get_with_defaults(arr, coords, nodata):
coords, shp = np.array(coords), np.array(arr.shape)
# Get values from arr, clipping to valid ranges
res = arr[tuple(np.clip(c, 0, s-1) for c, s in zip(coords, shp))]
# Set any output where one of the coords was out of range to nodata
res[np.any(~((0 <= coords) & (coords < shp[:len(coords), None])), axis=0)] = nodata
return res
import numpy as np
if __name__ == '__main__':
A = np.array([[1,2,3],[4,5,6]])
B = np.array([[[1, -9],[2, -8],[3, -7]],[[4, -6],[5, -5],[6, -4]]])
coords1 = [[0, 0, 1, 1, 2, 2], [1, 2, 2, 3, 3, 5]]
coords2 = [[0, 0, 1, 1, 2, 2], [1, 2, 2, 3, 3, 5], [1, 1, 1, 1, 1, 1]]
out1 = get_with_defaults(A, coords1, 13)
out2 = get_with_defaults(B, coords1, 13)
out3 = get_with_defaults(B, coords2, 13)
print(out1)
# [2, 3, 6, 13, 13, 13]
print(out2)
# [[ 2 -8]
# [ 3 -7]
# [ 6 -4]
# [13 13]
# [13 13]
# [13 13]]
print(out3)
# [-8, -7, -4, 13, 13, 13]
Suppose I have a 8-direction freeman chain code as follows, in a python list:
freeman_code = [3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5]
Where directions would be defined as follows:
I need to convert this to an image matrix of variable dimensions with valules of 1s and 0s where 1s would depict the shape, as follows, for example:
image_matrix = [
[0, 0, 1, 0, 0, 1],
[0, 0, 0, 1, 0, 1],
[0, 0, 0, 0, 1, 1]
]
Of course, the above is not an exact implementation of the above freeman code. Is there any implementation in python, or in any language that achieves this?
My idea (in python):
Use a defaultdict of defaultdicts with 0 as default:
ImgMatrixDict = defaultdict(lambda: defaultdict(lambda:0))
and then start at a midpoint, say ImgMatrixDict[25][25], and then change values to 1 depending on the freeman code values as I traverse. Afte tis I would convert ImgMatrixDict to a list of lists.
Is this a viable idea or are there any existing libraries or suggestions to implement this? Any idea/pseudo-code would be appreciated.
PS: On performance, yes it would not be important as I won't be doing this in realtime, but generally a code would be around 15-20 charactors in length. I assumed a 50*50 by matrix would suffice for this purpose.
If I am understanding your question correctly:
import numpy as np
import matplotlib.pyplot as plt
freeman_code = [3, 3, 3, 6, 6, 4, 6, 7, 7, 0, 0, 6]
img = np.zeros((10,10))
x, y = 4, 4
img[y][x] = 1
for direction in freeman_code:
if direction in [1,2,3]:
y -= 1
if direction in [5,6,7]:
y += 1
if direction in [3,4,5]:
x -= 1
if direction in [0,1,7]:
x += 1
img[y][x] = 1
plt.imshow(img, cmap='binary', vmin=0, vmax=1)
plt.show()
Here is a solution in python. A dictionary is not adapted to this problem, you would better use a list of list to simulate the table.
D = 10
# DY, DX
FREEMAN = [(0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1), (1, 0), (1, 1)]
freeman_code = [3, 3, 3, 3, 6, 6, 6, 6, 0, 0, 0, 0]
image = [[0]*D for x in range(D)]
y = D/2
x = D/2
image[y][x] = 1
for i in freeman_code:
dy, dx = FREEMAN[i]
y += dy
x += dx
image[y][x] = 1
print("freeman_code")
print(freeman_code)
print("image")
for line in image:
strline = "".join([str(x) for x in line])
print(strline)
>0000000000
>0100000000
>0110000000
>0101000000
>0100100000
>0111110000
>0000000000
>0000000000
>0000000000
>0000000000
Note that the image creation is a condensed expression of:
image = []
for y in range(D):
line = []
for x in range(D):
line.append(0)
image.append(line)
If one day, you need better performance for bigger images, there are solutions using numpy Library but requiring a good knowledge of basic python. Here is an example:
import numpy as np
D = 10
# DY, DX
FREEMAN = [(0, 1), (-1, 1), (-1, 0), (-1, -1), (0, -1), (1, -1), (1, 0), (1, 1)]
DX = np.array([1, 1, 0, -1, -1, -1, 0, 1])
DY = np.array([0, -1, -1, -1, 0, 1, 1, 1])
freeman_code = np.array([3, 3, 3, 3, 6, 6, 6, 6, 0, 0, 0, 0])
image = np.zeros((D, D), int)
y0 = D/2
x0 = D/2
image[y0, x0] = 1
dx = DX[freeman_code]
dy = DY[freeman_code]
xs = np.cumsum(dx)+x0
ys = np.cumsum(dy)+y0
print(xs)
print(ys)
image[ys, xs] = 1
print("freeman_code")
print(freeman_code)
print("image")
print(image)
Here, all loops built with 'for' on previous solution are fast-processed in C.
Say, I have a layer with output dims (4, x, y). I want to split this into 4 separate (1, x, y) tensors, which I can use as input for 4 other layers.
What I'm essentially looking for is the opposite of the Merge layer. I know that there's no split layer in keras, but is there a simple way to do this in keras?
Are you looking for something like this?
import keras.backend as K
import numpy as np
val = np.random.random((4, 2, 3))
t = K.variable(value=val)
t1 = t[0, :, :]
t2 = t[1, :, :]
t3 = t[2, :, :]
t4 = t[3, :, :]
print('t1:\n', K.eval(t1))
print('t2:\n', K.eval(t2))
print('t3:\n', K.eval(t3))
print('t4:\n', K.eval(t4))
print('t:\n', K.eval(t))
It gives the following output:
t1:
[[ 0.18787734 0.1085723 0.01127671]
[ 0.06032621 0.14528386 0.21176969]]
t2:
[[ 0.34292713 0.56848335 0.83797884]
[ 0.11579451 0.21607392 0.80680907]]
t3:
[[ 0.1908586 0.48186591 0.23439431]
[ 0.93413448 0.535191 0.16410089]]
t4:
[[ 0.54303145 0.78971165 0.9961108 ]
[ 0.87826216 0.49061012 0.42450914]]
t:
[[[ 0.18787734 0.1085723 0.01127671]
[ 0.06032621 0.14528386 0.21176969]]
[[ 0.34292713 0.56848335 0.83797884]
[ 0.11579451 0.21607392 0.80680907]]
[[ 0.1908586 0.48186591 0.23439431]
[ 0.93413448 0.535191 0.16410089]]
[[ 0.54303145 0.78971165 0.9961108 ]
[ 0.87826216 0.49061012 0.42450914]]]
Note that, now t1, t2, t3, t4 is of shape(2,3).
print(t1.shape.eval()) # prints [2 3]
So, if you want to keep the 3d shape, you need to do the following:
t1 = t[0, :, :].reshape((1, 2, 3))
t2 = t[1, :, :].reshape((1, 2, 3))
t3 = t[2, :, :].reshape((1, 2, 3))
t4 = t[3, :, :].reshape((1, 2, 3))
Now, you get the spitted tensors in correct dimension.
print(t1.shape.eval()) # prints [1 2 3]
Hope that it will help you to solve your problem.
You can define Lambda layers to do the slicing for you:
from keras.layers import Lambda
from keras.backend import slice
.
.
x = Lambda( lambda x: slice(x, START, SIZE))(x)
For your specific example, try:
x1 = Lambda( lambda x: slice(x, (0, 0, 0), (1, -1, -1)))(x)
x2 = Lambda( lambda x: slice(x, (1, 0, 0), (1, -1, -1)))(x)
x3 = Lambda( lambda x: slice(x, (2, 0, 0), (1, -1, -1)))(x)
x4 = Lambda( lambda x: slice(x, (3, 0, 0), (1, -1, -1)))(x)
You can just simply use tf.split.