I have np.ndarray A of shape (N, M, D).
I'd like to create np.ndarray B of shape (N, M, D, D) such that for every pair of fixed indices n, m along axes 0 and 1
B[n, m] = np.eye(A[n, m])
I understand how to solve this problem using cycles, yet I'd like to write code performing this in vectorized manner. How can this be done using numpy?
import numpy as np
A = ... # Your array here
n, m, d = A.shape
indices = np.arange(d)
B = np.zeros((n, m, d, d))
B[:, :, indices, indices] = A
Related
I have a tensor phi = np.random.rand(n, n, 3) and a matrix D = np.random.rand(3, 3). I want to multiply the matrix D along the last axis of phi so that the output has shape (n, n, 3). I have tried this
np.einsum("klj,ij->kli", phi, D)
But I am not confident in this notation at all. Basically I want to do
res = np.zeros_like(phi)
for i in range(n):
for j in range(n):
res[i, j, :] = D.dot(phi[i, j, :])
You are treating phi as an n, n array of vectors, each of which is to be left-multiplied by D. So you want to keep the n, n portion of the shape exactly as-is. The last (only) dimension of the vectors should be multiplied and summed with the last dimension of the matrix (the vectors are implicitly 3x1):
np.einsum('ijk,lk->ijl', phi, D)
OR
np.einsum('ij,klj->kli', D, phi)
It's likely much simpler to use broadcasting with np.matmul (the # operator):
np.squeeze(D # phi[..., None])
You can omit the squeeze if you don't mind the extra unit dimension at the end.
I have a tensor [a, b, c, d, e, f, g, h, i] with dimension 9 X 1536. I need to create a new tensor which is like [(a,b), (a,c), (a,d), (a,e),(a,f),(a,g), (a,h), (a,i)] with dimension [8 x 2 x 1536]. How can I do it with tensorflow ?
I tried like this
x = tf.zeros((9x1536))
x_new = tf.stack([(x[0],x[1]),
(x[0], x[2]),
(x[0], x[3]),
(x[0], x[4]),
(x[0], x[5]),
(x[0], x[6]),
(x[0], x[7]),
(x[0], x[8])])
This seems to work but I would like to know if there is a better solution or approach which can be used instead of this
You can obtain the desired output with a combination of tf.concat, tf.tile and tf.expand_dims:
import tensorflow as tf
import numpy as np
_in = tf.constant(np.random.randint(0,10,(9,1536)))
tile_shape = [(_in.shape[0]-1).value] + [1]*len(_in.shape[1:].as_list())
_out = tf.concat([
tf.expand_dims(
tf.tile(
[_in[0]],
tile_shape
)
,
1),
tf.expand_dims(_in[1:], 1)
],
1
)
tf.tile repeats the first element of _in creating a tensor of length len(_in)-1 (I compute separately the shape of the tile because we want to tile only on the first dimension).
tf.expand_dims adds a dimension we can then concat on
Finally, tf.concat stitches together the two tensors giving the desired result.
EDIT: Rewrote to fit the OP's actual use-case with multidimensional tensors.
I have two 2-D tensors of shape say m X d and n X d. What is the optimized(i.e. without for loops) or the tensorflow way of evaluating the pairwise euclidean distance between these two tensors so that I get an output tensor of shape m X n. I need it for creating the squared term of a Gaussian kernel for ultimately having a covariance matrix of size m x n.
The equivalent unoptimized numpy code would look like this
difference_squared = np.zeros((x.shape[0], x_.shape[0]))
for row_iterator in range(difference_squared.shape[0]):
for column_iterator in range(difference_squared.shape[1]):
difference_squared[row_iterator, column_iterator] = np.sum(np.power(x[row_iterator]-x_[column_iterator], 2))
I found the answer by taking help from here. Assuming the two tensors are x1 and x2, and their dimensions are m X d and n X d, their pair-wise Euclidean distance is given by
tile_1 = tf.tile(tf.expand_dims(x1, 0), [n, 1, 1])
tile_2 = tf.tile(tf.expand_dims(x2, 1), [1, m, 1])
pairwise_euclidean_distance = tf.reduce_sum(tf.square(tf.subtract(tile_1, tile_2)), 2))
My apologies if this has been answered many times, but I just can't find a solution.
Assume the following code:
import numpy as np
A,_,_ = np.meshgrid(np.arange(5),np.arange(7),np.arange(10))
B = (rand(7,10)*5).astype(int)
How can I slice A using B so that B represent the indexes in the first and last dimensions of A (I.e A[magic] = B)?
I have tried
A[:,B,:] which doesn't work due to peculiarities of advanced indexing.
A[:,B,np.arange(10)] generates 7 copies of the matrix I'm after
A[np.arange(7),B,np.arange(10)] gives the error:
ValueError: shape mismatch: objects cannot be broadcast to a single shape
Any other suggestions?
These both work:
A[0, B, 0]
A[B, B, B]
Really, only the B in axis 1 matters, the others can be any range that will broadcast to B.shape and are limited by A.shape[0] (for axis 1) and A.shape[2] (for axis 2), for a ridiculous example:
A[range(7) + range(3), B, range(9,-1, -1)]
But you don't want to use : because then you'll get, as you said, 7 or 10 (or both!) "copies" of the array you want.
A, _, _ = np.meshgrid(np.arange(5),np.arange(7),np.arange(10))
B = (rand(7,10)*A.shape[1]).astype(int)
np.allclose(B, A[0, B, 0])
#True
np.allclose(B, A[B, B, B])
#True
np.allclose(B, A[range(7) + range(3), B, range(9,-1, -1)])
#True
I have two 2-D arrays with the same first axis dimensions. In python, I would like to convolve the two matrices along the second axis only. I would like to get C below without computing the convolution along the first axis as well.
import numpy as np
import scipy.signal as sg
M, N, P = 4, 10, 20
A = np.random.randn(M, N)
B = np.random.randn(M, P)
C = sg.convolve(A, B, 'full')[(2*M-1)/2]
Is there a fast way?
You can use np.apply_along_axis to apply np.convolve along the desired axis. Here is an example of applying a boxcar filter to a 2d array:
import numpy as np
a = np.arange(10)
a = np.vstack((a,a)).T
filt = np.ones(3)
np.apply_along_axis(lambda m: np.convolve(m, filt, mode='full'), axis=0, arr=a)
This is an easy way to generalize many functions that don't have an axis argument.
With ndimage.convolve1d, you can specify the axis...
np.apply_along_axis won't really help you, because you're trying to iterate over two arrays. Effectively, you'd have to use a loop, as described here.
Now, loops are fine if your arrays are small, but if N and P are large, then you probably want to use FFT to convolve instead.
However, you need to appropriately zero pad your arrays first, so that your "full" convolution has the expected shape:
M, N, P = 4, 10, 20
A = np.random.randn(M, N)
B = np.random.randn(M, P)
A_ = np.zeros((M, N+P-1), dtype=A.dtype)
A_[:, :N] = A
B_ = np.zeros((M, N+P-1), dtype=B.dtype)
B_[:, :P] = B
A_fft = np.fft.fft(A_, axis=1)
B_fft = np.fft.fft(B_, axis=1)
C_fft = A_fft * B_fft
C = np.real(np.fft.ifft(C_fft))
# Test
C_test = np.zeros((M, N+P-1))
for i in range(M):
C_test[i, :] = np.convolve(A[i, :], B[i, :], 'full')
assert np.allclose(C, C_test)
for 2D arrays, the function scipy.signal.convolve2d is faster and scipy.signal.fftconvolve can be even faster (depending on the dimensions of the arrays):
Here the same code with N = 100000
import time
import numpy as np
import scipy.signal as sg
M, N, P = 10, 100000, 20
A = np.random.randn(M, N)
B = np.random.randn(M, P)
T1 = time.time()
C = sg.convolve(A, B, 'full')
print(time.time()-T1)
T1 = time.time()
C_2d = sg.convolve2d(A, B, 'full')
print(time.time()-T1)
T1 = time.time()
C_fft = sg.fftconvolve(A, B, 'full')
print(time.time()-T1)
>>> 12.3
>>> 2.1
>>> 0.6
Answers are all the same with slight differences due to different computation methods used (e.g., fft vs direct multiplication, but i don't know what exaclty convolve2d uses):
print(np.max(np.abs(C - C_2d)))
>>>7.81597009336e-14
print(np.max(np.abs(C - C_fft)))
>>>1.84741111298e-13
Late answer, but worth posting for reference. Quoting from comments of the OP:
Each row in A is being filtered by the corresponding row in B. I could
implement it like that, just thought there might be a faster way.
A is on the order of 10s of gigabytes in size and I use overlap-add.
Naive / Straightforward Approach
import numpy as np
import scipy.signal as sg
M, N, P = 4, 10, 20
A = np.random.randn(M, N) # (4, 10)
B = np.random.randn(M, P) # (4, 20)
C = np.vstack([sg.convolve(a, b, 'full') for a, b in zip(A, B)])
>>> C.shape
(4, 29)
Each row in A is convolved with each respective row in B, essentially convolving M 1D arrays/vectors.
No Loop + CUDA Supported Version
It is possible to replicate this operation by using PyTorch's F.conv1d. We have to imagine A as a 4-channel, 1D signal of length 10. We wish to convolve each channel in A with a specific kernel of length 20. This is a special case called a depthwise convolution, often used in deep learning.
Note that torch's conv is implemented as cross-correlation, so we need to flip B in advance to do actual convolution.
import torch
import torch.nn.functional as F
#torch.no_grad()
def torch_conv(A, B):
M, N, P = A.shape[0], A.shape[1], B.shape[1]
C = F.conv1d(A, B[:, None, :], bias=None, stride=1, groups=M, padding=N+(P-1)//2)
return C.numpy()
# Convert A and B to torch tensors + flip B
X = torch.from_numpy(A) # (4, 10)
W = torch.from_numpy(np.fliplr(B).copy()) # (4, 20)
# Do grouped conv and get np array
Y = torch_conv(X, W)
>>> Y.shape
(4, 29)
>>> np.allclose(C, Y)
True
Advantages of using a depthwise convolution with torch:
No loops!
The above solution can also run on CUDA/GPU, which can really speed things up if A and B are very large matrices. (From OP's comment, this seems to be the case: A is 10GB in size.)
Disadvantages:
Overhead of converting from array to tensor (should be negligible)
Need to flip B once before the operation