I have two tensors, t1 is a 200-dim vector with dynamic axes [batch_len, seq1_len], and t2 is also a 200-dim vector with dynamic axes [batch_len, seq2_len]. I want to get a matrix M of dimension (seq1_len, seq2_len), in which M_ij is the inner product of the two 200-dim vectors t1_i and t2_j. Is there a way to achieve this?
Currently (February 2017) there's no way to have two sequence axes. One possibility is to treat the sequence with the shortest range as a tensor with with only static axes.
Related
I have a matrix with dimension (22,2) and I want to decompose it using SVD. SVD in numpy doesn't return the correct dimensions though.I'd expect dimensions like (22,22), (22),(22,2)?
The returned dimensions are correct. The uu and vvh matrices are always square matrices, while depending on the software s can be an array with just the singular values (as in numpy) or a diagonal matrix with the dimension of the original matrix (as in MATLAB, for instance).
The dimensions of the uu matrix is the number of rows of the original matrix, while the dimension of the vvh matrix is the number of columns of the original matrix. This can never change or you would be computing something else instead of the SVD.
To reconstruct the original matrix from the decomposition in numpy we need to make s into a matrix with the proper dimension. For square matrices it's easy, just np.diag(s) is enough. Since your original matrix is not square and it has more rows than columns, then we can use something like
S = np.vstack([np.diag(s), np.zeros((20, 2))])
Then we get a S matrix which is a diagonal matrix with the singular values concatenated with a zero matrix. In the end, uu is 22x22, S is 22x2 and vvh is 2x2. Multiplying uu # S # vvh will give the original matrix back.
I have a tensorflow/keras CNN. It has layers and some are Conv2D. In a given layer I want to efficiently find the two filters in the Conv2D that are most similar.
The layer.weights is a list of shape (height, width, depth) filter_count long.
I want to compare by the difference or maybe the sqrt(diff^2) between each element in (height,width,depth) then sum so the difference is a single float value.
If T1 is thelayer.weights[idx1] and T2 is thelayer.weights[idx2]
then the comparison is tf.sqrt(tf.reduce_sum(tf.squared_difference(T1, T2)))
I want to compare every filter to every other filter and take the 3 lowest differences. (The first one will always be zero where it T1 and T2 are the same tensor, self)
Obviously I can do nested loops but that is not functional and nifty.
Is there some built in tensorflow or keras function to do this fast and possibly in the GPU?
Its not quite clear from your description, but I assume the shape of weights is [filter_count, height,width,depth]. If filter_count is along a different axis the arguments to "reduce_sum" will have to be modified accordingly.
You can use broadcasting to parallelize this process.
differences = tf.sqrt(
tf.reduce_sum(
tf.squared_difference(
tf.expand_dims(thelayer.weights,0),
tf.expand_dims(thelayer.weights,1),
),
(-1,-2,-3)
)
)
This will result in a tensor of shape [filter_count, filter_count] where element differences[i, j] measure differences between filter weights i and j.
You can then filter to find the desired elements.
Let T be a tensor of shape [n,f], which represents a batch. Now I want to slice T into m tensors along axis=0. The value of m depends on the current batch. I have another tensor I of shape [m,2] which stores pairs of indices which indicate where the slices should occur.
I am not really sure how to "iterate" over the indices to apply tf.slice. Any ideas?
Can this somehow be achieved using tf.scan?
I suppose you are looking for the split function.
Given two matrixes A and B with dimension of (x,y,z) and (y,x,z) respectively, how to dot product on the first two dimension of the two matrices? The result should have dimension of (x,x,z).
Thanks!
Use np.einsum with literally the same string expression -
np.einsum('xyz,yiz->xiz',a,b) # a,b are input arrays
Note that we have used yiz as the string notation for the second array and not yxz, as that i is supposed to be a new dimension in the output array and is not to be aligned with the first axis of the first array for which we have already assigned x. The dimensions that are to be aligned are given the same string notation.
I'm making the transition from MATLAB to Numpy and feeling some growing pains.
I have a 3D array, lets say it's 3x3x3 and I want the scalar sum of each plane.
In matlab, I would use:
sum_vec = sum(3dArray,3);
TIA
wbg
EDIT: I was wrong about my matlab code. Matlab only vectorizes in one dim, so a loop wold be required. So numpy turns out to be more elegant...cool.
MATLAB
for i = 1:3
sum_vec(i) = sum(sum(3dArray(:,:,i));
end
You can do
sum_vec = np.array([plane.sum() for plane in cube])
or simply
sum_vec = cube.sum(-1).sum(-1)
where cube is your 3d array. You can specify 0 or 1 instead of -1 (or 2) depending on the orientation of the planes. The latter version is also better because it doesn't use a Python loop, which usually helps to improve performance when using numpy.
You should use the axis keyword in np.sum. Like in many other numpy functions, axis lets you perform the operation along a specific axis. For example, if you want to sum along the last dimension of the array, you would do:
import numpy as np
sum_vec = np.sum(3dArray, axis=-1)
And you'll get a resulting 2D array which corresponds to the sum along the last dimension to all the array slices 3dArray[i, k, :].
UPDATE
I didn't understand exactly what you wanted. You want to sum over two dimensions (a plane). In this case you can do two sums. For example, summing over the first two dimensions:
sum_vec = np.sum(np.sum(3dArray, axis=0), axis=0)
Instead of applying the same sum function twice, you may perform the sum on the reshaped array:
a = np.random.rand(10, 10, 10) # 3D array
b = a.view()
b.shape = (a.shape[0], -1)
c = np.sum(b, axis=1)
The above should be faster because you only sum once.
sumvec= np.sum(3DArray, axis=2)
or this works as well
sumvec=3DArray.sum(2)
Remember Python starts with 0 so axis=2 represent the 3rd dimension.
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.sum.html
If you're trying to sum over a plane (and avoid loops, which is always a good idea) you can use np.sum and pass two axes as a tuple for your argument.
For example, if you have an (nx3x3) array then using
np.sum(a, (1,2))
Will give an (nx1x1), summing over a plane, not a single axis.