numpy-einsum

cross products with einsums

由 喜欢而已 提交于 2019-12-11 06:54:17
问题 I'm trying to compute the cross-products of many 3x1 vector pairs as fast as possible. This n = 10000 a = np.random.rand(n, 3) b = np.random.rand(n, 3) numpy.cross(a, b) gives the correct answer, but motivated by this answer to a similar question, I thought that einsum would get me somewhere. I found that both eijk = np.zeros((3, 3, 3)) eijk[0, 1, 2] = eijk[1, 2, 0] = eijk[2, 0, 1] = 1 eijk[0, 2, 1] = eijk[2, 1, 0] = eijk[1, 0, 2] = -1 np.einsum('ijk,aj,ak->ai', eijk, a, b) np.einsum('iak,ak-

Generating np.einsum evaluation graph

由 我的梦境 提交于 2019-12-11 06:22:48
问题 I was planning to teach np.einsum to colleagues, by hoping to show how it would be reduced to multiplications and summations. So, instead of numerical data, I thought to use alphabet chars. in the arrays. Say, we have A (2X2) as [['a', 'b'], ['c', 'd']] and B (2X1) as [['e'], ['f']] We could use einsum to create a matrix C, say like: np.einsum('ab , bc -> ac', A, B) . What I'd like to see is: it return the computation graph: something like: a*c + ..., etc. Ofcourse, np.einsum expects

Einsum optimize fails for basic operation

由 廉价感情. 提交于 2019-12-11 00:57:36
问题 With the recent update to Numpy (1.14), I found that it breaks my entire codebase. This is based on changing the default numpy einsum optimize argument from False to True. As a result, the following basic operation now fails: a = np.random.random([50, 2, 2]) b = np.random.random([50, 2]) np.einsum('bdc, ac -> ab', a, b, optimize=True) with the following error trace: ValueError Traceback (most recent call last) <ipython-input-71-b0f9ce3c71a3> in <module>() ----> 1 np.einsum('bdc, ac -> ab', a,

Theano version of a numpy einsum for two 3dim matrices

由 笑着哭i 提交于 2019-12-11 00:00:52
问题 I have two 3dim numpy matrices and I want to do a dot product according to one axis without using a loop in theano. a numpy solution with sample data would be like: a=[ [[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0], [ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0], [ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1], [ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0]], [[ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0], [ 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0], [ 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,

Can I use more than 26 letters in `numpy.einsum`?

由 谁说我不能喝 提交于 2019-12-08 17:43:37
问题 I am using np.einsum to multiply probability tables like: np.einsum('ijk,jklm->ijklm', A, B) The issue is that I am dealing with more than 26 random variables (axes) overall, so if I assign each random variable a letter I run out of letters. Is there another way I can specify the above operation to avoid this issue, without resorting to a mess of np.sum and np.dot operations? 回答1: The short answer is, you can use any of the 52 letters (upper and lower). That's all the letters in the English

Batch convolution 2d in numpy without scipy?

由 风格不统一 提交于 2019-12-07 18:41:18
问题 I have a batch of b m x n images stored in an array x , and a convolutional filter f of size p x q that I'd like to apply to each image (then use sum pooling and store in an array y ) in the batch, i.e. all(np.allclose(y[i][j][k], (x[i, j:j+p, k:k+q] * f).sum()) for i in range(b) for j in range(m-p+1) for k in range(n-q+1)) is true. Adapting this answer, I could write the following: b, m, n, p, q = 6, 5, 4, 3, 2 x = np.arange(b*m*n).reshape((b, m, n)) f = np.arange(p*q).reshape((p, q)) y = []

Batch convolution 2d in numpy without scipy?

由 久未见 提交于 2019-12-06 05:48:16
I have a batch of b m x n images stored in an array x , and a convolutional filter f of size p x q that I'd like to apply to each image (then use sum pooling and store in an array y ) in the batch, i.e. all(np.allclose(y[i][j][k], (x[i, j:j+p, k:k+q] * f).sum()) for i in range(b) for j in range(m-p+1) for k in range(n-q+1)) is true. Adapting this answer , I could write the following: b, m, n, p, q = 6, 5, 4, 3, 2 x = np.arange(b*m*n).reshape((b, m, n)) f = np.arange(p*q).reshape((p, q)) y = [] for i in range(b): shape = f.shape + tuple(np.subtract(x[i].shape, f.shape) + 1) strides = x[i]

How do I calculate all pairs of vector differences in numpy?

由 China☆狼群 提交于 2019-12-04 03:54:35
问题 I know I can do np.subtract.outer(x, x) . If x has shape (n,) , then I end up with an array with shape (n, n) . However, I have an x with shape (n, 3) . I want to output something with shape (n, n, 3) . How do I do this? Maybe np.einsum ? 回答1: You can use broadcasting after extending the dimensions with None/np.newaxis to form a 3D array version of x and subtracting the original 2D array version from it, like so - x[:, np.newaxis, :] - x Sample run - In [6]: x Out[6]: array([[6, 5, 3], [4, 3,

numpy einsum: nested dot products

由 北城余情 提交于 2019-12-02 11:31:05
问题 I have two n -by- k -by- 3 arrays a and b , e.g., import numpy as np a = np.array([ [ [1, 2, 3], [3, 4, 5] ], [ [4, 2, 4], [1, 4, 5] ] ]) b = np.array([ [ [3, 1, 5], [0, 2, 3] ], [ [2, 4, 5], [1, 2, 4] ] ]) and it like to compute the dot-product of all pairs of "triplets", i.e., np.sum(a*b, axis=2) A better way to do that is perhaps einsum, but I can't seem to get the indices straight. Any hints here? 回答1: You are loosing the third axis on those two 3D input arrays with that sum-reduction,

Using numpy einsum to compute inner product of column-vectors of a matrix

由 跟風遠走 提交于 2019-12-02 03:19:41
问题 Suppose I have a numpy matrix like this: [[ 1 2 3] [ 10 100 1000]] I would like to compute the inner product of each column with itself, so the result would be: [1*1 + 10*10 2*2 + 100*100 3*3 + 1000*1000] == [101, 10004, 1000009] I would like to know if this is possible using the einsum function (and to better understand it). So far, the closest result I could have is: import numpy as np arr = np.array([[1, 2, 3], [10, 100, 1000]]) res = np.einsum('ij,ik->jk', arr, arr) # [[ 101 1002 10003] #
订阅 numpy-einsum