Lets say we have a 3 Channel (RGB) image given by some matrix A A = [[[198 218 227] [196 216 225] [196 214 224] … … [185 201 217] [176 192 208] [162 178 194]] and a blur kernal as K = [[0.1111, 0.1111, 0.1111], [0.1111, 0.1111, 0.1111], [0.1111, 0.1111, 0.1111]] #which is actually … Read more
You can try this code: public class MyMatrix { Double[][] A = { { 4.00, 3.00 }, { 2.00, 1.00 } }; Double[][] B = { { -0.500, 1.500 }, { 1.000, -2.0000 } }; public static Double[][] multiplicar(Double[][] A, Double[][] B) { int aRows = A.length; int aColumns = A[0].length; int bRows = B.length; … Read more
The simplest code to do this relies on the broadcasting behavior of tf.multiply()*, which is based on numpy’s broadcasting behavior: x = tf.constant(5.0, shape=[5, 6]) w = tf.constant([0.0, 1.0, 2.0, 3.0, 4.0, 5.0]) xw = tf.multiply(x, w) max_in_rows = tf.reduce_max(xw, 1) sess = tf.Session() print sess.run(xw) # ==> [[0.0, 5.0, 10.0, 15.0, 20.0, 25.0], # … Read more
If you really don’t want to use numpy you can do something like this: def matmult(a,b): zip_b = zip(*b) # uncomment next line if python 3 : # zip_b = list(zip_b) return [[sum(ele_a*ele_b for ele_a, ele_b in zip(row_a, col_b)) for col_b in zip_b] for row_a in a] x = [[1,2,3],[4,5,6],[7,8,9],[10,11,12]] y = [[1,2],[1,2],[3,4]] import numpy … Read more
Using linear algebra, there exist algorithms that achieve better complexity than the naive O(n3). Solvay Strassen algorithm achieves a complexity of O(n2.807) by reducing the number of multiplications required for each 2×2 sub-matrix from 8 to 7. The fastest known matrix multiplication algorithm is Coppersmith-Winograd algorithm with a complexity of O(n2.3737). Unless the matrix is … Read more
logsumexp works by evaluating the right-hand side of the equation log(∑ exp[a]) = max(a) + log(∑ exp[a – max(a)]) I.e., it pulls out the max before starting to sum, to prevent overflow in exp. The same can be applied before doing vector dot products: log(exp[a] ⋅ exp[b]) = log(∑ exp[a] × exp[b]) = log(∑ exp[a … Read more
Normal matrix multiplication works as long as the vectors have the right shape. Remember that * in Numpy is elementwise multiplication, and matrix multiplication is available with numpy.dot() (or with the @ operator, in Python 3.5) >>> numpy.dot(numpy.array([[1], [2]]), numpy.array([[3, 4]])) array([[3, 4], [6, 8]]) This is called an “outer product.” You can get it … Read more