dot-product

Does this Python code actually find the dot product of two vectors? import operator vector1 = (2,3,5) vector2 = (3,4,6) dotProduct = reduce( operator.add, map( operator.mul, vector1, vector2)) You can also use the numpy implementation of dot product which has large array optimizations in native code to make computations slightly faster. Even better unless you are specifically trying to write a dot product routine or avoid dependencies, using a tried tested widely used library is much better than rolling your own. Yes it does. Here is another way >>> sum(map( operator.mul, vector1, vector2)) 48

I am having 2900 images in this path G:\newdatabase\ It is taking too much time to read images.For dot product also it is taking too much time. Questions: 1.Is there any alternative for imread command which increases performance? 2.Is there any alternative for dot command which increases performance? Source code i tried: srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists for b = 1 : length(srcFiles) filename = strcat('G:\newdatabase\',srcFiles(b).name); Imgdata = imread(filename); Source code i tried: for i = 1:aa pare = dot(NormImage,u(:,i)); p = [p; pare]; end 来源：

I have 2 matrices = X in R^(n*m) and W in R^(k*m) where k<<n . Let x_i be the i-th row of X and w_j be the j-th row of W. I need to find, for each x_i what is the j that maximizes <w_j,x_i> I can't see a way around iterating over all the rows in X, but it there a way to find the maximum dot product without iterating every time over all of W? A naive implementation would be: n = 100; m = 50; k = 10; X = rand(n,m); W = rand(k,m); Y = zeros(n, 1); for i = 1 : n max_ind = 1; max_val = dot(W(1,:), X(i,:)); for j = 2 : k cur_val = dot(W(j,:),X(i,:)); if cur_val > max_val max_val = cur_val; max_ind =

sorry for so many questions. I am running Mac OSX 10.6 on Intel core 2 Duo. I am running some benchmarks for my research and I have run into another thing that baffles me. If I run python -mtimeit -s 'import numpy as np; a = np.random.randn(1e3,1e3)' 'np.dot(a,a)' I get the following output: 10 loops, best of 3: 142 msec per loop However, if I run python -mtimeit -s 'import numpy as np; a = np.random.randint(10,size=1e6).reshape(1e3,1e3)' 'np.dot(a,a)' I get the following output: 10 loops, best of 3: 7.57 sec per loop Then I ran python -mtimeit -s 'import numpy as np; a = np.random.randn(1e3

I am trying to vectorize a loop, computing dot product of a large float vectors. I am computing it in parallel, utilizing the fact that CPU has large amount of XMM registers, like this: __m128* A, B; __m128 dot0, dot1, dot2, dot3 = _mm_set_ps1(0); for(size_t i=0; i<1048576;i+=4) { dot0 = _mm_add_ps( dot0, _mm_mul_ps( A[i+0], B[i+0]); dot1 = _mm_add_ps( dot1, _mm_mul_ps( A[i+1], B[i+1]); dot2 = _mm_add_ps( dot2, _mm_mul_ps( A[i+2], B[i+2]); dot3 = _mm_add_ps( dot3, _mm_mul_ps( A[i+3], B[i+3]); } ... // add dots, then shuffle/hadd result. I heard that using prefetch instructions could help

I calculated tf/idf values of two documents. The following are the tf/idf values: 1.txt 0.0 0.5 2.txt 0.0 0.5 The documents are like: 1.txt = > dog cat 2.txt = > cat elephant How can I use these values to calculate cosine similarity? I know that I should calculate the dot product, then find distance and divide dot product by it. How can I calculate this using my values? One more question: Is it important that both documents should have same number of words? a * b sim(a,b) =-------- |a|*|b| a*b is dot product some details: def dot(a,b): n = length(a) sum = 0 for i in xrange(n): sum += a[i] * b

I have two lists, one is named as A, another is named as B. Each element in A is a triple, and each element in B is just an number. I would like to calculate the result defined as : result = A[0][0] * B[0] + A[1][0] * B[1] + ... + A[n-1][0] * B[n-1] I know the logic is easy but how to write in pythonic way? Thanks! import numpy result = numpy.dot( numpy.array(A)[:,0], B) http://docs.scipy.org/doc/numpy/reference/ If you want to do it without numpy, try sum( [a[i][0]*b[i] for i in range(len(b))] ) Henri Andre Python 3.5 has an explicit operator @ for the dot product, so you can write a = A @ B