numpy-broadcasting

Given two matrices, I want to compute the pairwise differences between all rows. Each matrix has 1000 rows and 100 columns so they are fairly large. I tried using a for loop and pure broadcasting but the for loop seem to be working faster. Am I doing something wrong? Here is the code: from numpy import * A = random.randn(1000,100) B = random.randn(1000,100) start = time.time() for a in A: sum((a - B)**2,1) print time.time() - start # pure broadcasting start = time.time() ((A[:,newaxis,:] - B)**2).sum(-1) print time.time() - start The broadcasting method takes about 1 second longer and it's

I have the following code. It is taking forever in Python. There must be a way to translate this calculation into a broadcast... def euclidean_square(a,b): squares = np.zeros((a.shape[0],b.shape[0])) for i in range(squares.shape[0]): for j in range(squares.shape[1]): diff = a[i,:] - b[j,:] sqr = diff**2.0 squares[i,j] = np.sum(sqr) return squares You can use np.einsum after calculating the differences in a broadcasted way , like so - ab = a[:,None,:] - b out = np.einsum('ijk,ijk->ij',ab,ab) Or use scipy's cdist with its optional metric argument set as 'sqeuclidean' to give us the squared

I have a vector [x,y,z,q] and I want to create a matrix: [[x,y,z,q], [x,y,z,q], [x,y,z,q], ... [x,y,z,q]] with m rows. I think this could be done in some smart way, using broadcasting, but I can only think of doing it with a for loop. Certainly possible with broadcasting after adding with m zeros along the columns, like so - np.zeros((m,1),dtype=vector.dtype) + vector Now, NumPy already has an in-built function np.tile for exactly that same task - np.tile(vector,(m,1)) Sample run - In [496]: vector Out[496]: array([4, 5, 8, 2]) In [497]: m = 5 In [498]: np.zeros((m,1),dtype=vector.dtype) +

I need to create a 2D array where each row may start and end with a different number. Assume that first and last element of each row is given and all other elements are just interpolated according to length of the rows In a simple case let's say I want to create a 3X3 array with same start at 0 but different end given by W below: array([[ 0., 1., 2.], [ 0., 2., 4.], [ 0., 3., 6.]]) Is there a better way to do this than the following: D=np.ones((3,3))*np.arange(0,3) D=D/D[:,-1] W=np.array([2,4,6]) # last element of each row assumed given Res= (D.T*W).T Here's an approach using broadcasting -