X = np.asfortranarray( X ) %memit np .dot( X , Y) # maximum of 1: 905.093750 MB per loop If copying is a problem (e.g. when X is very large), what can you do about it? The best option would probably be to upgrade to a newer version of numpy – as @perimosocordiae points out, this performance issue was addressed in this pull request.
def EDM(self): ”’ Computes the EDM corresponding to the marker set ”’ if self.X is None: raise ValueError(‘No marker set’) G = np.dot(self.X.T, self.X) return np.outer(np.ones(self.m), np.diag(G)) – 2*G + np.outer(np.diag(G), np.ones(self.m)), Xt = np.transpose(X) XtX = np.dot(Xt,X) Xty = np.dot(Xt,y) beta = np.linalg.solve(XtX,Xty) The last line uses np.linalg.solve to compute ?, since the equation. ? = (X T X)-1 X T y. is mathematically equivalent to the system of equations (X T X) ? = X T y, X = X self. y = y # fit XtX = np. dot (X. T, X ) / sigma_squared I = np. eye ( X . shape [1]) / tau inverse = np. linalg. inv (XtX + I) Xty = np. dot ( X. T , y) / sigma_squared self. beta_hats = np. dot (inverse, Xty) # fitted values self. y_hat = np. dot ( X , self. beta_hats) Let’s fit a Bayesian regression model on.
12/15/2020 · import numpy as np from multiprocessing import Pool, Process def test( x ): arr = np.dot(x.T,x ) # On large matrices, this calc will use BLAS. if __name__ == ‘__main__’: x = np.random.random(size=((2000,500))) # Random matrix test( x ) evaluations = [ x for _ in range(5)] p = Pool() p.map_async(test,evaluations) # This is where Python will quit …
Sigmoid function def sigmoid(z): return 1 / (1 + np.exp(-z)) z = np .dot( X , weight) h = sigmoid(z) LR is also a transformation of a linear regression using the sigmoid function. If we compare with …
def ols( X , y): ”’returns parameters based on Ordinary Least Squares.”’ xtx = np.dot(X.T, X ) ## x -transpose times x inv_xtx = np.linalg.inv(xtx) ## inverse of x -transpose times x xty = np .dot( X.T , y) ## x -transpose times y return np .dot(inv_xtx, xty) Finally, let’s push the observational data through the function to find the thetas. …
In numpy, what’s the most efficient way to compute x.T * x , where x is a large (200,000 x 1000) dense float32 matrix and .T is the transpose operator?. For the avoidance of doubt, the result is 1000 x 1000. edit: In my original question I stated that np.dot(x.T, x ) was taking hours. It turned out that I had some NaNs sneak into the matrix, and for some reason that was completely killing the …
11/15/2018 · Sourced from Coursera dZ2 = A2 – Y dW2 = (1 / m) * np .dot(dZ2, A1.T) db2 = (1 / m) * np.sum(dZ2, axis=1, keepdims=True) dZ1 = np.multiply( np .dot(W2.T, dZ2), 1 – np.power(A1, 2)) dW1 = (1 / m) * np .dot(dZ1, X.T ) db1 = (1 / m) * np.sum(dZ1, axis=1, keepdims=True). 6. Update the parameters: Once we have computed our gradients, we multiply them with a factor called learning …
