Re: [Numpy-discussion] Improvement of performance

If your x data are equispaced I would do something like this def derive( func, x): """ Approximate the first derivative of function func at points x. """ # compute the values of y = func(x) y = func(x) # compute the step dx = x[1] - x[0] # kernel array for second order accuracy centered first derivative kc = np.array([-1.0, +0.0, +1.0]) / 2 / dx # kernel array for second order accuracy left and right first derivative kl = np.array([-3.0, +4.0, -1.0]) / 2 / dx kr = np.array([+1.0, -4.0, +3.0]) / 2 / dx # correlate it with the original array, # note that only the valid computation are performed derivs_c = np.correlate( y, kc, mode='valid' ) derivs_r = np.correlate( y[-3:], kr, mode='valid' ) derivs_l = np.correlate( y[:+3], kl, mode='valid' ) return np.r_[derivs_l, derivs_c, derivs_r] This is actually quite fast: on my machine (1.7GHz) i have this.

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:x = np.linspace(0,2*np.pi, 1e6) :func = lambda x: np.sin(x) :timeit derive(func, x) 10 loops, best of 3: 177 ms per loop

I'm curious if someone comes up with something faster. Regards, Davide On 4 May 2010 22:17, wrote:

On Tue, May 4, 2010 at 4:06 PM, gerardob wrote:

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Hello, I have written a very simple code that computes the gradient by

finite

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differences of any general function. Keeping the same idea, I would like modify the code using numpy to make it faster. Any ideas? Thanks.

def grad_finite_dif(self,x,user_data = None): assert len(x) == self.number_variables points=[] for j in range(self.number_variables): points.append(x.copy())

points[len(points)-1][j]=points[len(points)-1][j]+0.0000001

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delta_f = [] counter=0 for j in range(self.number_variables):

delta_f.append((self.eval(points[counter])-self.eval(x))/0.0000001)

it looks like your are evaluating the same point several times self.eval(x)

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counter = counter + 1 return array(delta_f)

That's what I used as a pattern for a gradient function

#from scipy.optimize def approx_fprime(xk,f,epsilon,*args): f0 = f(*((xk,)+args)) grad = np.zeros((len(xk),), float) ei = np.zeros((len(xk),), float) for k in range(len(xk)): ei[k] = epsilon grad[k] = (f(*((xk+ei,)+args)) - f0)/epsilon ei[k] = 0.0 return grad

Josef

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