Sorry, I guess I wasn't clear, I meant:

for n in xrange(1,N): y[n] = A*x[n] + B*y[n-1]

So y[n-1] is the result from the previous loop iteration.

Gerry

-----Original Message----- From: numpy-discussion-bounces@scipy.org [mailto:numpy-discussion-bounces@scipy.org] On Behalf Of Keith Goodman Sent: Wednesday, May 06, 2009 9:53 AM To: Discussion of Numerical Python Subject: Re: [Numpy-discussion] Recurrence relationships

On Wed, May 6, 2009 at 6:44 AM, Talbot, Gerry Gerry.Talbot@amd.com wrote:

Does anyone know how to efficiently implement a recurrence relationship in numpy such as:

y[n] = A*x[n] + B*y[n-1]

On an intel chip I'd use a Monte Carlo simulation. On an amd chip I'd use:

...

x = np.array([1,2,3]) y = np.array([4,5,6]) y = x[1:] + y[:-1] y

array([6, 8]) _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion

On Wed, May 6, 2009 at 10:21 AM, josef.pktd@gmail.com wrote:

On Wed, May 6, 2009 at 10:00 AM, Talbot, Gerry Gerry.Talbot@amd.com wrote:

...

Sorry, I guess I wasn't clear, I meant:

for n in xrange(1,N):          y[n] = A*x[n] + B*y[n-1]

So y[n-1] is the result from the previous loop iteration.

I was using scipy.signal for this but I have to look up what I did exactly. I think either signal.correlate or using signal.lti.

No, its signal.lfilter, below is a part of a script I used to simulate and estimate an AR(1) process, which is similar to your example.

I haven't looked at it in a while but it might give you the general idea.

Josef

# Simulate AR(1) #-------------- # ar * y = ma * eta ar = [1, -0.8] ma = [1.0]

# generate AR data eta = 0.1 * np.random.randn(1000) yar1 = signal.lfilter(ar, ma, eta)

etahat = signal.lfilter(ma, ar, y) np.all(etahat == eta)

# find error for given filter on data print 'AR(2)' for rho in [0.01, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.79, 0.8, 0.81, 0.9]: etahatr = signal.lfilter(ma, [1, --rho], yar1) print rho,np.sum(etahatr*etahatr) print 'AR(2)' for rho2 in np.linspace(-0.4,0.4,9): etahatr = signal.lfilter(ma, [1, -0.8, -rho2], yar1) print rho2,np.sum(etahatr*etahatr)

def errfn(rho): etahatr = signal.lfilter(ma, [1, -rho], yar1) #print rho,np.sum(etahatr*etahatr) return etahatr

def errssfn(rho): etahatr = signal.lfilter(ma, [1, -rho], yar1) return np.sum(etahatr*etahatr)

resultls = optimize.leastsq(errfn,[0.5]) print 'LS ARMA(1,0)', resultls

resultfmin = optimize.fmin(errssfn, 0.5) print 'fminLS ARMA(1,0)', resultfmin

On 5/6/2009 10:00 AM Talbot, Gerry apparently wrote:

for n in xrange(1,N): y[n] = A*x[n] + B*y[n-1]

So, x is known before you start? How big is N? Also, is y.shape (N,)? Do you need all of y or only y[N]?

Alan Isaac