Performance

Profiling Results for GaussianProcess.predict() (1st)

Call	Percentage (elapse time)	function
1	4.8	a = dist.cdist ()
1	8.6	a = expX[self.D]np.exp(-0.5a)
1	0.2	mu = np.dot( a.T, self.invQt)
1	5.2	var=sum(dot())
10	19.6	aa = self.inputs[:,d].flatten()[None,:] - testing[:,d].flatten()[:,None]
10	58.5	c = a*aa.T
10	2.0	deriv[:, d] = expX[d]*np.dot(c.T, self.invQt)

computational costs

function	matrix size	computation costs
cdist	(N,10),(250,10)	N* 250* 10
em	(N,250),(1)	N*250
matmul(dot)
aa	(250),(N)	250*N
c	(250,N)(250,N)	250*N

memory

problem size = N, parameter numbers P = 10 , M=250

matrix size of 250 * N: aa, a

matrix size of 10 * N: testing, deriv

N	single	double
1e6	1.9G	3.8G
1e7	19.3G	38.7G

Line profiler result for python code 1e5, 250, 10

Timer unit: 1e-06 s

Total time: 7.39544 s
File: test.py
Function: predict at line 14

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
    14                                               @profile
    15                                               def predict ( self, testing, do_unc=True ):
    16         1            4      4.0      0.0          ( nn, D ) = testing.shape
    17         1            1      1.0      0.0          assert D == self.D
    18
    19         1            7      7.0      0.0          expX = np.exp ( self.theta )
    20
    21         1       367538 367538.0      5.0          a = dist.cdist ( np.sqrt(expX[:(self.D)])*self.inputs, np.sqrt(expX[:(self.D)])*testing, 'sqeuclidean')
    22         1       621799 621799.0      8.4          a = expX[self.D]*np.exp(-0.5*a)
    23         1            6      6.0      0.0          b = expX[self.D]
    24
    25         1        18835  18835.0      0.3          mu = np.dot( a.T, self.invQt)
    26         1            2      2.0      0.0          if do_unc:
    27         1       454542 454542.0      6.1              var = b - np.sum (  a * np.dot(self.invQ,a), axis=0)
    28                                                   # Derivative and partial derivatives of the function
    29         1          518    518.0      0.0          deriv = np.zeros ( ( nn, self.D ) )
    30
    31        11           21      1.9      0.0          for d in xrange ( self.D ):
    32        10      1418347 141834.7     19.2              aa = self.inputs[:,d].flatten()[None,:] - testing[:,d].flatten()[:,None]
    33
    34        10      4342587 434258.7     58.7              c = a*aa.T
    35        10       163693  16369.3      2.2              deriv[:, d] = expX[d]*np.dot(c.T, self.invQt)
    36        10         7535    753.5      0.1              print deriv[0:10,d]
    37
    38         1            1      1.0      0.0          if do_unc:
    39         1            1      1.0      0.0              return mu, var, deriv
    40                                                   else:
    41                                           	    return mu, deriv