Ejemplo n.º 1
0
def sgs(data, model, hs, bw, xs, ys=None, pad=0.0):
    '''
    Input:  (data)   <N,3> NumPy array of data
            (hs)     NumPy array of distances
            (bw)     bandwidth of the semivariogram
            (xs)     number of cells in the x dimension
            (ys)     number of cells in the y dimension
    Output: (M)      <xsteps,ysteps> NumPy array of data
                     representing the simulated distribution
                     of the variable of interest 
    '''
    # check for meshsize in second dimension
    if ys == None:
        ys = xs
    # create path
    xdim = (data[:, 0].min() - pad, data[:, 0].max() + pad, xs)
    ydim = (data[:, 1].min() - pad, data[:, 1].max() + pad, ys)
    path = gridpath(xdim, ydim)
    # create array for the output
    M = np.zeros((xs, ys))
    # for each cell in the grid..
    for step in path:
        # grab the index, the cell address, and the physical location
        idx, cell, loc = step
        # perform the kriging
        kv = k.krige(data, model, hs, bw, loc, 4)
        # add the kriging estimate to the output
        M[cell[0], cell[1]] = kv
        # add the kriging estimate to a spatial location
        newdata = [loc[0], loc[1], kv]
        # add this new point to the data used for kriging
        data = np.vstack((data, newdata))
    return M
Ejemplo n.º 2
0
def sgs( data, model, hs, bw, xs, ys=None, pad=0.0 ):
    '''
    Input:  (data)   <N,3> NumPy array of data
            (hs)     NumPy array of distances
            (bw)     bandwidth of the semivariogram
            (xs)     number of cells in the x dimension
            (ys)     number of cells in the y dimension
    Output: (M)      <xsteps,ysteps> NumPy array of data
                     representing the simulated distribution
                     of the variable of interest 
    '''
    # check for meshsize in second dimension
    if ys == None:
	      ys = xs
    # create path
    xdim = ( data[:,0].min()-pad, data[:,0].max()+pad, xs )
    ydim = ( data[:,1].min()-pad, data[:,1].max()+pad, ys )
    path = gridpath( xdim, ydim )
    # create array for the output
    M = np.zeros((xs,ys))
    # for each cell in the grid..
    for step in path :
        # grab the index, the cell address, and the physical location
        idx, cell, loc = step
        # perform the kriging
        kv = k.krige( data, model, hs, bw, loc, 4 )
        # add the kriging estimate to the output
        M[cell[0],cell[1]] = kv
        # add the kriging estimate to a spatial location
        newdata = [ loc[0], loc[1], kv ]
        # add this new point to the data used for kriging
        data = np.vstack(( data, newdata ))
    return M
Ejemplo n.º 3
0
    def test_simple_kriging_gaussian_model(self):
        '''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
        kv = k.krige(data, k.gaussian, hs, bw, u, N)
        ans = 2.14052910511
        self.assertTrue(abs(ans - kv) < eps)
Ejemplo n.º 4
0
    def test_simple_kriging_exponential_model(self):
        '''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
        kv = k.krige(data, k.exponential, hs, bw, u, N)
        ans = 2.42879523605
        self.assertTrue(abs(ans - kv) < eps)
Ejemplo n.º 5
0
    def test_simple_kriging_spherical_model(self):
        '''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
        kv = k.krige(data, k.spherical, hs, bw, u, N)
        ans = 2.29891949337
        self.assertTrue(abs(ans - kv) < eps)
Ejemplo n.º 6
0
	def test_simple_kriging_gaussian_model( self ):
		'''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
		kv = k.krige( data, k.gaussian, hs, bw, u, N )
		ans = 2.14052910511
		self.assertTrue( abs( ans - kv ) < eps )
Ejemplo n.º 7
0
	def test_simple_kriging_exponential_model( self ):
		'''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
		kv = k.krige( data, k.exponential, hs, bw, u, N )
		ans = 2.42879523605
		self.assertTrue( abs( ans - kv ) < eps )
Ejemplo n.º 8
0
	def test_simple_kriging_spherical_model( self ):
		'''
		Does krige() return a result within
		eps=0.0001 of a benchmarked value?
		'''
		kv = k.krige( data, k.spherical, hs, bw, u, N )
		ans = 2.29891949337
		self.assertTrue( abs( ans - kv ) < eps )
Ejemplo n.º 9
0
covfct = model.covariance(model.spherical, (4000, sill))

# <markdowncell>

# We can then krige the data, using the covariance model, the point we are interested in, (2000,47000), and `N=6` signifying that we only want to use the six nearest points. The output of the simple and ordinary kriging functions below is the krigin estimate, and the standard deviation of the kriging estimate.

# <codecell>

kriging.simple(P, covfct, pt, N=6)

# <codecell>

kriging.ordinary(P, covfct, pt, N=6)

# <codecell>

est, kstd = kriging.krige(
    P,
    covfct, [[2000, 4700], [2100, 4700], [2000, 4800], [2100, 4800]],
    'simple',
    N=6)

# <codecell>

est

# <codecell>

kstd
Ejemplo n.º 10
0
# The covariance modeling function function will return a spherical covariance model that takes a distance as input, and returns an covariance estimate. We've used the global variance of the porosity in `ZoneA.dat` as the sill.

# <codecell>

covfct = model.covariance( model.spherical, ( 4000, sill ) )

# <markdowncell>

# We can then krige the data, using the covariance model, the point we are interested in, (2000,47000), and `N=6` signifying that we only want to use the six nearest points. The output of the simple and ordinary kriging functions below is the krigin estimate, and the standard deviation of the kriging estimate.

# <codecell>

kriging.simple( P, covfct, pt, N=6 )

# <codecell>

kriging.ordinary( P, covfct, pt, N=6 )

# <codecell>

est, kstd = kriging.krige( P, covfct, [[2000,4700],[2100,4700],[2000,4800],[2100,4800]], 'simple', N=6 )

# <codecell>

est

# <codecell>

kstd