コード例 #1
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def make_knots(size=30, num_knots=4, order=7):

    knots = np.round(-1 +
                     np.logspace(np.log10(1), np.log10(size - 1), num_knots))
    knots = np.round(np.sqrt(knots) * (size - 1) / max(np.sqrt(knots)))
    knots = np.unique(knots)
    knots = splinelab.augknt(knots, order)

    return knots
コード例 #2
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def create_bspline_basis(xmin, xmax, p=3, nknots=5):
    """ compute a Bspline basis set where:
        
        :param p: order of spline (3 = cubic)
        :param nknots: number of knots (endpoints only counted once)
    """

    knots = np.linspace(xmin, xmax, nknots)
    k = splinelab.augknt(knots, p)  # pad the knot vector
    B = bspline.Bspline(k, p)
    return B
コード例 #3
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ファイル: hbr.py プロジェクト: Hesterhuijsdens/nispat
def bspline_fit(X, order, nknots):

    X = X.squeeze()
    if len(X.shape) > 1:
        raise ValueError('Bspline method only works for a single covariate.')

    knots = np.linspace(X.min(), X.max(), nknots)
    k = splinelab.augknt(knots, order)
    bsp_basis = bspline.Bspline(k, order)

    return bsp_basis
コード例 #4
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def test():
    ########################
    # config
    ########################

    p = 3  # order of spline basis (as-is! 3 = cubic)

    nknots = 5  # for testing: number of knots to generate (here endpoints count only once)

    tau = [0.1, 0.33]  # collocation sites (i.e. where to evaluate)

    ########################
    # usage example
    ########################

    knots = np.linspace(
        0, 1, nknots)  # create a knot vector without endpoint repeats
    k = splinelab.augknt(
        knots, p)  # add endpoint repeats as appropriate for spline order p
    B = bspline.Bspline(k, p)  # create spline basis of order p on knots k

    # build some collocation matrices:
    #
    A0 = B.collmat(tau)  # function value at sites tau
    A2 = B.collmat(tau, deriv_order=2)  # second derivative at sites tau

    print(A0)
    print(A2)

    ########################
    # tests
    ########################

    # number of basis functions
    n_interior_knots = len(knots) - 2
    n_basis_functions_expected = n_interior_knots + (p + 1)
    n_basis_functions_actual = len(
        B(0.))  # perform dummy evaluation to get number of basis functions
    assert n_basis_functions_actual == n_basis_functions_expected, "something went wrong, number of basis functions is incorrect"

    # partition-of-unity property of the spline basis
    assert np.allclose(
        np.sum(A0, axis=1), 1.0
    ), "something went wrong, the basis functions do not form a partition of unity"
コード例 #5
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ファイル: test_blr.py プロジェクト: mariam186/PCNtoolkit
Phips = np.zeros((Xs.shape[0], dimpoly))
colid = np.arange(0, 1)
for d in range(1, dimpoly + 1):
    Phips[:, colid] = np.vstack(Xs**d)
    colid += 1

# generative model
b = [0.5, -0.1, 0.005]  # true regression coefficients
s2 = 0.05  # noise variance
y = Phip.dot(b) + np.random.normal(0, s2, N)

# cubic B-spline basis (used for regression)
p = 3  # order of spline (3 = cubic)
nknots = 5  # number of knots (endpoints only counted once)
knots = np.linspace(0, 10, nknots)
k = splinelab.augknt(knots, p)  # pad the knot vector
B = bspline.Bspline(k, p)
Phi = np.array([B(i) for i in X])
Phis = np.array([B(i) for i in Xs])

hyp0 = np.zeros(2)
#hyp0 = np.zeros(4) # use ARD
B = BLR(hyp0, Phi, y)
hyp = B.estimate(hyp0, Phi, y, optimizer='powell')

yhat, s2 = B.predict(hyp, Phi, y, Phis)
plt.fill_between(Xs,
                 yhat - 1.96 * np.sqrt(s2),
                 yhat + 1.96 * np.sqrt(s2),
                 alpha=0.2)
plt.scatter(X, y)
コード例 #6
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start_n = 1000  # sample size at first analysis
typ_list = [2, 3]  # phi function type list
nk_list = [(200, 5), (20, 50)]  #
delta_list = [0.0, 0.10, 0.15, 0.20, 0.25, 0.30]  # effect size typ three
alpha = 0.05
e = 0.5  # error variance
ms = [1, 2, 3]  # alpha spending functions
lm = len(ms)

lb, ub = -2, 2

order = 3  # order of spline (as-is; 3 = cubic)
nknots = 4  # number of knots to generate
knots = np.linspace(lb, ub,
                    nknots)  # create a knot vector without endpoint repeats
knots = splinelab.augknt(
    knots, order)  # add endpoint repeats as appropriate for spline order
bases = bspline.Bspline(knots,
                        order)  # create spline basis of order p on knots k
nfeatures = 3
p = nfeatures * (order + nknots - 1) + 1

rho = 0.5
cov_mat = (1 - rho) * np.eye(nfeatures) + rho * np.ones((nfeatures, nfeatures))
L = np.linalg.cholesky(cov_mat).T

for (typ, (n, k), delta) in product(typ_list, nk_list, delta_list):

    name = 'result/HTE/' + 'typ' + str(typ) + 'n' + str(n) + 'k' + str(
        k) + 'delta' + str(delta) + '_nonlinear_adaptive_HTE.npz'

    #     if os.path.exists(name):
コード例 #7
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def main():
    """Demonstration: plot a B-spline basis and its first three derivatives."""

    #########################################################################################
    # config
    #########################################################################################

    # Order of spline basis.
    #
    p = 3

    # Knot vector, including the endpoints.
    #
    # For convenience, endpoints are specified only once, regardless of the value of p.
    #
    # Duplicate knots *in the interior* are allowed, with the standard meaning for B-splines.
    #
    knots = [0., 0.25, 0.5, 0.75, 1.]

    # How many plotting points to use on each subinterval [knots[i], knots[i+1]).
    #
    # Only intervals with length > 0 are actually plotted.
    #
    nt_per_interval = 101

    #########################################################################################
    # the demo itself
    #########################################################################################

    # The evaluation algorithm used in bspline.py uses half-open intervals  t_i <= x < t_{i+1}.
    #
    # This causes the right endpoint of each interval to actually be the start point of the next interval.
    #
    # Especially, the right endpoint of the last interval is the start point of the next (nonexistent) interval,
    # so the basis will return a value of zero there.
    #
    # We work around this by using a small epsilon to avoid evaluation exactly at t_{i+1} (for each interval).
    #
    epsrel = 1e-10
    epsabs = epsrel * (knots[-1] - knots[0])

    original_knots = knots
    knots = splinelab.augknt(
        knots, p)  # add repeated endpoint knots for splines of order p

    # treat each interval separately to preserve discontinuities
    #
    # (useful especially when plotting the highest-order nonzero derivative)
    #
    B = bspline.Bspline(knots, p)
    xxs = []
    for I in zip(knots[:-1], knots[1:]):
        t_i = I[0]
        t_ip1 = I[1] - epsabs
        if t_ip1 - t_i > 0.:  # accept only intervals of length > 0 (to skip higher-multiplicity knots in the interior)
            xxs.append(np.linspace(t_i, t_ip1, nt_per_interval))

    # common settings for all plotted lines
    settings = {"linestyle": 'solid', "linewidth": 1.0}

    # create a list of unique colors for plotting
    #
    # http://stackoverflow.com/questions/8389636/creating-over-20-unique-legend-colors-using-matplotlib
    #
    NUM_COLORS = nbasis = len(
        B(0.))  # perform dummy evaluation to get number of basis functions
    cm = plt.get_cmap('gist_rainbow')
    cNorm = matplotlib.colors.Normalize(vmin=0, vmax=NUM_COLORS - 1)
    scalarMap = matplotlib.cm.ScalarMappable(norm=cNorm, cmap=cm)
    colors = [scalarMap.to_rgba(i) for i in range(NUM_COLORS)]

    labels = [
        r"$B$", r"$\mathrm{d}B\,/\,\mathrm{d}x$",
        r"$\mathrm{d}^2B\,/\,\mathrm{d}x^2$",
        r"$\mathrm{d}^3B\,/\,\mathrm{d}x^3$"
    ]

    # for plotting the knot positions:
    unique_knots_xx = np.unique(original_knots)
    unique_knots_yy = np.zeros_like(unique_knots_xx)

    # plot the basis functions B(x) and their first three derivatives
    plt.figure(1)
    plt.clf()
    for k in range(4):
        ax = plt.subplot(2, 2, k + 1)

        # place the axis label where it fits
        if k % 2 == 0:
            ax.yaxis.set_label_position("left")
        else:
            ax.yaxis.set_label_position("right")

        # plot the kth derivative; each basis function gets a unique color
        f = B.diff(order=k)  # order=0 is a passthrough
        for xx in xxs:
            yy = np.array([
                f(x) for x in xx
            ])  # f(scalar) -> rank-1 array, one element per basis function
            for i in range(nbasis):
                settings["color"] = colors[i]
                plt.plot(xx, yy[:, i], **settings)
            plt.ylabel(labels[k])

        # show knot positions
        plt.plot(unique_knots_xx, unique_knots_yy, "kx")

    plt.suptitle(r"$B$-spline basis functions, $p=%d$" % (p))