具有NaN值或掩模的大数组的双变量结构插值

Question

我正在尝试使用Scipy的windstress类来定期插入网格RectBivariateSpline数据。在某些网格点，输入数据包含无效数据条目，这些条目设置为NaN值。首先，我使用了关于二维插值的Scott's question解决方案。使用我的数据，插值返回仅包含NaN的数组。假设我的数据是非结构化的并使用SmoothBivariateSpline类，我也尝试了一种不同的方法。显然我有太多的数据点来使用非结构化插值，因为数据数组的形状是（719 x 2880）。

为了说明我的问题，我创建了以下脚本：

from __future__ import division
import numpy
import pylab

from scipy import interpolate


# The signal and lots of noise
M, N = 20, 30  # The shape of the data array
y, x = numpy.mgrid[0:M+1, 0:N+1]
signal = -10 * numpy.cos(x / 50 + y / 10) / (y + 1)
noise = numpy.random.normal(size=(M+1, N+1))
z = signal + noise


# Some holes in my dataset
z[1:2, 0:2] = numpy.nan
z[1:2, 9:11] = numpy.nan
z[0:1, :12] = numpy.nan
z[10:12, 17:19] = numpy.nan


# Interpolation!
Y, X = numpy.mgrid[0.125:M:0.5, 0.125:N:0.5]
sp = interpolate.RectBivariateSpline(y[:, 0], x[0, :], z)
Z = sp(Y[:, 0], X[0, :])

sel = ~numpy.isnan(z)
esp = interpolate.SmoothBivariateSpline(y[sel], x[sel], z[sel], 0*z[sel]+5)
eZ = esp(Y[:, 0], X[0, :])


# Comparing the results
pylab.close('all')
pylab.ion()

bbox = dict(edgecolor='w', facecolor='w', alpha=0.9)
crange = numpy.arange(-15., 16., 1.)

fig = pylab.figure()
ax = fig.add_subplot(1, 3, 1)
ax.contourf(x, y, z, crange)
ax.set_title('Original')
ax.text(0.05, 0.98, 'a)', ha='left', va='top', transform=ax.transAxes, 
    bbox=bbox)

bx = fig.add_subplot(1, 3, 2, sharex=ax, sharey=ax)
bx.contourf(X, Y, Z, crange)
bx.set_title('Spline')
bx.text(0.05, 0.98, 'b)', ha='left', va='top', transform=bx.transAxes, 
    bbox=bbox)

cx = fig.add_subplot(1, 3, 3, sharex=ax, sharey=ax)
cx.contourf(X, Y, eZ, crange)
cx.set_title('Expected')
cx.text(0.05, 0.98, 'c)', ha='left', va='top', transform=cx.transAxes, 
    bbox=bbox)

其中包含以下结果：

该图显示了构建的数据图（a）以及使用Scipy的RectBivariateSpline（b）和SmoothBivariateSpline（c）类的结果。第一次插值产生仅具有NaN的阵列。理想情况下，我会期望类似于第二次插值的结果，这是更加计算密集的。我不一定需要在域区域之外进行数据外推。

Answer 1

你可以使用griddata，唯一的问题是它不能很好地处理边缘。这可以通过反映来帮助，具体取决于您的数据如何... 这是一个例子：

from __future__ import division
import numpy
import pylab
from scipy import interpolate


# The signal and lots of noise
M, N = 20, 30  # The shape of the data array
y, x = numpy.mgrid[0:M+1, 0:N+1]
signal = -10 * numpy.cos(x / 50 + y / 10) / (y + 1)
noise = numpy.random.normal(size=(M+1, N+1))
z = signal + noise


# Some holes in my dataset
z[1:2, 0:2] = numpy.nan
z[1:2, 9:11] = numpy.nan
z[0:1, :12] = numpy.nan
z[10:12, 17:19] = numpy.nan

zi = numpy.vstack((z[::-1,:],z))
zi = numpy.hstack((zi[:,::-1], zi))
y, x = numpy.mgrid[0:2*(M+1), 0:2*(N+1)]
y *= 5 # anisotropic interpolation if needed.

zi = interpolate.griddata((y[~numpy.isnan(zi)], x[~numpy.isnan(zi)]),
                zi[~numpy.isnan(zi)], (y, x), method='cubic')
zi = zi[:(M+1),:(N+1)][::-1,::-1]

pylab.subplot(1,2,1)
pylab.imshow(z, origin='lower')
pylab.subplot(1,2,2)
pylab.imshow(zi, origin='lower')
pylab.show()

如果内存不足，您可以按以下方式拆分数据：

def large_griddata(data_x, vals, grid, method='nearest'):
    x, y = data_x
    X, Y = grid
    try:
        return interpolate.griddata((x,y),vals,(X,Y),method=method)
    except MemoryError:
        pass

    N = (np.min(X)+np.max(X))/2.
    M = (np.min(Y)+np.max(Y))/2.

    masks = [(x<N) & (y<M),
             (x<N) & (y>=M),
             (x>=N) & (y<M),
             (x>=N) & (y>=M)]

    grid_mask = [(X<N) & (Y<M),
             (X<N) & (Y>=M),
             (X>=N) & (Y<M),
             (X>=N) & (Y>=M)]

    Z = np.zeros_like(X)
    for i in range(4):
        Z[grid_mask[i]] = large_griddata((x[masks[i]], y[masks[i]]),
                    vals[masks[i]], (X[grid_mask[i]], Y[grid_mask[i]]), method=method)

    return Z