RegularGridInterpolatorを使った例

[stackoverflow: Scipy interpolation with masked data?]
(https://stackoverflow.com/questions/35807321/scipy-interpolation-with-masked-data)
が分かりやすい．
MaskedArrayの場合は，（マスクされた部分をnp.nanで埋めるよりも）マスク自体を変換する方が高速．

import numpy as np
from scipy import interpolate

def conv_resol(arr, newshape, *args, **kwargs):
    '''
    Args:
        arr [2d ndarray or MaskedArray]
        newshape [tuple of int]
        *args, **kwargs: for interpolate.RegularGridInterpolator
    '''
    nx0, ny0 = arr.shape
    nx1, ny1 = newshape
    x0 = np.linspace(0, 1, nx0)
    y0 = np.linspace(0, 1, ny0)
    x1 = np.linspace(0, 1, nx1)
    y1 = np.linspace(0, 1, ny1)
    x1, y1 = np.meshgrid(x1, y1) # x1 [ny1,nx1], y1 [ny1, nx1]
    xy1 = np.array((x1, y1)).T   # xy1 [nx1, ny1, 2]
    arr1 = interpolate.RegularGridInterpolator((x0, y0), arr, *args, **kwargs)(xy1)
    if isinstance(arr, np.ma.MaskedArray):
        mask1 = interpolate.RegularGridInterpolator((x0, y0), arr.mask.astype('float'), *args, **kwargs)(xy1) > 0.0
        return np.ma.masked_array(arr1, mask=mask1)
    return arr1

テスト

import matplotlib.pyplot as plt


nx0, ny0 = 10, 20
arr0 = np.arange(nx0 * ny0).reshape(nx0, ny0) * (1.0 / (nx0 * ny0))

nx1, ny1 = 5, 10
arr1 = conv_resol(arr0, (nx1, ny1))

nx2, ny2 = 20, 40
arr2 = conv_resol(arr0, (nx2, ny2))

plt.figure(figsize=(12,8))
plt.subplot(2,2,1)
plt.imshow(arr0, vmin=0.0, vmax=1.0)
plt.subplot(2,2,3)
plt.imshow(arr1, vmin=0.0, vmax=1.0)
plt.subplot(2,2,4)
plt.imshow(arr2, vmin=0.0, vmax=1.0)