Filling gaps in a numpy array

Question

I just want to interpolate, in the simplest possible terms, a 3D dataset. Linear interpolation, nearest neighbour, all that would suffice (this is to start off some algorith

Answer 1

Using scipy.ndimage, your problem can be solved with nearest neighbor interpolation in 2 lines :

from scipy import ndimage as nd

indices = nd.distance_transform_edt(invalid_cell_mask, return_distances=False, return_indices=True)
data = data[tuple(ind)]

---

Now, in the form of a function:

import numpy as np
from scipy import ndimage as nd

def fill(data, invalid=None):
    """
    Replace the value of invalid 'data' cells (indicated by 'invalid') 
    by the value of the nearest valid data cell

    Input:
        data:    numpy array of any dimension
        invalid: a binary array of same shape as 'data'. 
                 data value are replaced where invalid is True
                 If None (default), use: invalid  = np.isnan(data)

    Output: 
        Return a filled array. 
    """    
    if invalid is None: invalid = np.isnan(data)

    ind = nd.distance_transform_edt(invalid, 
                                    return_distances=False, 
                                    return_indices=True)
    return data[tuple(ind)]

Exemple of use:

def test_fill(s,d):
     # s is size of one dimension, d is the number of dimension
    data = np.arange(s**d).reshape((s,)*d)
    seed = np.zeros(data.shape,dtype=bool)
    seed.flat[np.random.randint(0,seed.size,int(data.size/20**d))] = True

    return fill(data,-seed), seed

import matplotlib.pyplot as plt
data,seed  = test_fill(500,2)
data[nd.binary_dilation(seed,iterations=2)] = 0   # draw (dilated) seeds in black
plt.imshow(np.mod(data,42))                       # show cluster

result: