矢量化 2D 插值

Question

I am using SciPy's 2D interpolation function ( scipy.interpolate.interp2d ). I have the field which is stored as Z = [x, y, p1, p2] . I would like to interpolate at every x and y point using the 2D matrix [x, y, :, :] . The shape of Z is (40, 40, 6, 6) .

The only way I could do this was by using a for loop and going over each point. Below is the code I am using for it. a and b are arrays which have the shape (6,) and (6,) .

for i in range(40):
    for j in range(40):
        ff = interpolate.interp2d(a, b, z[i,j,:,:]) 
        zi[i,j] = ff(atest,btest)

Is there a way I can do this without using for loop?

Answer 1

Since you are essentially interpolating a bunch of different functions over the same 2d points, it's reasonable that there's no built-in way to do this. Although some of the computation can be spared since for every function you have the same 2d grid underlying the interpolation, so the expectation isn't impossible either. In any case I couldn't find a built-in way to do this without looping.

Fortunately, with some elbow grease we can implement a bilinear interpolator (just like the loopy one) that use vectorization to work on all the points at once. Here's a class that implements this:

import numpy as np 

class VectorizedBilinearInterpolator: 
    def __init__(self, x0, y0, z0): 
        """Create a bilinear interpolator for gridded input data 

        Inputs: 
            x0: shape (ngridx,)
            y0: shape (ngridy,)
            z0: shape batch_shape + (ngridy, ngridx) (viewed as batches) 
        """

        if z0.shape[-2:] != y0.shape + x0.shape: 
            raise ValueError("The last two dimensions of z0 must match that of y0 and x0, respectively!") 

        ind_x = np.argsort(x0) 
        self.x0 = x0[ind_x] 

        ind_y = np.argsort(y0) 
        self.y0 = y0[ind_y] 

        self.batch_shape = z0.shape[:-2] 
        indexer = np.ix_(ind_y, ind_x) 
        self.z0 = z0[..., indexer[0], indexer[1]].reshape(-1, y0.size, x0.size)  # shape (nbatch, ngridy, ngridx)

        # compute auxiliary coefficients for interpolation 
        # we have ngridx-1 boxes along x and ngridy-1 boxes along y
        # for each box we need 4 coefficients for bilinear interpolation
        # see e.g. https://en.wikipedia.org/wiki/Bilinear_interpolation#Alternative_algorithm
        # construct a batch of matrices with size (ngridy-1, ngridx-1, 4, 4) to invert
        x1 = self.x0[:-1]
        x2 = self.x0[1:]
        y1 = self.y0[:-1, None]
        y2 = self.y0[1:, None]
        x1,x2,y1,y2,one = np.broadcast_arrays(x1, x2, y1, y2, 1)  # all shaped (ngridy-1, ngridx-1)

        M = np.array([[one, x1, y1, x1*y1], [one, x1, y2, x1*y2],
                      [one, x2, y1, x2*y1], [one, x2, y2, x2*y2]]).transpose(2, 3, 0, 1)  # shape (ngridy-1, ngridx-1, 4, 4)
        zvec = np.array([self.z0[:, :-1, :-1], self.z0[:, 1:, :-1], self.z0[:, :-1, 1:], self.z0[:, 1:, 1:]])  # shape (4, nbatch, ngridy-1, ngridx-1)

        self.coeffs = np.einsum('yxab,bnyx -> nyxa', np.linalg.inv(M), zvec)  # shape (nbatch, ngridy-1, ngridx-1, 4) for "a0,a1,a2,a3" coefficients
        # for a given box (i,j) the interpolated value is given by self.coeffs[:,i,j,:] @ [1, x, y, x*y]

    def __call__(self, x, y): 
        """Evaluate the interpolator at the given coordinates 

        Inputs: 
            x: shape (noutx,) 
            y: shape (nouty,) 

        Output: 
            z: shape batch_shape + (nouty, noutx) (see __init__) 
        """

        # identify outliers (and mask at the end)
        out_x = (x < self.x0[0]) | (self.x0[-1] < x)
        out_y = (y < self.y0[0]) | (self.y0[-1] < y)

        # clip outliers, mask later
        xbox = (self.x0.searchsorted(x) - 1).clip(0, self.x0.size - 2)  # shape (noutx,) indices
        ybox = (self.y0.searchsorted(y) - 1).clip(0, self.y0.size - 2)  # shape (nouty,) indices
        indexer = np.ix_(ybox, xbox)

        xgrid,ygrid = np.meshgrid(x, y)  # both shape (nouty, noutx)

        coeffs_now = self.coeffs[:, indexer[0], indexer[1], :]  # shape (nbatch, nouty, noutx, 4)
        poly = np.array([np.ones_like(xgrid), xgrid, ygrid, xgrid*ygrid])  # shape (4, nouty, noutx)
        values =  np.einsum('nyxa,ayx -> nyx', coeffs_now, poly)  # shape (nbatch, nouty, noutx)

        # reshape final result and mask outliers
        z = values.reshape(self.batch_shape + xgrid.shape)
        z[..., out_y, :] = np.nan
        z[..., :, out_x] = np.nan

        return z

And here are some tests:

from scipy import interpolate  # only needed for the loopy comparison

# input points
x0 = np.linspace(0, 1, 6)
y0 = np.linspace(0, 1, 7)
z0 = np.random.rand(40, 41, y0.size, x0.size)

# output points
xtest = np.linspace(0, 1, 20)
ytest = np.linspace(0, 1, 21)

# the original loopy version
def loopy():
    zi = np.empty(z0.shape[:2] + (ytest.size, xtest.size))
    for i in range(z0.shape[0]):
        for j in range(z0.shape[1]):
            ff = interpolate.interp2d(x0, y0, z0[i,j,:,:])
            zi[i,j] = ff(xtest, ytest)
    return zi

# the new, vectorized version
def vector():
    f = VectorizedBilinearInterpolator(x0, y0, z0)
    zi = f(xtest, ytest)
    return zi

First, we should make sure that the two interpolators do the same thing:

>>> np.allclose(loopy(), vector())
True

Second, we can time this dummy example using ipython's %timeit magic due to laziness:

>>> %timeit loopy()
... %timeit vector()
216 ms ± 2.57 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
25.9 ms ± 1.08 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

So at least in this small example the vectorized version is 8 times faster. The actual speedup depends greatly on the size of your actual example. Although I've tried implementing outliers properly (ie setting outliers to np.nan rather than extrapolating nonsense), I didn't thoroughly check all edge cases, so make sure you test the implementation for your real data (by comparing the loop with the new interpolator for a few typical cases).

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If your question contains the real problem, ie a final array of shape (40,40,6,6) , then none of this work is worth any of your time (considering the very short runtimes), and you should just use the loop.