CODEWITHSUNDEEP
pythonnumpymatplotlibscipy
user1979961
user1979961

Reputation:

Scipy Fast 1-D interpolation without any loop

I have two 2D array, x(ni, nj) and y(ni,nj), that I need to interpolate over one axis. I want to interpolate along last axis for every ni.

I wrote

import numpy as np
from scipy.interpolate import interp1d

z = np.asarray([200,300,400,500,600])
out = []
for i in range(ni):
    f = interp1d(x[i,:], y[i,:], kind='linear')
    out.append(f(z))
out = np.asarray(out)

However, I think this method is inefficient and slow due to loop if array size is too large. What is the fastest way to interpolate multi-dimensional array like this? Is there any way to perform linear and cubic interpolation without loop? Thanks.

Upvotes: 7

Views: 7558

Answers (2)

Jaime
Jaime

Reputation: 67487

The method you propose does have a python loop, so for large values of ni it is going to get slow. That said, unless you are going to have large ni you shouldn't worry much.

I have created sample input data with the following code:

def sample_data(n_i, n_j, z_shape) :
    x = np.random.rand(n_i, n_j) * 1000
    x.sort()
    x[:,0] = 0
    x[:, -1] = 1000
    y = np.random.rand(n_i, n_j)
    z = np.random.rand(*z_shape) * 1000
    return x, y, z

And have tested them with this two versions of linear interpolation:

def interp_1(x, y, z) :
    rows, cols = x.shape
    out = np.empty((rows,) + z.shape, dtype=y.dtype)
    for j in xrange(rows) :
        out[j] =interp1d(x[j], y[j], kind='linear', copy=False)(z)
    return out

def interp_2(x, y, z) :
    rows, cols = x.shape
    row_idx = np.arange(rows).reshape((rows,) + (1,) * z.ndim)
    col_idx = np.argmax(x.reshape(x.shape + (1,) * z.ndim) > z, axis=1) - 1
    ret = y[row_idx, col_idx + 1] - y[row_idx, col_idx]
    ret /= x[row_idx, col_idx + 1] - x[row_idx, col_idx]
    ret *= z - x[row_idx, col_idx]
    ret += y[row_idx, col_idx]
    return ret

interp_1 is an optimized version of your code, following Dave's answer. interp_2 is a vectorized implementation of linear interpolation that avoids any python loop whatsoever. Coding something like this requires a sound understanding of broadcasting and indexing in numpy, and some things are going to be less optimized than what interp1d does. A prime example being finding the bin in which to interpolate a value: interp1d will surely break out of loops early once it finds the bin, the above function is comparing the value to all bins.

So the result is going to be very dependent on what n_i and n_j are, and even how long your array z of values to interpolate is. If n_j is small and n_i is large, you should expect an advantage from interp_2, and from interp_1 if it is the other way around. Smaller z should be an advantage to interp_2, longer ones to interp_1.

I have actually timed both approaches with a variety of n_i and n_j, for z of shape (5,) and (50,), here are the graphs:

enter image description here

enter image description here

So it seems that for z of shape (5,) you should go with interp_2 whenever n_j < 1000, and with interp_1 elsewhere. Not surprisingly, the threshold is different for z of shape (50,), now being around n_j < 100. It seems tempting to conclude that you should stick with your code if n_j * len(z) > 5000, but change it to something like interp_2 above if not, but there is a great deal of extrapolating in that statement! If you want to further experiment yourself, here's the code I used to produce the graphs.

n_s = np.logspace(1, 3.3, 25)
int_1 = np.empty((len(n_s),) * 2)
int_2 = np.empty((len(n_s),) * 2)
z_shape = (5,)

for i, n_i in enumerate(n_s) :
    print int(n_i)
    for j, n_j in enumerate(n_s) :
        x, y, z = sample_data(int(n_i), int(n_j), z_shape)
        int_1[i, j] = min(timeit.repeat('interp_1(x, y, z)',
                                        'from __main__ import interp_1, x, y, z',
                                        repeat=10, number=1))
        int_2[i, j] = min(timeit.repeat('interp_2(x, y, z)',
                                        'from __main__ import interp_2, x, y, z',
                                        repeat=10, number=1))

cs = plt.contour(n_s, n_s, np.transpose(int_1-int_2))
plt.clabel(cs, inline=1, fontsize=10)
plt.xlabel('n_i')
plt.ylabel('n_j')
plt.title('timeit(interp_2) - timeit(interp_1), z.shape=' + str(z_shape))
plt.show()

Upvotes: 15

Dave
Dave

Reputation: 8109

One optimization is to allocate the result array once like so:

import numpy as np
from scipy.interpolate import interp1d

z = np.asarray([200,300,400,500,600])
out = np.zeros( [ni, len(z)], dtype=np.float32 ) 
for i in range(ni):
    f = interp1d(x[i,:], y[i,:], kind='linear')
    out[i,:]=f(z)

This will save you some memory copying that occurs in your implementation, which occurs in the calls to out.append(...).

Upvotes: 2

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