Slower than scipy

[inigozubeldia]

Hello,

I've tried the following test code, which does 2d interpolation, and fast_interp is slower than scipy by three orders of magnitude. Is there something obvious I am doing wrong?

import fast_interp
import numpy as np
import time
import scipy.interpolate as interpolate

nx = 50
ny = 37
xv, xh = np.linspace(0, 1,       nx, endpoint=True,  retstep=True)
yv, yh = np.linspace(0, 2*np.pi, ny, endpoint=False, retstep=True)
x, y = np.meshgrid(xv, yv, indexing='ij')

test_function = lambda x, y: np.exp(x)*np.exp(np.sin(y))
f = test_function(x, y)

n = 128**2
test_x = np.random.rand(n)
test_y = np.random.rand(n)

t0 = time.time()

interpolater = fast_interp.interp2d([0,0], [1,2*np.pi], [xh,yh], f, k=1, p=[False,False], e=[0,0])
fe = interpolater(test_x, test_y)

t1 = time.time()

print(t1-t0)

cpdf = interpolate.RegularGridInterpolator((xv,yv),f,method="linear",fill_value=0.,bounds_error=False)((test_x,test_y))

t2 = time.time()

print(t2-t1)

I get t1-t0 = 2.1 s (fast_interp) and t2-t1 = 0.008 s (scipy).

[ReadingClouds]

One comment - fast_interp uses numba, so first time you call it the code gets 'compiled' (in the numba sense). This is a one-time overhead.

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From: inigozubeldia @.> Sent: Tuesday, October 24, 2023 6:46 PM To: dbstein/fast_interp @.> Cc: Subscribed @.***> Subject: [dbstein/fast_interp] Slower than scipy (Issue #7)

Hello,

I've tried the following test code, which does 2d interpolation, and fast_interp is slower than scipy by three orders of magnitude. Is there something obvious I am doing wrong?

import cnc import numpy as np import time import scipy.interpolate as interpolate

nx = 50 ny = 37 xv, xh = np.linspace(0, 1, nx, endpoint=True, retstep=True) yv, yh = np.linspace(0, 2*np.pi, ny, endpoint=False, retstep=True) x, y = np.meshgrid(xv, yv, indexing='ij')

test_function = lambda x, y: np.exp(x)*np.exp(np.sin(y)) f = test_function(x, y)

n = 128**2 test_x = np.random.rand(n) test_y = np.random.rand(n)

t0 = time.time()

interpolater = cnc.interp2d([0,0], [1,2*np.pi], [xh,yh], f, k=1, p=[False,False], e=[0,0]) fe = interpolater(test_x, test_y)

t1 = time.time()

print(t1-t0)

cpdf = interpolate.RegularGridInterpolator((xv,yv),f,method="linear",fill_value=0.,bounds_error=False)((test_x,test_y))

t2 = time.time()

print(t2-t1)

I get t1-t0 = 2.1 s (fast_interp) and t2-t1 = 0.008 s (scipy).

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[shashi4u]

For a large dataset of 4096x4096 regular grid fast_interp is around 2918.554 ms (~ 3 secs) whereas scipy's RectBivariateSpline is around 31405.037 (~31 secs). In this case fast_interp is 10x faster compared to scipy's RectBivariateSpline.

From my understanding fast_interp is not advisable if you want to interpolate small datasets only once.

When to use fast_interp:

• When you have a large datasets
• When you want reuse the same function against multiple datasets

When to avoid fast_interp:

• When you want to interpolate fewer and smaller datasets against a single function