NumPy: How to calulate piecewise linear interpolant on multiple axes - numpy

Question

Given the following ndarray t -
In [26]: t.shape
Out[26]: (3, 3, 2)
In [27]: t
Out[27]:
array([[[ 0, 1],
[ 2, 3],
[ 4, 5]],
[[ 6, 7],
[ 8, 9],
[10, 11]],
[[12, 13],
[14, 15],
[16, 17]]])
this piecewise linear interpolant for the points t[:, 0, 0] can evaluated for [0 , 0.66666667, 1.33333333, 2.] as follows using numpy.interp -
In [38]: x = np.linspace(0, t.shape[0]-1, 4)
In [39]: x
Out[39]: array([0. , 0.66666667, 1.33333333, 2. ])
In [30]: xp = np.arange(t.shape[0])
In [31]: xp
Out[31]: array([0, 1, 2])
In [32]: fp = t[:,0,0]
In [33]: fp
Out[33]: array([ 0, 6, 12])
In [40]: np.interp(x, xp, fp)
Out[40]: array([ 0., 4., 8., 12.])
How can all the interpolants be efficiently calculated and returned together for all values of fp -
array([[[ 0, 1],
[ 2, 3],
[ 4, 5]],
[[ 4, 5],
[ 6, 7],
[ 8, 9]],
[[ 8, 9],
[10, 11],
[12, 13]],
[[12, 13],
[14, 15],
[16, 17]]])

Answer 1

As the interpolation is 1d with changing y values it must be run for each 1d slice of t. It's probably faster to loop explicitly but neater to loop using np.apply_along_axis
import numpy as np
t = np.arange( 18 ).reshape(3,3,2)
x = np.linspace( 0, t.shape[0]-1, 4)
xp = np.arange(t.shape[0])
def interfunc( arr ):
""" Function interpolates a 1d array. """
return np.interp( x, xp, arr )
np.apply_along_axis( interfunc, 0, t ) # apply function along axis 0
""" Result
array([[[ 0., 1.],
[ 2., 3.],
[ 4., 5.]],
[[ 4., 5.],
[ 6., 7.],
[ 8., 9.]],
[[ 8., 9.],
[10., 11.],
[12., 13.]],
[[12., 13.],
[14., 15.],
[16., 17.]]]) """
With explicit loops
result = np.zeros((4,3,2))
for c in range(t.shape[1]):
for p in range(t.shape[2]):
result[:,c,p] = np.interp( x, xp, t[:,c,p])
On my machine the second option runs in half the time.
Edit to use np.nditer
As the result and the parameter have different shapes I seem to have to create two np.nditer objects one for the parameter and one for the result. This is my first attempt to use nditer for anything so it could be over complicated.
def test( t ):
ts = t.shape
result = np.zeros((ts[0]+1,ts[1],ts[2]))
param = np.nditer( [t], ['external_loop'], ['readonly'], order = 'F')
with np.nditer( [result], ['external_loop'], ['writeonly'], order = 'F') as res:
for p, r in zip( param, res ):
r[:] = interfunc(p)
return result
It's slightly slower than the explicit loops and less easy to follow than either of the other solutions.

Answer 2

As requested by #Tis Chris, here is a solution using np.nditer with the multi_index flag but I prefer the explicit nested for loops method above because it is 10% faster
In [29]: t = np.arange( 18 ).reshape(3,3,2)
In [30]: ax0old = np.arange(t.shape[0])
In [31]: ax0new = np.linspace(0, t.shape[0]-1, 4)
In [32]: tnew = np.zeros((len(ax0new), t.shape[1], t.shape[2]))
In [33]: it = np.nditer(t[0], flags=['multi_index'])
In [34]: for _ in it:
...: tnew[:, it.multi_index[0], it.multi_index[1]] = np.interp(ax0new, ax0old, t[:, it.multi_
...: index[0], it.multi_index[1]])
...:
In [35]: tnew
Out[35]:
array([[[ 0., 1.],
[ 2., 3.],
[ 4., 5.]],
[[ 4., 5.],
[ 6., 7.],
[ 8., 9.]],
[[ 8., 9.],
[10., 11.],
[12., 13.]],
[[12., 13.],
[14., 15.],
[16., 17.]]])

Answer 3

You could try scipy.interpolate.interp1d:
from scipy.interpolate import interp1d
import numpy as np
t = np.array([[[ 0, 1],
[ 2, 3],
[ 4, 5]],
[[ 6, 7],
[ 8, 9],
[10, 11]],
[[12, 13],
[14, 15],
[16, 17]]])
# for the first slice
f = interp1d(np.arange(t.shape[0]), t[..., 0], axis=0)
# returns a function which you call with values within range np.arange(t.shape[0])
# data used for interpolation
t[..., 0]
>>> array([[ 0, 2, 4],
[ 6, 8, 10],
[12, 14, 16]])
f(1)
>>> array([ 6., 8., 10.])
f(1.5)
>>> array([ 9., 11., 13.])