Scipy interpolate returns a 'dimensionless' array

Question

I understand that interp1d expects an array of values to interpolate, but the behavior when passing it a float is strange enough to ask what is going on and what exactly is being returned

import numpy as np
from scipy.interpolate import interp1d

x = np.array([1,2,3,4])
y = np.array([5,7,9,15])
f = interp1d(x,y, kind='cubic')
a = f(2.5)

print(repr(a))
print("type is {}".format(type(a)))
print("shape is {}".format(a.shape))
print("ndim is {}".format(a.ndim))
print(a)

Output:

array(7.749999999999992)
type is <class 'numpy.ndarray'>
shape is ()
ndim is 0
7.749999999999992

EDIT: To clarify, I would not expect numpy to even have a dimensionless, shapeless array much less a scipy function return one.

print("Numpy version is {}".format(np.__version__))
print("Scipy version is {}".format(scipy.__version__))

Numpy version is 1.10.4
Scipy version is 0.17.0

I can replicate this. Could you specify what your question is? I guess you expect that it should return a `float` and you are asking if this is a bug. — Fabian Rost, Jun 02 '16 at 13:11
This related question may help clarify things: http://stackoverflow.com/questions/773030/why-are-0d-arrays-in-numpy-not-considered-scalar — Ryan J. Smith, Jun 02 '16 at 13:14
I don't find it strange that the interpolated value for 2.5 is somewhere in between 7 and 9, given that f(2) = 7 and f(3) = 9. — Forzaa, Jun 02 '16 at 13:14
@Forzaa- the problem is not the interpolated value, it's the dimensionless, shapeless array returned. I agree the value itself is correct. — michael_j_ward, Jun 02 '16 at 13:26

hpaulj · Accepted Answer · 2016-06-02T17:52:04.957

The interp1d returns a value that matches the input in shape - after wrapping in np.array() if needed:

In [324]: f([1,2,3])
Out[324]: array([ 5.,  7.,  9.])

In [325]: f([2.5])
Out[325]: array([ 7.75])

In [326]: f(2.5)
Out[326]: array(7.75)

In [327]: f(np.array(2.5))
Out[327]: array(7.75)

Many numpy operations do return scalars instead of 0d arrays.

In [330]: np.arange(3).sum()
Out[330]: 3

though actually it returns a numpy object

In [341]: type(np.arange(3).sum())
Out[341]: numpy.int32

which does have a shape () and ndim 0.

Whereas interp1d returns an array.

In [344]: type(f(2.5))
Out[344]: numpy.ndarray

You can extract the value with [()] indexing

In [345]: f(2.5)[()]
Out[345]: 7.75

In [346]: type(f(2.5)[()])
Out[346]: numpy.float64

This may just be an oversight in the scipy code. How often do people want to interpolate at just one point? Isn't interpolating over a regular grid of points more common?

==================

The documentation for f.__call__ is quite explicit about returning an array.

Evaluate the interpolant

Parameters
----------
x : array_like
    Points to evaluate the interpolant at.

Returns
-------
y : array_like
    Interpolated values. Shape is determined by replacing
    the interpolation axis in the original array with the shape of x.

===============

The other side to the question is why does numpy even have a 0d array. The linked answer probably is sufficient. But often the question is asked by people who are used to MATLAB. In MATLAB nearly everything is 2d. There aren't any (true) scalars. Now MATLAB has structures and cells, and matrices with more than 2 dimensions. But I recall a time (in the 1990s) when it didn't have those. Everything, literal, was a 2d matrix.

The np.matrix approximates that MATLAB case, fixing its arrays at 2d. But it does have a _collapse method that can return a 'scalar'.

Scipy interpolate returns a 'dimensionless' array

1 Answers1