Interpolation in SciPy: Finding X that produces Y

Question

Is there a better way to find which X gives me the Y I am looking for in SciPy? I just began using SciPy and I am not too familiar with each function.

import numpy as np
import matplotlib.pyplot as plt
from scipy import interpolate

x = [70, 80, 90, 100, 110]
y = [49.7, 80.6, 122.5, 153.8, 163.0]
tck = interpolate.splrep(x,y,s=0)
xnew = np.arange(70,111,1)
ynew = interpolate.splev(xnew,tck,der=0)
plt.plot(x,y,'x',xnew,ynew)
plt.show()
t,c,k=tck
yToFind = 140
print interpolate.sproot((t,c-yToFind,k)) #Lowers the spline at the abscissa

Can you elaborate on what you want to be better? Performance, accuracy, conciseness? — Ryan, Dec 04 '09 at 04:17

ihuston · Accepted Answer · 2011-06-27T09:07:53.613

18

The UnivariateSpline class in scipy makes doing splines much more pythonic.

x = [70, 80, 90, 100, 110]
y = [49.7, 80.6, 122.5, 153.8, 163.0]
f = interpolate.UnivariateSpline(x, y, s=0)
xnew = np.arange(70,111,1)

plt.plot(x,y,'x',xnew,f(xnew))

To find x at y then do:

yToFind = 140
yreduced = np.array(y) - yToFind
freduced = interpolate.UnivariateSpline(x, yreduced, s=0)
freduced.roots()

I thought interpolating x in terms of y might work but it takes a somewhat different route. It might be closer with more points.

edited Jun 27 '11 at 09:07

answered Jun 23 '09 at 09:25

ihuston

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Wouldn't this require twice the amount of CPU calculations since are interpolating practically the same data set two times? – JcMaco Jun 23 '09 at 11:29
1

@JcMaco, the first use of UnivariateSpline is just to make a pretty plot. The second usage is what actually gives the values. – Theran Dec 09 '09 at 04:07
Craig is right, can you correct it on your example as it's otherwise great! – Mike Vella Jun 23 '11 at 23:50
Fixed the typo. Thanks Craig. – ihuston Jun 27 '11 at 09:09

score 3 · Answer 2 · answered Jun 23 '09 at 04:20

3

If all you need is linear interpolation, you could use the interp function in numpy.

answered Jun 23 '09 at 04:20

Vicki Laidler

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I'd prefer spline interpolation. How would the interp function help me solve my problem more easily? – JcMaco Jun 23 '09 at 11:23
Your question didn't specify what type of interpolation you needed -- if linear isn't good enough for your problem, then I don't think interp will help. – Vicki Laidler Jun 24 '09 at 03:00

score 0 · Answer 3 · answered Nov 25 '09 at 13:50

I may have misunderstood your question, if so I'm sorry. I don't think you need to use SciPy. NumPy has a least squares function.

#!/usr/bin/env python

from numpy.linalg.linalg import lstsq



def find_coefficients(data, exponents):
    X = tuple((tuple((pow(x,p) for p in exponents)) for (x,y) in data))
    y = tuple(((y) for (x,y) in data))
    x, resids, rank, s = lstsq(X,y)
    return x

if __name__ == "__main__":
    data = tuple((
        (1.47, 52.21),
        (1.50, 53.12),
        (1.52, 54.48),
        (1.55, 55.84),
        (1.57, 57.20),
        (1.60, 58.57),
        (1.63, 59.93),
        (1.65, 61.29),
        (1.68, 63.11),
        (1.70, 64.47),
        (1.73, 66.28),
        (1.75, 68.10),
        (1.78, 69.92),
        (1.80, 72.19),
        (1.83, 74.46)
    ))
    print find_coefficients(data, range(3))

This will return [ 128.81280358 -143.16202286 61.96032544].

>>> x=1.47 # the first of the input data
>>> 128.81280358 + -143.16202286*x + 61.96032544*(x**2)
52.254697219095988

0.04 out, not bad

The problem is finding which number will give me 52.21. Of course, there can be many solutions if the interpolation is quadratic (or higher power). — JcMaco, Nov 26 '09 at 18:27

Interpolation in SciPy: Finding X that produces Y

3 Answers3