反转插值以给出与所需插值函数值关联的变量 [英] Invert interpolation to give the variable associated with a desired interpolation function value

问题描述

我正在尝试使用scipy的插值函数反转插值函数.假设我创建了一个内插函数，

I am trying to invert an interpolated function using scipy's interpolate function. Let's say I create an interpolated function,

import scipy.interpolate as interpolate
interpolatedfunction = interpolated.interp1d(xvariable,data,kind='cubic')

当我指定a时，是否存在一些可以找到x的函数

Is there some function that can find x when I specify a:

interpolatedfunction(x) == a

换句话说，我希望内插函数等于a；使我的函数等于a的xvariable的值是多少?"

In other words, "I want my interpolated function to equal a; what is the value of xvariable such that my function is equal to a?"

我很高兴可以使用某些数字方案来实现此目的，但是有没有更简单的方法?如果插值函数在xvariable中是多值的怎么办?

I appreciate I can do this with some numerical scheme, but is there a more straightforward method? What if the interpolated function is multivalued in xvariable?

推荐答案

创建内插函数interp_fn后，您可以通过函数根查找x其中interp_fn(x) == a的值

After creating an interpolated function interp_fn, you can find the value of x where interp_fn(x) == a by the roots of the function

interp_fn2 = lambda x: interp_fn(x) - a

有许多选项可以找到scipy.optimize中的根.例如，要使用初始值从10开始的牛顿方法:

There are number of options to find the roots in scipy.optimize. For instance, to use Newton's method with the initial value starting at 10:

from scipy import optimize

optimize.newton(interp_fn2, 10)

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实际示例

创建一个内插函数，然后找到fn(x) == 5

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Actual example

Create an interpolated function and then find the roots where fn(x) == 5

import numpy as np
from scipy import interpolate, optimize

x = np.arange(10)
y = 1 + 6*np.arange(10) - np.arange(10)**2
y2 = 5*np.ones_like(x)
plt.scatter(x,y)
plt.plot(x,y)
plt.plot(x,y2,'k-')
plt.show()

# create the interpolated function, and then the offset
# function used to find the roots

interp_fn = interpolate.interp1d(x, y, 'quadratic')
interp_fn2 = lambda x: interp_fn(x)-5

# to find the roots, we need to supply a starting value
# because there are more than 1 root in our range, we need 
# to supply multiple starting values.  They should be 
# fairly close to the actual root

root1, root2 = optimize.newton(interp_fn2, 1), optimize.newton(interp_fn2, 5)

root1, root2
# returns:
(0.76393202250021064, 5.2360679774997898)

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