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A library for calculating the numerical inverse of a function

Project description

A module specialized on calculating the numerical inverse of any invertible function.

Requirements

Scipy Numpy

Install

In order to install this tool you’ll need pip:

pip install pynverse

Usage

Pynverse provides a central function inversefunc that calculates the numerical inverse of a function f passed as the first argument in the form of a callable.

>>> from pynverse import inversefunc

It can be used to calculate the inverse function at certain y_values points:

>>> cube = (lambda x: x**3)
>>> invcube = inversefunc(cube, y_values=3)
array(3.0000000063797567)

Or to obtain a callable that will calculate the inverse values at any other points if y_values is not provided:

>>> invcube = inversefunc(cube)
>>> invcube(27)
array(3.0000000063797567)

It requires the function to be continuous and strictly monotonic (i.e. purely decreasing or purely increasing) within the domain of the function. By default, the domain includes all real numbers, but it can be restricted to an inverval using the domain argument:

>>> import numpy as np
>>> inversefunc(np.cos, y_values=[1, 0, -1], # Should give [0, pi / 2, pi]
...             domain=[0, np.pi])
array([ 0.        ,  1.57079632,  3.14159265])

Additionally, the argument open_domain can be used to specify the open/closed characters of the end of the domain interval on one side:

>>> inversefunc(np.log10, y_values=-2, # Should give 0.01
...             domain=0, open_domain=[True, False])
array(0.0099999999882423)

Or on both sides:

>>> invtan = inversefunc(np.tan,
...                      domain=[-np.pi / 2, np.pi / 2],
...                      open_domain=True)
>>> invtan([1, 0, -1]) # Should give [pi / 4, 0, -pi / 4]
array([  7.85398163e-01,   1.29246971e-26,  -7.85398163e-01])

Additional parameters may be passed to the function for easier reusability using the args argument:

>>> invsquare = inversefunc(np.power, args=(2), domain=0)
>>> invsquare([4, 16, 64])
array([ 2.,  4.,  8.])

The image of the function in the interval may be also provided for cases where the function is non continuous right at the end of an open interval with the image argument:

>>> invmod = inversefunc(np.mod, args=(1), domain=[5,6],
...                      open_domain=[False,True], image=[0,1])
>>> invmod([0.,0.3,0.5])
array([ 5. ,  5.3,  5.5])

Additionally an argument can be used to check for the number of digis of accuracy of the results. Giving a warning in case it is not meet:

>>> inversefunc(np.log10, y_values=-8, # Should give 0.01
...             domain=0, open_domain=True, accuracy=6)
pynverse\inverse.py:195: RuntimeWarning: Results obtained with less than 6 decimal digits of accuracy
array(9.999514710830838e-09)

As it is compatible with arrays, it can very easily used to obtain the inverse for broad ranges. These are some examples of using the callables with vectors to make plots, and compare to the analytical inverse, each of them calculated as simply as:

log = lambda x: np.log10(x)
invlog = scipy.misc.inversefunc(log, domain=0, open_domain=True)
x1=np.linspace(0.00001,10,100)
x2=np.linspace(-5,1,100)
ax1.plot(x1,log(y1),'b-')
ax2.plot(x2,invlog(x2),'b-')

invlog_a = lambda x: 10**x
ax2.plot(x2,invlog_a(x2),'r--')

In particular, for the piecewise function case, there is piecewise util function provided that solves the issues of np.piecewise in order to make it work for both scalars and arrays. For example, the inverse for the last plot was obtained as:

from pynverse import inversefunc, piecewise

pw=lambda x: piecewise(x,[x<1,(x>=1)*(x<3),x>=3],[lambda x: x, lambda x: x**2, lambda x: x+6])
invpw =inversefunc(pw)

Disclaimer

Just to clarify, the problem of calculating the numerical inverse of an arbitrary funtion in unlimited or open intervals, is still an open question in applied maths. The main purpose of this package is not to be fast, or as accurate as it could be if the inverse was calculated specifically for a known function, using more specialised techniques. The current implementation essentially uses the existing tools in scipy to solve the particular problem of finding the inverse of a function meeting the continuity and monotonicity conditions, but while it performs really well it may fail under certain conditions. For example when inverting a log10 it is known to start giving inccacurate values when being asked to invert -10, which should correspond to 0.0000000001 (1e-10), but gives instead 0.0000000000978 (0.978e-10).

The advantage about estimating the inverse function is that the accuracy can be verified by checking if f(finv(x))==x..

Details about the implementation

The summarized internal strategy is the following:

  1. Homogenize and normalize the input parameters.

  2. Figure out if the function is increasing or decreasing. For this two reference points ref1 and ref2are need:

  • In a case of a closed interval, the points can 1/4 and 3/4 through the interval.

  • In an open interval any two values would work really.

  • if f(ref1)<(ref2), the function is increasing, otherwise is decreasing.

  1. Figure out the image of the function in the interval.

  • If the user gives values, then those are used.

  • In a closed interval just calculate f(a) and f(b).

  • In a closed interval try to calculate f(a) and f(b), if this works those are used. other wise it will be assume to be (-Inf, Inf).

  1. I build a function the following function: -bounded_f(x): -return -Inf if x below interval, and f is increasing. -return +Inf if x below interval, and f is decreasing. -return +Inf if x above interval, and f is increasing. -return -Inf if x above interval, and f is decreasing. -return f(x) otherwise

  2. If the required number y0 for the inverse is outside the image, raise an exception.

  3. I find roots for bounded_f(x)-y0, by minimizing (bounded_f(x)-y0)**2, using the Brent method, making sure that the algorithm for minimising starts in a point inside the original interval by setting ref1, ref2 as brackets. As soon as if goes outside the allowed intervals, bounded_f will return infinite, forcing the algorithm to go back to search inside the interval. This could be further improved, although the current performance, is quite good already.

  4. Check that the solutions are accurate and they meet f(x0)=y0 to some desired precision, raising a warning otherwise.

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