Reverse Box-Cox transformation

Question

I am using SciPy\'s boxcox function to perform a Box-Cox transformation on a continuous variable.

from scipy.stats import boxcox
import numpy as np
y = np.random         


        

Answer 1

SciPy has added an inverse Box-Cox transformation.

https://docs.scipy.org/doc/scipy/reference/generated/scipy.special.inv_boxcox.html

scipy.special.inv_boxcox scipy.special.inv_boxcox(y, lmbda) =

Compute the inverse of the Box-Cox transformation.

Find x such that:

y = (x**lmbda - 1) / lmbda  if lmbda != 0
    log(x)                  if lmbda == 0

Parameters: y : array_like

Data to be transformed.

lmbda : array_like

Power parameter of the Box-Cox transform.

Returns:
x : array

Transformed data.

Notes

New in version 0.16.0.

Example:

from scipy.special import boxcox, inv_boxcox
y = boxcox([1, 4, 10], 2.5)
inv_boxcox(y, 2.5)

output: array([1., 4., 10.])

Answer 2

In order to inverse the boxcox transformation from scipy.stats.boxcox using scipy.special.inv_boxcox you have to identify the lambda which was generated.

First apply the transformation and print the lambda (ie. param).

df[feature_boxcox], param = stats.boxcox(df[feature])
print('Optimal lambda', param)

Then in order to inverse the transformation you input the generated lambda.

inv_boxcox(df[feature_boxcox], param)

Answer 3

1. Here it is the code. It is working and just test. Scipy used neperian logarithm, i check the BoxCox transformation paper and it seens that they used log10. I kept with neperian, because it works with scipy

2. Follow the code:

   #Function
   def invboxcox(y,ld):
      if ld == 0:
         return(np.exp(y))
      else:
         return(np.exp(np.log(ld*y+1)/ld))
   
   # Test the code
   x=[100]
   ld = 0
   y = stats.boxcox(x,ld)
   print invboxcox(y[0],ld)
   

Answer 4

Thanks to @Warren Weckesser, I've learned that the current implementation of SciPy does not have a function to reverse a Box-Cox transformation. However, a future SciPy release may have this function. For now, the code I provide in my question may serve others to reverse Box-Cox transformations.