Yeo-Johnson Transformation

YeoJohnson.py

#!/usr/bin/env python
# -*- coding: UTF-8 -*-
import warnings
import numpy as np
import pandas as pd
import sys

__author__ = "Mohsen Mesgarpour"
__copyright__ = "Copyright 2016, https://github.com/mesgarpour"
__credits__ = ["Mohsen Mesgarpour"]
__license__ = "GPL"
__version__ = "1.0"
__maintainer__ = "Mohsen Mesgarpour"
__email__ = "mohsen.mesgarpour@gmail.com"


class YeoJohnson:
    """
    Computing Yeo-Johnson transofrmation, which is an extension of Box-Cox transformation
    but can handle both positive and negative values.

    References:
    Weisberg, S. (2001). Yeo-Johnson Power Transformations.
    Department of Applied Statistics, University of Minnesota. Retrieved June, 1, 2003.
    https://www.stat.umn.edu/arc/yjpower.pdf

    Adapted from CRAN - Package VGAM
    """
    def fit(self, y, lmbda, derivative=0, epsilon=np.finfo(np.float).eps, inverse=False):
        """
        :param y: The variable to be transformed (numeric array).
        :param lmbda: The function's Lambda value (numeric value or array).
        :param derivative: The derivative with respect to lambda
        (non-negative integer; default: ordinary function evaluation).
        :param epsilon: The lambda's tolerance (positive value).
        :param inverse: The inverse transformation option (logical value).
        :return: The Yeo-Johnson transformation or its inverse, or its derivatives with respect to lambda, of y.
        """
        # Validate arguments
        self.__validate(y, lmbda, derivative, epsilon, inverse)

        # initialise
        y = np.array(y, dtype=float)
        result = y
        if not (isinstance(lmbda, list) or isinstance(lmbda, np.ndarray)):
            lmbda, y = np.broadcast_arrays(lmbda, y)
            lmbda = np.array(lmbda, dtype=float)
        l0 = np.abs(lmbda) > epsilon
        l2 = np.abs(lmbda - 2) > epsilon

        # Inverse
        with warnings.catch_warnings():  # suppress warnings
            warnings.simplefilter("ignore")
            if inverse is True:
                mask = np.where(((y >= 0) & l0) == True)
                result[mask] = np.power(np.multiply(y[mask], lmbda[mask]) + 1, 1 / lmbda[mask]) - 1

                mask = np.where(((y >= 0) & ~l0) == True)
                result[mask] = np.expm1(y[mask])

                mask = np.where(((y < 0) & l2) == True)
                result[mask] = 1 - np.power(np.multiply(-(2 - lmbda[mask]), y[mask]) + 1, 1 / (2 - lmbda[mask]))

                mask = np.where(((y < 0) & ~l2) == True)
                result[mask] = -np.expm1(-y[mask])

            # Derivative
            else:
                if derivative == 0:
                    mask = np.where(((y >= 0) & l0) == True)
                    result[mask] = np.divide(np.power(y[mask] + 1, lmbda[mask]) - 1, lmbda[mask])

                    mask = np.where(((y >= 0) & ~l0) == True)
                    result[mask] = np.log1p(y[mask])

                    mask = np.where(((y < 0) & l2) == True)
                    result[mask] = np.divide(-(np.power(-y[mask] + 1, 2 - lmbda[mask]) - 1), 2 - lmbda[mask])

                    mask = np.where(((y < 0) & ~l2) == True)
                    result[mask] = -np.log1p(-y[mask])

                # Not Derivative
                else:
                    p = self.fit(y, lmbda, derivative=derivative - 1, epsilon=epsilon, inverse=inverse)

                    mask = np.where(((y >= 0) & l0) == True)
                    result[mask] = np.divide(np.multiply(np.power(y[mask] + 1, lmbda[mask]),
                                                         np.power(np.log1p(y[mask]), derivative)) -
                                             np.multiply(derivative, p[mask]), lmbda[mask])

                    mask = np.where(((y >= 0) & ~l0) == True)
                    result[mask] = np.divide(np.power(np.log1p(y[mask]), derivative + 1), derivative + 1)

                    mask = np.where(((y < 0) & l2) == True)
                    result[mask] = np.divide(-(np.multiply(np.power(-y[mask] + 1, 2 - lmbda[mask]),
                                                                    np.power(-np.log1p(-y[mask]), derivative)) -
                                                        np.multiply(derivative, p[mask])), 2 - lmbda[mask])

                    mask = np.where(((y < 0) & ~l2) == True)
                    result[mask] = np.divide(np.power(-np.log1p(-y[mask]), derivative + 1), derivative + 1)

        return result

    @staticmethod
    def __validate(y, lmbda, derivative, epsilon, inverse):
        try:
            if not isinstance(y, (list, np.ndarray, pd.Series)):
                raise Exception("Argument 'y' must be a list!")
            if not isinstance(lmbda, (int, float, np.int, np.float)):
                if not isinstance(lmbda, (list, np.ndarray, pd.Series)) or len(lmbda) != len(y):
                    raise Exception("Argument 'lmbda' must be a number "
                                    "or a list, which its length matches 'y' argument!")
            if not isinstance(derivative, (int, float, np.int, np.float)) or derivative < 0:
                raise Exception("Argument 'derivative' must be a non-negative integer!")
            if not isinstance(epsilon, (int, float, np.int, np.float)) or epsilon <= 0:
                raise Exception("Argument 'epsilon' must be a positive number!")
            if not isinstance(inverse, bool):
                raise Exception("Argument 'inverse' must be boolean!")
            if inverse is True and derivative != 0:
                raise Exception("Argument 'derivative' must be zero "
                                "when argument 'inverse' is 'True'!")
        except ():
            sys.exit()

Could you tell me how to extract lambda from R using rpy2? This value should be somewhere in r_scale.rx('yj'), right?
Thanks!

Sorry @croessert, I was not monitoring the comments.

The Lambda value indicates the power to which all data should be raised to transform it to normal distribution.

You have three principal options:

1. Try different set of values for lambda, and assess the feature distribution (e.g. use Q-Q plot).
2. If your feature has positive values only, then use boxcox transformation to maximizes the log-likelihood function using a simple linear regression (y~1). Under the hood, box-cox searches for lambda from within a range of values.
3. Use an automated grid-search approach in combination with a normality test performed on the transformed data to determine optimal lambda, such that the p-value from the normality test is the highest. For instance, you can use "scipy.optimize.brute" and "scipy.stats.mstats.normaltest" functions to achieve that.

Some general notes:

• Always check for outliers and missing values, to make sure that some artifact is not driving the transformation.
• Consider the confidence interval around the optimal lambda, and whether a particular transformation makes sense.