Solved – Determine when time-series should be logged (or any other transformation) and applied automatically

by

Is there any way to test whether a series should be logged or transformed in another way?

I have a code of which i use to run lots of different data through to forecast. Some of the data definitely need transforming however some don't. As the code has been written to be fully automatic it will be used by non-statisticians within the company so they will have no idea whether they should change the code to transform the data depending on the series. So i need tests which will check that for them and apply the transformation accordingly.

Here is a example data set that you can use:

M <- matrix(c("08Q1", "08Q2", "08Q3", "08Q4", "09Q1", "09Q2", "09Q3", "09Q4", "10Q1", "10Q2", "10Q3", "10Q4", "11Q1", "11Q2", "11Q3", "11Q4", "12Q1", "12Q2", "12Q3", "12Q4", "13Q1", "13Q2", "13Q3", "13Q4", "14Q1", "14Q2", "14Q3",  5403.676,  6773.505,  7231.117,  7835.552,  5236.710, 5526.619,  6555.782, 11464.727,  7210.069,  7501.610,  8670.903, 10872.935,  8209.023,  8153.393, 10196.448, 13244.502,  8356.733, 10188.442, 10601.322, 12617.821, 11786.526, 10044.987, 11006.005, 15101.946, 10992.273, 11421.189, 10731.312),ncol=2,byrow=FALSE) Nu <- M[, length(M[1,])]

I have found boxcoxfit() from the package geoR finds the lambda for transformation….does anyone know how accurate this is for transforming the data?

ml <- boxcoxfit(Nu)   Fitted parameters:     lambda    beta  sigmasq       0.59  375.43  3649.39 N<- ((Nu^(ml$lambda))-1)/ml$lambda
Contents hide
1 Best Answer
2 Similar Posts:

Best Answer

As @Irishstat points out you could use boxcox power transformation, which is a more general transformation function which also includes log transformation. R's forecast package has a function called BoxCox.lambda and BoxCox, you could use these two functions and determine if your data needs transformation. if lambda is close to 1 then your data needs no transformation, else you data needs appropriate power transformation.

Using your data

x <- ts(c(5403.676,  6773.505,  7231.117,  7835.552,  5236.710, 5526.619,  6555.782, 11464.727,  7210.069,  7501.610,  8670.903, 10872.935,  8209.023,  8153.393, 10196.448, 13244.502,  8356.733, 10188.442, 10601.322, 12617.821, 11786.526, 10044.987, 11006.005, 15101.946, 10992.273, 11421.189, 10731.312),frequency =4)  lambda <- BoxCox.lambda(x, method=c("guerrero")) lambda 0.3855427  x.transform <- BoxCox(x,lambda) plot(x.transform)

Using the box.cox lambda yielded a lambda value 0.3855. You could use this in the BoxCox function as shown above.

Let us know if you find this post useful.

Similar Posts:

Rate this post

Leave a Comment