similar to: Inverse BoxCox transformation

I've got a vector of data (hours to drive from a to b) y. After a qqplot I know, that they don't fit the normal probability. I would like to transform these data with the boxcox transformation (MASS), that they fit the model. When I try ybx<-boxcox(y~1,0) qqnorm(ybx) the plot is different from library (TeachingDemos) ybct<-bct(y,0) // qqnorm(ybct) How can I transform

Dear r-helpers, Prior to analysis of variance, I ran the Boxcox function (MASS library) to find the best power transformation of my data. However, reading the Boxcox help file, I cannot figure out if this function (through its associated log-likelihood function) corrects for * normality only * or if it also induces * homogeneity of variances *. I found in Biometry (Sokal and Rohlf, p. 419)

Dear R-helpers, Apologies in advance for this (probably) simple question. I've searched the R Archive and can't seem to find a solution to my problem. I have a data frame of vegetation quadrat data with the following format: Q S C 1 A 5 1 B 10 1 C 50 1 D 10 2 A 20 2 E 10 2 C 40 3 D 5 3 F 1 3 G 5 3 B 75 Where Q is the sample (vegetation quadrats), S is the species and C is the

dear members, I have a heteroscedastic time series which I want to transform to make it homoscedastic by a box cox transformation. I am using Otexts by RJ hyndman and George Athanopolous as my textbook. They discuss transformation and also say the fpp3 and the fable package automatically back transforms the point forecast. they also discuss the process which I find to be

Hi there, I wrote a function that wraps MASS::boxcox as: bc <- function(vec) { lam <- boxcox(lm(vec ~ 1)) lam <- lam$x[which.max(lam$y)] (vec^lam - 1)/lam } When I invoke it as: > x <- runif(20) > bc(x) Error in eval(predvars, data, env) : object 'vec' not found I have googled, and rewrote the above function as: bc <- function(vec) { dat <<-

Hi there, I wrote a function that wraps MASS::boxcox as: bc <- function(vec) { lam <- boxcox(lm(vec ~ 1)) lam <- lam$x[which.max(lam$y)] (vec^lam - 1)/lam } When I invoke it as: > x <- runif(20) > bc(x) Error in eval(predvars, data, env) : object 'vec' not found I have googled, and rewrote the above function as: bc <- function(vec) { dat <<-

Dear Ron and Bert, First (and without considering why one would want to do this, e.g., adding a start of 1 to the data), the following works for me: ------ snip ------ > library(MASS) > BoxCoxLambda <- function(z){ + b <- boxcox(z + 1 ~ 1, + lambda = seq(-5, 5, length.out = 101), + plotit = FALSE) + b$x[which.max(b$y)] + } > mrow <- 500

Hi, Thakns all for your help I am doing the next in my dataframe tabla, column pend1, because the Lilliefors (Kolmogorov-Smirnov) test give me a pvalue < alfa. (data no normal distribution). I need do a transformation with box-cox or other: > bc <- boxcox.fit(tabla$pend1) R send to me: Error in boxcox.fit(tabla$pend1) : Transformation requires positive data The summary for my data

Hi, I'm doing some work with linear models, and I've scaled my data using the scale(dataset) function. This was great at removing the skew, but I now can't perform the Box Cox transformation on the data set (using the boxcox(dataset) function), as the scaling has returned negative values. So my question is: how can I get the scale function to return a positive set of data (so I can

Hi, I am new to R. I need help with regards to box cox transformation. I have phenotypic data for e.g. plant height. data is non-normal. Skewness is 0.34. Could you please help me? Regards, Yogi -- View this message in context: http://r.789695.n4.nabble.com/boxcox-transformations-tp4682077.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]]

Hi, Firstly, apologies as I have posted this on community.rstudio.com too. I want to optimise a Box-Cox transformation on columns of a matrix (ie, a unique lambda for each column). So I wrote a function that includes the call to MASS::boxcox in order that it can be applied to each column easily. Except that I'm getting an error when calling the function. If I just extract a column of the

Thanks John. ?boxcox says: ************************* Arguments object a formula or fitted model object. Currently only lm and aov objects are handled. ************************* I read that as saying that boxcox(lm(z+1 ~ 1),...) should run without error. But it didn't. And perhaps here's why: BoxCoxLambda <- function(z){ b <- MASS:::boxcox.lm(lm(z+1 ~ 1), lambda = seq(-5, 5,

Hi Bert, On 2023-07-08 3:42 p.m., Bert Gunter wrote: > Caution: This email may have originated from outside the organization. Please exercise additional caution with any links and attachments. > > > Thanks John. > > ?boxcox says: > > ************************* > Arguments > > object > > a formula or fitted model object. Currently only lm and aov objects

No, I'm afraid I'm wrong. Something went wrong with my R session and gave me incorrect answers. After restarting, I continued to get the same error as you did with my supposed "fix." So just ignore what I said and sorry for the noise. -- Bert On Sat, Jul 8, 2023 at 8:28?AM Bert Gunter <bgunter.4567 at gmail.com> wrote: > Try this for your function: > >

Try this for your function: BoxCoxLambda <- function(z){ y <- z b <- boxcox(y + 1 ~ 1,lambda = seq(-5, 5, length.out = 61), plotit = FALSE) b$x[which.max(b$y)] # best lambda } ***I think*** (corrections and clarification strongly welcomed!) that `~` (the formula function) is looking for 'z' in the GlobalEnv, the caller of apply(), and not finding it. It finds

Hi, Hope this does not sound too ignorant . I am trying to detrend and transform variables to achieve normality and stationarity (for time series use, namely spectral analysis). I am using the boxcox transformations. As my dataset contains zeros, I found I need to add a constant to it in order to run "boxcox". I have ran tests adding several types of constants, from .0001

Help page for boxcox function in MASS library says that the transformation is y^lambda, which is different from the Y' = log(Y) if lambda = 0 , Y' = ((Y ^ lambda) - 1)/lambda otherwise I'm used to. Is this just a help page typo ? Thanks. ---------------------------------------------------------------- This message was sent using IMP, the Internet Messaging Program.

Greetings R-ians: I know what doesn?t work but I don?t know why, nor how to remedy things. I have a character string containing "." which I want to replace with " " gsub(".", " ", file.label) replaces the every character with a blank. However gsub(".xls", " ", file.label) replaces ".xls" with a blank as expected. It appears

Greetings, R-ians: I would like to know which version or R was used to create a given package. I think I remember seeing that topic discussed recently but cannot find it among my notes. Can anyone tell me how to determine which version of R created a package? Thanks. Charles Annis, P.E. <mailto:Charles.Annis@StatisticalEngineering.com> Charles.Annis@StatisticalEngineering.com

R-ians: After some effort I coerced my code to do what I want but my syntax is a kludge. Suggestions on more elegant syntax? par <- NIM.results$par do.call("Draw.NIM.POD.curve", list(par[1], par[2], par[3], par[4], par[5], a.hat.decision, .... et cetera ... It seems that I should be able to avoid defining the variable "par" and then specifying each of its elements,