similar to: Getting an error calling MASS::boxcox in a function

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

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

Estimados Le consulto porque estoy intentando presentar unos datos de alguna forma un poco distinta, salir un poco de la costumbre. Primero describo los datos: Suponiendo una secuencia: golpe al auto -> torura de vidrio -> rotura de aciento otro auto golpe al auto -> -> rotura de aciento , a este por alguna causa no se le rompio el vidio o esa información es desconocida. Esta

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

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 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)

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, 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

Hi, I can't seem to find a function in R that will reverse a BoxCox transformation. Can somebody help me locate one please? Thanks in advance. Best wishes, Des [[alternative HTML version deleted]]

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]]

I am trying to model the observed leaching of wood preservative chemicals from treated wood during an outdoor experiment where leaching is caused by rainfall events. For each rainfall event, the amount of rainfall was recorded as well as the amount of preservative chemical leached. A number of climatic variables were measured, but the most important is the amount of rainfall. I have tried a

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.

Dear R People: In the DASL library, there is a story about hot dogs. Here are the data: Beef 186 495 Beef 181 477 Beef 176 425 Beef 149 322 Beef 184 482 Beef 190 587 Beef 158 370 Beef 139 322 Beef 175 479 Beef 148 375 Beef 152 330 Beef 111 300 Beef 141 386 Beef 153 401 Beef 190 645 Beef 157 440 Beef 131 317 Beef 149 319 Beef 135 298 Beef 132 253 Meat 173 458 Meat 191 506 Meat 182 473 Meat 190

Full_Name: Chris Hane Version: 2.10.1 OS: Windows Submission from: (NULL) (198.203.181.181) When using barchart in the flexcust package, setting mcol=NULL to avoid the lollipops causes an error. Each panel shows the text message "Error using packet n replacement has length zero." where n is the panel number. > data(iris) > cl <- cclust(iris[,-5], k=3) > barplot(cl,

Buenas tardes, He construido la función “myfun” al objeto de considerar aquellas persones que a partir de una determinada fecha de Apertura tienen como mínimo 65 años. Se tiene su fecha de nacimiento, su fecha de inicio en la institución y su fecha de salida de la misma. Doy vueltas al script y no acabo se saber cómo poder aplicar de un modo eficiente las instrucciones “if” ó bien “ifelse”, y me