similar to: boxcox transformations

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

t<--4:4 y<-c(5,7,10,13,15,16,14,12,11) plot(t,y,type="l") how can i add a curve y=0.83*t-0.44*t^2 in the graph? [[alternative HTML version deleted]]

Hi there! My problem is the following: I am experimenting with virtualization with XEN. I am running WinXP in a HVM host with a Debian Gnu/Linux (Etch) dom0. I am also using audio emulation (sb16). This way, sounds from WinXP work fine. And here comes the problem: The HVM guest completely reserves the audio device; other software can not use it. (mplayer says this: alsa-init: using ALSA

Hi, I tried with library(MASS) methods('boxcox') # [1] boxcox.default* boxcox.formula* boxcox.lm* Then > boxcox.lm Error: object 'boxcox.lm' not found How to look into the codes of boxcox.lm? Thanks for your time [[alternative HTML version deleted]]

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)

Hmm, this looks like a buglet/infelicity in update.lm rather than MASS::boxcox per se. Moving to R-devel. I think the story is that update.lm eventually does eval(call, parent.frame()) where the call is extracted from the lm object, but call$formula is unevaluated, and does not contain environment information like formula(obj) would do. Then when the call is evaluated and parent.frame()

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

Hello, I am implementing the BoxCox transformation for multiple regression using the function boxcox() from library MASS (this seems to be the best one). To avoid negative values in the variable to be transformed however, one must specify a shift parameter. However different choices for the value of the shift parameter result in different estimates for the exponent (lambda) and in different

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 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, 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.

Hi all I?ve got a problem using boxcox.nls function in nlrwr packagge. I?m fitting several non linear models to these data: > x [1] 2 1 1 5 4 6 13 11 13 101 101 101 > y [1] 1.281055090 1.563609934 0.001570796 2.291579783 0.841891853 [6] 6.553951324 14.243274230 14.519899320 15.066473610 21.728809880 [11] 18.553054450 23.722637370 I used nls function with self

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 <<-

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

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

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: > >

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