similar to: Potential bug in fitted.nls

Doesn't nls() expect that the lengths of vectors on both sides of the formula match (if both are supplied)? Perhaps it should check for that. -Bill On Thu, Jan 26, 2023 at 12:17 AM Dave Armstrong <darmst46 at uwo.ca> wrote: > Dear Colleagues, > > I recently answered [this question]() on StackOverflow that identified > what seems to be unusual behaviour with

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

On Tue, Oct 13, 2015 at 4:43 PM, Hadley Wickham <h.wickham at gmail.com> wrote: > Seems easy enough to write yourself: > > where <- function(x, env = parent.frame()) { > if (identical(env, emptyenv())) > return(NULL) > if (exists(x, envir = env, inherits = FALSE)) > return(env) > where(x, parent.env(env)) > } > > sample2 <-

Hi all, I am running into problems using `summary.lm` as calling `var` on a `units` object does not retain the `units` attribute. The units package addresses this issue for `mean` by implementing `mean.units`: units:::mean.units function (x, ...) { .as.units(NextMethod(), units(x)) } This is, however, not possible for `var` (as it's not a generic), which breaks `summary.lm`: ```

Gracias Emilio. Si, son asimétricos porque se mueven en rangos distintos. De hecho, miden cosas distintas, pero una muestra no puede ser positiva y negativa a la vez. Me planteo multiplicar las positivas por 2 y al editar el mapa corregirlo para que indique los valores reales. Eso funcionaría, pero son muchos mapas y quizás haya una función en ggplot que evite el problema directamente. Manuel El

No sé muy bien, Carlos, como hacerlo, pero no parece más sencillo que lo de multiplicar por 2 los positivos y después editar el mapa. De lo que decías, Emilio, de cambiar el color del punto central creo que haría más confusa la interpretación. El vie., 20 sept. 2019 a las 10:33, Carlos Ortega (<cof en qualityexcellence.es>) escribió: > Hola, > > Una alternativa que suele

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: > >

To David W.'s point about lack of a suitable reprex ("reproducible example"), Bill's solution seems to be for only one station. Here is a reprex and modification that I think does what was requested for multiple stations, again using base R and data frames, not dplyr and tibbles. First the reprex with **two** stations: > d <- data.frame( station =

I think a simple reproducible example ("reprex") may be necessary for you to get a useful reply. Questions with vague specifications such as yours often result in going round and round with attempts to clarify what you mean without a satisfactory answer. Clarification at the outset with a reprex may save you and others a lot of frustration. Cheers, Bert On Sat, Nov 4, 2023 at 1:41?AM

Bert wrote ... Bill's solution seems to be for only one station. No, it works for any number of stations. Bill Dunlap TIBCO Software wdunlap tibco.com On Sun, Nov 26, 2017 at 11:10 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > To David W.'s point about lack of a suitable reprex ("reproducible > example"), Bill's solution seems to be for only one

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

Dear All, Here is a reprex: > x<- 1:100 > x[-which(x>100)] integer(0) In words, I am finding out which indices correspond to values in x which are greater than 100 ( there are no such items ) . Then I remove those indices. I should get back the x that I started with since there are no items in x which are bigger than 100 . Instead, it is returning an empty vector. Why is this ?

try this: ########################################## library(dplyr) input <- tribble( ~station, ~from, ~to, ~record, "07EA001" , 1960 , 1960 , "QMS", "07EA001" , 1961 , 1970 , "QMC", "07EA001" , 1971 , 1971 , "QMM", "07EA001" , 1972 , 1976 , "QMC", "07EA001" , 1977 ,

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

Hi All, I realize that this is not a really nice reprex, but anyone has an idea why a background R session would "remember" an interrupt (SIGINT) on Unix? rs <- callr::r_session$new() rs$interrupt() # just sends a SIGINT #> [1] TRUE rs$run(function() 1+1) #> Error: interrupt rs$run(function() 1+1) #> [1] 2 It seems that the main loop somehow stores the SIGINT it

Before responding to Jeff's posting, let me reiterate my question: Why does a function using m1*m1 produce an integer overflow, but m1^2 does not? As for Jeff's 'response': > a) Numeric values may be either integers (signed 32 bit) or double precision (53 bit mantissa). > b) Double precision constants are numeric with no decoration (e.g. 61224). Integer constants have an L

a) Numeric values may be either integers (signed 32 bit) or double precision (53 bit mantissa). b) Double precision constants are numeric with no decoration (e.g. 61224). Integer constants have an L (e.g. 61224L). c) 61224*61224 > 2^31-1 so that answer cannot fit into an integer. d) Exponentiation is a floating point operation so the result of 61224L^2L is a floating point answer that CAN

On 11/26/2017 08:42 PM, jim holtman wrote: > try this: > > ########################################## > > library(dplyr) > > input <- tribble( > ~station, ~from, ~to, ~record, > "07EA001" , 1960 , 1960 , "QMS", > "07EA001" , 1961 , 1970 , "QMC", > "07EA001" , 1971 , 1971 ,