similar to: recursive lapply and keeping track of data

You could replace your 'depth' argument with one that shows where in the original data you are at: leaf.func <- function(data, where) { if(is.null(data)) stop("Null data at ", deparse(where)) return(mean(data)) } visit.level <- function(data, where = integer()) { if (length(where) == 2) { return(leaf.func(data, where)) } else

Hello, I'm having troubles defining a list where names are variables (of type character). Like this, which gives "foo" instead of "world" (the way I meant it is that "world" is the value of the variable foo). Any hint? > f <- function(foo, bar) { list(foo = bar) } > x <- f("hello", "world") > names(x) [1] "foo"

Hi Giovani, I would create an unnamed list and set the names after. Best, Ulrik On Fri, 4 Aug 2017 at 12:08 Giovanni Gherdovich <g.gherdovich at gmail.com> wrote: > Hello, > > I'm having troubles defining a list where names are variables (of type > character). Like this, which gives "foo" instead of "world" (the way I > meant it is that

Do you mean like this? > f <- function(foo, bar) { + result <- list(bar) + names(result) <- foo + result + } > (x <- f("hello", "world")) $hello [1] "world" > names(x) [1] "hello" -- Thomas Mailund On 4 August 2017 at 12.08.28, Giovanni Gherdovich (g.gherdovich at gmail.com) wrote: Hello, I'm having troubles defining

You can wrap the list-creating function call (e.g. lapply) in a call to ?setNames, or you can use the ?map function from the purrr package. -- Sent from my phone. Please excuse my brevity. On August 4, 2017 3:14:44 AM PDT, Ulrik Stervbo <ulrik.stervbo at gmail.com> wrote: >Hi Giovani, > >I would create an unnamed list and set the names after. > >Best, >Ulrik > >On

Hello Thomas, Ulrik, thanks for your suggestions. Giovanni On Fri, Aug 4, 2017 at 12:13 PM, Thomas Mailund <thomas.mailund at gmail.com> wrote: > Do you mean like this? > > >> f <- function(foo, bar) { > + result <- list(bar) > + names(result) <- foo > + result > + } > >> (x <- f("hello", "world")) > $hello >

Hello: When I use SPSS I execute the AGGREGATE DATA comand for the next data: 2112141123212213212213334 3143244113442312121213344 2114141123112214212113344 2112211122212413421213221 3114444123442414343413344 2312231223212222323223322 2143241123212313131213234 2113241113212313222213333 2113141123112214212113344 2114141123412111114413344 2113211122342314222313234 2114141123112414212113344

Dear List, the advent of multicore machines in the consumer segment makes me wonder whether it would, at least in principle, be possible to divide a computational task into more slave R processes running on the different cores of the same processor, more or less in the way package SNOW would do on a cluster. I am thinking of simple 'embarassingly parallel' problems, just like inverting

Hello. I am having a problem setting up a self-starting function for use in nonlinear regression (and eventually in the mixed model version). The function is a non-rectangular hyperbola - called "NRhyperbola" - which is used for fitting leaf photosynthetic rate to light intensity. It has one independent variable (Irr) and four parameters (theta, Am, alpha and Rd). I have created this

Buenas a todos, Ayer realizamos un Meetup conjunto con el grupo de "Machine Learning Spain". Parte del material ya está subido a nuestro portal. http://madrid.r-es.org/50-miercoles-14-de-marzo-2018/ Estamos a la espera de que nos compartan los videos. Por si queréis ir viendo la parte de Santiago, mi parte es la misma que la que presenté en la reunión del Grupo hace unas semanas.

> f <- function(foo,bar) structure(list(bar),names =foo) > f("hello","world") $hello [1] "world" Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Fri, Aug 4, 2017 at 6:12 AM, Jeff

Hello, just a question, I can't see a core "release-2.3.6" branch, nor a pigeonhole "release-0.5.6" branch on the GIT repository, it is outdated? or do you have changed your policy to create a new branch for each new release of dovecot and pigeonhole? Thanks and regards -- +-------------------------------------------------------------------+ | Juan C. Blanco

Gracias Carlos. No uso caret, pero lo miraré. Quoting Carlos Ortega <cof en qualityexcellence.es>: > Hola, > > Creo que si utilizas "caret" y en la función "trainControl()" defines "oob" > como criterio de randomización, puedes luego recuperar del objeto del > modelo, las predicciones individuales... > > Saludos, > Carlos Ortega >

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hello: I'm having problems with this line of code: X.lm <- lapply(names(d), function(x) lm(d["cls"] ~ d[x], data=d)) d[x] is what is giving trouble here, but I don't know exactly how to solve it. What I'm trying to do is to create a linear model from each column of the data frame 'd' to apply ANOVA later. Thanks

OK, thanks!! Juan C. Blanco On 06/05/2019 10:25, Aki Tuomi wrote: > Hi , > > sorry they were accidentically not pushed. Will do so soon. > > Aki >> On 6 May 2019 11:11 Juan C. Blanco via dovecot < dovecot at dovecot.org >> <mailto:dovecot at dovecot.org>> wrote: >> >> >> Hello, just a question, I can't see a core

Gracias Jorge. No entiendo bien; la variable objetivo es ya factor. El árbol me la predice bien, como factor, también. Es al ir construyendo el vector que lo anota con un nº, según de cuál de las 4 categorías se trate. Quoting Jorge I Velez <jorgeivanvelez en gmail.com>: > Estimado Manuel, > > Debes definir ecsta como factor usando, por ejemplo, > > factor(ecsta,

Bueno, creo que no contesté tu pregunta. Con training <- data[-i, ] crea una df llamada training, sin la muestra i, que después utiliza para entrenar el algoritmo. Quoting Javier Marcuzzi <javier.ruben.marcuzzi en gmail.com>: > Estimado Manuel Mendoza > > Con sus datos y a modo de curiosidad, ¿que pasa en training <- data[-i, ]? > > Javier Rubén Marcuzzi >

Gracias Carlos J., sale bien, pero me transforma las 6 categorías en números del 1 al 6 ¿sabes cómo evitarlo? Quoting "Carlos J. Gil Bellosta" <cgb en datanalytics.com>: > apply(data, 1, function(x) which.max(table(x))) > > El sáb., 14 abr. 2018 a las 19:54, Manuel Mendoza (<mmendoza en mncn.csic.es>) > escribió: > >> >> Buenas tardes de

Hi everyone, I am interested in estimating this type of random effects panel: y_it = x'_it * beta + u_it + e_it u_it = rho * u_it-1 + d_it rho belongs to (-1, 1) where: u and e are independent and normally zero-mean distributed. d is also independently normally zero-mean distributed. So, I want random effects for group i to be correlated in t, following an AR(1) process. I am

Hola de nuevo. Se me olvidaba la principal razón para utilizar gbm.step del paquete dismo. Como sabéis, los boosted si sobreajustan (a diferencia de los random forest o cualquier otro bootstrap) pero gbm.step hace validación cruzada para determinar el nº óptimo de árboles y evitarlo. Es fundamental. La opción que me queda, Carlos, es hacerlo con gbm, pero muchas veces, y usar el