similar to: Thanks to Mark, Henrique, and Jorge

I am writing a simple R program to execute a t-test repeatedly on data contained in a data frame. My data looks like this: Category Value1 Value2 1 .5 .8 1 .3 .9 . . . . . . . . . 2

Dear list, I came up with a two functions that flatten arbitrary deeply nested lists (as long as they're named; not tested for unnamed) and environments (see attachment; 'flatten_examples.txt' contains some examples). The paradigm is somewhat similar to that implemented in 'unlist()', yet extends it. I would have very much liked to build upon the superfast functionality

Daniel, en vez de mean utilicé table. Entonces me genera una lista y quiero que quede una matriz. Unlist no me sirve porque me pasa todos los valores a un vector y me pierdo de saber a que caso corresponde. Saludos, Sebastián. El 26 de septiembre de 2013 16:50, daniel <daniel319@gmail.com> escribió: > Sebastián, tengo dificultades para entender si sigues con el ejemplo > anterior

It looks from str(SA) that Response IPS1 is a data.frame of class "anova", which probably cannot be coerced to vector. Maybe you can use unlist() instead of as.vector() Or something like SA[["Response IPS1"]]["as.factor(WSD)",] ## to select the first row only, even maybe with unlist() Without a better REPRODUCIBLE example, I cannot tell more (maybe some others

Hi All, I don't know if this has been reported before, but according to Henrique Dallazuanna's program (below) the number of R packages has exceeded the 3,000 mark. The count today is 3,175. I ran this just a couple of months ago & the number was still in the high 2,000s, so it must be fairly recent. I think this represents about 50% growth in the last year. Not bad! Does

Sebastián, tengo dificultades para entender si sigues con el ejemplo anterior o esto es algo nuevo. Tú descripción se debe entender que dentro de aggregate(cbind(X1,X2) ~ B + C, t, function(x) mean(x < 2)) utilizas un table? Aparentemente ese no es el problema sino que el resultado que obtienes es una lista, ¿te sirve ?unlist ? De cualquier manera no estoy frente a tu ordenador y solo estoy

Muchas gracias Jorge. La función va bien, pero las columnas las completa con NAs y no con un dígito (99 o cualquier otro, pero dígito). Saludos, Manuel On Wed, 31 Aug 2011 12:00:11 +0200, r-help-es-request@r-project.org wrote: > Envíe los mensajes para la lista R-help-es a > r-help-es@r-project.org [3] > > Para subscribirse o anular su subscripción a través de la WEB >

Hello Everyone, I have recently been introduced to the ggplot package by Hadley Wickham and must say I am quite impressed so far at how easy it is to make attractive plots, but one thing I am struggling over is how to consolidate legends. I have 3 plots that I would like to put on a single page and all 3 map the same dimension of the data to the colour aesthetic. Right now, when I plot all

Ryan, from the console what does "zap show channel 1" or 2/3/4 in your case say. I have X100P's and I seem to be having similar sounding problems. I noticed that the above command shows the channel to be off-hook at all times when a phone line is plugged in. I don't know why or if it is a bug in the application reporting the status. dbc. Ryan Courtnage wrote: > On July 8,

Thank you very much for your useful suggestions. These are exactly what I was looking for. foo <- list(foo1, foo2, foo3) lapply(foo, function(x) matrix(unlist(x), nrow = length(x), byrow = TRUE)) or lapply(foo, function(x) do.call('rbind', x)) Best, Muhammad Subianto On 4/11/06, Muhammad Subianto <msubianto at gmail.com> wrote: > Dear all, > I have a result my experiment

I am not sure if something is wrong with my programming or is a bug of chron or something else. In the following script, db is a large dataframe (dim(db)=c(60698,14)), then I select a very small part for a specific date and compute some basic statistics. date.base is a chron object foo<-function(db,date.base){ date.base<-as.numeric(date.base) #convert to number

I don't know what do you (devels) fixed (maybe that what i read in WWN 287), but alsa works NICE on my system. So, I step in here to wish you luck with the project. I'm deeply impressed that this works :P -- Pawe? R??a?ski

Hi, I'm new to R and would like to know, how one can populate the list with array data. I'm reading a tab separated table in R. The data in the table looks something like this. #Table Data Comp    A    B    C Extracellular    103    268    535759 Nucleus    45603    47783    442744 #R code myData <- read.table("table.data",                 header=T,                

Congratulations for the baby !! I would like to thank those lots of people who have made such a nice work and who have been so generous with your efforts. Jorge Luis Ojeda Cabrera Dep. M?todosEstad?sticos Fac. De Ciencias, U. de Zaragoza Pedro Crbuna,12 50009 Zaragoza Espa?a -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

How do I concatenate two vectors of factors? --8<---------------cut here---------------start------------->8--- > a <- factor(5:1,levels=1:9) > b <- factor(9:1,levels=1:9) > str(c(a,b)) int [1:14] 5 4 3 2 1 9 8 7 6 5 ... > str(unlist(list(a,b),use.names=FALSE)) Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...

Hola, primero de todo muchas gracias. He probado a ejecutar el programa en una máquina virtual y efectivamente funciona. La única diferencia con mi ordenador está en locale, que me aparece esto: > sessionInfor() locale: [1] LC_CTYPE=es_ES.UTF-8 LC_NUMERIC=C [3] LC_TIME=es_ES.UTF-8 LC_COLLATE=es_ES.UTF-8 [5] LC_MONETARY=es_ES.UTF-8

Hi there, I have a problem about lapply, strsplit, and accessing list elements, which I don't understand or cannot solve: I have e.g. a character vector with three elements: x = c("349/077,349/074,349/100,349/117", "340/384.2,340/513,367/139,455/128,D13/168", "600/437,128/903,128/904") The task I want to perform, is to generate a list,

Copiando tus códigos llego a esto (adjunto) En mi caso veo bien las fórmulas excepto en el índice. Estas son mis versiones de los paquetes: R version 3.0.2 (2013-09-25) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=Spanish_Spain.1252 LC_CTYPE=Spanish_Spain.1252 [3] LC_MONETARY=Spanish_Spain.1252 LC_NUMERIC=C [5] LC_TIME=Spanish_Spain.1252 attached base packages: [1] stats

Thanks for your help, Rui! That works and will save me a lot of trouble. --Kelly -----Original Message----- From: Rui Barradas [mailto:ruipbarradas at sapo.pt] Sent: Tuesday, July 10, 2012 2:24 AM To: Vining, Kelly Cc: r-help at r-project.org Subject: Re: [R] boxplot with "cut" Hello, Maybe this iss what you're looking for. GD is your data.frame. multi.boxplot <-

I have a nested list l like: l <- list(A=c(1,2,3), B=c("a", "b")) l <- list(l,l, list(l,l)) I want the list to be unlisted, but not on the lowest level of each "branch". I want the lowest level of each list branch to remain as it is. So unlist or unlist(rec=F) do not work here as the level of nesting may differ on the elements. The result should look like: