similar to: Named rows in a table (data frame) read from a file

Buenas tardes a todos: Primero que nada un saludo y desearles un feliz 2009. Escribo para que me asesoren con la instalación de un editor de R, estube siguiendo los pasos de un manual 1. Instalar R 2. Instalar WinEdt 5 (V. 5.2 o superior) 3. Instalar *SWinRegistry *(disponible en http://www.omegahat.org/SWinRegistry): *Packages ! Install package(s) from local zip ¯les... * 4. Instalar

I'm studying lists and I came to an example where > L $name [1] "Fred" $wife [1] "Mary" $no.children [1] 4 $child.ages [1] 4 7 9 then following the instructions to extend the list with a new component, I executed: > L[5] <-list(NewName="something") and the new list I got was: > L $name [1] "Fred" $wife [1] "Mary"

All, I have some data on parasites on apple leaves and want to do a goodness of fit test to a Poisson distribution. This seems to do it: mites <- c(rep(0,70), rep(1,38), rep(2,17), rep(3,10), rep(4,9), rep(5,3), rep(6,2), rep(7,1)) tab <- table(mites) NSU <- length(mites) N <-

Dear all i?m working with a program i?ve made in R (using functions that others created) to run my program i need a sample. if i generate the sample using for example, rnorm(n, mu, sigma) i have no problem but if i obtain a sample from a column in excel and i copy it, the program says that there is a mistake: it says "Error en `[.data.frame`(data, indices) : undefined columns

I am running a binomial glm with response variable the no of mites of two species y->cbind(mitea,miteb) against two continuous variables (temperature and predatory mites) - see below. My model shows overdispersion as the residual deviance is 48.81 on 5 degrees of freedom. If I use quasibinomial to account for overdispersion the dispersion parameter estimate is 2501139, which seems

When I was studying the function cut I found this example: > x <- rep(0:8, tx0) > x [1] 0 0 0 0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5 5 5 5 5 5 5 5 6 [39] 6 6 6 6 7 7 7 8 8 8 8 8 > cut(x, b = 8) [1] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] [6] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (0.994,2] [11] (0.994,2]

Hello All, A colleague of mine started working with R and out of curiosity I did some research on the language. Very nice. In my opinion this is one of the best languages I've found for getting tasks I'm interested in done. I wrote this simple die roller and was curious to know if it is R enough. ############################################################################## #

I thought I've understood the 'order' function, using simple examples like: order(c(5,4,-2)) [1] 3 2 1 However, I arrived to the following example: order(c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045)) [1] 8 9 10 7 11 6 5 4 3 2 1 and I was completely perplexed! Shouldn't the output vector be 11 10 9 8 7 6 4 1 2 3 5 ? Do I have a damaged

I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply

No soy experto en data mining con R, sé que existe algún paquete al respecto, pero no sé si R sería lo más adecuado para cantidades ingentes de datos… Saludos, Juan Carmona. De: Eduardo Bieñkowski [mailto:edukoski@gmail.com] Enviado el: miércoles, 04 de junio de 2014 17:07 Para: Joan Carmona CC: Isidro Hidalgo; r-help-es Asunto: Re: [R-es] Cargar csv de 16GB en R Si pero que pasaría

Hello All: Using the below dataset how can I make a barplot with Date(X) and NumEggs(Y) by Site. Then plot Temp(lineplot) It seems really simple, but I am having a hard time trying to do it by Site. Thanks Date NumEggs Site Temp 1 2008-04-22 0 Massacre Flat (RK424.5) 51.20 2 2008-04-23 0 Massacre Flat (RK424.5) 50.80 3 2008-04-24

I'm trying to figure out about factors, however the on-line documentation is rather sparse. I guess, factors are intended for grouping arrays members into categories, which R names "Levels". And so we have: * state <- c("tas", "sa", "qld", "nsw", "nsw", "nt", "wa", "wa",

Is there any way to issue operating system commands and geting back the results, in R? I mean, for instance, in Linux, to execute from R the 'ls' command and getting back a list of files in the current directory, or, equivalently, in Windows/DOS, the 'dir' command? I'm not interested in the 'ls' or 'dir' commands it is just an example. Do you have any

I'm trying to read a data file that contains characters from the Spanish language: > Station <- read.fwf("LosDatos.txt",widths=c(7,7,25,8,8,5),header=FALSE, + skip=3,n=separ[1]-4) Then the R interpreter issues the following message: Error en substring(x, first, last) : invalid multibyte string at '<d1>A, S.' Calls: read.fwf ->

I'm trying to write a Rscript program capable of reading several tables from the standard input. The problem is that the tables aren't in files because they are coming from another process that is generating them. From the R-console the following works pretty well: |> f <- stdin() |> t <- read.table(f) |> t2 <- read.table(f) |> t uno dos 01 3 4

I'm trying to produce a series of powers of a number as follows: |> 0.05^0:5 [1] 1 2 3 4 5 This is not the result I expected. I guess some kind of coercion happened, since, |> class(0.05^0:5) [1] "integer" Could anyone explain me what is happening here? Thanks, -Sergio.

I have an ordered "set" of numbers, represented by a vector, say > X <- c(10:13, 17,18) > X [1] 10 11 12 13 17 18 then I have a "sub-set" of X, say > Y <- c(11,12,17,18) Is there a simple way in R to have a logical vector (parallel to X) indicating what elements of X are in Y, i.e., > Inclusion [1] FALSE TRUE TRUE FALSE TRUE TRUE I'm

I try to display the grouping variable in coplot. It work, but it's special solution and rather ugly. Any better idea? # Simulate my data frame data(state) x77<-data.frame(state.x77) x77$region<-state.region coplot(Life.Exp ~ Income | region, data=x77, show.given=F, subscripts=T, panel = function(x, y,subscripts, ...) { panel.smooth(x, y, span = 1., ...)

On 6/15/20 6:19 AM, Leon Fauster via CentOS wrote: > Am 15.06.20 um 05:38 schrieb Strahil Nikolov via CentOS: >> Working with different Linux Distributions makes the life harder. >> So far I have found out that 'poweroff' & 'reboot' has the same >> behaviour on? Linux/Unix/BSDs. >> > > Yeah, poweroff seems the appropriate command instead of

I learnt that functions can be handled as objects, the same way the variables are. So, the following is perfectly valid: > f = function(a, b) { + print(a) + print(b) + } > > f1 = function(foo) { + foo(1,2) + } > > f1(f) [1] 1 [1] 2 > I also know that operators are functions, so, I can call: > '+'(1,2) [1] 3 > However, when I want to pass the