similar to: categorical variables

Hi, I'm doing kolmogorv-smirnov test but I don't know what conclusions to take, I want to know if my data has a normal distribution: > ks.test(dados1,"pnorm") One-sample Kolmogorov-Smirnov test data: dados1 D = 0.972, p-value < 2.2e-16 alternative hypothesis: two-sided Warning message: cannot compute correct p-values with ties in: ks.test(dados1,

Hola Agustin Creo que esto te podría valer Tiempos<-rnorm(100) Caudales<-rnorm(100) qmed<-0 #Tiempos<-c(1,3,5) #Caudales<-c(-1,1,-1) zeros<-function(tiempo,valor,qmed){ orden<-order(tiempo) tiempo<-tiempo[orden] valor<-valor[orden] zeros<-c() for(i in 1:(length(tiempo)-1)){ if((valor[i]-qmed)*(valor[i+1]-qmed)<0){

Hi all, I'm having this error, since I'm working with a data matrix I don't understand what's happening; I've tried several ways to solve this, even working with sparse matrix, but nothing seems to solve it, I've also tried svm (with a simple matrix 3*3 and still got the same error. > dados<-read.table("b.txt",sep="",nrows=30000) >

Hi, I have 35 data files for reading. I would like get a program for performing reading of 35 files at once. All are of the type: Dados1.raw, Dados2.raw and so on. If the files have the same number of columns, I can read with the following commands: rm(list=ls()) filenames = list.files(path="~/Silvano/Arq", pattern="Dados+.*raw") names = substr(filenames, 1, 7) for(i in

Buenas a todos, no encuentro la forma de hacer lo que quiero hacer, a ver si alguien me muestra el camino. La cuestión es así: estoy graficando una serie temporal en la cual tengo un único valor por mes del caudal de un río. La serie es de unos 4 años. Lo grafico con plot(..., type="l") y le agrego una abline() horizontal con un valor de caudal ficticio para mi serie temporal pero que

Dear R Community, I'm trying to write a loop to split my data into different series. I need to make a new matrix (or series) according to the series code. For instance, every time the "code" column assumes the value "433" I need to save "date", "value", and "code" into the "dados433" matrix. Please take a look at the following

Hi Second question in a day, i'm beginnning to feel incompetent... This time i'm having a weird problem, i'm importing the next data base: >car<-read.csv2("Historicos.csv") 'data.frame': 1818 obs. of 6 variables: $ Dpto : Factor w/ 11 levels "ANTIOQUIA","ATL?NTICO",..: 2 2 2 2 2 1 1 1 1 5 ... $ Rio : Factor w/ 43 levels

Hola buenos días, me presento, me llamo Miguel y 'soy de' y 'vivo en' Galicia. Soy profesor de secundaria (Bachillerato Adultos) y llevo 15 días estudiando R a un buen ritmo, pero todavía me faltan miles de cosas. He visto que R facilita, no solo el análisis de datos y que posee una potencia en cálculos estadísticos a cualquier nivel, sino gran caudal de recursos para Data Mining,

*Thanks* all those who took the time to help me (even if the "question" was not related to - the use of - R). Now I think I can soundly make my point w/ the referee (can I use your replies? If so I intend to properly cite its use?!?). Regards, Eduardo Esteves ps - Sorry for not explaining the "biological details" of my posting: RNA/DNA is the ratio of RNA content to

En relación con lo que comenta Carlos, por ejemplo para el caso de las Variaciones sin Repetición, puede ser instructivoenseñar como se construye como por ejemplo: VsinR <- function(m, n){ return (factorial(m)/factorial(m-n))} VsinR(9,3) ------------------------- Creo que con la función factorial que viene por defecto en R puedes construir siguiendo este modelo rápidadmentecasi cualquier

Hi, This is my first post in this page althougt i've been using R for several months (by the way english is not my native language so sorry about the grammar). I have a problem with a xyplot on lattice, it's supossed to print different lines for each river (Rio) conditioned by a chemical parameter (Var) and the state at which it's located (Dpto) against the year the sample were taken

Hola Miguel, Sí se pueden obtener las variaciones con y sin repetición en R. Eso sí están un poco escondidas... Se pueden calcular de esta forma: #---------------------- > #Cargar el paquete gtools > library(gtools) > #Definir el conjunto sobre el que se hará el cálculo > x <- c('rojo', 'azul', 'verde') > #Utilizar la función "permutations()"

Hi, I try to make a model using lmer, but the weigths is not accept. m1<-lmer(ocup/total~tempo+(tempo|estacao),family=binomial,weights=total) Erro em lmer(ocup/total ~ tempo + (tempo | estacao), family = binomial, : object `weights' of incorrect type I dont understand why this error, with glm this work. the total object is a vector. Any idea? Thanks Ronaldo -- God is subtle, but

Hi: I am going through some of the xtable examples and I can't make the one below work. I need to create a longtable on the fly keeping the column headers for all the pages and I thought this example could give some ideas on how to do it. I am using Sweave and xtable to create my tables and graphics. I wonder if someone could tell me what's wrong. Thanks ## Not run: \begin{small}

Sí, también... Para las permutaciones, n=r. Y con el parámetro "repeats.allowed" controlas si son con o sin repetción: #---------------------- > #Permutaciones *con repetición* > permutations(n=3, r=3, v=x, repeats.allowed=TRUE) [,1] [,2] [,3] [1,] "azul" "azul" "azul" [2,] "azul" "azul" "rojo" [3,]

Dear all, Does anyone knows how I could extract the p-value in: > survdiff(Surv(tempo,status) ~ grupo,data=dados1,rho=1) Call: survdiff(formula = Surv(tempo, status) ~ grupo, data = dados1, rho = 1) N Observed Expected (O-E)^2/E (O-E)^2/V grupo=1 21 5.12 12.00 3.94 14.5 grupo=2 21 14.55 7.68 6.16 14.5 Chisq= 14.5 on 1 degrees of freedom,

Hola amigos, muchas gracias por vuestra ayuda. Entonces veo que mi sorpresa era legítima. Por todos vuestros mails la conclusión es que: - En el módulo base de R no incluye combinatoria elemental, ni siquiera el número combinatorio Cm,n hay que cargar el paquete *combinat* - Y para las variaciones con repetición el paquete* gtools* - Y aún así no tenemos ni las combinaciones ni las

Hello, in the data.frame "resultsfuzzy" I would like to replace the characters in the second column ("5a", "5b", ... "5e") with numbers from 1 to 5. The data.frame has 39150 entries. I seems to work on samples that are << nrow(resultsfuzzy) but it takes suspicously long. Do you have any suggestions how to make the character replacing more

Hi all I am having trouble with my SMBFS and it is the following Every time I try to connect to other machine in my network, throught the command MOUNT, the folowing ERROR appears. I've already tried to see the manpage but i had not success. [root@backup_sp bin] mount -t smbfs //sarq/c /mnt/windows Password: ERROR: smbfs filesystem not supported by the kernel Please refer to the smbnt (8)

Hi, There's anyone here who go to "Estacao Voip" in Brazil??????? http://www.estacaovoip.com.br/ I was think to go.... Anyone here ?? -- Felipe Amaral Vento Livre Internet -------------- next part -------------- An HTML attachment was scrubbed... URL: http://lists.digium.com/pipermail/asterisk-users/attachments/20061109/7a64d327/attachment.htm