similar to: Regressão linear

Hi! With the following script, I'm trying to make a demonstration of a Confidence Interval, but I'm observing some differences on tails. # Teste de média entre uma amostra e uma população normal # Autor: Raphael de Freitas Saldanha # Agosto de 2008 n <- 200 # Sample size xbar <- 100 # Sample mean s <- 2 # Sample SD nc <- 0.95 # Confidence level (95%

Hola a todos. Tengo un proyecto entre manos que consiste en lo siguiente: Un suario manda a un correo electrónico un mail con un fichero adjunto, este fichero se descarga de forma automática en una carpeta de un sistema que corre bajo Linux (Ubuntu), una vez descargado quisiera generar un script de R de forma automática seleccionase el último fichero descargado e hiciese con él una serie de

Hi, I'm trying to use the function "error.bars", but the program don't find it, and I dont't found any package with this function. Is there some another functin to draw barplots with error bars? Sueli Rodrigues Eng. Agr?noma - UNESP Mestranda - USP/ESALQ PPG-Solos e Nutri??o de Plantas Fones (19)93442981 (19)33719762

Hi, I have the following file, and I need to work out the linear regression for each sample. I tried the model(*) and receive the error message (**): > data=split(mydata,rep(1:(nrow(mydata)/6),each=6)) > arrang.linear=lapply(data,lm,formula=KA~PA) Erro em storage.mode(y) <- "double" : invalid to change the storage mode of a factor Al?m disso: Warning message: In

Oi F?bio. Pelo IP tamb?m n?o acessa. A minha suspeita de que era um problema ligado a interface est? errada. Aleat?riamente usu?rios nas duas interfaces perdem o acesso. Ainda assim, somente usu?rios com Windows XP tem esse problema. Vou testar suas indica??es. F?bio Rabelo wrote: > > > 2010/3/23 Leonardo Carneiro - Veltrac <lscarneiro at veltrac.com.br > <mailto:lscarneiro

Hello again! :-) Mine he doubts he is with relation to smbpasswd, would like to add to an user and a password standard from one script (*in botton), but I did not obtain to eliminate prompt requesting the new password and password verification, like make in useradd. This is possible? I use Red Hat 8.0 box with a Samba 3.0 Alpha running a PDC. Me he seems that alone it functions compiling with

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Hi, When i put the command pdbedit plus any user name show the message user not found. Only user root is exibited. Why the others not show? the service smbd winb nmbd slapd is running without problem. Example *root at AbramisSRV smbldap-tools]# pdbedit -Lv teste* Username not found! *[root at AbramisSRV smbldap-tools]# pdbedit -Lv root* Unix username: root NT username: root

Dear R users, I?m trying to optimize simultaneously two binomials inequalities used to acceptance sampling, which are nonlinear solution, so there is no simple direct solution. The 'n' represents the sample size and the 'c' an acceptance number or maximum number of defects (nonconforming) in sample size. The objective is to obtain the smallest value of 'n' (sample size)

Pessoal, Bom dia a Todos!!! Sou novo aqui na lista e tamb?m no uso do Asterisk. Estou implementando um ambiente de laborat?rio para entender melhor todo a quest?o de uso e configura??o da ferramenta. Bom o meu problema esta no seguinte:--------------------------------------------------- Tenho uma placa FXO com 4 canais anal?gicos (Troco), onde estou conectando minhas 4 linhas da anal?gica da

Dear all, This is the first time I am sending mail to the mailing list, so I hope I do not make a mistake... The last months I have been working on my MSc thesis project on performing data mining techniques on user logs of a software-as-a-service application. The main problem I am experiencing is how to process the huge amount of data. More specifically: I am using R 2.10.1 in a laptop with

Hi. How do I plot the straight line and r? in a scatter plot using a simple file x~y? Sueli Rodrigues Eng. Agr?noma - UNESP Mestranda - USP/ESALQ PPG-Solos e Nutri??o de Plantas Fones (19)93442981 (19)33719762

Hi R users, I have a question about fitting a cosine curve. I don't know how to set the approximate starting values. Besides, does the method work for sine curve as well? Thanks. Part of the dataset is in the following: y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96,

Using a more stable nonlinear modeling tool will also help, but key is to get the periodicity right. y=c(16.82, 16.72, 16.63, 16.47, 16.84, 16.25, 16.15, 16.83, 17.41, 17.67, 17.62, 17.81, 17.91, 17.85, 17.70, 17.67, 17.45, 17.58, 16.99, 17.10) t=c(7, 37, 58, 79, 96, 110, 114, 127, 146, 156, 161, 169, 176, 182, 190, 197, 209, 218, 232, 240) lidata <- data.frame(y=y, t=t) #I use the

If you know the period and want to fit phase and amplitude, this is equivalent to fitting a * sin + b * cos > >>> > I don't know how to set the approximate starting values. I'm not sure what you meant by that, but I suspect it's related to phase and amplitude. > >>> > Besides, does the method work for sine curve as well? sin is the same as cos with

I'm trying the different parameters, but don't know what the error is: Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates Thanks for any suggestions. On Tue, Jun 20, 2017 at 7:37 PM, Don Cohen <don-r-help at isis.cs3-inc.com> wrote: > > If you know the period and want to fit phase and amplitude, this is > equivalent to

Hi lily, You can get fairly good starting values just by eyeballing the curves: plot(y) lines(supsmu(1:20,y)) lines(0.6*cos((1:20)/3+0.6*pi)+17.2) Jim On Wed, Jun 21, 2017 at 9:17 AM, lily li <chocold12 at gmail.com> wrote: > Hi R users, > > I have a question about fitting a cosine curve. I don't know how to set the > approximate starting values. Besides, does the method

Hi, Your mistake is related with "read.table(n)". I haven`t got what you want but I put an example below (where I used loop to read many files). Hope it helps. Anyway, I thougth should have sent your question to <r-help@r-project.org>. Bye Nilza =====example ==================== nomesout <- dir(dat.dir,pattern="^[s]") #obtem no diretorio de atual todos arquivo

Thanks to all. I summarized (in order to thank the list) the solutions to help future workers searching subjects like this at R help.   # Number of rows nr = 10 # Data set dataf = as.data.frame(matrix(c(rnorm(nr),rnorm(nr)*2,runif(nr),sort(c(1,1,2,2,3,3,sample(1:3,nr-6,replace=TRUE)))),ncol=4)) names(dataf)[4] = "class" #----------------------------------------------------- #Solution 1:

I am trying to use lm for a simple linear fit with weights. The results I get from IDL (which I am more familiar with) seem correct and intuitive, but the "lm" function in R gives outputs that seem strange to me. Unweighted case: > x<-1:4 > y<-(1:4)^2 > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: 1 2 3 4 1 -1 -1 1 Coefficients: