similar to: matrix subscripts in replacement

Platform: WIN2000 Version of R: 1.6.2 I'm interested in plotting fitted values in a trellis xyplot. I believe the following should work; however, I only get the points (not the fitted lines). library(lattice) trellis.device(bg="white") xyplot(MULTDV~TIME|SUBNUM,data=TEMP, panel=function(x,y){ panel.xyplot(x,y) lines(x,fitted(lm(y~poly(x,1),na.action=na.omit)))

I apologize if this is a FAQ -- I kind of recall seeing something along these lines before, but I couldn't find the message when I searched the archives. Problem: 1. I have hundreds of small files in a subdirectory ("c:\\temp") and I would like to combine the files into a single data frame. 2. Individually, it is easy to read each file

Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,40000), var4 = c(100000,20000,30000,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1)))

Hi all, I have been trying to solve this problem and have had no luck so far. I have numeric vectors VAR1, VAR2, and VAR3 which I am trying to cbind. I also have a character vector "VAR1,VAR2,VAR3". How do I manipulate this character vector such that I can input a transformed version of the character vector into cbind and have it recognize that I'm trying to refer to my numeric

Buenas noches Javier y José, Estoy en contra de usar attach(), asi que propongo la siguiente alternativa con with(): # paquete require(epicalc) # los argumentos en ... pasan de epicalc:::cc # ver ?cc para mas informacion foo <- function(var1, var2, var3, ...){ or1 <- cc(var1, var2, ...) or2 <- cc(var1, var3, ...) list(or1 = or1, or2 = or2) } # datos x <-

Looking to the wonderful statistical advice that this group can offer. In behavioral science applications of stats, we are often introduced to coefficients for orthogonal polynomials that are nice integers. For instance, Kirk's experimental design book presents the following coefficients for p=4: Linear -3 -1 1 3 Quadratic 1 -1 -1 1 Cubic -1 3 -3 1 In R orthogonal

HI R help, I was trying to get identical data frame from a list using two methods. #Suppose my list is: listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2)) #Creating dataframe using cbind dat1<-data.frame(do.call("cbind",listdat1)) colnames(dat1)<-c("Var1","Var2","Var3") #Second dataframe conversion

Can someone please give me a clue how to 're'write this so I dont need to use loops. a<-ts(matrix(c(1,1,1,10,10,10,20,20,20),nrow=3),names=c('var1','var2','var3')) b<-ts(matrix(c(2,2,2,11,11,11,21,21,21),nrow=3),names=c('var1','var2','var3'))

Dear all, I am having trouble creating summary tables using aggregate function. given the following table: Var1 Var2 Var3 dummy S1 T1 I 1 S1 T1 I 1 S1 T1 D 1 S1 T1 D 1 S1 T2 I 1 S1 T2 I 1 S1 T2 D 1 S1 T2 D 1 S2

Hi, I have a table like that: > datatest var1 var2 var3 1 1 1 1 2 3 1 2 3 8 1 3 4 6 1 4 5 10 1 5 6 2 2 1 7 4 2 2 8 6 2 3 9 8 2 4 10 10 2 5 I need to create another table based on that with the rules: take a random sample by var2==1 (2 sample rows for example): var1 var2 var3 1 1

Hi, I am using R 1.9.1 on on i686 PC with SuSE Linux 9.0. I have a data.frame, e.g.: > myData <- data.frame( var1 = c( 1:4 ), var2 = c (5:8 ) ) If I add a new column by > myData$var3 <- myData[ , "var1" ] + myData[ , "var2" ] everything is fine, but if I omit the commas: > myData$var4 <- myData[ "var1" ] + myData[ "var2" ] the name

Hi your initial ds > str(ds) 'data.frame': 2 obs. of 3 variables: $ var1: num 1 1 $ var2: logi TRUE FALSE $ var3: logi NA NA first result > str(ds) 'data.frame': 2 obs. of 6 variables: $ var1 : num 1 1 $ var2 : logi TRUE FALSE $ var3 : logi NA NA $ value_and_logical: logi TRUE TRUE $ logical_and_na : logi TRUE NA

Carlos, en principio si sería algo así, sólo que en vez de quedarme con todas las columnas var1 a var4 tuviera sólo 3, ya que en mis datos no hay ningún caso que tenga el valor 1 en más de 3 variables.. Había llegado a una solución (mucho menos elegante que usando reshape), que implicaba un for sobre las filas. Jorge, creo que tu solución me vale. Muchas gracias a los dos.. Saludos El

Estimados Estoy entrenando hacer funciones que respondan a comandos, en esta caso en la salida gráfica se observa que dice : Exposure=var3 y outcome=var 1 quisiéramos que se reflejan los nombres de la base de datos : var1=estado, var2=cake, var3=chocolate Espero haberme explicado adecuadamente Adjunto tabla con datos #################################### #Comando que llama

Hello! # I have 3 vectors of values: values1<-rnorm(10) values2<-rnorm(10) values3<-rnorm(10) # In real life, all 3 vectors have a length of 25 # I create all possible combinations of 4 based on 10 elements: mycombos<-expand.grid(1:10,1:10,1:10,1:10) dim(mycombos) # Removing rows that contain pairs of identical values in any 2 of these columns: mycombos<-mycombos[!(mycombos$Var1

Gentlemen, I have googled, searched the mailing list archives, and even spoke on the IRC channel, but have not found an answer to the following problem. I am attempting to retrieve multiple columns in an ODBC query using ARRAY per the solutions offered by many individuals. My dialplan code is as follows: exten => _.,n,Set(ARRAY(var1,var2,var3)=${ODBC_LOOKUP(${KEYVAL})}) exten =>

Hi, this is a simple question I have this data.frame: > test <- data.frame(var1=c(1,1,1,1,1,1),var2=c("a","a","b","c","d","e"),var3=c("a1","a1","b1","a1","c1","d1")) > test var1 var2 var3 1 1 a a1 2 1 a a1 3 1 b b1 4 1 c a1 5 1

Dear R-Helpers Given a function like foo <- function(data,var1,var2,var3) { f <- formula(paste(var1,'~',paste(var2,var3,sep='+'),sep='')) linmod <- lm(f) return(linmod) } By typing foo(mydata,'a','b','c') I get the result of the linear model a~b+c. How can I rewrite the function so that the formula can be updated inside the function,

Dear community, I've plotted data and coloured depending on the factor variable v3. In the legend, I'd like to assign properly the same colors than in the factor (the factor has 5 levels). I've been trying this but it doesn't work. plot(var1, var2, xlab = "var1", ylab = "var2", col =var3 , bty='L') legend(locator(1),c("level 1 var3",

Dear list, I'm looking for an inverse function of melt(which is in package reshape).Namely, I had a data frame like this (Table1) YEAR VAR1 VAR2 VAR3 1995 7 3 45 1996 5 6 32 1997 6 10 15 I transformed my data by using the melt function and my data was reshaped in the following format: (Table2) YEAR variable