similar to: Putting column names in some automated way

Hi, I have one question on expand.grid() function. When I write following syntax :expand.grid(c("u", "l"), c("u", "l"), c("u", "l")) I get following as desired : Var1 Var2 Var3 1 u u u 2 l u u 3 u l u 4 l l u 5 u u l 6 l u l 7 u l l 8 l l l However

Dear list, I'm trying to concatenate the values of two columns but im not able to do it: i have a dataframe with the following two columns: X VAR1 VAR2 1 2 2 1 3 2 4 3 5 4 6 4 what i would like to

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

Hello I have a data frame d with columns "var1", "var2", "var3" Then I have two vectors: columns <- c("var2", "var3") values <- c(0, 1) Is there a compact way to subset the data frame using these two vectors and get the result equivalent to: select (d, var2==0 & var3==1) ? Thank You Blazej

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 <-

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

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

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'))

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

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

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

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",

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

Buenos días. Hoy ando un poco (o bastante) espeso y no doy con la tecla de una cosa que seguro que es muy simple.. Pongo un ejemplo. var1 <- c(rep(0,3),rep(1,2)) var2 <- c(rep(1,2),0,0,1) var3 <- c(rep(1,2),rep(0,3)) var4 <- c(rep(1,2),rep(0,3)) datos <- data.frame(fila=1:5,var1, var2, var3, var4) datos datos fila var1 var2 var3 var4 1 1 0 1 1 1 2 2 0

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

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

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 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

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

Dear R community members I have been struggling on this simple question, but never get appropriate solution. So please help. # my data, though I have a large number of variables var1 <- rnorm(500, 10,4) var2 <- rnorm(500, 20, 8) var3 <- rnorm(500, 30, 18) var4 <- rnorm(500, 40, 20) datafr1 <- data.frame(var1, var2, var3, var4) # my unsuccessful codes nvar <- ncol(datafr1)