similar to: matching observations and ranking

Dear all, Suppose I have a data frame like this: [code] var1 <- c(1,999,2) var2 <- c(999,1,2) var3 <- c(1,2,999) example <- data.frame(var1,var2,var3) [/code] I want to replace all 999 to NA in all observations in all columns. I know how to do it in each individual column. [code] example$var1[example$var1==999] <- NA [/code] I think it can be done with a for loop. [code] for

G'day, I'm?calculating LSM for the following model, and am finding that R and SAS give different answers. Whilst the error is at the second or third decimal, the percentage error can be quite large. I'm using the effects library (Version: 2.0-10) on R?2.11.1 in the following manner: options(contrasts=c("contr.helmert","contr.poly"))

Hola.. yo lo haría de la siguiente manera ... En excel:genero una columna con la serie de fechas continuas ... con la función buscarv, agrego las variables que tienen dato a esta serie.. en base a la fecha muy largo y mecánico para mi gusto.... En R ..de igual manera genero el vector de fechas require(chron)# crear el vector continuo de fechasfch01 <- data.frame(''fch'' =

Hi all, I analysed my data with lme and after that I spent a lot of time for mean separation of treatments (post hoc). But still I couldn’t make through it. This is my data set and R scripts I tried. replication fertilizer variety plot height 1 level1 var1 1504 52 1 level1 var3 1506 59 1 level1 var4 1509 54 1 level1 var2 1510 48 2 level1 var1 2604 47 2 level1 var4 2606 51 2 level1 var3

Hi everybody, I'm trying to analyse a set of data with a non-normal response, 2 fixed effects and 1 nested random effect with strong heteroscedasticity in the model. I planned to use the function lmer : lmer(resp~var1*var2 + (1|rand)) and then use permutations based on the t-statistic given by lmer to get p-values. 1/ Is it a correct way to obtain p-values for my variables ? (see below)

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

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

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 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 problem converting a XML file, via the XML package, to a data.frame. The XML file looks like this: <Transaction> <ID value="0044"/> <Var1 value="XYZ159"/> <Var2 value="_"/> <Var3 value="AMR1.0-INT-1005"/> <Var4 value="2010-05-25 10:44:16:673"/> <Var5 value="1"/>

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 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,

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 plotting line chart using ggplot and want to use geom_line and geom_point simultaneously. I get the plot but now I have two legends. None of the legend is representing the true values. I need the legend with shape and color both. Thanks > con = textConnection("inputs var1 var2 var3+ 100 10 5 2+ 1000 20 10 4+ 5000 30 15 8+ 10000 40 20 16+ 30000 50 25 32")> data =

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",

I'm reluctant to draw the S-PLUS and R comparison (these are different programs after all), but could someone tell me why the following matrix substitution works in S-PLUS, but not R. I'm curious because matrix substitution is a really slick way to "cleaning up" columns of data in data frames. For example, in the following I change values of 1 to values of 10, but only for

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

Dear Help, I'm trying to calculate a weighted correlation matrix from a data frame with 6 columns (variables) and 297 observations extracted from the regression. The last column is a weight column which I want to apply. $ model :'data.frame': 297 obs. of 6 variables: ..$ VAR1 : num [1:297] 5.21 9.82 8.08 0.33 8.7 6.82 3.94 4 0 5 ... ..$ VAR2 : num [1:297]

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 =>