search for: x1

...'fct'. How can I remove these double-quotes? I tried as.name() but it did not work (because of size?). These are creating trouble with subsequent programs, which I tested with strings that for some reason do not have these double quotes (see very bottom). > cum1 [1] "A11*(X11*x1+X21*x2)+1*sqrt(B11*(X11*x1+X21*x2)^2+C11)A12*(X12*x1+X22*x2)+1*sqrt(B12*(X12*x1+X22*x2)^2+C12)A13*(X13*x1+X23*x2)+-1*sqrt(B13*(X13*x1+X23*x2)^2+C13)A14*(X14*x1+X24*x2)+-1*sqrt(B14*(X14*x1+X24*x2)^2+C14)A15*(X15*x1+X25*x2)+1*sqrt(B15*(X15*x1+X25*x2)^2+C15)A16*(X16*x1+X26*x2)+1*sqrt(B16*(X16*x1+X...

Hi Elisa, I hope this is what you wanted. dat1<-read.csv("peaks.csv",sep=",") #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x); abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2....

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1] "8.49*x1*z1 + 6.13*x1*z2 + 6.4*x1*z3 + 6.9*x2*z1 + 4.54*x2*z2 + 3.99*x2*z3 + 19.31*x3*z1 + 12.49*x3*z2 + 3.86*x3*z...

...ons with categorical variables when the full hierarchy of effects is not present. Depending on which lower level interactions are specified, the factor encoding changes for a higher level interaction. Consider the following minimal reproducible example: -------------- > runmatrix = expand.grid(X1=c(1,-1),X2=c(1,-1),X3=c("A","B","C"))> model.matrix(~(X1+X2+X3)^3,data=runmatrix) (Intercept) X1 X2 X3B X3C X1:X2 X1:X3B X1:X3C X2:X3B X2:X3C X1:X2:X3B X1:X2:X3C 1 1 1 1 0 0 1 0 0 0 0 0 0 2 1 -1 1 0...

Hi R-helpers! I have the following dataframe: firm<-c(rep(1:3,4)) year<-c(rep(2001:2003,4)) X1<-rep(c(10,NA),6) X2<-rep(c(5,NA,2),4) data<-data.frame(firm, year,X1,X2) data So I want to obtain the same dataframe with a variable X3 that is: X1, if X2=NA X2, if X1=NA X1+X2 if X1 and X2 are not NA So my final data is X3<-c(15,NA,12,5,10,2,15,NA,12,5,10,2) finaldata<-data.frame...

Dear expeRts, I would simplify following code. --------------------------------------------- youtput <- function(x1, x2){ n <- length(x1) y <- vector(mode="numeric", length=n) for(i in 1:n){ if(x1[i] >=5 & x1[i] <= 10 & x2[i] >=5 & x2[i] <=10) y[i] <- 0.631 * x1[i]^0.55 * x2[i]^0.65 if(x1[i] >=10 & x1[i] <= 15 & x2[i] >=5 &...

Hi everyone! #I need a loop or a function that creates a X2 variable that is X1 without the extreme values (or X1 winsorized) by industry and year. #My reproducible example: firm<-sort(rep(1:1000,10),decreasing=F) year<-rep(1998:2007,1000) industry<-rep(c(rep(1,10),rep(2,10),rep(3,10),rep(4,10),rep(5,10),rep(6,10),rep(7,10),rep(8,10),rep(9,10), rep(10,10)),1000) X1&...

Hello Tyler, You write that you understand what I am saying. However, I am now at loss about what exactly is the problem with the behavior of R. Here is a script which reproduces your experiments with three variables (excluding the full model): m=expand.grid(X1=c(1,-1),X2=c(1,-1),X3=c("A","B","C")) model.matrix(~(X1+X2+X3)^3-X1:X3,data=m) model.matrix(~(X1+X2+X3)^3-X2:X3,data=m) model.matrix(~(X1+X2+X3)^3-X1:X2,data=m) Below are the three results, similar to your first mail. (The first two are basically the same, of course.)...

Greetings I want to take a fitted regression and replace all uses of a variable in a formula. For example, I'd like to take m1 <- lm(y ~ x1, data=dat) and replace x1 with something else, say x1c, so the formula would become m1 <- lm(y ~ x1c, data=dat) I have working code to finish that part of the problem, but it fails when the formula is more complicated. If the formula has log(x1) or x1:x2, the update code I'm testing doesn...

...1) What am I getting an error message for # 5 and # 7 ? 2) How to fix the code? I would appreciate receiving your help. Thanks, Pradip Muhuri ###### Reproducible Example ##### N <- 100 set.seed(13) df<-data.frame(matrix(sample(c(1:10),N, replace=TRUE),ncol=5)) keep_var <- c("X1", "X2") drop_var <- c("X3", "X4", "X5") df[df$X1>=8,] [,1:2] #1 df[df$X1>=8,] [,-c(3,4,5)] #2 df[df$X1>=8,] [,c(-3,-4,-5)] #3 df[df$X1>=8,] [,c("X1", "X2")] #4 df[df$X1&...

hi, I have following data and code; cov <- c (1.670028 ,-1.197685 ,-2.931445,-1.197685,1.765646,3.883839,-2.931445,3.883839,12.050816) cov.matrix <- matrix(cov, 3, 3, dimnames=list(c("y1","x1","x2"), c("y1","x1","x2"))) path.model <- specify.model() x1 -> y1, x1-y1 x2 <-> x1, x2-x1 x2 <-> x2, x2-x2 x1 <-> x1, x1-x1 y1 <-> y1, y1-y1 x2 -> y1, x2-y1 summary(sem(path.model, cov.matrix, N =...

...discussed there, in Stata and in R the drop of the continuous variable has no effect on the degrees of freedom here: it is just a reparameterisation of the full model, protecting you against losing marginality... Hence the model.matrix 'mm' is still square and nonsingular after the drop of X1, unless of course when a row is removed from the matrix 'design' when before creating 'mm'. Arie On Sun, Oct 15, 2017 at 7:05 PM, Tyler <tylermw at gmail.com> wrote: > You could possibly try to explain away the behavior for a missing main > effects term, since with...

Hello Tyler, I rephrase my previous mail, as follows: In your example, T_i = X1:X2:X3. Let F_j = X3. (The numerical variables X1 and X2 are not encoded at all.) Then T_{i(j)} = X1:X2, which in the example is dropped from the model. Hence the X3 in T_i must be encoded by dummy variables, as indeed it is. Arie On Thu, Nov 2, 2017 at 4:11 PM, Tyler <tylermw at gmail.com&g...

Dear all, I have one dependent variable y and two independent variables x1 and x2 which I would like to use to explain y. x1 and x2 are design factors in an experiment and are not correlated with each other. For example assume that: x1 <- rbind(1,1,1,2,2,2,3,3,3) x2 <- rbind(1,2,3,1,2,3,1,2,3) cor(x1,x2) The problem is that I do not only want to analyze the effect...

On 24/02/2018 1:53 PM, William Dunlap via R-help wrote: > x1 = rbind(unique(preval),mydat) > x2 <- x1[is.na(x1)] <- 0 > x2 # gives 0 > > Why introduce the 'x2'? x1[...] <- 0 alters x1 in place and I think that > altered x1 is what you want. > > You asked why x2 was zero. The value of the expression >...

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4 = 2, x5 = 1 x2 x1...

Hi, final<-data.frame() ?? for (m1 in 4:10) { ?????? for (n1 in 4:10){? ?????????? for (x1 in 0: m1) { ????????????? for (y1 in 0: n1) { final<- rbind(final,c(m1,n1,x1,y1)) res}}}} ?final1<-within(final,{flag<-ifelse(x1/m1>y1/n1, 1,0)}) ?head(final1) #? m1 n1 x1 y1 flag #1? 4? 4? 0? 0??? 0 #2? 4? 4? 0? 1??? 0 #3? 4? 4? 0? 2??? 0 #4? 4? 4? 0? 3??? 0 #5? 4? 4? 0? 4??? 0 #6? 4?...

Hi all, I still have problems with the predict function by setting up the values on which I want to predict ie: original df: p1 (193 obs) variates y x1 x2 rm(list=ls()) x1<-rnorm(193) x2<-runif(193,-5,5) y<-rnorm(193)+x1+x2 p1<-as.data.frame(cbind(y,x1,x2)) p1 y x1 x2 1 -0.6056448 -0.1113607 -0.5859728 2 -4.2841793 -1.0432688 -3.3116807 ...... 192 -1.3228239 1.0263013 -2.7801324 193 1.8736683 1.0480...

...quot;... F_j is coded by contrasts if T_{i(j)} has appeared in the formula and by dummy variables if it has not" You find: "However, the example I gave demonstrated that this dummy variable encoding only occurs for the model where the missing term is the numeric-numeric interaction, ~(X1+X2+X3)^3-X1:X2." We have here T_i = X1:X2:X3. Also: F_j = X3 (the only factor). Then T_{i(j)} = X1:X2, which is dropped from the model. Hence the X3 in T_i must be encoded by dummy variables, as indeed it is. Arie On Tue, Oct 31, 2017 at 4:01 PM, Tyler <tylermw at gmail.com> wrote:...

Hello r-help, I am using the excellent np package to conduct a nonparametric kernel regression and am having some trouble plotting the results. I have 2 covariates, x1 and x2, and a continuous outcome variable y. I am conducting a nonparametric regression of y on x1 and x2. The one somewhat unusual feature of these data is that, to be included in the dataset, x1 must be at least as large as x2. The basics of the analysis are to calculate the correct bandwidth us...