search for: x2

...ct'. 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+X26*x2)^...

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*z3 + 5.2...

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(firm, year,X1,X2,X3)...

...tegorical 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 0 -1...

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 & x2[...

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 = 422)) a...

libvorbis does not currently have any simd/vectorization. Following patches add generic framework for simd/vectorization and on top, add ARM-NEON simd vectorization using intrinsics. I was able to get over 34% performance improvement on my Beaglebone Black which is single Cortex-A8 based CPU. You can find more information on metrics and procedure I used to measure at

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 of x1...

The code below works as expected but: Using the proto package, is this the best way to 1) make a parameter persist if the parameter is passed in with a value, 2) allow for calling the bias() function without a parameter assignment, 3) have the x2 value initialize as 5? Thanks for your feedback. Giving the proto package a test beat and establishing some templates for myself. > oo <- proto(expr = {x = c(10, 20, 15, 19, 17) x2 = 5 # so x2 initializes as 5, but can be overwritten with param assignment...

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...

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.) Please pic...

...ing variable. More precisely: for all values of that variable. In other words: you have a three-way interaction, with all values of that variable. An even smaller example is the following script with only two variables, each being a factor: df <- expand.grid(X1=c("p","q"), X2=c("A","B","C")) print(model.matrix(~(X1+X2)^2 ,data=df)) print(model.matrix(~(X1+X2)^2 -X1,data=df)) print(model.matrix(~(X1+X2)^2 -X2,data=df)) The result is: (Intercept) X1q X2B X2C X1q:X2B X1q:X2C 1 1 0 0 0 0 0 2 1 1...

...results on R 4.0.0 and R-devel (2020-05-22 r78545). I haven't found any info about that change in NEWS. Was the change intentional? Sys.setlocale("LC_CTYPE","C") Sys.setlocale("LC_COLLATE","C") x1 = "fa\xE7ile" Encoding(x1) = "latin1" x2 = iconv(x1, "latin1", "UTF-8") base::order(c(x2,x1,x1,x2)) Encoding(x2) = "unknown" base::order(c(x2,x1,x1,x2)) # R 4.0.0 base::order(c(x2,x1,x1,x2)) #[1] 1 4 2 3 Encoding(x2) = "unknown" base::order(c(x2,x1,x1,x2)) #[1] 2 3 1 4 # R-devel base::order(c(x2,x...

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 1.0480632...

hello: I am very confused when it comes to list operations in R. I seek help in the following problem. I have two different vectors myIDs - a character vector with no names to their elements x2 - another character vector derived from unlisting a list where an element has both a name and value. I am not happy the way x2 is with names and values to every element. > myIDs[1:10] [1] "54898" "7083" "7031" "7083" "4175" "4605&q...

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 using npr...

...quot;Chandler Carruth" <chandlerc at google.com>, "LLVM Developers Mailing List" <llvmdev at cs.uiuc.edu> > Sent: Friday, November 14, 2014 5:31:24 PM > Subject: Re: [LLVMdev] Upcoming Changes/Additions to Scoped-NoAlias metadata > > You don't have x1 and x2 in your example, assuming you mean: > > int i = 0; > T A; > T * y2 = ... > { > T * x1 = &A; > a = x1[i]; > } > { > T * restrict x2 = y2; > b = x2[i]; > } > > then the answer is no, the fact that x2 is restrict qualified does > not allow us to concl...

Dear all, I have a question about how to set arguments in my own function. For example, I have a function that looks like this: my.f <- function(a = x1, b = x2) { x1 = equation 1 x2 = equation 2 x3 = equation 3 y = a + b } x1, x2, and x3 are temporary variables (intermediate results) calculated from other variables within the funciton. I want to use two of these three variables to calculate y, and write R script as below: my.f(a = x1, b =...

...obtengo que los coeficientes estimados no se asemejan mucho a los originales. ¿Se le ocurre a alguien cuál es el motivo de esta discrepancia? ¿y cómo solucionarlo? Muchas gracias Emilio logisticsimulation <- function(n){ dat <- data.frame(x1=sample(0:1, n,replace=TRUE), x2=sample(0:1, n,replace=TRUE)) odds <- exp(-1 - 4 * dat$x1 + 7*dat$x2 - 1 *dat$x1* dat$x2 ) pr <- odds/(1+odds) res <- replicate(100, { dat$y <- rbinom(n,1,pr) coef(glm(y ~ x1*x2, data = dat, family = binomial())) }) t(res) } res <- logisticsimulation(100) apply(res,...

...t;- cbind(mixAndMatch, modelcv[ ,!colnames(modelcv) %in% colnames(mixAndMatch)]) newdf } predictOMatic <- function(model = NULL, fl = NULL, ...){ nd <- newdataMaker(model, fl) fit <- predict(model, newdata=nd, ...) cbind(fit, nd) } set.seed(12345) x1 <- rnorm(100) x2 <- rnorm(100) x3 <- rnorm(100) x4 <- rnorm(100) x5 <- rpois(100, 5) x6 <- rgamma(100, 2,1) xcat1 <- gl(2,50, labels=c("M","F")) xcat2 <- cut(rnorm(100), breaks=c(-Inf, 0, 0.4, 0.9, 1, Inf), labels=c("R", "M", "D", "P",...