search for: formula

...e name of the data frame for extracting weights or subsets within a function call. The problems can be reproduced as follows: ********************** datai <- data.frame( y = rnorm(10), x1 = rnorm(10), x2 = abs(rnorm(10)), x3 = rep(seq(1,5),2), counter = rep(c(1,2),c(5,5))) formula <- as.formula(y ~ x1) # the following fails in R-1.2.0 but not in R-1.1.1 # > Error in eval(expr, envir, enclos) : Object "datos" not found lapply(split(datai,datai$counter), function(datos,formula) {lm(formula = formula, data = datos, wei...

...r below, it's very much like what went on in the more complex example. best Niels Debugging levelplot: > library(lattice) > debug(levelplot) > source("lptest.r") [1] "test" debugging in: levelplot(matrix(11:14, 2, 2)) debug: { ocall <- match.call() formula <- ocall$formula if (!is.null(formula)) { warning("The 'formula' argument has been renamed to 'x'. See ?xyplot") ocall$formula <- NULL if (!("x" %in% names(ocall))) ocall$x <- formula else warning("...

...**************************************************************************************** I need to store the coefficients using lm() for different combinations of the 4 factors, or different combinations of 3 factors or different combinations of 2 factors or differennt combinations of 1 factor. The formula remains fixed as: > Formula UncDmd ~ M1 + M2 + M3 + M4 + M5 + M6 + M7 + M8 + M9 + M10 + M11 So, different models I want to solve in R are : 1) Community : lm(Formula,TEST1[ as.logical( (TEST1[[1]]=="20232") ) , ]) 2) WT : lm(Formula,TEST...

I'm trying to create a formula object to pass on to a function that applies the function terms.formula() to it. f <- function(formula, ...) { ... mf <- match.call() term <- terms.formula(mf$formula) ... } However, my code below gives an error. form <- as.formula("y~x") f(form, ...) The error message w...

I've run into a scoping problem in R. I'm calling a function that * creates a formula * calculates a weight vector * calls lm with that formula and weights This fails. Here's a simplified reproduce example: # f works, g doesn't, h is a workaround rm(w) data <- data.frame(y=runif(20), x=runif(20), z=runif(20)) f <- function(k){ w <- data$z^k coef(lm(y~x, d...

Until I get a fix that works, a work-around would be to rename the 'y1' column, used a fixed formula, and rename it back afterwards. Thanks for your help. SGO. -----Original Message----- From: Bert Gunter [mailto:bgunter.4567 at gmail.com] Sent: 22 August 2017 20:38 To: Stephen O'hagan <SOhagan at manchester.ac.uk> Cc: r-help at r-project.org Subject: Re: [R] boot.stepAIC fails with c...

Feature request: I want to use update.formula to subtract an intercept (or other) term from a formula with a dot on the RHS. However, as this causes an error, I propose a patch below. Thus, I want: > update.formula(y ~ ., ~ . -1) [1] y ~ . - 1 Instead I get this error: Error in terms.formula(tmp, simplify = TRUE) : '.' in formu...

I am writing a program in which I would like to take in a formula, change the response (Y) variable into something else, and then pass the formula, with the new Y variable to another function. That is, I am starting with formula <- Y~X1+X2+X3 and I'd like to do something like Y <- formula$Y newY <- f(Y) lm(newY~X...

...x, data, ...) UseMethod("xyplot") <environment: namespace:lattice> > library(plm) Loading required package: bdsmatrix Attaching package: ‘bdsmatrix’ The following object(s) are masked from ‘package:base’: backsolve Loading required package: nlme Loading required package: Formula Loading required package: MASS Loading required package: sandwich Loading required package: zoo Attaching package: ‘zoo’ The following object(s) are masked from ‘package:base’: as.Date, as.Date.numeric > ?plm > plm function (formula, data, subset, na.action, effect = c("individua...

It seems that if you build the formula as a character string, and postpone the "as.formula" into the lm call, it works. instead of frm1 <- as.formula(paste(trg,"~1")) use frm1a <- paste(trg,"~1") and then strt <- lm(as.formula(frm1a),dat) regards, Heinz Stephen O'hagan wrote/hat geschriebe...

The formula attribute of the builtin CO2 dataset seems a bit strange: > formula(CO2) Plant ~ Type + Treatment + conc + uptake What is one supposed to do with that? Certainly its not suitable for input to lm and none of the examples in ?CO2 use the above.

With the recent changes in R 1.9.x, it is now impossible to properly call plot on a formula() where the formula is provided via a named first argument. On today's R-1.9.1-alpha: > x <- 1:10 > y <- rnorm(x,0.25) > > plot(x~y) > > plot(x=x~y) Error in terms.formula(formula, data = data) : argument is not a valid model > > plot(formula=x~y) Error in p...

Hi, i 'dont understand how to take a general formula, view this: x<-1:5 y<-c(0,1,1.7,2,2.1.4) dummy<-data.frame(x=x,y=y) formula<-"y~A*log(x)/log(2)" formu<-as.formula(formula) fm<-lm(formu,data=dummy) Error in eval(expr, envir, enclos) : Object "A" not found but A is the parameter of fitting, why is this?Than...

Hi, I am very new to R. I am hoping to create formulas and assign them to locations within an array (or matrix, if it will work). Here's a simplified example of what I'm trying to do: form.arr <- array(31,5,3) for (i in seq(from=1, to=31, by=1)) { for (j in seq(from=1, to=5, by=1)) { form.arr[i,j,] <- as.formula(y~1+2) } } which r...

Dear all I have a work-flow issue with lm(). When I use > lm(y1~x1, anscombe) Call: lm(formula = y1 ~ x1, data = anscombe) Coefficients: (Intercept) x1 3.0001 0.5001 I get as expected the formula, "y1 ~ x1", in the print()ed results or summary(). However, if I pass through a formula object > (form <- formula(y1~x1)) y1 ~ x1 > lm(form, anscombe) Cal...

...art automatically generated by your mailer, which isn't always readable. ? Wed, 30 Oct 2024 17:45:29 +0100 Elys?e Aristide <ariel92and at gmail.com> ?????: > I am using the gmm function from the gmm package and encountered an > unexpected error. No model can be estimated if I load formula.tools?I > need to restart R each time. I can reproduce the problem: library(gmm) data(Finance) r <- Finance[1:300, 1:10] rm <- Finance[1:300, "rm"] rf <- Finance[1:300, "rf"] z <- as.matr...

Hi all, I want to write a wrapper for an analysis of variance and I face a curious problem. Here are two different wrappers: fun.1 <- function(formula) { summary(aov(formula)) } fun.2 <- function(formula) { oneway.test(formula) } values <- c(15, 8, 17, 7, 26, 12, 8, 11, 16, 9, 16, 24, 20, 19, 9, 17, 11, 8, 15, 6, 14) group <- rep(1:3, each=7) # While the first wrapper works just fine: fun.1(values ~ group) # the s...

In R 2.12.0 I get > d <- data.frame(x=1:10, y=log(1:10), f3=LETTERS[rep(1:3,c(3,3,4))]) > m <- model.frame(y~.^2, data=d) > formula(m) y ~ x + f3 In S+ formula(m) gives formula given to model.frame(), but in R you have to do the following get that formula: > formula(attr(m, "terms")) y ~ (x + f3)^2 Would it break anything to add to the top of formula.data.frame something like if (!is.null(tms <- attr(x,...

Hi everyone, I am using the gmm function from the gmm package and encountered an unexpected error. No model can be estimated if I load formula.tools?I need to restart R each time. Here is a simple reproducible example: *library(gmm)data(Finance)r <- Finance[1:300, 1:10]rm <- Finance[1:300, "rm"]rf <- Finance[1:300, "rf"]z <- as.matrix(r-rf)zm <- rm-rfres <- gmm(z ~ zm, x = ~ zm)summa...

I just noticed the following in update.formula and I'm wondering if this behavior is the intention of the developers. Here's an example: update(`a: b` ~ x, ~ . + y) Note now that the response has no backticks and is interpreted as a:b (i.e. ":" is now an operator). This is because in update.formula the call to terms.formula...