search for: formulae

Dear All, A library (PHYLOGR) that passed the usual tests in R-1.1.1 gives errors with R-devel; my (mis?)understanding of scoping rules is that it should have worked in both. The problems seem related to using the 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 =

I have been using levelplot but have had trouble calling it inside functions - something seems to go wrong when it's not called directly from the R command prompt. Simplest reproducible example: $ R --vanilla > library(lattice) > levelplot(matrix(1:4,2,2)) - This gives a nice plot in soothing pastel colors. Now, with a file lptest.r containing 2 lines: library(lattice)

Hi, I am trying to find an efficient way of applying a linear regression model to different factor combinations in a data frame. I want to obtain the output with minimal or no use of loops if possible. Please let me know if this query is unclear. Thanks, Murtaza

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$formula) ... } However, my code below gives an error. form <- as.formula("y~x") f(form, ...) The error message was: Error in terms.formula(mf$formula): argument is not a valid

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, data

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

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 formula and no 'data' argument While the error

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~X1+X2+X3) So far, it seems that my

Hi, How could I see the source code of functions in an R package? If we type ?function_name , we will see documentations of the function_name. If we type function_name, is what returns just the source code? Could we just save it in an .R file and modify as we want? However, it seems that sometimes the source code is hidden (or stored elsewhere?) As an example, could we see the source

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 geschrieben on/am 23.08.2017 12:07: > Until I get a fix that works, a

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

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?Thanks Ruben

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 results in this error: Error in form.arr[i, j,

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) Call:

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

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

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 uses simplify = TRUE. I'm working with data that

Dear,?? The script'formula("x")' loops infinitely. More specifically, it throws thefollowing error:? ? Error: evaluationnested too deeply: infinite recursion / options(expressions=)?? ? As a side effect,this makes an IDE like RStudio to crash. On the other hand,the script 'formula("y")'?works as expected : if ?y? does notexist in the global environment,

According to the man page for formula, "a formula object has an associated environment". However, update.default doesn't use this environment, which creates problems like the following: make.model <- function(x) { lm(medv~.,x) } library(MASS) data(Boston) fit = make.model(Boston) fit = update(fit,".~.-crim") # Object "x" not found Here is a