search for: formulas

Displaying 20 results from an estimated 7309 matches for "formulas".

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2000 Oct 12
2
works in R-1.1.1 but not in R-development; why?
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 =
2006 Jun 13
2
levelplot and source() problems
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)
2008 Dec 26
3
lm() with same formula but different column/factor combinations in data frame
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
2011 Jul 13
1
question on formula and terms.formula()
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
2008 May 30
2
scoping problem when calling lm(precomputed formula, weights) from function (PR#11540)
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$z^k coef(lm(y~x, data
2017 Aug 23
3
boot.stepAIC fails with computed formula
Until I get a fix that works, a work-around would be to rename the 'y1' column, used a fixed formula, and rename it back afterwards. Thanks for your help. SGO. -----Original Message----- From: Bert Gunter [mailto:bgunter.4567 at gmail.com] Sent: 22 August 2017 20:38 To: Stephen O'hagan <SOhagan at manchester.ac.uk> Cc: r-help at r-project.org Subject: Re: [R] boot.stepAIC
2017 May 23
0
Allow dot in RHS of update.formula's old formula
Feature request: I want to use update.formula to subtract an intercept (or other) term from a formula with a dot on the RHS. However, as this causes an error, I propose a patch below. Thus, I want: > update.formula(y ~ ., ~ . -1) [1] y ~ . - 1 Instead I get this error: Error in terms.formula(tmp, simplify = TRUE) : '.' in formula and no 'data' argument While the error
2007 Oct 01
4
Disentagling formulas
I am writing a program in which I would like to take in a formula, change the response (Y) variable into something else, and then pass the formula, with the new Y variable to another function. That is, I am starting with formula <- Y~X1+X2+X3 and I'd like to do something like Y <- formula$Y newY <- f(Y) lm(newY~X1+X2+X3) So far, it seems that my
2013 May 17
2
How could I see the source code of functions in an R package?
...uot;pht") ht <- eval(ht, parent.frame()) return(ht) } if (!is.null(dots$instruments)) { as.Formula(formula, dots$instruments) deprec.instruments <- paste("the use of the instruments argument is deprecated,", "use two-part formulas instead") warning(deprec.instruments) } if (inherits(data, "pdata.frame") && !is.null(index)) warning("the index argument is ignored because data is a pdata.frame") if (!inherits(data, "pdata.frame")) data <- pdata.fr...
2017 Aug 23
0
boot.stepAIC fails with computed formula
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
2007 Jul 16
5
formula(CO2)
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.
2004 May 25
3
problems with plot.formula
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
2004 Jul 15
5
formula
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
2010 Jul 06
4
Assign Formulas to Arrays or Matrices?
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 re...
2011 Nov 24
2
proper work-flow with 'formula' objects and lm()
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:
2024 Nov 01
1
Invalid term in model formula with gmm after formula.tools is loaded
Hi Aristide and welcome to R-help! Your message was a bit mangled [*]. It's best to compose messages to this mailing list in plain text. Otherwise (when composed in HTML), the mailing list eats the HTML part and we're left with the plain text part automatically generated by your mailer, which isn't always readable. ? Wed, 30 Oct 2024 17:45:29 +0100 Elys?e Aristide <ariel92and at
2007 Jan 09
5
a question of substitute
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
2011 Jan 07
1
formula(model.frame(y~.^2, data=d)) does not return formula from terms attribute of the model.frame
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
2024 Oct 30
1
Invalid term in model formula with gmm after formula.tools is loaded
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
2009 Jul 30
3
update.formula and backticked colons
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