Displaying 20 results from an estimated 7255 matches for "formulas".
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formula
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:
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
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
2019 Jun 14
1
Bug report: 'formula("x")' loops infinitely ('formula("y")' does not)
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,
2003 Jun 18
3
update.default bugfix (PR#3288)
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