I need to use mixed model lme function many times in my code. But I do not know how to use it within a function. If used otherwise, the lme function works just well but when used within the function, it throws errors:
myfunc<- function(cc, x, y, z)
{
model <- lme(fixed = x ~1 , random = ~ 1|y/z,
data=cc,
method="REML")
}
on calling this function:
myfunc (dbcon2, birthweight, sire, dam)
I get the error :
Error in model.frame.default(formula = ~x + y + z, data = list(animal
= c("29601/9C1", : invalid type (list) for variable 'x'
I think, there is a different procedure for using this which I am unaware of. Any help would be greatly appreciated.
Thanks in advance
Not sure if you are looking for this, you may try to use this, as correctly pointed out by #akrun, you may use paste, I am using paste0 however(its a special case of paste), paste concatenates two strings:
Here the idea is to concatenate the variable names with the formula, but since paste converts it to a string hence you can't refer that as formula to build a model,so you need to convert that string to a formula using as.formula which is wrapped around paste0 statement.
To understand above, Try writing a formula like below using paste:
formula <-paste0("mpg~", paste0("hp","+", "am"))
print(formula)
[1] "mpg~hp+am"
class(formula)
[1] "character" ##This should ideally be a formula rather than character
formula <- as.formula(formula) ##conversion of character string to formula
class(formula)
[1] "formula"
To work inside a model, you would always require a formula object, also please also try to learn about collapse and sep option in paste they are very handy.
I don't have your data , hence I have used mtcars data to represent the same.
library("nlme")
myfunc<- function(cc, x, y, z)
{
model <- lme(fixed = as.formula(paste0(x," ~1")) , random = as.formula(paste0("~", "1|",y,"/",z)),
data=cc,
method="REML")
}
models <- myfunc(cc=mtcars, x="hp", y="mpg", z="am")
summary(models)
You can read more about paste by typing ?paste in your console.
Related
I am trying to create my own function that contains 1.) the mgcv gamm function and 2.) a nested autocorrelation (ARMA) argument. I am getting an error when I try to run the function like this:
df <- AirPassengers
df <- as.data.frame(df)
df$month <- rep(1:12)
df$yr <- rep(1949:1960,each=12)
df$datediff <- 1:nrow(df)
try_fxn1 <- function(dfz, colz){gamm(dfz[[colz]] ~ s(month, bs="cc",k=12)+s(datediff,bs="ts",k=20), data=dfz,correlation = corARMA(form = ~ 1|yr, p=2))}
try_fxn1(df,"x")
Error in eval(predvars, data, env) : object 'dfz' not found
I know the issue is with the correlation portion of the formula, as when I run the same function without the correlation structure included (as seen below), the function behaves as expected.
try_fxn2 <- function(dfz, colz){gamm(dfz[[colz]] ~ s(month, bs="cc",k=12)+ s(datediff,bs="ts",k=20), data=dfz)}
try_fxn2(df,"x")
Any ideas on how I can modify try_fxn1 to make the function behave as expected? Thank you!
You are confusing a vector with the symbolic representation of that vector when building a formula.
You don't want dfz[[colz]] as the response in the formula, you want x or whatever you set colz too. What you are getting is
dfz[[colz]] ~ ...
when what you really want is the variable colz:
colz ~ ...
And you don't want a literal colz but whatever colz evaluates to. To do this you can create a formula by pasting the parts together:
fml <- paste(colz, '~ s(month, bs="cc", k=12) + s(datediff,bs="ts",k=20)')
This turns colz into whatever it was storing, not the literal colz:
> fml
[1] "x ~ s(month, bs=\"cc\", k=12) + s(datediff,bs=\"ts\",k=20)"
Then convert the string into a formula object using formula() or as.formula().
The final solution then is:
fit_fun <- function(dfz, colz) {
fml <- paste(colz, '~ s(month, bs="cc", k=12) + s(datediff,bs="ts",k=20)')
fml <- formula(fml)
gamm(fml, data = df, correlation = corARMA(form = ~ 1|yr, p=2))
}
This really is not an issue with corARMA() part, other than that triggers somewhat different evaluation code for the formula. The guiding mantra here is to always get a formula as you would type it if not programming with formulas. You would never (or should never) write a formula like
gamm(df[[var]] ~ x + s(z), ....)
While this might work in some settings, it will fail miserably if you ever want to use predict()` and it fails when you have to do something a little more complicated.
loess.smooth <- function(dat) {
dat <- dat[complete.cases(dat),]
## response
vars <- colnames(dat)
## covariate
id <- 1:nrow(dat)
## define a loess filter function (fitting loess regression line)
loess.filter <- function (x, span) loess(formula = paste(x, "id", sep = "~"),
data = dat,
degree = 1,
span = span)$fitted
## apply filter column-by-column
new.dat <- as.data.frame(lapply(vars, loess.filter, span = 0.75),
col.names = colnames(dat))
}
When I try to apply loess.smooth to a dataframe, I get the error:
Error in model.frame.default(formula = paste(x, "id", sep = "~"), data = dat) :
invalid type (closure) for variable 'id'
I don't understand why this is a problem since id is not a function, which is implied by the error.
When I run through these lines of code outside of the function, it works perfectly fine and does exactly what I want it to do.
It is a scoping issue involving passing a vector of strings to the loess function instead of passing a vector of formulas. The problem is that the environment returns NULL for the former, so loess doesn't know where to find it. If you wrap the formula in as.formula it works. This variable will be assigned the local environment inside the function call by default.
As to the cryptic error, it happens when you name a variable the same name of a given function from another package that is loaded, since if a function doesn't find a variable in the local environment, it will scope in the loaded packages for the function. In my case, the id function was loaded by the dplyr library.
I am trying to run a t-test within a custom function, and am running into a quosure misapplication (I believe). Any help would be greatly appreciated.
library(tidyverse)
tp_pull <- function(mydata, dv, iv){
dv <- enquo(dv)
iv <- enquo(iv)
t.test(!!dv ~ !!iv, mydata)
}
tp_pull(mydata = mtcars, dv = mpg, iv = vs)
My error message reads:
numerical expression has 2 elements: only the first usedNAs introduced by
coercion
Show Traceback
Error in quo_name(dv):~!(!iv) : NA/NaN argument
For context this t-test will be part of a larger custom function.
Quosures are unique to tidyeval and are not supposed by the base R language. Right now they only work with dplyr. It is very unlikely these will ever work with base functions such as t.test.
If you want to do this with base R, you can use G. Grothendieck's suggestion of
tp_pull <- function(mydata, dv, iv){
t.test(formula(substitute(dv ~ iv)), mydata)
}
tp_pull(mydata = mtcars, dv = mpg, iv = vs)
The substitute captures the un-evaulated symbol names from the promise passed to the call and allows you to re-assemble them into a new expression. The formula() call helps coerce the un-evaulated expression returned by substitute() into a proper R formula object.
I get Error: $ operator not defined for this S4 class when I try to run a ctree from the party package, but only when the formula is writen as a string that I transform using as.formula().
Below the example :
#This works fine :
y <- ctree(formula = quotation ~ minute + temp, data=test[[1]], controls = ctree_control(mincriterion = 0.99))
#While this doesn't :
x <- "ctree(formula = quotation ~ minute + temp, data=test[[1]], controls = ctree_control(mincriterion = 0.99))"
y <- as.formula(x)
Error: $ operator not defined for this S4 class
My ultimate purpose is to create a function that iterates through the list test to create multiple trees.
Any idea ?
ctree is a function and not a formula. formula is the class of the object resulting from the function '~' (tilde). You can learn more about formulas from help('~') and help('formula').
The most common way to use as.formula is to convert a string that represents the formula syntax to an object of class formula. Something like as.formula('y ~ x'). Also, check class(as.formula(y~x)).
In your case you saved a string representing function ctree to variable x. Function ctree only contains a string representing a formula syntax (quotation ~ minute + temp) but it cannot be coerced to formula (it does not represent a formula, it just contains a formula syntax string) because it does not follow the formula syntax.
If you want to execute a function from text you need to use eval(parse(text = x)) although this technique is not encouraged..
I want to run a multiple comparisons analysis for the different variables of a model. My idea is as follows:
library(multcomp)
set.seed(123)
x1 <- gl(4,10)
x2 <- gl(5,2,40)
y <- rnorm(40)
fm1 <- lm(y ~ x1 + x2)
for(var in c('x1', 'x2'))
{
mc1 <- glht(fm1, linfct=mcp(var='Tukey'))
print(summary(mc1))
}
When I run, I get the following error:
Error en mcp2matrix(model, linfct = linfct) :
Variable(s) ‘var’ have been specified in ‘linfct’ but cannot be found in ‘model’!
That is, it is not possible to use a character to specify an attribute of the mcp function.
Anyone knows a solution?
It's generally better to avoid working with strings representing code wherever possible - it prevents errors that are hard to debug, and aesthetically is much more elegant. This problem turns out to be fairly easy to solve if you use do.call and the setNames function:
var <- "x1"
cmp <- do.call(mcp, setNames(list("Tukey"), var))
glht(fm1, linfct = cmp)
You can't use substitute here because it does not allow you modify the names of function parameters. I have some intuition for why this is reasonable, but not enough to explain it :/
If you're a package author, it's a good idea to provide an alternative version of functions that use unusual syntax so they can be accessed programmatically without jumping through hoops.
(Update: Make sure to see Hadley's answer for the better way of doing this, without resorting to string-pasting. My answer will still be useful for explaining why that is harder-than-usual in this case.)
The peculiarities of mcp() require you to use the relatively brute force approach of pasting together the expression you'd like to evaluate and then passing it through eval(parse()).
The tricky bit is that mcp() interprets its first argument in a nonstandard way. Within mcp(), x1 = 'Tukey' does not (as it normally would) mean "assign a value of 'Tukey' to the argument x1". Instead, the whole thing is interpreted as a symbolic description of the intended contrasts. (In this, it is much like more familiar formula objects such as the y ~ x1 + x2 in your lm() call).
for(var in c('x1', 'x2')) {
# Construct a character string with the expression you'd type at the command
# line. For example : "mcp(x1 = 'Tukey')"
exprString <- paste("mcp(", var, "='Tukey')")
# eval(parse()) it to get an 'mcp' object.
LINFCT <- eval(parse(text = exprString))
mc1 <- glht(fm1, linfct = LINFCT)
print(summary(mc1))
}
Have you tried: eval(parse(text='variable'))
or assign ?