I'm optimising a simple function in r using 'nloptr' and I'm having difficulty passing arguments in to the objective function. Here is the code I'm using:
require("nloptr")
Correl <- matrix(c(1,-.3,-.3,1), nrow=2, ncol=2)
Wghts <- c(.01,.99)
floor <- .035
expret <- c(.05,.02)
pf.return <- function(r, x, threshold=0){
return(r * x - threshold)
}
pf.vol <- function(x, C){
return(sqrt(x %*% C %*% x))
}
res <- nloptr(x0=Wghts,eval_f = pf.vol,eval_g_ineq=pf.return,opts=list(algorithm="NLOPT_GN_ISRES"), x=Wghts,C=Correl)
(I know I'm missing parameters here but I'm trying to highlight a behaviour I don't understand)
Running this gives the following error:
Error in .checkfunargs(eval_f, arglist, "eval_f") :
x' passed to (...) in 'nloptr' but this is not required in the eval_f function.
Just to see what happens I can try running it without the x argument:
res <- nloptr(x0=Wghts,eval_f = pf.vol,eval_g_ineq=pf.return,opts=list(algorithm="NLOPT_GN_ISRES"), C=Correl)
which gives the error:
Error in .checkfunargs(eval_g_ineq, arglist, "eval_g_ineq") :
eval_g_ineq requires argument 'x' but this has not been passed to the 'nloptr' function.
So including x throws an error that it is unnecessary and omitting it throws an (at least understandable) error that it has been omitted.
Ok for posterity:
I rewrote the functions so that they had the same set of arguments, in the same order.
I also omitted the x=Wghts bit as this is the parameter I'm trying to search over.
Related
I am getting a Error in 1:T : argument of length 0 when running the Performance Analytics package in R. am I missing a package? Below is my code with error.
#clean z, all features, alpha = .01, run below
setwd("D:/LocalData/casaler/Documents/R/RESULTS/PLOTS_PCA/CLN_01")
PGFZ_ALL <- read.csv("D:/LocalData/casaler/Documents/R/PG_DEUX_Z.csv", header=TRUE)
options(max.print = 100000) #Sets ability to view all dealer records
pgfzc_all <- PGFZ_ALL
#head(pgfzc_all,10)
library("PerformanceAnalytics")
library("RGraphics")
Loading required package: grid
pgfzc_elev <- pgfzc_all$ELEV
#head(pgfzc_elev,5)
#View(pgfzc_elev)
set.seed(123) #for replication purposes; always use same seed value
cln_elev <- clean.boudt(pgfzc_elev, alpha = 0.01) #set alpha .001 to give the most extreme outliers
Error in 1:T : argument of length 0
It's hard to answer your question without knowing what your data looks like. But I can tell you what throws that error. Looking into the source code of the clean.boudt function I find the following cause of your error:
T = dim(R)[1]
...
for (t in c(1:T)) {
d2t = as.matrix(R[t, ] - mu) %*% invSigma %*% t(as.matrix(R[t,
] - mu))
vd2t = c(vd2t, d2t)
}
...
The dim(R)[1] extracts the number of rows in the data supplied to the R argument in the function. It appears that your data has no rows, so check the data type of pgfzc_elev
The cause of the error is likely from your use of $ to subset pgfzc_all.
pgfzc_elev <- pgfzc_all$ELEV
I reckon it is of class integer, which is why dim(R)[1] does not work in the function.
Rather subset your object like this:
pgfzc_elev <- pgfzc_all[, ELEV, drop = F]
Try that and see if it works.
I have a difficulty in learning how to use eval() to evaluate a function,
suppose i have a function:
sq <- function(y){ y**2 }
u can evaluate this function like this:
call <- match.call(expand.dots = FALSE)
call[[1]] <- as.name('sq')
call$y <- 0.2
call <- call[c(1,3)]
eval(call)
and it will give u 0.2^2 = 0.04
But if i want to calculate sth like sq(y), where y = sin(x), i may write:
call <- match.call(expand.dots = FALSE)
call[[1]] <- as.name('sq')
call$y <- as.name('sin')
call$x <- 0.2
call <- call[c(1,3:4)]
eval(call)
it will give me this error:
Error in sq(y = sin, x = 0.2) : unused argument (x = 0.2)
Seems that R cannot recognize x as an argument of sin, but an argument of sq instead. how can we tell R that x is an argument of sin?
Also, it seems that R is the only language i have learned that uses eval() to evaluate a function (i know C++ and Python, but havent seen that syntax before), what is the different (or advantage) to evaluate a function in this way instead of calling sq(y=sin(x=0.2))?
Is there a good book or tutorial talking about its usage, and when to use between the two ways? Thanks!
PS: the example above is actually a simplified version of the code in mlogit package im studying, in which the log likelihood is returned by calling 'lnl.slogit' and is passed to 'mlogit.optim' and get optimized (Line 407 of https://github.com/cran/mlogit/blob/master/R/mlogit.R). I used the same method as the code in the package to call two functions, but i got the error above.
The code is trying to pass:
an argument x to sq but sq has no x argument
the function sin in argument y but a number is required, not a function.
Try this:
x <- 0.2
cl <- call("sq", y = quote(sin(x)))
cl
## sq(y = sin(x))
eval(cl)
## [1] 0.0394695
or maybe what you want is:
x <- 0.2
cl <- call("sq", y = sin(x))
cl
## sq(y = 0.198669330795061)
eval(cl)
## [1] 0.0394695
or
match.fun("sq")(sin(x))
## [1] 0.0394695
or just:
sq(sin(x))
## [1] 0.0394695
Note that ordinarily you do not have to use eval. Just listing the function with its arguments is enough to evaluate it as the in last line of code.
The regression functions in R internally use non-standard code due to considerations related to environments but ordinarily that would not be needed in other contexts.
I'm just learning to create functions in R so I'm trying to make a function which graphs residual lines for a linear regression. I've already tried it and the code works outside of the function, but once I put it all into a function I get the 'x' and 'y' lengths differ error.
Here is my function:
`reslines <- function(x,y) {
abline(lm(y~x))
for(k in 1: length(y)) lines(c(x[k],x[k]), c(y[k], predict(lm(y~x))))
}`
The tracebook shows that the error occurs here:
6 stop("'x' and 'y' lengths differ")
5 xy.coords(x, y)
4 plot.xy(xy.coords(x, y), type = type, ...)
3 lines.default(c(x[k], x[k]), c(y[k], predict(lm(y ~ x))))
2 lines(c(x[k], x[k]), c(y[k], predict(lm(y ~ x))))
1 reslines(a, b)
I've checked the lengths of each data set I've tried using the length() function, and they all match, so something is happening inside the function which appears to change the length or 'x' or 'y' or both.
Can anyone tell me what the error is and how to fix it? Thanks.
I think I fixed it, it was not super easy, mainly the problem was in your predict where you used y, x instead of y[k], x[k]. But there was a little bit more:
reslines <- function(x,y) {
plot(y~x)
abline(lm(y~x))
lm.xy <- lm(y~x)
for(k in 1: length(y)) {
lines(c(x[k],x[k]), c(y[k], predict(lm.xy, data.frame(x=x[k], y = y[k]))))
}
}
Now a test
set.seed(123)
reslines(rnorm(10), rnorm(10))
How does R dispatch plot functions? The standard generic is defined as
plot <- function (x, y, ...) UseMethod("plot")
So usually, all plot methods need arguments x and y. Yet, there exists a variety of other functions with different arguments. Some functions only have argument x:
plot.data.frame <- function (x, ...)
others even have neither x nor y:
plot.formula <- function(formula, data = parent.frame(), ..., subset,
ylab = varnames[response], ask = dev.interactive())
How does this work and where is this documented?
Background
In my package papeR (see GitHub) I want to replace the function plot.data.frame, which is defined in the R package graphics with my own version. Yet, this is usually not allowed
Do not replace registered S3 methods from base/recommended packages,
something which is not allowed by the CRAN policies and will mean that
everyone gets your method even if your namespace is unloaded.
as Brian Ripley let me know last time I tried to do such a thing. A possible solution is as follows:
If you want to change the behaviour of a generic, say predict(), for
an existing class or two, you could add such as generic in your own
package with default method stats::predict, and then register modified
methods for your generic (in your own package).
For other methods I could easily implement this (e.g. toLatex), yet, with plot I am having problems. I added the following to my code:
## overwrite standard generic
plot <- function(x, y, ...)
UseMethod("plot")
## per default fall back to standard generic
plot.default <- function(x, y, ...)
graphics::plot(x, y, ...)
## now specify modified plot function for data frames
plot.data.frame <- function(x, variables = names(x), ...)
This works for data frames and plots with x and y. Yet, it does not work if I try to plot a formula etc:
Error in eval(expr, envir, enclos) :
argument "y" is missing, with no default
I also tried to use
plot.default <- function(x, y, ...)
UseMethod("graphics::plot")
but then I get
Error in UseMethod("graphics::plot") :
no applicable method for 'graphics::plot' applied to an object of class "formula"
So the follow up question is how I can fix this?
[Edit:] Using my solution below fixes the problems within the package. Yet, plot.formula is broken afterwards:
library("devtools")
install_github("hofnerb/papeR")
example(plot.formula, package="graphics") ## still works
library("papeR")
example(plot, package = "papeR") ## works
### BUT
example(plot.formula, package="graphics") ## is broken now
Thanks to #Roland I solved part of my problem.
It seems that the position of the arguments are used for method dispatch (and not only the names). Names are however partially used. So with Rolands example
> plot.myclass <- function(a, b, ...)
> print(b)
> x <- 1
> y <- 2
> class(x) <- "myclass"
we have
> plot(x, y)
[1] 2
> plot(a = x, b = y)
[1] 2
but if we use the standard argument names
> plot(x = x, y = y)
Error in print(b) (from #1) : argument "b" is missing, with no default
it doesnt't work. As one can see x is correctly used for the dispatch but b is then "missing". Also we cannot swap a and b:
> plot(b = y, a = x)
Error in plot.default(b = y, a = x) :
argument 2 matches multiple formal arguments
Yet, one could use a different order if the argument one wants to dispatch for is the first (?) element without name:
> plot(b = y, x)
[1] 2
Solution to the real problem:
I had to use
plot.default <- function(x, y, ...)
graphics::plot(x, y, ...)
The real issue was internally in my plot.data.frame method where I was using something along the lines of:
plot(x1 ~ y1)
without specifying data. Usually this works as data = parent.frame() per default. Somehow in this case this wasn't working. I now use plot(y1, x1) which works like a charm. So this was my sloppiness.
I have a weird problem with R that I can't seem to work out.
I've tried to write a function that performs K-fold cross validation for a model chosen by the stepwise procedure in R. (I'm aware of the issues with stepwise procedures, it's purely for comparison purposes) :)
Now the issue is, that if I define the function parameters (linmod,k,direction) and run the contents of the function, it works flawlessly. BUT, if I run it as a function, I get an error saying the datas.train object can't be found.
I've tried stepping through the function with debug() and the object clearly exists, but R says it doesn't when I actually run the function. If I just fit a model using lm() it works fine, so I believe it's a problem with the step function in the loop, while inside a function. (try commenting out the step command, and set the predictions to those from the ordinary linear model.)
#CREATE A LINEAR MODEL TO TEST FUNCTION
lm.cars <- lm(mpg~.,data=mtcars,x=TRUE,y=TRUE)
#THE FUNCTION
cv.step <- function(linmod,k=10,direction="both"){
response <- linmod$y
dmatrix <- linmod$x
n <- length(response)
datas <- linmod$model
form <- formula(linmod$call)
# generate indices for cross validation
rar <- n/k
xval.idx <- list()
s <- sample(1:n, n) # permutation of 1:n
for (i in 1:k) {
xval.idx[[i]] <- s[(ceiling(rar*(i-1))+1):(ceiling(rar*i))]
}
#error calculation
errors <- R2 <- 0
for (j in 1:k){
datas.test <- datas[xval.idx[[j]],]
datas.train <- datas[-xval.idx[[j]],]
test.idx <- xval.idx[[j]]
#THE MODELS+
lm.1 <- lm(form,data= datas.train)
lm.step <- step(lm.1,direction=direction,trace=0)
step.pred <- predict(lm.step,newdata= datas.test)
step.error <- sum((step.pred-response[test.idx])^2)
errors[j] <- step.error/length(response[test.idx])
SS.tot <- sum((response[test.idx] - mean(response[test.idx]))^2)
R2[j] <- 1 - step.error/SS.tot
}
CVerror <- sum(errors)/k
CV.R2 <- sum(R2)/k
res <- list()
res$CV.error <- CVerror
res$CV.R2 <- CV.R2
return(res)
}
#TESTING OUT THE FUNCTION
cv.step(lm.cars)
Any thoughts?
When you created your formula, lm.cars, in was assigned its own environment. This environment stays with the formula unless you explicitly change it. So when you extract the formula with the formula function, the original environment of the model is included.
I don't know if I'm using the correct terminology here, but I think you need to explicitly change the environment for the formula inside your function:
cv.step <- function(linmod,k=10,direction="both"){
response <- linmod$y
dmatrix <- linmod$x
n <- length(response)
datas <- linmod$model
.env <- environment() ## identify the environment of cv.step
## extract the formula in the environment of cv.step
form <- as.formula(linmod$call, env = .env)
## The rest of your function follows
Another problem that can cause this is if one passes a character (string vector) to lm instead of a formula. vectors have no environment, and so when lm converts the character to a formula, it apparently also has no environment instead of being automatically assigned the local environment. If one then uses an object as weights that is not in the data argument data.frame, but is in the local function argument, one gets a not found error. This behavior is not very easy to understand. It is probably a bug.
Here's a minimal reproducible example. This function takes a data.frame, two variable names and a vector of weights to use.
residualizer = function(data, x, y, wtds) {
#the formula to use
f = "x ~ y"
#residualize
resid(lm(formula = f, data = data, weights = wtds))
}
residualizer2 = function(data, x, y, wtds) {
#the formula to use
f = as.formula("x ~ y")
#residualize
resid(lm(formula = f, data = data, weights = wtds))
}
d_example = data.frame(x = rnorm(10), y = rnorm(10))
weightsvar = runif(10)
And test:
> residualizer(data = d_example, x = "x", y = "y", wtds = weightsvar)
Error in eval(expr, envir, enclos) : object 'wtds' not found
> residualizer2(data = d_example, x = "x", y = "y", wtds = weightsvar)
1 2 3 4 5 6 7 8 9 10
0.8986584 -1.1218003 0.6215950 -0.1106144 0.1042559 0.9997725 -1.1634717 0.4540855 -0.4207622 -0.8774290
It is a very subtle bug. If one goes into the function environment with browser, one can see the weights vector just fine, but it somehow is not found in the lm call!
The bug becomes even harder to debug if one used the name weights for the weights variable. In this case, since lm can't find the weights object, it defaults to the function weights() from base thus throwing an even stranger error:
Error in model.frame.default(formula = f, data = data, weights = weights, :
invalid type (closure) for variable '(weights)'
Don't ask me how many hours it took me to figure this out.