I am searching for a fast and simple way to give comprehensible names to the columns of a VAR-coefficient matrix.
What I would like to use is the function VAR.names, which is used in the function VAR.est() in the VAR.etp-package. When I use the function VAR.est(), this works perfectly, but as soon as I modify VAR.est (by adding another element to the list of values which are returned), I receive an error message stating "could not find function VAR.names".
I could not find any information on the function VAR.names.
Example:
library(VAR.etp)
data(dat)
M=VAR.est(dat,p=2,type="const")
M$coef
Another possibility would be to use a loop as in the function VAR() from the vars package, but if VAR.names would actually work, this would be a lot more elegant!
Related
bmi<-function(x,y){
(x)/((y/100)^2)
}
bmi(70,177) it can work
but with apply() it does't work
apply(Student,1:2,bmi(Student$weight,Student$height))
Error in match.fun(FUN) :
'bmi(Student$weight, Student$height)' is not a function, character or symbol
It's a bit unclear what the goal is. If it's just to get an answer, then the comments do answer it. If on the other hand, the goal is to understand what you are doing wrong, then read on. I'd say the first error going from left to right is passing the whole dataframe. I would have only passed the 'height' and 'weight' columns.
The next error, again going from left to right, is the use of 1:2 as the second argument to apply. You obviously want to do this "by rows" which mean you should use only 1, i.e. the first dimension of the dataframe.
And the third error is using a function call rather than the function name. Functions with arguments in parentheses don't work when an R function (meaning apply in this case) is expecting a function name or an anonymous function as illustrated in comments.
Fourth error is not assigning the value to a column in your dataframe. So this probably would have succeeded in making the desired extra column via the apply method. But, as noted in comments this is not the most efficient method.:
Student$bmi_val <- apply(Student[ ,c("weight", "height")], bmi)
# didn't want my column name to be the same as the function name
The apply function was actually designed to work with matrices and arrays, so for many purposes it is ill-suited when used with dataframes. In this case where all the arguments to the bmi function are numeric and you can control the order of argument in the first argument to match the x and y positions, it's arguably an acceptable strategy, but not most R-ish method. When working with dates or factor variables, you should definitely avoid apply.
How do I remove an object from the current function environment?
I'm trying to achieve this:
foo <- function(bar){
x <- bar
rm(bar, envir = environment())
print(c(x, is.null(bar)))
}
Because I want the function to be able to handle multiple inputs.
Specifically I'm trying to pass either a dataframe or a vector to the function, and if I'm passing a dataframe I want to set the vector to NULL for later error handling.
If you want, you can watch my DepthPlotter script, where I want to let the second function check if depth is a dataframe, and if so, assign it to df in stead and remove depth from the environment.
Here is a very brief sketch of how to set this up using S3 method dispatch.
First, you define your generic:
DepthPlotter <- function(depth,...){
UseMethod("DepthPlotter", depth)
}
Then you define methods for specific classes of the argument depth. As a very basic example in your case, you might create only two, a data.frame method and a default method to handle the vector case:
DepthPlotter.default <- function(depth, variable, ...){
#Here you write a function assuming that depth is
# anything but a data frame
}
DepthPlotter.data.frame <- function(depth,...){
#Here you'd write a function that assumes
# that depth is a data frame
}
And then you can call DepthPlotter() using either type of argument and the correct function will be run based upon the result of class(depth).
The example I've sketched out here is a little crude, since I've used a default method to handle the vector case. You could write .numeric and .integer methods to handle numeric or integer vectors more specifically. In my example, the .default method will be called for any case other than data.frame, so if you go this route you'd want to write some code in there that checks for strange cases like depth being a complicated list, or other odd object, if you think there's a chance something like that might be passed to the function.
I am new to R and mostly working with old code written by someone else. And I am trying to create my own R functions.
I found some of the following code used for eigenvalue decomposition.
eigenMatrix = eigen(myMatrix)[[2]]
eigenVals = eigen(myMatrix)[[1]]
Here there is single function that can output 2 different data structures, being, a vector and a matrix depending of the value in the brackets.
When I search of functions with multiple outputs, they usually use lists to output multiple variables at once which does not work, possibly because of different types.
I don't understand why there are two setts of brackets and how the underlying function would work.
The posted code takes the eigen function, which returns a list with 2 values.
Then the [[]] are use to extract the first and second items from the list.
The [[]] is needed to return the underlying structure, and is better explained here: How to Correctly Use Lists in R?
Also, since the eigen function is run twice the code in the question is inefficient.
resultList = eigen(myMatrix)
eigenMatrix = resultList[[2]]
eigenVals = resultList[[1]]
This code is better since eigen is run only once and saves the result of the function as a list and then reads the values from the list.
For the function itself can be coaded as any function with multiple outputs such as here: https://stat.ethz.ch/pipermail/r-help/2007-March/126851.html or here: How to assign from a function with multiple outputs?
The list values can hold any structure and [[]] can be used to return the underlying structure of each value.
I have this function
ANN<-function (x,y){
DV<-rep(c(0:1),5)
X1<-c(1:10)
X2<-c(2:11)
ANN<-neuralnet(x~y,hidden=10,algorithm='rprop+')
return(ANN)
}
I need the function run like
formula=X1+X2
ANN(DV,formula)
and get result of the function. So the problem is to say the function USE the object which was created during the run of function. I need to run trough lapply more combinations of x,y, so I need it this way. Any advices how to achieve it? Thanks
I've edited my answer, this still works for me. Does it work for you? Can you be specific about what sort of errors you are getting?
New response:
ANN<-function (y){
X1<-c(1:10)
DV<-rep(c(0:1),5)
X2<-c(2:11)
dat <- data.frame(X1,X2)
ANN<-neuralnet(DV ~y,hidden=10,algorithm='rprop+',data=dat)
return(ANN)
}
formula<-X1+X2
ANN(formula)
If you want so specify the two parts of the formula separately, you should still pass them as formulas.
library(neuralnet)
ANN<-function (x,y){
DV<-rep(c(0:1),5)
X1<-c(1:10)
X2<-c(2:11)
formula<-update(x,y)
ANN<-neuralnet(formula,data=data.frame(DV,X1,X2),
hidden=10,algorithm='rprop+')
return(ANN)
}
ANN(DV~., ~X1+X2)
And assuming you're using neuralnet() from the neuralnet library, it seems the data= is required so you'll need to pass in a data.frame with those columns.
Formulas as special because they are not evaluated unless explicitly requested to do so. This is different than just using a symbol, where as soon as you use it is evaluated to something in the proper frame. This means there's a big difference between DV (a "name") and DV~. (a formula). The latter is safer for passing around to functions and evaluating in a different context. Things get much trickier with symbols/names.
I'm trying to subset a dataframe within a function using a mixture of fixed variables and some variables which are created within the function (I only know the variable names, but cannot vectorise them beforehand). Here is a simplified example:
a<-c(1,2,3,4)
b<-c(2,2,3,5)
c<-c(1,1,2,2)
D<-data.frame(a,b,c)
subbing<-function(Data,GroupVar,condition){
g=Data$c+3
h=Data$c+1
NewD<-data.frame(a,b,g,h)
subset(NewD,select=c(a,b,GroupVar),GroupVar%in%condition)
}
Keep in mind that in my application I cannot compute g and h outside of the function. Sometimes I'll want to make a selection according to the values of h (as above) and other times I'll want to use g. There's also the possibility I may want to use both, but even just being able to subset using 1 would be great.
subbing(D,GroupVar=h,condition=5)
This returns an error saying that the object h cannot be found. I've tried to amend subset using as.formula and all sorts of things but I've failed every single time.
Besides the ease of the function there is a further reason why I'd like to use subset.
In the function I'm actually working on I use subset twice. The first time it's the simple subset function. It's just been pointed out below that another blog explored how it's probably best to use the good old data[colnames()=="g",]. Thanks for the suggestion, I'll have a go.
There is however another issue. I also use subset (or rather a variation) in my function because I'm dealing with several complex design surveys (see package survey), so subset.survey.design allows you to get the right variance estimation for subgroups. If I selected my group using [] I would get the wrong s.e. for my parameters, so I guess this is quite an important issue.
Thank you
It's happening right as the function is trying to define GroupVar in the beginning. R is looking for the object h by itself (not within the dataframe).
The best thing to do is refer to the column names in quotes in the subset function. But of course, then you'd have to sidestep the condition part:
subbing <- function(Data, GroupVar, condition) {
....
DF <- subset(Data, select=c("a","b", GroupVar))
DF <- DF[DF[,3] %in% condition,]
}
That will do the trick, although it can be annoying to have one data frame indexing inside another.