When I try:
x=c(1,2,3)
...I get the message unused argument (3)
When I type x=c(1,2) it takes the function but when I type x it gives me a value of [3]. I want to create a simple vector of numbers but I don't know why it is giving me this message.
I presume you created a user-defined function called c, thereby replacing the built-in base-r function. Try rm("c") and then everything should work as expected again.
Related
This is the code in R and I'm having trouble understanding the role of function(x) and qdata[[x]] in this line of code. Can someone elaborate me this piece by piece? I didn't write this code. Thank you
outs=lapply(names(qdata[,12:35]), function(x)
hist(qdata[[x]],data=qdata,main="Histogram of Quality Trait",
xlab=as.character(x),las=1.5)$out)
This code generate a series of histograms, one for each of columns 12 to 35 of dataframe qdata. The lapply function iterates over the columns. At each iteraction, the name of the current column is passed as argument "x" to the anonymous function defined by "function(x)". The body of the function is a call to the hist() function, which creates the histogram. qdata[[x]] (where x is the name of a column) extracts the data from that column. I am actually confused by "data=qdata".
We don't have the data object named qdata so we cannot really be sure what will happen with this code. It appears that the author of this code is trying to pass the values of components named outs from function calls to hist. If qdata is an ordinary dataframe, then I suspect that this code will fail in that goal, because the hist function does not have an out component. (Look at the output of ?hist. When I run this with a simple dataframe, I do get histogram plots that appear in my interactive plotting device but I get NULL values for the outs components. Furthermore the 12 warnings are caused by the lack of a data parameter to hte hist function.
qdata <- data.frame(a=rnorm(10), b=rnorm(10))
outs=lapply(names(qdata), function(x)
hist(qdata[[x]],data=qdata,main="Histogram of Quality Trait",
xlab=as.character(x),las=1.5)$out)
#There were 12 warnings (use warnings() to see them)
> str(outs)
List of 2
$ : NULL
$ : NULL
So I think we need to be concerned about the level of R knowledge of the author of this code. It's possible I'm wrong about this presumption. The hist function is generic and it is possible that some unreferenced package has a function designed to handle a data object and retrun an outs value when delivered a vector having a particular class. In a typical starting situation with only the base packages loaded however, there are only three hist.* functions:
methods(hist)
#[1] hist.Date* hist.default hist.POSIXt*
#see '?methods' for accessing help and source code
As far as the questions about the role of function and [[x]]: the keyword function returns a language object that can receive parameter values and then do operations and finally return results. In this case the names get passed to the anonymous function and become, each in turn, the local name, x and the that value is used by the '[['-function to look-up the column in what I am presuming is the ‘qdata’-dataframe.
I want to minimize the first 11 (or any large number) parameters of a function, which takes 15 (or any number greater than 11 / the first number stated) parameters in total. The problem is that when I use the optim function with par=c(...) for the first 11 values, the function cannot be evaluated as it takes all 11 values for its first argument only. Is there any way to force each input argument to be a scalar / single number?
To be more specific:
fun_to_optim <- function(m1,m2,m3,m4,m5,m6,m7,m8,m9,m10,m11,p1,p2,p3,p4)
{
# I want to make sure m1 to m11 are single numbers and no vectors
...
}
# optimization
optim(par=c(0,0,0,0,0,0,0,0,0,0,0),fn=fun_to_optim,p1=data,p2=opt,p3=data,p4=1)
Trying to run this code always ends with the error message: Error in fn(par, ...) : argument "m1" is missing, with no default. By printing out the arguments, I figured that the function assumes that all values from par are for the input variable m1. So m1 is taken to be a vector with the initial values given through par.
Please note: I am aware of the possibility to simply put the first 11 parameters into a vector (let's say para_to_optim) and then call the optim function as I did. However, I explicitly do not want to do this because I need to optimize the parameters iteratively, i.e. starting with just 1 parameter and ending will all 11. Writing the function 11 times with different input argument structures seems a bit inefficient.
I have already checked similar post for function arguments such as Forcing specific data types as arguments to a function and How to define argument types for R functions?, but it would not be possible in my case to check the type of the argument within the function and wrapper or S3 classes are also kind of problematic. Is there any other way to solve this?
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.
Having come from Matlab I am struggling to work out why the following does not work:
plot(x=rand(10),y=rand(10))
Produces a graph correctly.
x=rand(10)
y=rand(10)
plot(x,y)
produces error:
ERROR: plot has no method matching plot(::Array(Float64,1),::Array(Float64,1))
I would be very grateful if someone coould explain to me why embeding the code within the plot line produces a result, but defining the variables beforehand results in an error. Logic says they should produce the same result.
I am using Julia v 0.3.1 and have loaded Gadfly as charting tool.
In the first case, you are using keyword argument syntax, not assigning to variables x and y (the meaning of = inside function calls is special). To get the same effect in the second case, you should use
x=rand(10)
y=rand(10)
plot(x=x,y=y)
which passes the value in the variable x in the keyword argument x to plot, and the value in the variable y in the keyword argument y.
In case you didn't. Write this before your code:
using plots
plyplot()
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!