I am putting together an R function that takes some undefined input through the ... argument described in the docs as:
"..." the special variable length argument ***
The idea is that the user will enter a number of column names here, each belonging to a dataset also specified by the user. These columns will then be cross-tabulated in comparison to the dependent variable by tapply. The function is to return a table (independent variable x indedependent variable).
Thus, I tried:
plotter=function(dataset, dependent_variable, ...)
{
indi_variables=list(...); # making a list of the ... input as described in the docs
result=with (dataset, tapply(dependent_variable, indi_variables, mean); # this fails
}
I figured this should work as tapply can take a list as input.
But it does not in this case ('Error in tapply...arguments must have same length') and I think it is because indi_variables is a list of strings.
If I input the contents of the list by hand and leave out the quotation marks, everything works just fine.
However, if the user feeds the function the column names as non-strings, R will interpret them as variable names; and I cannot figure out how to transform the list indi_variables in the right way, unsuccessfully trying things like this:
indi_variables=lapply(indi_variables, as.factor)
So I am wondering
What causes the error described above? Is my interpretation correct?
How would one go about transforming the list created through ... in the right way?
Is there an overall better way of doing this, in the input or the implementation of tapply?
Any help is much appreciated!
Thanks to Joran's helpful reading, I have come up with these improvements than make things work out...
indi_variables=substitute(list(...));
result=with (dataset, tapply(dependent_variable, eval(indi_variables, dataset), FUN=mean));
Related
im super newbie on R and i have been learning for myself for a few weeks already due my work degree.
Im almost done with the statistical analysis that i need, but it is through an ugly and messy code, that is, repeating lot of codes for several data frames, to apply different statistical tests, save results, etc.
Well now, for personal interest, want to write this better, but im totally trapped in my ignorance and really need a push to get the idea, please.
For example, i want to create a function that measure the correlation on all the data tables im using and save those results as a tables using the input name as part of the output name.
I mean, if we had the iris data but measured on different seasons, e.g. iris_fall, iris_winter, iris_spring and iris_summer, after apply cor(X) method to each one, i want to save those results as tables called like "mCoriris_fall.txt", "mCoriris_winter.txt", "mCoriris_spring.txt" and "mCoriris_summer.txt" respectively.
My useless code for now say:
cor_PQ<-function(X) {
cor_PQ<-cor(X, use="pairwise.complete.obs")
return(cor_PQ)
}
savecor<-function(t) {
outputname<-(paste0("mCor",t)) #HOW DO I CALL THE NAME OF THE INPUT? t is cor_PQ result matrix.
savecor<-write.table(t, file=paste0(outputname,".txt"))
return(savecor)
}
cor_PQ(Iris_fall)
I expect to get cor result and save it as a table in my workspace, using the input name as part of the output name.
Im aware this are 2 separates functions and the one to write table should be inside the function for cor(x), but i cant understand how.
I have been reading a lot but i just cant fit all in my head.
Thanks to anyone who can help me.
Regards.
UNTIL HERE IT HAS BEEN SOLVED...
But after making a list with my 14 data frames to apply cor and other methods, the write.table function overwrite the 14 cor results on 1 single doc. This is my code.
PQ_files<-list.files(path="C:/Users/Sol/Documents/ProyectoTÃtulo/CalidadAgua/Matrices/Regs",pattern="\\_PQ.txt")
PQ_data<-lapply(PQ_files, read.table)
names(PQ_data)<-gsub("\\_PQ.txt","", PQ_files)
PQ_data
cor_PQ<-function(X) {
cor_PQ<-cor(X, use="pairwise.complete.obs")
outputname.txt<-paste0("mCor",deparse(substitute(X)),".txt")
write.table(cor_PQ, file=outputname.txt)
outputname.pdf<-paste0("Cor",deparse(substitute(X)),".pdf")
pdf(outputname.pdf)
plot(X)
dev.off()
return(cor_PQ)
}
for (i in seq_along(PQ_data)){
Correlaciones<-lapply(PQ_data,cor_PQ)
}
Correlaciones
On SUM: seems to work almost good, until the write.table and plot(x) overwrite the outputs from the 14 dataframes on my PQ_data withe the name mCor[[i]] and CorX[[i]], respectively.
Should i define [i] somehow to have each results with the right name?
Also, when i run Correlaciones at the end, i can see the cor result for the 14 dataframes in one single dataframe, but i dont know how to split them correctly.
I guess almost there.
THANKS AGAIN!
You can combine the two functions and use deparse substitute to get input names as string
cor_PQ <- function(X) {
cor_PQ<-cor(X, use="pairwise.complete.obs")
outputname<- paste0("mCor",deparse(substitute(X)), ".txt")
write.table(t, file=outputname)
return(cor_PQ)
}
and then call
cor_PQ(Iris_fall)
Help files call attributes() a function. Its syntax looks like a function call. Even class(attributes) calls it a function.
But I see I can assign something to attributes(myobject), which seems unusual. For example, I cannot assign anything to log(myobject).
So what is the proper name for "functions" like attributes()? Are there any other examples of it? How do you tell them apart from regular functions? (Other than trying supposedfunction(x)<-0, that is.)
Finally, I guess attributes() implementation overrides the assignment operator, in order to become a destination for assignments. Am I right? Is there any usable guide on how to do it?
Very good observation Indeed. It's an example of replacement function, if you see closely and type apropos('attributes') in your R console, It will return
"attributes"
"attributes<-"
along with other outputs.
So, basically the place where you are able to assign on the left sign of assignment operator, you are not calling attributes, you are actually calling attributes<- , There are many functions in R like that for example: names(), colnames(), length() etc. In your example log doesn't have any replacement counterpart hence it doesn't work the way you anticipated.
Definiton(from advanced R book link given below):
Replacement functions act like they modify their arguments in place,
and have the special name xxx<-. They typically have two arguments (x
and value), although they can have more, and they must return the
modified object
If you want to see the list of these functions you can do :
apropos('<-$') and you can check out similar functions, which has similar kind of properties.
You can read about it here and here
I am hopeful that this solves your problem.
bmi<-function(x,y){
(x)/((y/100)^2)
}
bmi(70,177) it can work
but with apply() it does't work
apply(Student,1:2,bmi(Student$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.
I am reading Wickham's Advanced R book. This question is relating to solving Question 5 in chapter 12 - Functionals. The exercise asks us to:
Implement a version of lapply() that supplies FUN with both the name and value of each component.
Now, when I run below code, I get expected answer for one column.
c(class(iris[1]),names(iris[1]))
Output is:
"data.frame" "Sepal.Length"
Building upon above code, here's what I did:
lapply(iris,function(x){c(class(x),names(x))})
However, I only get the output from class(x) and not from names(x). Why is this the case?
I also tried paste() to see whether it works.
lapply(iris,function(x){paste(class(x),names(x),sep = " ")})
I only get class(x) in the output. I don't see names(x) being returned.
Why is this the case? Also, how do I fix it?
Can someone please help me?
Instead of going over the data frame directly you could switch things around and have lapply go over a vector of the column names,
data(iris)
lapply(colnames(iris), function(x) c(class(iris[[x]]), x))
or over an index for the columns, referencing the data frame.
lapply(1:ncol(iris), function(x) c(class(iris[[x]]), names(iris[x])))
Notice the use of both single and double square brackets.
iris[[n]] references the values of the nth object in the list iris (a data frame is just a particular kind of list), stripping all attributes, making something like mean(iris[[1]]) possible.
iris[n] references the nth object itself, all attributes intact, making something like names(iris[1]) possible.
I have been trying to figure out how to apply the apply functions plyr is out there. I will learn that later. But, I need help. I can get output with actually typing the object name in, but I am trying to loop a list through it. The code is as follows:
list<-noquote(c("T","AAVL"))
lapply(list,function(i) xts(l.df$i[,-1:-5],order.by=as.POSIXct(rownames(l.df$i))))
If I just do xts(l.df$T[,-1:-5],order.by=as.POSIXct(rownames(l.df$T))
I get the xts file that I need. Could someone please help me loop the names without quotes into the lapply(), so that I could have this work for numerous elements in my list? Thank you!
There are a number of ways to subset a list in R. See https://ramnathv.github.io/pycon2014-r/learn/subsetting.html or http://adv-r.had.co.nz/Subsetting.html for more detailed discussion.
However, in your case the issue is that the dollar operator $ takes a fixed string rather than a variable name. So myList[["item"]] and myList$item are equivalent. In the example you gave, you're trying to find the member of the data.frame called "i", not the one referenced by the variable i. The noquote class you used purely affects printing of a character vector; it has no effect on subsetting.
The version of your code that works doesn't work as you explain in your comment. It works because you're now subsetting the column whose name is stored in i not the one called "i".
I want to add 106 new columns to a dataframe that are the length of the df ofcourse and filled with zeros (0). How would I loop over i in this case:
geo <- unique(df$geo)
geo
[1] "AL" "AT1" "AT2" "AT3" "BE1" "BE2" "BE3"
for(i in geo) {
df$i <- v(0,length(df)
}
Emil Krabbe 2 mins ago Edit