[R] Tying to underdressed the magic of lm redux

[Sorkin, John]

Bert,
Thank you for your reply. You are correct that your code will print the contents of the data frame. While it works, it is not as elegant as the lm function. One does not have to pass the independent and dependent variables to lm In parentheses.

Fit1<-lm(y~x,data=mydata)

None of the parameters to lm are passed in quotation marks. Somehow, using deparse(substitute()) and other magic lm is able to get the data in the dataframe mydata. I want to be able to do the same magic in functions I write; pass a dataframe and column names, all without quotation marks and be able to write code that will provide access to the columns of the dataframe without having to pass the column names in quotation marks.
Thank you,
John

John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine
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On May 29, 2019, at 9:59 PM, Bert Gunter <bgunter.4567 using gmail.com<mailto:bgunter.4567 using gmail.com>> wrote:

Basically, huh?

> df <- data.frame(a = 1:3, b = letters[1:3])
> nm <- names(df)
> print(df[,nm[1]])
[1] 1 2 3
> print(df[,nm[2]])
[1] a b c
Levels: a b c

This can be done within a function, of course:

> demo <- function(df, colnames){
+    print(df[,colnames])
+ }
> demo(df,c("a","b"))
  a b
1 1 a
2 2 b
3 3 c

Am I missing something? (Apologies, if so).

Bert Gunter



On Wed, May 29, 2019 at 6:40 PM Sorkin, John <jsorkin using som.umaryland.edu<mailto:jsorkin using som.umaryland.edu>> wrote:
Thanks to several kind people, I understand how to use deparse(substitute(paramter)) to get as text strings the arguments passed to an R function. What I still can't do is put the text strings recovered by deparse(substitute(parameter)) back together to get the columns of a dataframe passed to the function. What I want to do is pass a column name to a function along with the name of the dataframe and then, within the function access the column of the dataframe.

I want the function below to print the columns of the dataframe testdata, i.e. testdata[,"FSG"] and testdata[,"GCM"]. I have tried several ways to tell the function to print the columns; none of them work.

I thank everyone who has helped in the past, and those people who will help me now!

John

testdata <- structure(list(FSG = c(271L, 288L, 269L, 297L, 311L, 217L, 235L,

                                   172L, 201L, 162L), CGM = c(205L, 273L, 226L, 235L, 311L, 201L,

                                   203L, 155L, 182L, 163L)), row.names = c(NA, 10L), class = "data.frame")

cat("This is the data frame")

class(testdata)

testdata



BAPlot <- function(first,second,indata){

  # these lines of code work

    col1 <- deparse(substitute(first))

    col2 <- deparse(substitute(second))

    thedata <- deparse(substitute(third))

    print(col1)

    print(col2)

    print(thedata)

    cat("This gets the data, but not as a dataframe\n")

    zoop<-paste(indata)

    print(zoop)

    cat("End This gets the data, but not as a dataframe\n")

     # these lines do not work

    print(indata[,first])

    print(indata[,"first"])

    print(thedata[,col1])

    paste(zoop[,paste(first)])

    paste(zoop[,first])

    zap<-paste(first)

    print(zap)

}




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