access a column of a dataframe without qualifying the name of the column

Here is an alternaive approach that is closer to that used by lm and friends.
df <- data.frame(x=1:10,y=11:20)
test <- function(col, dat) eval(substitute(col), envir = dat)
test(x, df)
[1]  1  2  3  4  5  6  7  8  9 10
test(y, df)
[1] 11 12 13 14 15 16 17 18 19 20

There is a slight added bonus this way
test(x+y+1, df)
[1] 13 15 17 19 21 23 25 27 29 31

(Well, I did say 'slight'.)

Bill Venables.
On Dec 29, 2010, at 7:11 PM, John Sorkin wrote:

> I am trying to write a function that will access a column of a data
> frame without having to qualify the name of the data frame column as
> long as the name of the dataframe is passed to the function. As can
> be seen from the code below, my function is not working:

Not sure what the verb "qualify" means in programming. Quoting?

>
> df <- data.frame(x=1:10,y=11:20)
> df
>
> test <- function(column,data) {
>  print(data$column)
> }
>
> test(x,df)
>
> I am trying to model my function after the way that lm works where
> one needs not qualify column names, i.e.

 > df <- data.frame(x=1:10,y=11:20)
 > test <- function(column,dat) { print(colname <-
deparse(substitute(column)))
+  dat[[colname]]
+ }
 >
 > test(x,df)
[1] "x"
  [1]  1  2  3  4  5  6  7  8  9 10
 >

--
David.

>
>
> fit1<- lm(y~x,data=df)
>
>
> John David Sorkin M.D., Ph.D.
> Chief, Biostatistics and Informatics
> University of Maryland School of Medicine Division of Gerontology
> Baltimore VA Medical Center
> 10 North Greene Street
> GRECC (BT/18/GR)
> Baltimore, MD 21201-1524
> (Phone) 410-605-7119
> (Fax) 410-605-7913 (Please call phone number above prior to faxing)
>
> Confidentiality Statement:
> This email message, including any attachments, is for th...{{dropped:
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>
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David Winsemius, MD
West Hartford, CT

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access a column of a dataframe without qualifying the name of the column

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