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

Here is an alternaive
Now that's a rather interesting novel word you got there, Bill....
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.

Yes, and similar stuff goes on in subset.data.frame(), with(), etc. However, there is a trend towards preferring to do non-standard evaluation in a more standard (less non-standard?) way, using formulas:
test <- function(f, dat) eval(f[[2]], envir = dat, enclos=environment(f))
df <- data.frame(x=1:10,y=11:20)
test(~x,df)
[1]  1  2  3  4  5  6  7  8  9 10
test(~x+y,df)
[1] 12 14 16 18 20 22 24 26 28 30

Or, you could pass expression(x) or quote(x) explicitly, but it's a bit more long-winded and you lose the ability to pass along the environment of the formula (and typically use enclos=parent.frame() instead).

________________________________________
From: r-help-bounces at r-project.org [r-help-bounces at r-project.org] On Behalf Of David Winsemius [dwinsemius at comcast.net]
Sent: 30 December 2010 10:44
To: John Sorkin
Cc: r-help at r-project.org
Subject: Re: [R] access a column of a dataframe without qualifying the name     of the column

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)

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______________________________________________
R-help at r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd.mes at cbs.dk  Priv: PDalgd at gmail.com

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

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