On Dec 30, 2010, at 07:12 , <bill.venab...@csiro.au> <bill.venab...@csiro.au>
wrote:
> 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-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf
> Of David Winsemius [dwinsem...@comcast.net]
> Sent: 30 December 2010 10:44
> To: John Sorkin
> Cc: r-help@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)
>>
>> Confidentiality Statement:
>> This email message, including any attachments, is for th...{{dropped:
>> 6}}
>>
>> ______________________________________________
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>
> David Winsemius, MD
> West Hartford, CT
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> 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....@cbs.dk Priv: pda...@gmail.com
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