R help - Jun 2012 - passing function parameters into a 'with' statement to dynamically pick out columns

Hi,

I've built a function to generate plots and would like to be able pass in
column names as a parameter. Here's a boiled down version of what I have
right now:

pmts <- data.frame(date=c(1,2,3), all=c(5,6,7),maj=c(4,5,6),ind=c(3,4,5))
perc.mktshare <- function(df){  
  range1 <- floor(min(with(df, 100*ind/all)))
  range2 <- ceiling(max(with(df, 100*ind/all)))  
  with(df,plot(date,100*ind/all,ylim=c(range1,range2),ylab="%
Marketshare"))
}
perc.mktshare(pmts)

What I'd like to do is something like this:

perc.mktshare <- function(df, scaling.column){  
  range1 <- floor(min(with(df, 100*scaling.column/all)))
  range2 <- ceiling(max(with(df, 100*scaling.column/all)))  
 
with(df,plot(date,100*"scaling.column"/all,ylim=c(range1,range2),ylab="%
Marketshare"))
}
perc.mktshare(pmts,"ind")
perc.mktshare(pmts,"maj")

I've tried going about this a couple of different ways but I can't make
it work. My suspicion is that there's probably no way to do this using the
'with' statement and I'll have to trim the input dataframe to the
pertinent columns first using subset or something.

Thanks for your help!

David

"With" is designed for interactive use, so you are right that it is
not the answer.

You can use list indexing with column names within functions, though, like
df[[scaling.column]] or df[["all"]].
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David Marx <dmarx at SOUNDEXCHANGE.COM> wrote:
>Hi,
>
>I've built a function to generate plots and would like to be able pass
>in column names as a parameter. Here's a boiled down version of what I
>have right now:
>
>pmts <- data.frame(date=c(1,2,3),
>all=c(5,6,7),maj=c(4,5,6),ind=c(3,4,5))
>perc.mktshare <- function(df){  
>  range1 <- floor(min(with(df, 100*ind/all)))
>  range2 <- ceiling(max(with(df, 100*ind/all)))  
>with(df,plot(date,100*ind/all,ylim=c(range1,range2),ylab="%
>Marketshare"))
>}
>perc.mktshare(pmts)
>
>What I'd like to do is something like this:
>
>perc.mktshare <- function(df, scaling.column){  
>  range1 <- floor(min(with(df, 100*scaling.column/all)))
>  range2 <- ceiling(max(with(df, 100*scaling.column/all)))  
>with(df,plot(date,100*"scaling.column"/all,ylim=c(range1,range2),ylab="%
>Marketshare"))
>}
>perc.mktshare(pmts,"ind")
>perc.mktshare(pmts,"maj")
>
>I've tried going about this a couple of different ways but I can't
make
>it work. My suspicion is that there's probably no way to do this using
>the 'with' statement and I'll have to trim the input dataframe
to the
>pertinent columns first using subset or something.
>
>Thanks for your help!
>
>David
>
>______________________________________________
>R-help at 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.

On Jun 20, 2012, at 16:45 , David Marx wrote:
> Hi,
> 
> I've built a function to generate plots and would like to be able pass
in column names as a parameter. Here's a boiled down version of what I have
right now:
> 
> pmts <- data.frame(date=c(1,2,3),
all=c(5,6,7),maj=c(4,5,6),ind=c(3,4,5))
> perc.mktshare <- function(df){  
>  range1 <- floor(min(with(df, 100*ind/all)))
>  range2 <- ceiling(max(with(df, 100*ind/all)))  
>  with(df,plot(date,100*ind/all,ylim=c(range1,range2),ylab="%
Marketshare"))
> }
> perc.mktshare(pmts)
> 
> What I'd like to do is something like this:
> 
> perc.mktshare <- function(df, scaling.column){  
>  range1 <- floor(min(with(df, 100*scaling.column/all)))
>  range2 <- ceiling(max(with(df, 100*scaling.column/all)))  
> 
with(df,plot(date,100*"scaling.column"/all,ylim=c(range1,range2),ylab="%
Marketshare"))
> }
> perc.mktshare(pmts,"ind")
> perc.mktshare(pmts,"maj")
> 
> I've tried going about this a couple of different ways but I can't
make it work. My suspicion is that there's probably no way to do this using
the 'with' statement and I'll have to trim the input dataframe to
the pertinent columns first using subset or something.
> 
Something with eval(bquote(.....)) should do it, e.g.

foo <- function(col) eval(bquote(with(airquality, plot(.(col), Ozone))))
foo(quote(Wind))
foo(quote(Solar.R))

Or, use .(as.name(col)) inside bquote, and foo("Wind") in the call.

Possibly, substitute() can be used as well, but it might get tricky.


> Thanks for your help!
> 
> David
> 
> ______________________________________________
> R-help at 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, Professor
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