Thank you for the help. Much appreciated. On Mon, Dec 6, 2010 at 9:12 PM, Hadley Wickham wrote: > It's easiest to see what's going on if you use eval.quoted directly: > > eval.quoted(.(cyl), mtcars) > eval.quoted(.("cyl"), mtcars) > eval.quoted(.(as.name("cyl")), mtcars) > > But you shouldn't need to do any syntactic hackery because the default > method automatically parses the string for you: > > eval.quoted(as.quoted("cyl"), mtcars) > > Hadley > > On Mon, Dec 6, 2010 at 6:22 PM, Sunny Srivastava > wrote: > > Hi Hadley: > > I was trying to use ddply using the format . (var1) for splitting. > > I thought . ( as.name(grp) ) would do the same thing. But it does not. I > was > > just trying to know my mistake. I am sorry if it is a basic question. > > Thank you and others for your reply. > > Best Regards, > > S. > > > > On Mon, Dec 6, 2010 at 5:28 PM, Hadley Wickham wrote: > >> > >> On Mon, Dec 6, 2010 at 3:58 AM, Sunny Srivastava > >> wrote: > >> > Dear R-Helpers: > >> > > >> > I am using trying to use *ddply* to extract min and max of a > particular > >> > column in a data.frame. I am using two different forms of the > function: > >> > > >> > > >> > ## var_name_to_split is a string -- something like "var1" which is the > >> > name > >> > of a column in data.frame > >> > > >> > ddply( df, .(as.name(var_name_to_split)), function(x) c(min(x[ , 3] , > >> > max(x[ > >> > , 3]))) ## fails with an error - case 1 > >> > ddply( df, var_name_to_split , function(x) c(min(x[ , 3] , max(x[ , > >> > 3]))) > >> > ## works fine - case 2 > >> > > >> > I can't understand why I get the error in case 1. Can someone help me > >> > please? > >> > >> Why do you expect case 1 to work? > >> > >> Hadley > >> > >> -- > >> Assistant Professor / Dobelman Family Junior Chair > >> Department of Statistics / Rice University > >> http://had.co.nz/ > > > > > > > > -- > Assistant Professor / Dobelman Family Junior Chair > Department of Statistics / Rice University > http://had.co.nz/ > [[alternative HTML version deleted]]