[plyr] Question regarding ddply: use of .(as.name(varname)) and varname in ddply function
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
<research.baba at gmail.com> 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 <hadley at rice.edu> wrote:
On Mon, Dec 6, 2010 at 3:58 AM, Sunny Srivastava <research.baba at gmail.com> 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/