CODEWITHSUNDEEP
r
jessica
jessica

Reputation: 1355

String Loop to Form a Variable name in R

calld=data.frame(matrix(rnorm(100*50,0,1),1000,50))

for (x in names(calld)) {
   assign(paste("calld$",x,sep=""),pnorm(get(paste("calld$",x,sep="")),0,1,lower.tail=T,log.p=F))
    }
Error in get(paste("calld$", x, sep = "")) : object 'calld$X1' not found

Am I using the get function correctly?? I am trying to concatenate the names of the data set via a loop and paste of it's existing valued by passing the values through a pnorm (cumulative normal distribution function). But I keep getting an error. The function works when I call the variable names in the "calld" dataframe. The problem is the concentration process of creating the loop. Where am I going wrong? I appreciate your help

Update::

I took your advice guys and reedited the loop, to. 

for (n in names(calld)) {

get("calld")[[n]]=pnorm(get("calld")[[n]],0,1,lower.tail=T,log.p=F)

}

Error in get("calld")[[n]] = pnorm(get("calld")[[n]], 0, 1, lower.tail = T,  : 
  target of assignment expands to non-language object

But now I am getting this new error. So everything on the right hand side of the equation in the loop when I tested it it works. The error arises when I set it the value equal to itself, replacing the prior values.

Upvotes: 0

Views: 147

Answers (1)

mnel
mnel

Reputation: 115485

Have mercy on kittens!

You can't use assign this way, nor get. 

 calld[] <- lapply(calld, pnorm, mean = 0, sd = 1)

Explanantion: calld[]<- replaces all existing columns of calld (whilst retaining the structure as a data.frame) with the results of lapply(calld, pnorm, mean = 0, sd = 1) which cycles through all columns of calld, applying pnorm on each one.

 library(fortunes)
 fortune(312)

The problem here is that the $ notation is a magical shortcut and like any other magic if used incorrectly is likely to do the programmatic equivalent of turning yourself into a toad.

-- Greg Snow (in response to a user that wanted to access a column whose name is stored in y via x$y rather than x[[y]])

R-help (February 2012)

Upvotes: 4

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