Re: [R] simple problem, but not for me

From: Rolf Turner <rolf_at_math.unb.ca>
Date: Sun 20 Mar 2005 - 04:38:52 EST

alexbri@netcabo.pt wrote:

> Hello, I'm new in R and I want to do one thing that is very easy in
> excel, however, I cant do it in R.

	Well, if you've deadened your brain by using Excel,
	no wonder.

> Suppose we have the data frame:
>
> data<- data.frame(A=c("a1","a2","a3","a4","a5"))

	Oh, for Pete's sake!  This makes ``data'' (NOT a good name
	for an object!) into a data frame with a single column named
	``A''.  That column will be a factor with 5 entries (an 5
	levels) with these levels being (the character strings)
	"a1","a2","a3","a4", and "a5".
	
	Nothing to do with what you actually want.
 

> I need to obtain another column in the same data frame (lets say
> B=c(b1,b2,b3,b4,b5) in the following way:

	This would make B a ***vector*** equal to the concatenation
	of b1, ..., b5.

	Perhaps you mean:

		B <- cbind(b1,b2,b3,b4,b5)

> b1=a1/(a1+a2+a3+a4+a5)
>
> b2=a2/(a2+a3+a4+a5)
>
> b3=a3/(a3+a4+a5)
>
> b4=a4/(a4+a5)
>
> b5=a5/a5
>
> a1..a5 and b1...b5 are always numeric values
>
> (this is just an example, what I really want is apply this kind of
> formula to a much larger data frame)
>

	You are very confused.  Your notation and your use of
	the function c() are all wrong.

	If you are going to use R, get the basic syntax straight.

	You probably should be using matrices rather than data frames
	given that the entries are all numeric.

	Be that as it may, if A is a (numeric) matrix then

		B <- A/t(apply(A,1,function(x){rev(cumsum(x))}))

	will give what you appear to want.

				cheers,

					Rolf Turner
					rolf@math.unb.ca

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