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Creating a factor from a combination of vectors

2 messages · Richard O'Keefe, Yves Brostaux

Original
Richard O'Keefe
#
Yves Brostaux <brostaux.y at fsagx.ac.be> wrote:
I want to produce a factor from a subset of the combination of two 
	vectors. I have the vectors a et b in a data-frame :
	
	 > df <- expand.grid(a=c(0, 5, 10, 25, 50), b=c(0, 25, 50, 100, 200))
	...
	and want to create a factor which levels correspond to particular 
	combinations of a and b (let's say Low for a=0 & b=0, Medium for a=10 & 
	b=50, High for a=50 & b=200, others levels set to NA), reading them from 
	a data-frame which describes the desired subset and corresponding levels.
	
	Here's my own solution (inputs are data-frames df and cas, output is the 

Why not do it the obvious way?

	ifelse(a == 0 & b == 0, "Low",
	  ifelse(a == 10 & b == 50, "Medium",
	    ifelse(a == 50 & b == 200, "High", 
	      "Other")))

gives you the mapping from vectors a and b to strings you want.
To get at the vectors locally, you need

	with(df, ...)

To convert the vector of strings you get to an ordered factor,
with "Other" mapped to NA, just do

	ordered(..., levels = c("Low","Medium","High"))

because any string not listed in levels= will be mapped to NA.
Put these pieces together, and you get

    output <- ordered(with(df,
		ifelse(a == 0 & b == 0, "Low",
		  ifelse(a == 10 & b == 50, "Medium",
		    ifelse(a == 50 & b == 200, "High",
		      "Other")))),
		levels = c("Low","Medium","High"))
Yves Brostaux
# 
Thank you Richard, but I dismissed the 'ifelse' solution because it 
needs explicit manual definition of the factor levels and corresponding 
vectors' combinations and does not define it automaticaly  from the 
'cas' data-frame (from which values, number of levels and rownames can 
vary).

Eric Lecoutre's code does exactly what I want, many thanks to both of you.

Richard A. O'Keefe a ??crit :