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 :
Yves Brostaux <[EMAIL PROTECTED]> 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"))
--
Ir. Yves BROSTAUX
Unit� de Statistique et Informatique
Facult� universitaire des Sciences agronomiques de Gembloux (FUSAGx)
8, avenue de la Facult�
B-5030 Gembloux
Belgique
T�l: +32 81 62 24 69
Email: [EMAIL PROTECTED]
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