Read in the data noting that the row names are in column "name" and
interpreting character data as character type.
The sapply loops over columns creating factors with the same levels (in
the same order), orders them and uses that to index the row.names. Finally
we create row names for the result out of the levels.
d <- read.table("clipboard", header=T, row.names="name", as.is=T)
lev <- sort(d[,1])
d2 <- sapply( d, function(x) row.names(d)[ order( factor(x,levels=lev) ) ] )
row.names(d2) <- lev
Douglas Federman <dfederman <at> mco.edu> writes:
:
: I have the following dataframe "s"
:
: name jul aug sep
: 1 Joe m1 m2 m3
: 2 Bill m2 m3 m2
: 3 Sue m3 m1 m1
:
: I am interested in transforming the data into the following:
:
: jul aug sep
: m1 Joe Sue Sue
: m2 Bill Joe Bill
: m3 Sue Bill Joe
:
: Is there a simple way to do this?
:
: --
: "It has often been remarked that an educated man has probably forgotten most
of the facts he acquired in
: school and university. Education is what survives when what has been learned
has been forgotten." -- B.F.
: Skinner, New Scientist, 31 May 1964, p. 484
:
:
:
: BEGIN:VCARD
: VERSION:2.1
: X-GWTYPE:USER
: FN:Douglas Federman
: ORG:;1331
: EMAIL;WORK;PREF;NGW:dfederman <at> mco.edu
: N:Federman;Douglas
: END:VCARD
:
:
:
: ______________________________________________
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