Hi,
If you only need "aa" to be replaced by "a", "bb" by "b" and so on, try
the following:
Dat$att <- as.factor(substr(Dat$att,1,1))
HTH,
Pascal
Le 15/01/2013 18:48, Christofer Bogaso a écrit :
Dear all,
Let say I have following data-frame:
Dat <- structure(list(dat = c(-0.387795842956327, -0.23270882099043,
-0.89528973290562, 0.95857175595512, 1.61680582493783, -1.17738110289352,
0.210601060411423, -0.827369747447338, -0.36896112964414, 0.440288648776096,
1.28018410608809, -0.897113649961341, 0.342216546981718, -1.17288066266219,
-1.57994101992621, -0.913655547602414, -2.54753726314408, -0.617703410989815,
-0.443272763891558, 0.359181170918489), att = structure(c(2L,
2L, 2L, 4L, 2L, 2L, 1L, 3L, 1L, 4L, 1L, 2L, 2L, 2L, 2L, 4L, 2L,
2L, 4L, 2L), .Label = c("aa", "bb", "cc", "dd"), class = "factor")),
.Names = c("dat",
"att"), row.names = c(NA, -20L), class = "data.frame"); Dat
Now I want to replace the 2nd column with:
Replace <- letters[1:20]
The rule is as follows:
Consider the 1st element of the 2nd colume: 'bb'. Now I need to see
which element of 'Replace' is contained in 'bb'? Obviously, this will
be 'b'. Therefore I need to replace "bb" with "b". And so on for all
elements of 2nd column.
This is obviously an straightforward example. However I have some more
complex problem of replacing like that. I want to get the idea how to
achieve that programmatically.
Can somebody help me?
Thanks and regards,
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