Hi,
Try:
dat1 <- read.table(text="a b c d e
1 2 3 4 5
10 9 8 7 6",sep="",header=TRUE)
Names1<- read.table(text="Original New
e ee
b bb
a aa
c cc
d dd",sep="",header=TRUE,stringsAsFactors=FALSE)
It is better to dput() your dataset. For example:
dput(Names1)
structure(list(Original = c("e", "b", "a", "c", "d"), New = c("ee",
"bb", "aa", "cc", "dd")), .Names = c("Original", "New"), class = "data.frame",
row.names = c(NA,
-5L))
names(dat1) <- Names1[,2][match(names(dat1), Names1[,1])] ##
dat1
# aa bb cc dd ee
#1 1 2 3 4 5
#2 10 9 8 7 6
A.K.
On Thursday, November 21, 2013 1:45 PM, Nitisha jha <nitisha...@gmail.com>
wrote:
Hi,
Thanks. I used as.character() and got the right strings.
Btw, I have lots of handicaps regarding R.
I have to rename the columns(I have 22 columns here). I have the new names
along with the original names in another dataset. Right now, I am going
hardcoding all the 19 name changes(tedious and not optimum). 1st 3 names
remain the same. I will give u a sample dataset. Let me know if there is any
easy way of doing this. Pardon the displaced column labels.
Original dataset.
a b c d
e
1 2 3 4 5
10 9 8 7 6
Dataset for name change
Original New
e ee
b bb
a aa
c cc
d dd
I want my final dataset to be like this:
aa bb cc dd ee
1 2 3 4 5
10 9 8 7 6
Could u tell me an optimal way to do it. My method is tedious and not good.
Also, is there a way to import .xls without perl (windows)?
Thanks for being patient. :)
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