assuming that the row entries for the columns with the same name are not
all zero, you can try something in the following lines:
dfrm <- data.frame(a = 1:4, a = 1:4, b = 1:4,
b = 1:4, b = 1:4, check.names = FALSE)
dfrm[3, 1:3] <- NA
dfrm
vals <- unlist(dfrm)
res <- tapply(vals, names(vals), sum, na.rm = TRUE)
res[res == 0] <- NA
as.data.frame(matrix(res, ncol = 2))
I hope it helps.
Best,
Dimitris
On 5/16/2011 4:25 PM, Assu wrote:
Hi all
I have a data frame with duplicate columns and i want to remove duplicates
by adding rows in each group of duplicates, but have lots of NA's.
Data:
dfrm<- data.frame(a = 1:4, b= 1:4, cc= 1:4, dd=1:10, ee=1:4)
names(dfrm)<- c("a", "a", "b", "b", "b")
dfrm[3,2:3]<-NA
dfrm
a a b b b
1 1 1 1 1 1
2 2 2 2 2 2
3 NA NA NA 3 3
4 4 4 4 4 4
I did: sapply(unique(names(dfrm)),function(x){
rowSums(dfrm[ ,grep(x, names(dfrm)),drop=FALSE])})
which works. However, I want rowSums conditional:
1) if there is at least one value non NA in a row of each group of
duplicates, apply rowSums to get the value independently of the existence of
other NA's in the group row.
2) if all values in a row of duplicates are NA, I get NA
In my data dfrm I would get
a b
1 2 3
2 4 6
3 NA 6
4 8 12
Can't use na.rm=TRUE or FALSE.
I tried: sapply(unique(names(dfrm)),function(x) ifelse(any(!is.na(dfrm[
,grep(x, names(dfrm))])), rowSums(dfrm[ ,grep(x,
names(dfrm)),drop=FALSE],na.rm=TRUE),NA))
and it doesn't work.
Can someone please help me?
Thanks in advance.
--
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
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