[R] Looking for a quick way to combine rows in a matrix

Rocko22 rock.ouimet at gmail.com
Wed May 13 14:37:22 CEST 2009


I reviewed my code and this will work now for any number of successive "TA",
I hope:

b=matrix(1:64, ncol=4)
key <- rownames(b)
key[key == "AT"] <- "TA"
c <- b

for(i in 2:I(nrow(c))) {
   if(rownames(c)[i]=="TA" & rownames(c)[i-1]=="TA") { c[i,] <-
              c[i-1,]<-NA}} # sums the rows and replace the used rows by NA
c <- c[apply(c,1,function(x)any(!is.na(x))),] # removes the rows with NA


Rocko22 wrote:
> In the first reply, what was calculated was the overall means by group
> (amino acids). It does not work for a larger database.
> I am quite really new to R, and I worked on your question just to learn
> how to manipulate data with R.
> The following seems to work. The code could be made a lot more elegant and
> straightforward, but it works only when there is no more than two
> successive "TA":
> Let's try with a matrix "b" that contains more rows than in your example:
> b=matrix(1:32, ncol=4)
> rownames(b)=rep(c("AA","AT","TA","TT"),2)
> key <- rownames(b)
> key[key == "AT"] <- "TA"
> rownames(b)=key
> for(i in 1:I(nrow(b)-1)) {
>    if(rownames(b)[i]=="TA" & rownames(b)[i+1]=="TA") { b[i,] <-
> colSums(b[i:I(i+1),])
>               b[i+1,]<-NA}} # sums the rows and replace the used rows by
> NA values
> b <- b[order(b[,1],na.last=NA),] # removes the rows with NA values
> Of course, the rows are reordered, and that may be not wanted. The
> ordering was just to remove the NA rows.
> Rock :-D

View this message in context: http://www.nabble.com/Looking-for-a-quick-way-to-combine-rows-in-a-matrix-tp23491348p23520900.html
Sent from the R help mailing list archive at Nabble.com.

More information about the R-help mailing list