Hello,
I reviewed my code and this will work now for any number of successive "TA",
I hope:
b=matrix(1:64, ncol=4)
rownames(b)=rep(c("AA","AT","TA","TT"),each=4)
key <- rownames(b)
key[key == "AT"] <- "TA"
c <- b
rownames(c)=key
for(i in 2:I(nrow(c))) {
if(rownames(c)[i]=="TA" & rownames(c)[i-1]=="TA") { c[i,] <-
colSums(c[i:I(i-1),])
c[i-1,]<-NA}} # sums the rows and replace the used rows by NA
values
c <- c[apply(c,1,function(x)any(!is.na(x))),] # removes the rows with NA
values
c
Rock
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
>
>
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