You can automate this step
>key[key == "AT"] <- "TA"
## create a function to reverse a string -- see strsplit help page for this
strReverse function
reverse <- function(x) sapply(lapply(strsplit(x, NULL), rev), paste,
collapse="")
key <- rownames(a)
# combine rownames with reverse (rownames)
n<-
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)[
Dear Jacy,
If AT and TA always one after the other, you might consider the following as
an alternative:
res <- apply(a, 2, function(x) c(x[1], sum(x[2:3]), x[4] ))
rownames(res) <- rownames(a)[-3]
res
#[,1] [,2] [,3] [,4]
#AA159 13
#AT5 13 21 29
#TT48 12 16
HTH
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 elegan
jim holtman schrieb:
Try this:
key <- rownames(a)
key[key == "AT"] <- "TA"
do.call(rbind, by(a, key, colSums))
something like
paste(sort(strsplit(key, split="")[[1]]), "")
might be more general.
__
R-help@r-project.org mailing list
https:/
Try this:
> key <- rownames(a)
> key[key == "AT"] <- "TA"
> do.call(rbind, by(a, key, colSums))
V2 V3 V4 V5
AA 1 5 9 13
TA 5 13 21 29
TT 4 8 12 16
On Mon, May 11, 2009 at 4:53 PM, Crosby, Jacy R
wrote:
> I'm working with genotype data in a frequency table:
>
> > a=matrix(1:16, nrow=4)
I'm working with genotype data in a frequency table:
> a=matrix(1:16, nrow=4)
> rownames(a)=c("AA","AT","TA","TT")
> a
[,1] [,2] [,3] [,4]
AA159 13
AT26 10 14
TA37 11 15
TT48 12 16
'AT' and 'TA' are essentially the same, and I'd like to combine
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