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,] -
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-cbind(key,
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
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
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,
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 (add) the
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
jacy.r.cro...@uth.tmc.eduwrote:
I'm working with genotype data in a frequency table:
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