Jim & Michael: I really appreciate your guidance in creating the function I
wanted. I took suggestions from both of you and was able to complete this
function. I had to split the process into two functions as listed below.
I just thought to send the results to the list in case someone might
Right now, using `method = "cv"` or `method = "repeatedcv"` does stratified
sampling. Depending on what you mean by "ensure" and the nature of your
outcome (categorical?), it probably already does.
On Mon, Nov 23, 2015 at 7:04 PM, TJUN KIAT TEO wrote:
> In the caret train
Thank you for your suggestions. I am quite grateful to understand that
plotting is reliable and consistent in R. I had believed that this was
based on a built-in dataset within the R programming language, just as the
New Zealand volcano is.
I look forward to further participation in R as I
... or read about set.seed() and use it.
B.
On Nov 24, 2015, at 2:58 PM, William Dunlap wrote:
> rpois(100, 5) gives a different set of random numbers each time it is
> called, so if you want repeatable results compute it once and use its
> value in the calls to plot.
But please spend some time with an R tutorial or two (An Intro to R
ships with R; there are many more on the Web) before you post further
here. Many such elementary confusions and time wasted -- both yours
and ours -- will be avoided if you do so.
Cheers,
Bert
Bert Gunter
"Data is not
Hello,
I am quite new to R and have high expectations for my future with it.
R version 3.0.2 (2013-09-25) -- "Frisbee Sailing"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-pc-linux-gnu (64-bit)
I have stepped back to an earlier tutorial and found an odd
rpois(100, 5) gives a different set of random numbers each time it is
called, so if you want repeatable results compute it once and use its
value in the calls to plot. E.g.,
r <- rpois(100, 5)
plot(table(r), type="h", col="red", lwd=10, main="hello")
Bill Dunlap
TIBCO Software
wdunlap
I need to fit a sinusoidal curve to
x-y data that exhibits a sinusoidal
pattern. The curve will be:
y = a*sin(w*x +p) ;
where I need to get the best
fit choice for the parameters
a, w, and p. Could anyone
suggest which package and
routine I should use? I have
less than 1000 data
> On 24 Nov 2015, at 13:32, Duncan Murdoch wrote:
>
>> Perhaps it would make sense always to use the well-known standard XQuartz
>> paths on Mac and only consider other locations if explicitly asked for by
>> the user?
>
> If rgl is using non-standard features in
Wrong list (mostly). You would do much better at the Bioconductor
list, https://support.bioconductor.org/
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Tue, Nov 24, 2015 at 9:04 AM, debra
One way would be to use the biostrings and msa packages from bioconductor.
Use AAStringSet() to collect your sequences, and (e.g.) msaMuscle() to align
them. Then use BStringset() to extract the subrange you need, and e.g. table to
look at the distributions...
table(BStringSet(, start=50,
I have 15 protein sequences of 99 amino acids each. After doing some looking
around I have found that there are several ways you can read sequences into R
and do pairwise or multiple alignments. I, however, do not know how to probe
changes at specific positions. For instance, I would like to
Well ... that depends on what the best solution will turn out to be.
filling a matrix with the characters, or even substr() would be alternative
"lightweight" appraoches.
B.
On Nov 24, 2015, at 12:43 PM, Bert Gunter wrote:
> Wrong list (mostly). You would do much
I am giving to the community two examples
My code currently look like:
breaks<-c(0,0.01,0.05,0.08,0.1,0.15,0.2,0.3,0.5,0.7,0.9,1)
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude, colour
=Divergence), data = PlotPoints, size =10)+
Hi Nilesh,
I simplified your code a bit:
fun1<-function (dataset, plot.id, ranges2use, control) {
m1 <- strsplit(as.character(ranges2use), ",")
dat1 <- data.frame()
row_check_mean <- NA
row_check_adj_yield <- NA
x <- length(plot.id)
for (i in 1:x) {
cat(i,"\n")
dat1 <-
Dear Dennis, it would be better if not plotting the lon and lat. Keeping it
blank is better for the aesthetics of my map.
I am not sure how I can give a reproducible example herebut I want to
ggmap(mp, darken = 0) + geom_point(aes(Longitude, Latitude,
colour=Divergence), data = PlotPoints,
Some comments on the second part of your message.
On 23/11/2015 7:45 AM, Stefan Evert wrote:
On 23 Nov 2015, at 11:50, Duncan Murdoch wrote:
The OSX binary version of rgl on CRAN is ancient. You'll need to reinstall
it from source for a current one.
Since you
Buenas tardes,
Estoy extrayendo un dato de una web y el separador de miles es el ".",
que es el separador decimal en R. He intentado reemplazaro por el
caracter, para convertirlo en ´numérico de R, vacío pero no me deja.
Lo que he intentado es lo siguiente:
> a<-"17.691.700" # valor que obtengo
Muchas gracias, con esto ya funciona perfectamente.
Un saludo.
El día 24 de noviembre de 2015, 17:44, Carlos J. Gil Bellosta
escribió:
> a <- "17.691.700" # valor que obtengo de la web
> a <- as.numeric(gsub("\\.", "", a))
>
> "sub and gsub perform replacement of the
a <- "17.691.700" # valor que obtengo de la web
a <- as.numeric(gsub("\\.", "", a))
"sub and gsub perform replacement of the first and all matches respectively."
Además, ten cuidado con ".": es el comodín para cualquier caracter. Por eso
a <- "17.691.700"
sub(".", "", a)
da lo que da (y no es
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