> On Feb 11, 2016, at 7:41 PM, Harun Rashid via R-help
> wrote:
>
> Hello,
> I am having trouble with putting subscript in facet_grid label. Here is
> an example of the work I have been trying to do.
>
>>
> df<-data.frame(species=gl(2,10,labels=c('sp1','sp2')),age=sample(3:12,40,replace=T),v
I would like to have a legend given in the manner
legend("topleft",pch=c(20,8,1),lty=1:3,bty="n",
legend=c("clyde","irving","melvin"))
but with the point symbol *NOT* being superimposed on the line segments
that are plotted.
I saw that I can specify "merge=FALSE" in the call to legend
Hello,
I am having trouble with putting subscript in facet_grid label. Here is
an example of the work I have been trying to do.
>
df<-data.frame(species=gl(2,10,labels=c('sp1','sp2')),age=sample(3:12,40,replace=T),variable=gl(2,20,labels=c('N1P1
var','N2P1 var')),value=rnorm(40))
>
test.pl
Thank you very much Dan!
I want go with the second one, because the data very huge (>25,000
columns) and > 3,000 row.
The data is loaded as "testdat"
Can you help me to fit in the following code please,
# faster but a little more difficult to see what is going on:
outdat<-indat %*%
array(c(r
Hi Val,
There are probably more elegant ways to do it, but the following is fairly
transparent:
# input data arranged as an array:
indat<-cbind(c(1,2,2,1),c(1,2,1,1),c(2,2,2,2),c(2,2,2,2),c(2,2,2,1),c(2,2,2,2),c(2,2,2,1),c(2,2,2,2),c(1,2,1,1),c(1,2,1,2))
indat
outdat<-array(dim=c(dim(indat)[1],di
Hi all,
I have SNP data set: the first column is the ID and the the
subsequent pair of columns are the alleles for each
SNP1, SNP2 and So on. Each SNP has two columns. Based on the alleles
I want make phenotype
if the alleles are 1 1then genotype is 0
2 2then
> On Feb 11, 2016, at 11:30 AM, C W wrote:
>
> Hi David,
>
> My real function is actually a multivariate normal, the simple toy 1-d normal
> won't work.
>
> But, you gave me an idea about restricting the bounds, and focus integrating
> on that. I will get back to you if I need any further a
Hi David,
My real function is actually a multivariate normal, the simple toy 1-d
normal won't work.
But, you gave me an idea about restricting the bounds, and focus
integrating on that. I will get back to you if I need any further
assistance.
Thank you so much!
On Thu, Feb 11, 2016 at 2:06 PM,
> On Feb 11, 2016, at 9:20 AM, C W wrote:
>
> I want to do numerical integration w.r.t. mu: P(mu) × N(mu, 0.1)
>
> Because the variance is small, it results in density like: 7.978846e+94
>
> Is there any good suggestion for this?
So what's the difficulty? It's rather like the Dirac functi
Thanks to everybody for trying to help me with this, I think there are
a few workable options here. However, I think the most efficient
option that I've found was to avoid the join/aggregate in R
altogether. I've joined them at the database level to accomplish the
same thing. This may not be a h
I should have said that R-3.2.3 requires the $ to be backslashed even when
it
is not at the end of the pattern:
> gsub("$[[:digit:]]*", "", c("$VAR", "$20/oz."))
[1] "$VAR""$20/oz."
> gsub("\\$[[:digit:]]*", "", c("$VAR", "$20/oz."))
[1] "VAR" "/oz."
Modern Linuxen's tools like sed d
In certain programs (not current R), a pattern with stuff after a naked
dollar
sign would not match anything because dollar meant end-of-string.
In any case I prefer simple rules like 'backslash a dollar sign' instead of
'backslash a dollar sign at the end of the pattern but not elsewhere'.
Bill
I want to do numerical integration w.r.t. mu: P(mu) × N(mu, 0.1)
Because the variance is small, it results in density like: 7.978846e+94
Is there any good suggestion for this?
Thanks so much!
On Thu, Feb 11, 2016 at 9:14 AM, C W wrote:
> Wow, thank you, that was very clear. Let me give
The "end of string" special meaning only applies when the dollar sign is at the
right end of the string (as it was in the OP attempt). That is, it is NOT
generally necessary to wrap it in brackets to remove the special meaning unless
it would otherwise be at the end of the pattern string.
--
Hi,
So working with docker and rJava i have a freeze at the Xrs check
there is a solution that is to use the --disable-Xrs option
https://github.com/s-u/rJava/issues/63
but as i'am a beginner : install.packages("rJava", repos='
http://cran.us.r-project.org', configure.args ="--disable-Xrs")
does
The R logo has been revised in to be more compatible with the principles
of flat design followed by some recent user interfaces such as Microsoft
Windows >= 8 and Mac OS X >= 10.10 (Yosemite). The new logo is
available for download from the R project web site in SVG and
high-resolution PNG formats:
R Users,
My sincere thanks to all those who have been coming forward to test my GPU
package and provide bug reports. I want to followup on my initial request
with a few qualifiers.
1. I neglected to tell users to also use my github version of 'RViennaCL'
instead of the CRAN version. I have made
Wow, thank you, that was very clear. Let me give it some more runs and
investigate this.
On Thu, Feb 11, 2016 at 12:31 AM, William Dunlap wrote:
> Most of the mass of that distribution is within 3e-100 of 2.
> You have to be pretty lucky to have a point in sequence
> land there. (You will get
> On Feb 10, 2016, at 7:10 PM, Val wrote:
>
> Hi all,
>
> I want create a frequency table using this :
>
> xc1<- sample(c(1:10), 100, replace = TRUE)
> xc2<- sample(c(0,1), 100, replace = TRUE)
>
> xc3<- cbind(xc1,xc2)
>
> tab1<- xc3[,list( d1=sum(xc2==0), d2=sum(xc2==1)),by=xc1]
>
> but no
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