I imagine you want the ggplot2 package.
something like:
ggplot(dataframe, aes(x = yourxvar, y = youryvar)) +
geom_point() +
facet_wrap(~ ProbeSet.ID)
Or facet_grid(), either of which makes a different panel for each unique level
of ProbeSet.ID
see gggplot help here: http://had.co.nz/ggplot2/
Hi Ivan,
thanks for your replay!
but the problem is there that the dataframe has 2 rows and ca. 2000
groups, but i dont have the column with the groupnames, because the groups
are depending on 2 onother columns ...
any other idea or i didnt understand waht are you posted ... :(
--
View th
> Date: Thu, 24 Feb 2011 13:28:18 -0800
> From: dannyb...@gmail.com
> To: r-help@r-project.org
> Subject: Re: [R] Group rows by common ID and plot?
>
does this do what you want?
library("lattice")
df<-data.frame(x=1:100,y=1.0/(1:100),f=floor
Apologies, I'm really new with R, Can you help me with the syntax?
here is my data.frame in which I introduce independent variables:
> varind <-
> data.frame(datpos$hdom2,datpos$NumPies,datpos$InHart,datpos$CV,datpos$CA,datpos$FCC)
varind has dimensions(194, 6), in case that's necessary. Then
hi,
I would like to find the x value (independent variable) for a certain dependent
value using the fitted model with nls.
with (predict) I can find y that corresponds to a list of x. I need the other
way around. can it be done?
thanks,
afadda
__
R-hel
Apologies, I'm really new with R, Can you help me with the syntax?
here is my data.frame in which I introduce independent variables:
> varind <-
> data.frame(datpos$hdom2,datpos$NumPies,datpos$InHart,datpos$CV,datpos$CA,datpos$FCC)
varind has dimensions(194, 6), in case that's necessary. Then I
Here's the way I would explore this, and some of the code is made more tidy.
Note that also you could vectorize your simulation. I have used set.seed
multiple times to make bootstrap samples the same across runs. -Frank
. . .
if (data[i, 3] == 4) data[i, 5] <- sample(c(0, 1), 1, prob=c(.06, .
That's interesting. You might also like:
http://en.wikipedia.org/wiki/Von_Mises%E2%80%93Fisher_distribution
I'm not sure how to plot the wireframe sphere, but you can visualize the
points by transforming to Cartesian coordinates like so:
u <- runif(1000,0,1)
v <- runif(1000,0,1)
theta <- 2 * pi *
On 25/02/2011 8:21 AM, Lorenzo Isella wrote:
Dear All,
I need to plot some points on the surface of a sphere, but I am not sure
about how to proceed to achieve this in R (or if it is suitable for this
at all).
In any case, I am not looking for really fancy visualizations; for
instance you can con
Brian Tsai gmail.com> writes:
>
> Hi all,
>
> I'm trying to figure out the effective differences between BFGS and L-BFGS-B
> are, besides the obvious that L-BFGS-B should be using a lot less memory,
> and the user can provide box constraints.
>
> 1) Why would you ever want to use BFGS, if L-BF
And in that vein, the recently released MethComp package by Bendix
Carstensen may be of service.
HTH,
Dennis
On Fri, Feb 25, 2011 at 5:39 AM, Marc Schwartz wrote:
> On Feb 24, 2011, at 4:50 PM, Denis Kazakiewicz wrote:
>
> > Dear R people
> > Could you please help with following
> >
> > Trying
On Feb 24, 2011, at 4:50 PM, Denis Kazakiewicz wrote:
> Dear R people
> Could you please help with following
>
> Trying to compare accuracy of tumor size evaluation by different
> methods. So data looks like
>
> id true metod1 method2 ...
> 1 2 2 2.5
> 2 1.52 2
> 3 2 2
Ha... it was way too simple!
I thought it would be like system.time()... my bad. Thanks for the tip!
As we thought, foo_reg() takes most of the computing time, and I cannot
improve that.
Any ideas of how to improve the rest?
Thanks again for your help
Ivan
Le 2/25/2011 14:29, jim holtman a é
You invoke Rprof, run your code and then terminate it:
Rprof()
... code you want to profile
Rprof(NULL) # generate output
summaryRprof()
example:
> Rprof()
> for (i in 1:1e6) sin(i) + cos(i) + sqrt(i)
> Rprof(NULL)
> summaryRprof()
$by.self
self.time self.pct total.time total.pct
sin
Dear All,
I need to plot some points on the surface of a sphere, but I am not sure
about how to proceed to achieve this in R (or if it is suitable for this
at all).
In any case, I am not looking for really fancy visualizations; for
instance you can consider the images between formulae 5 and 6 a
I've rolled up R-2.12.2.tar.gz a short while ago. This is an update release,
which fixes a number of mostly minor issues, and one major issue in which
complex arithmetic was being messed up on some compiler platform.
You can get it from
http://cran.r-project.org/src/base/R-2/R-2.12.2.tar.gz
or
Dear Sacha,
Do you revisit the same locations per site? If so, use (1|site/location) as
random effect. Otherwise use just (1|site). You might want to add a crossed
random effect (1|date) if you can expect an effect of phenology.
Best regards,
Thierry
PS R-sig-mixed-models is a better list for
Hi,
I have two data set of normalized Affymetrix CEL files, wild type vs Control
type.(each set have further three replicates).
> wild.fish
AffyBatch object
size of arrays=712x712 features (10 kb)
cdf=Zebrafish (15617 affyids)
number of samples=3
number of genes=15617
annotation=zebrafish
notes=
hello dear list! I wonder about the layout of my csv for my study design:
i have 11 different sites.
each site had been visited 9 times.
on each visit, 6 distinctive water parameters had been taken ONCE on
each visit (as continuous variables).
on each visit, the fish abundance was counted us
Hi,
I think ave() might do what you want:
df <- data.frame(a=rep(c("this","that"),5), b1=rnorm(10), b2=rnorm(10))
ave(df[,2], df[,1], FUN=mean)
For all columns, you could do that:
d <- lapply(df[,2:3], FUN=function(x)ave(x,df[,1],FUN=mean))
df2 <- cbind(df, d)
HTH,
Ivan
Le 2/25/2011 12:11, zem
You need to use "==" instead of "=" for testing equality. While you're at it,
you should check for positive values, not just screening out 0s. This works
for me:
R> mydata = data.frame(x=0:10, y=runif(11))
R> fm = lm(y ~ log(x), mydata, subset=x>0)
Andy
> -Original Message-
> From:
Dear Jim,
I've tried to use Rprof() as you advised me, but I don't understand how
it works.
I've done this:
Rprof(for (i in seq_along(seq.yvar)){
all_my_commands
})
summaryRprof()
But I got this error:
Error in summaryRprof() : no lines found in ‘Rprof.out’
I couldn't really understand from
Hi Jessica,
try this: Q[k:c(k+3)]
--
View this message in context:
http://r.789695.n4.nabble.com/select-element-from-vector-tp3323725p3324286.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://s
Hi all,
i have a little problem, and i think it is really simple to solve, but i
dont know exactly how to.
here is the "challange":
i have a data.frame with n colum, i have to group 2 of them and calculate
the mean value of the 3. one. so far so good, that was easy - i used
aggregate function to
I want to do a lm regression, some of the variables are going to be affected
with log, I would like not no take into account the values which imply doing
log(0)
for just one variable I have done the following but it doesn't work:
lmod1.lm <- lm(log(dat$inaltu)~log(dat$indiam),subset=(!(dat$in
On 02/25/2011 09:33 PM, Lucia Rueda wrote:
Hi,
I am using this loop
par(mfrow=c(3,3))
annos<-c(2001:2007,2009)
for (i in annos) {
t<-subset(masia,YEAR==i)
t$FAMILIA<-drop.levels(t$FAMILIA)
pie(table(t$FAMILIA),main=i)
}
To make piecharts of species composition among years (my data frame is
use Rprof to find where time is being spent. probably in 'plot' which might
imply it is not the 'for' loop and therefore beyond your control.
Sent from my iPad
On Feb 25, 2011, at 6:19, Ivan Calandra wrote:
> Thanks Nick for your quick answer.
> It does work (no missed bracket!) but unfortuna
Thanks Nick for your quick answer.
It does work (no missed bracket!) but unfortunately doesn't really speed
up anything: with my real data, it takes 82.78 seconds with the double
lapply() instead of 83.59s with the double loop (about 0.8 s).
It looks like my double loop was not that bad. Does
Hi all,
I am modelling a time series with missing data.
*Q1)* However, I am not sure if I should use the next *graphics* to
understand my data:
*a)* ACF & PACF (original series)
*b)* ACF & PACF (residuals)
* *
*Q2)* I am using *tsdiag*, so I obtain a graphic with 3 plots: stand.
residuals vs time
Hi,
Take a look at ?points, ?legend and ?par (specifically col and pch)
HTH,
Ivan
Le 2/25/2011 11:58, amir a écrit :
Hi,
I have two X1,X2 and Y1,Y2 and I want to draw them ((X1,Y1), (X2,Y2)) in
a scatter graph.
How can I draw both of them in a same graph with different legends?
And is there
Hi,
I am using this loop
par(mfrow=c(3,3))
annos<-c(2001:2007,2009)
for (i in annos) {
t<-subset(masia,YEAR==i)
t$FAMILIA<-drop.levels(t$FAMILIA)
pie(table(t$FAMILIA),main=i)
}
To make piecharts of species composition among years (my data frame is
called "masia"). So I get 1 piechart of
Hi,
I have two X1,X2 and Y1,Y2 and I want to draw them ((X1,Y1), (X2,Y2)) in
a scatter graph.
How can I draw both of them in a same graph with different legends?
And is there any way to show different labels on each point?
Regards,
Amir
__
R-help@r-pr
On Fri, Feb 25, 2011 at 4:11 AM, zbynek.jano...@gmail.com
wrote:
>
> I am having following problem:
> I´m constructing model for calculation of area of triangle.
> I know sides a, b, and gamma angle.
> I wish to calculate the area using heron´s formula:
> S <- sqrt(s*(s-a)*(s-b)*(s-c))
> where
> s
Simply avoiding the for loops by using lapply (I may have missed a bracket
here or there cause I did this without opening R)...
Haven't checked the speed up, though.
lapply(seq.yvar, function(k){
plot(mydata1[[k]]~mydata1[[ind.xvar]], type="p",
xlab=names(mydata1)[ind.xvar], ylab=names(mydata1)
Hi,
If I follow you correctly, you could write a function:
foo <- function(a,b,gamma){
c <- sqrt(a^2 + b^2 -2*a*b*cos(gamma))
s <- (a+b+c)/2
A <- sqrt(s*(s-a)*(s-b)*(s-c))
return(A)
}
I hope I didn't make mistakes, but it can still help you, I guess.
Ivan
Le 2/25/2011 10:11, zbynek.jano..
Dear users,
I have a double for loop that does exactly what I want, but is quite
slow. It is not so much with this simplified example, but IRL it is slow.
Can anyone help me improve it?
The data and code for foo_reg() are available at the end of the email; I
preferred going directly into the
Try
my.date <- strptime("20/2/06 11:16:16.683", "%d/%m/%y %H:%M:%OS")
Then you can examine my.date$mon.
--
Robert Tirrell | r...@stanford.edu | (607) 437-6532
Program in Biomedical Informatics | Butte Lab | Stanford University
On Thu, Feb 24, 2011 at 14:12, Belle wrote:
>
> I think I got it,
I am having following problem:
I´m constructing model for calculation of area of triangle.
I know sides a, b, and gamma angle.
I wish to calculate the area using heron´s formula:
S <- sqrt(s*(s-a)*(s-b)*(s-c))
where
s <- (a+b+c)/2
and c is calculated using law of cosines:
c <- sqrt(a^2 + b^2 -2*a*
I can't think of a reason why they would...
Rob Tirrell
On Thu, Feb 24, 2011 at 23:56, Vedajit Boyd wrote:
> Hi,
>
> Please someone let me know that the installation of both R for Windows
> 2.12.2 and MS office 2010 on the same system will interfere each other or
> not.
> In short, are these t
Hi Celine,
GLM outputs usually give the null deviance and residual deviance in the
summary() term - so you can work out % deviance explained for a variable/model
from this. Hope this helps.
Best wishes,
Clare
Dr Clare B Embling
Visiting Research Fellow
Marine Inst
On Fri, 25 Feb 2011, Vedajit Boyd wrote:
Hi,
Please someone let me know that the installation of both R for Windows
2.12.2 and MS office 2010 on the same system will interfere each other or
not.
In short, are these two tools compatible to each other?
There is nothing special about R, but you
One of my hypotheses of what you want is:
for(code in codes) {
get(code)
}
The other one is:
for(code in codes) {
as.name(code)
}
On 25/02/2011 06:55, Noah Silverman wrote:
How can I dynamically use a variable as the name for another variable?
I realize this sounds cryptic,
Hi,
Please someone let me know that the installation of both R for Windows
2.12.2 and MS office 2010 on the same system will interfere each other or
not.
In short, are these two tools compatible to each other?
Thanks in advance.
Best Regards,
Vedajit
[[alternative HTML version deleted]]
Hello,
I was searching online to find more info about Biometics
and I came across your information.
Can you tell me, are you still involved with Biometics?
If you are, how are things going for you?
Please let me know.
Sincerely,
Will Hammack
__
I think I got it, I post it here see if you have better way, please let me
know.
index <- rep(0, length(mydata[,1]))
index[as.Date(mydata3$Date) < as.Date("2006-11-30 23:29:29 PM") &
as.Date(mydata3$Date) > as.Date("2006-09-01 00:00:00 AM")] <- 1
index[as.Date(mydata3$Date) < as.Date("2007-11-30
Hi:
I want to give an index with all the dates between Sept. to Nov. as 1, and
anything else is 0. It doesn't matter which year it is, as long as it is
between Sept. to Nov, then set up to 1, otherwise is 0.
My data frame looks like below:
ID Date
201 1/1/05 6:07 AM
201 3/27/09 9:4
Hello,
I came across a package called cubature (
http://cran.r-project.org/web/packages/cubature/index.html) to perform
multivariate integration. I was not able to understand few stuff:
What is the need for package flags under src/Makevars?
What is the purpose of fWrapper in the rcubature.c file?
On Feb 24, 2011, at 3:38 PM, Kent Alleman wrote:
Hello,
I'm fairly new to R. I'm a chemist, not a programmer so please bear
with me.
I have a large data.frame that I want to break down (subset) into
smaller data.frames for analysis. I would like to give the
data.frames descriptive na
On Feb 25, 2011, at 1:55 AM, Noah Silverman wrote:
How can I dynamically use a variable as the name for another variable?
I realize this sounds cryptic, so an example is best:
#Start with an array of "codes"
codes <- c("a1", "b24", "q99")
Is there some reason not to use list(a1, b24, q99)?
Dear All,
I'd like to trim the output produced in a Sweave code chunk. For
instance, in
fit <- lm(conc ~ . - Plant, data = CO2)
summary(fit)
I'd like, skip the info after the coefficients' table, and possibly
replace it with '...'.
I've created this small function to do this, which is base
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