Hello,
I have been struggling to do a plot in ggplot(2) that's of lattice equivalent.
The following code shows the lattice plot.
dd<-rbind(data.frame(rbind(c("V1","A",0.3),c("V2","A",0.5),c("V3","A",0.2))),data.frame(rbind(c("V1","B",0.3),c("V2","B",0.4),c("V3","B",0.8))),data.frame(rbind(c("V1"
Hi all,
Forgive me if this question has been addressed, but I was unable to find
anything in the r-help list or in cyberspace. My question is this: is there a
function, or set of functions, that will enable a script to detect its own
path? I have tried file.path() but that was not what I was lo
Try this:
PATH <- dirname(sys.frame(1)$ofile)
On Wed, Sep 29, 2010 at 5:00 PM, Stu Field wrote:
> Hi all,
>
> Forgive me if this question has been addressed, but I was unable to find
> anything in the r-help list or in cyberspace. My question is this: is there
> a function, or set of functions,
Hi:
The deal with replicate() is that its second argument is a *function*; more
specifically, a function *call*. That's why Henrique's solution worked and
your attempt didn't. Inside replicate(), if the function has arguments, they
need to be supplied. This works:
testdat <- function(df, n) df[sa
Mike,
Without completely knowing your end game with these questions and this
procedure it does seem like you are "re-inventing the wheel" here. If that is
true and given the nls() fit that you are using I would suggest that you look
at boot.case() in the alr3 package or nlsBoot() in the nlstoo
Try this:
growthBoot <- replicate(3, growth[sample(9,12,replace=T),], simplify =
FALSE)
lapply(growthBoot, nls, formula = Length ~ Linf * (1 - exp(-K * (Age -
to))), start = par)
On Wed, Sep 29, 2010 at 4:56 PM, Michael Larkin wrote:
> I apologize if this comes across as confusing. I will try t
HI:
I've seen a few threads about this topic but
still can't find a straightforward way on this.
Is there a package that can control R within an access form. For example,
I want to send a query to R, perform some statistics in R and send the output or
summary back to Access and display it on a f
Hello All,
I am drawing a graph having 18 small graphs inside using par(mfrow = c(6,3))
command. My problem is how to specify the margins of the whole 18 graphs. I
used par(mar=c(6.5, 6.5, 1.5, 1.5)) for each graph separately already but it
does not left any margins for the 'mtext()' for the margi
Use par(oma=c(1,1,1,1)) # oma = "outer margin area"
Is this what your looking for?
On Wed, Sep 29, 2010 at 5:03 PM, Mohsen Jafarikia wrote:
> Hello All,
>
> I am drawing a graph having 18 small graphs inside using par(mfrow =
> c(6,3))
> command. My problem is how to specify the margins of the
Hi:
There's no way you could produce a loess plot based on the data supplied
below. V1, your purported x-variable, is a factor; moreover, you have one
point per V1 * V2 factor combination. (BTW, you might also consider using
the carriage return when demarcating individual lines of code.)
The reas
Dear R users,
I am leaning MCMC sampling, and have a problem while trying to sample
exponential r.v.'s via the following code:
samp <- MCMCmetrop1R(dexp, theta.init=1, rate=2,
mcmc=5000, burnin=500,
thin=10, verbose=500, logfun=FALSE)
I tried o
for example, when I am calculating a posterior density, I need to calculate
gamma(75*3+5)=gamma(220) which is out of the bound of gamma function. what
shall I do for this condition>
Thank you
[[alternative HTML version deleted]]
__
R-help@r-pro
>
> Forgive me if this question has been addressed, but I was unable to find
> anything in the r-help list or in cyberspace. My question is this: is there a
> function, or set of functions, that will enable a script to detect its own
> path? I have tried file.path() but that was not what I was l
I used split() to split a variable by 3 years, but am wondering how to call up
that split data and use it in further analyses. Can I make separate columns for
the 3 resulting year groups?
[[alternative HTML version deleted]]
__
R-help
Thanks much for all the help, R-helpers. Ended up getting the counts of the
categories of the matching variable in both x and y and then limiting the
sample from there. No longer really random, but I think it's fine for my
purposes.
Thanks again.
LB
On 28 September 2010 18:40, Michael Bedward wro
Hi,
I've been reading quite a bit about the proper way of analyzing repeated
measured data and understand the advantages/pitfalls of doing it using
either a MANCOVA or linear mixed model approach. But I was wondering, for
the sake of really understanding, if anyone has some data to show how a
stan
Ali -
A reproducible example would be very helpful, but I'll
try to guess what you mean.
mydat = data.frame(year=rep(2008:2010,each=5),var=1:15)
sdat = split(mydat$var,mydat$year)
do.call(cbind,sdat)
2008 2009 2010
[1,]16 11
[2,]27 12
[3,]38 13
[4,]4
Hi Dennis,
Sorry for not being considerate. I should have at least mentioned I was
using MCMCpack.
I have no idea about traceback() though.
I appreciate your suggestion.
Best,
Zhongyi
On Wed, Sep 29, 2010 at 5:00 PM, Dennis Murphy wrote:
> Hi:
>
> It might be helpful to inform the list whic
Can you send your code and data as separate files so we can get it into R
easily?
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 5
I don't have enough RAM for this problem, so I need a work around. This is
what I want to do:
y<- sample(2^32, 10, replace=FALSE)
but my machine won't let me do that. so I now do this:
x<- seq(1,2^32, by=100)
y<- sample(x, 10, replace=FALSE)
this works fine, but by selecting every 100
On Thu, 30 Sep 2010, Matthew Finkbeiner wrote:
I don't have enough RAM for this problem, so I need a work around. This is
what I want to do:
y<- sample(2^32, 10, replace=FALSE)
y <- trunc(runif( 10, 1, 2^32+1))
while( any( dup.y <-duplicated(y) ) ) y[dup.y] <-
trunc(
Hadley,
I'm not sure this will solve the issue because if I move the script, I would
still have to go into the script and edit the "/path/to/my/script.r", or do I
misunderstand your workaround?
I'm looking for something like:
file.path.is.here("myscript.r")
and which would return something li
THanks for all the replies.
I guess I should have been clearer from the beginning.
When I said I don't want to write my code, I meant I don't want to CREATE a new
function.
Not because I can't, but because I don't want to.
The cmh.test in the {lawstat} package still doesn't look like the cocrahn
On Wed, Sep 29, 2010 at 6:55 AM, hairryharry wrote:
> Hi,
>
> Fairly new to R - have done basic plots but now faced with plotting a
> matrix/table of results -I know what I want but cannot find out how to do
> it.
>
> Basically have individual questions ( x) to which an organization can rate
> the
On 30 September 2010 02:48, Michael Larkin wrote:
> >
> > testdat <- replicate( 50, growth[ sample(nrow(growth), 8, rep=TRUE) ] )
>
> I can't seem to get it to work. I keep getting the error message of
> "undefined columns selected"
>
> Any advice?
I'd need to know the dimensions of your matrix
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Michael Bedward
> Sent: Wednesday, September 29, 2010 5:34 PM
> To: Michael Larkin; Rhelp
> Subject: Re: [R] repeat a function
>
> On 30 September 2010 02:48, Michael Larkin
> w
Perhaps use lgamma?
> lgamma(220)
[1] 964.8206
Jonathan
On Wed, Sep 29, 2010 at 3:22 PM, song song wrote:
> for example, when I am calculating a posterior density, I need to calculate
> gamma(75*3+5)=gamma(220) which is out of the bound of gamma function. what
> shall I do for this condition>
I have a vector that looks like this:
> foo
[1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
How can we find the percentage of "o" and "x" in
that vector in R?
- G.V
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-he
On Wed, Sep 29, 2010 at 6:43 PM, Gundala Viswanath wrote:
> I have a vector that looks like this:
>
>> foo
> [1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
>
> How can we find the percentage of "o" and "x" in
> that vector in R?
table(foo)/length(foo)
Peter
Dear R Users,
I have model simulated data for 240 days.
The day 1 is Jan 1, 2009, 00:00 hrs and then with 3-hourly interval and so
on.
The time axis is 1,2,3,4..1920; so the total rows in the data are
1920.
How to convert above time axis in "year" "month" "day" "hour" format
Great Thank
Hi,
I am have a data set of around 43000 probes(rows), and have to calculate
correlation matrix. When I run cor function in R, its throwing an error message
of RAM shortage which was obvious for such huge number of rows. I am not
getting a logical way to cut off this huge number of entities,
Try this:
prop.table(table(foo))
On Wed, Sep 29, 2010 at 10:43 PM, Gundala Viswanath wrote:
> I have a vector that looks like this:
>
> > foo
> [1] "o" "o" "o" "x" "o" "o" "o" "o" "o" "x" "x" "o" "x
>
> How can we find the percentage of "o" and "x" in
> that vector in R?
>
> - G.V
>
> _
sum(foo=="o")/length(foo)
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-get-a-proportion-of-a-Vector-Member-tp2720060p2720067.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hello,
My apologies, it was the hotmail writer which ate my post (hopefully this will
get there, intact).
dd<-rbind(data.frame(rbind(c("V1","A",0.3),c("V2","A",0.5),c("V3","A",0.2))),
data.frame(rbind(c("V1","B",0.3),c("V2","B",0.4),c("V3","B",0.8))),
data.frame(rbind(c("V1","C",0.9)
Here's my cooked up example:
# Faked data
x <- sample(1:100, 300, replace = TRUE)
# y = a + bx + cx^2 + noise, where a, b, c differ in each group
y <- rep(c(2, 5, 3), each = 100) + rep(c(-0.5, 0.5, 1), each = 100) * x +
rep(c(0.01, -0.01, 0.02), each = 100) * x^2 + rnorm(300, 0, 10)
g <- rep
Hi
Thanks for the wonderful package! I have a question on plotting.
How do I control the dodging (like the position_dodge in ggplot2) of
the points after creating a plot with ezPlot? when plotting a 2-way
ANOVA the error bars cover each other...
Thanks!
shai
On Aug 31, 2:53 pm, Mike Lawrence
Hello
I am trying to use the correlog function to estimate a spatial correlogram
for the residuals of a logistic regression and I have run accross the
following error.
> summary(binom1 <- glm(Use~X20mslop+X20mdem+soilsst, family=binomial,
+ data=M60m2000NE_1.df))
> correlog1.1 <- correlog(M60
On Wed, Sep 29, 2010 at 1:27 PM, Jyotasana Gulati wrote:
> Hi,
>
> I am have a data set of around 43000 probes(rows), and have to calculate
> correlation matrix. When I run cor function in R, its throwing an error
> message of RAM shortage which was obvious for such huge number of rows. I am
>
hi R-users!
does anyone know how I can access/print only the first two digits of a
number? if i have the number 23732, i would like to get 23. if i have
355 i would like to get 35. if i have 4 i would like to get 40.
thanks for your help!
christian
_
On Wed, Sep 29, 2010 at 7:18 PM, Christian Schoder
wrote:
> hi R-users!
>
> does anyone know how I can access/print only the first two digits of a
> number? if i have the number 23732, i would like to get 23. if i have
> 355 i would like to get 35. if i have 4 i would like to get 40.
It's a stran
Try this:
trunc(x / 10 ^ nchar(x) * 100)
On Wed, Sep 29, 2010 at 11:18 PM, Christian Schoder
wrote:
> hi R-users!
>
> does anyone know how I can access/print only the first two digits of a
> number? if i have the number 23732, i would like to get 23. if i have
> 355 i would like to get 35. if i
I have a variable that looks like this:
> print(pred$posterior)
ox
1 2.356964e-03 9.976430e-01
2 8.988153e-01 1.011847e-01
3 9.466137e-01 5.338627e-02
4 2.731429e-11 1.00e+00
Now what I want to do is to access "o" and "x"
How come this approach fa
Give us more information to work with. What does
str(pred$posterior)
show so that we can see the structure of the data. Is it a matrix, if
so then you would do
pred$posterior[, 'o']
On Wed, Sep 29, 2010 at 10:41 PM, Gundala Viswanath wrote:
> I have a variable that looks like this:
>
>> prin
Thank you for your help
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-rotate-the-x-axis-lable-to-an-interested-angle-tp2718591p2720118.html
Sent from the R help mailing list archive at Nabble.com.
[[alternative HTML version deleted]]
___
Hi List,
I am trying to create a spatial representation of some wind data.
I have the season, frequency, strength and direction of the wind from 10
different locations, the coverage of the area that I am interested in is
not 100% there are small gaps in my coverage due to the location of the
we
Hello,
I am struggling with computing nodes in Unix.
I have the use of a Unix server that has 30 nodes and I would like to
batch scripts.
Here is an R example that results in 72 repeated tasks based on the 2
loops. If I wanted to send these out to the different nodes, each node
has 1 task and th
101 - 146 of 146 matches
Mail list logo