Dear All,
I am given a time series such at, at every time t_i, I am given a set
of data (every element of the set is just an integer number).
What I need is an injective function able to map every set into a
number (possibly an integer number, but that is not engraved in the
stone). Does anybody
I apologize if this comes across as confusing. I will try to explain my
situation as best I can.
I have R bootstrapping my growth data for fish. It's resampling my database
of age and length data and then produces several new datasets for me. In
this case, it's resampling my data to
Hello,
I have been struggling to do a plot in ggplot(2) that's of lattice equivalent.
The following code shows the lattice plot.
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
Try this:
PATH - dirname(sys.frame(1)$ofile)
On Wed, Sep 29, 2010 at 5:00 PM, Stu Field s...@colostate.edu 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
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)
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
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 mlar...@rsmas.miami.eduwrote:
I apologize if this comes across as
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
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
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 jafari...@gmail.comwrote:
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
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
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
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
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__
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
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?
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__
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
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
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 djmu...@gmail.com wrote:
Hi:
It might be helpful to
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)
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 100th
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] -
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 like:
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
On Wed, Sep 29, 2010 at 6:55 AM, hairryharry hairryha...@tesco.net 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
On 30 September 2010 02:48, Michael Larkin mlar...@rsmas.miami.edu 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
-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
Perhaps use lgamma?
lgamma(220)
[1] 964.8206
Jonathan
On Wed, Sep 29, 2010 at 3:22 PM, song song rprojecth...@gmail.com 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
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
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
On Wed, Sep 29, 2010 at 6:43 PM, Gundala Viswanath gunda...@gmail.com 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 Thanks,
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 gunda...@gmail.comwrote:
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)
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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),c(V2,C,0.2),c(V3,C,0.4
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 -
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 -
On Wed, Sep 29, 2010 at 1:27 PM, Jyotasana Gulati jgul...@ice.mpg.de 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
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
schoc...@newschool.edu 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
Try this:
trunc(x / 10 ^ nchar(x) * 100)
On Wed, Sep 29, 2010 at 11:18 PM, Christian Schoder
schoc...@newschool.eduwrote:
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
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 fail?
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 gunda...@gmail.com wrote:
I have a variable that looks
Thank you for your help
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