I have generated a multi-page pdf with lattice. Now I want to make an mpeg3
file animation from it. I have been experimenting with the 'animation' package.
The saveM ovie function will create an mp3 file, but it is not clear if it will
take a pdf as the 'expr' that generates animations.
S
On Thu, Jul 22, 2010 at 8:23 AM, Christopher David Desjardins
wrote:
> * Please cc me if you reply as I am a digest subscriber *
>
> Hi,
> I am wondering how I can run a multilevel survival model in R? Below is
> some of my data.
>
>> head(bi0.test)
> childid famid lifedxm sex age delta
Can't give more specific advice without more information (see the
posting guide), but I suspect you want to use nlme::lme, specifying a
weights argument to adjust for the heterogeneity.
Kingsford
On Wed, Jul 21, 2010 at 11:10 PM, Kline, Keith A wrote:
> I have heterogeneous variance in a two-way
not an xtable solution, but R2HTML::HTML has an impressive list of
methods, including HTML.infl:
lm.SR <- lm(sr ~ pop15 + pop75 + dpi + ddpi, data = LifeCycleSavings)
influencia <- influence.measures(lm.SR)
library(R2HTML)
methods(HTML)
HTML(influencia, 'influencia.html')
and open the html file i
All sorted
Thanks very much for your help Peter and Jannis.
Toni
Peter Alspach wrote:
Tena koe Toni
Assume your data is in a data.frame called toniData (naming a data.frame data
is not a good idea as data is a function in R - see ?data), and that Group is
of class character (try str(
hi,
Library DAAG has onet.permutation function for one-sample permutation test
and twot.permutation function for two-sample permutation test.
Anyway, p-value is a result of a test, what's your test?
-
A R learner.
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Hi Ethan,
I should clarify that you do not *need* to make size an integer, so
you technically do not need trunc(), but if you use trunc() or
round(), you get to be in control of what you get. To answer your
second question, look at this little function, it even lets you choose
different percentag
On Jul 22, 2010, at 9:06 PM, stephen sefick wrote:
Agian, Plead read the bottom of this email. Also, it looks like you
should read An Introduction to R.
It certainly does appear this poster needs to do both.
On Thu, Jul 22, 2010 at 7:52 PM, jd6688 wrote:
The dataframe is
id sala
?subset
HTH,
Jorge
On Thu, Jul 22, 2010 at 8:52 PM, jd6688 <> wrote:
>
> The dataframe is
>
> id salary
> 100 500
> 101 600
> 102 700
> 103 800
>
> how can i generate a subsets if salary>600?
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Filtering-i
Agian, Plead read the bottom of this email. Also, it looks like you
should read An Introduction to R.
On Thu, Jul 22, 2010 at 7:52 PM, jd6688 wrote:
>
> The dataframe is
>
> id salary
> 100 500
> 101 600
> 102 700
> 103 800
>
> how can i generate a subsets if salary>600?
>
>
>
The dataframe is
id salary
100500
101600
102700
103800
how can i generate a subsets if salary>600?
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make this reproducible!
On Thu, Jul 22, 2010 at 7:40 PM, jd6688 wrote:
>
> POS=sum(x[-1][x[-1]>0],na.rm=TRUE)
> is this the correct syntax?
> --
> View this message in context:
> http://r.789695.n4.nabble.com/na-rm-TRUE-tp2299596p2299596.html
> Sent from the R help mailing list archive at Nabble
On Thu, Jul 22, 2010 at 4:56 PM, Tyler Williamson wrote:
> Hello,
>
> Suppose one is interested in fitting a GLM with a log link to binomial data.
> How does R choose starting values for the estimation procedure? Assuming I
> don't supply them.
>
Assuming weights are not specified it uses thi
POS=sum(x[-1][x[-1]>0],na.rm=TRUE)
is this the correct syntax?
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__
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sum() has a na.rm argument
sum(x, na.rm=TRUE)
On Thu, Jul 22, 2010 at 7:19 PM, jd6688 wrote:
>
> I have a DF
>
> ID VALUE
> 100 120
> 101 100
> 102 100
> 103
> 104
> 105
>
>
> when i calculate the sum of the values, it returned NA. should I populate
> the blank value as 0?
>
> Thanks,
>
I have a DF
ID VALUE
100 120
101 100
102 100
103
104
105
when i calculate the sum of the values, it returned NA. should I populate
the blank value as 0?
Thanks,
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Sent from
On Thu, Jul 22, 2010 at 7:45 PM, Kevin T. Ryan wrote:
> Hi There -
>
> New to the list, so hopefully I don't mess this up too much ... I am
> hoping maybe some of you can help me. I have a problem I'm trying to
> solve regarding customer accounts, attrition and forecasting.
> Basically, here is w
It works well for the title command. Thanks.
t
--- On Thu, 7/22/10, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] , how to express bar(zeta) in main title in boxplot
To: "Peter Ehlers"
Cc: "Marcus Liu" , r-help@r-project.org
Date: Thursday, July 22, 2010, 1:40 PM
On Jul 22,
Tena koe Toni
Assume your data is in a data.frame called toniData (naming a data.frame data
is not a good idea as data is a function in R - see ?data), and that Group is
of class character (try str(toniData)) then:
toniData[toniData$Group %in% 'A','Group'] <- 'C'
will work. But from your mess
Toni Pitcher schrieb:
This works to some extent, all As have been changes to Cs, but the Bs
have also been changed, in this case to "2", instead of remaining as
their original values. How do I get the Bs to stay the same?
You should consult some introductory tutorial dealing with Rs way of
t
Hi There -
New to the list, so hopefully I don't mess this up too much ... I am
hoping maybe some of you can help me. I have a problem I'm trying to
solve regarding customer accounts, attrition and forecasting.
Basically, here is what I want to do (not sure of the best way):
1. Track accounts by
n.via...@libero.it schrieb:
I would like to Know if somewhere in xtable I can put the command \pagebreak .
There is an add.to.row argument in print.xtable, where you could add
this command. You have to specify a line where to add this though. How
about appending each printed xtable object
On Thu, Jul 22, 2010 at 4:11 PM, jd6688 wrote:
>
> Hi Joshua:
>
> how to do a permutation test on the following sample with the sample size in
> 2 :
Take a look at some of the results from
RSiteSearch("permutation test")
the "coin" package also comes to mind
Josh
>
> id weight
I have a sample in size of 2 rows, for instanse:
idweightp value
100 125
101160
102 150
,
how can I calculate the p value after the permutation and how the
permutation test cab be done?
Thanks,
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Hi Joshua:
how to do a permutation test on the following sample with the sample size in
2 :
idweight P VALUE
100 100
101 200
102 200
103 150
...
Thanks
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> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Xebar Saram
> Sent: Thursday, July 22, 2010 3:50 PM
> To: r-help@r-project.org
> Subject: [R] choosing a random sample by precentage
>
> hi all
>
> i have found the follwoing wa
Hi Ethan,
Sure with a bit of extra work:
x <- matrix(1:109, ncol = 1)
x[sample(nrow(x), trunc(nrow(x)*.1)), ]
you could also use round() in place of trunc(), you just need
something to get it to an integer. If you didn't like having to type
the name several times you could make a simple functio
Is this the kind of thing you mean?
> wData <- within(wData, {
+ z <- (weigth - mean(weigth))/sd(weigth)
+ "p-value" <- 2*pnorm(-abs(z)) ## 2-sided
+ rm(z) })
> wData
employee_id weigth p-value
1 100150 0.3763641
2 101200 0.9081403
3 102300 0.1547139
4
What is your null hypothesis? What is your alternate hypothesis? What
is the test statistic? Why do you want a p-value?
Hadley
On Thu, Jul 22, 2010 at 5:40 PM, jd6688 wrote:
>
> Here is my dataframe with 1000 rows:
>
> employee_id weigth p-value
>
> 100 150
> 1
Thank you so much! I think I had tried those two pieces separately
and obviously had no success. I also typed panel3pt1.fn in the
console without the () following it.
On Jul 22, 2010, at 5:57 PM, David Winsemius wrote:
On Jul 22, 2010, at 5:34 PM, David Winsemius wrote:
On Jul 22, 2
hi all
i have found the follwoing way to choose a random sample by sample size (200):
ten_per_T2000 <- F_T2000_All[sample(nrow(F_T2000_All), 200), ]
but i wondered if there is a way to choose a sample size by
precentage (10% etc..)
thx
ethan
__
R-
kdionisio gmail.com> writes:
>
>
> Hi,
> Is there any similar command to "predict" which can be used with a logistic
> random effects model?
Poke around on http://glmm.wikidot.com , let me know how you do.
Ben Bolker
__
R-help@r-project.org ma
Here is my dataframe with 1000 rows:
employee_id weigth p-value
100 150
101 200
102 300
103 180
.
My question:
how can I calculate the p-value in R for each employee? the
distributio
Peter Ehlers wrote:
> On 2010-07-22 15:21, David Winsemius wrote:
>> On Jul 22, 2010, at 5:01 PM, Peter Ehlers wrote:
>>
>>> On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
> On 2010-07-22 11:44, Marcus Liu wrote:
>> Hi everyone, I am p
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Peter Dalgaard
> Sent: Thursday, July 22, 2010 3:13 PM
> To: Pat Schmitz
> Cc: r-help@r-project.org
> Subject: Re: [R] Question about a perceived irregularity in R syntax
>
> Pat
Good day R-listers,
I'm currently working on a panel data analysis (N=17, T=5), in order
to check for the spurious regression problem, i have to test for
stationarity but i've read somewhere that i needn't to test for it as
my T<10 , what do you think? if yes is there any other test i have
to
Uwe Ligges wrote:
>
> On 22.07.2010 21:47, Filoche wrote:
>> Hi everyone.
>>
>> I'm trying to display the r^2 of a linear regression on a plot using
>> text(...). I first build the string to display. However, I can't find why
>> this is not working (it display literally r^2 instead of r superscri
Pat Schmitz wrote:
> Both vector query's can select the values from the data.frame as written,
> however in the first form assigning a value to said selected numbers fails.
> Can you explain the reason this fails?
>
> dat <- data.frame(index = 1:10, Value = c(1:4, NA, 6, NA, 8:10))
>
> dat$Value
On 2010-07-22 15:21, David Winsemius wrote:
On Jul 22, 2010, at 5:01 PM, Peter Ehlers wrote:
On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
It occurs to me that Hannah may simply be plotting to a plot window that
is overall too small.
Try setting the plot window to full-screen (or close to that), and then
running your plot & legend. Once the picture on the screen is the way
you like it, save or copy (clipboard) to your format of
Both vector query's can select the values from the data.frame as written,
however in the first form assigning a value to said selected numbers fails.
Can you explain the reason this fails?
dat <- data.frame(index = 1:10, Value = c(1:4, NA, 6, NA, 8:10))
dat$Value[dat$Value == "NA"] <- 1 #Why doe
Hi Brian,
On 07/21/2010 10:02 AM, Davis, Brian wrote:
[...]
Part 2)
My next step in the string processing is to take the characters in the output
of CleanRead and subtract 33 from the ascii value of the character to obtain an
integer. Again I have a solution that works, involving splitting the
On Jul 22, 2010, at 5:34 PM, David Winsemius wrote:
On Jul 22, 2010, at 5:27 PM, Daniel Hocking wrote:
I am sorry if this question is vague or uninformed. I am just
learning R and struggling. I am using the book Hierarchical Modeling
and Inference in Ecology and they provide examples of R
Sam,
I recommend taking a look at the ggplot2 package. This page from the
author's website contains an example of what I think you are trying to
achieve:
http://had.co.nz/ggplot2/geom_segment.html
Obviously, this would require doing the whole plot in ggplot2, but that's
not at all unpleasant. Th
Hi
I'm new to R and would like some help with a couple of problems. I
suspect the solutions are quite simple.
I have a data.frame (data) with 40 variables and 5238 observations
created from ~150 text files using read.table.
I would like to change some of the entries within two different col
On Jul 22, 2010, at 5:27 PM, Daniel Hocking wrote:
I am sorry if this question is vague or uninformed. I am just
learning R and struggling. I am using the book Hierarchical Modeling
and Inference in Ecology and they provide examples of R code. I have
the following code from the book but when
I am sorry if this question is vague or uninformed. I am just
learning R and struggling. I am using the book Hierarchical Modeling
and Inference in Ecology and they provide examples of R code. I have
the following code from the book but when I run it I don't get any
output. I cannot get
On Jul 22, 2010, at 5:01 PM, Peter Ehlers wrote:
On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), ": ", bar(zeta), "
And, surely, regardless of R version:
resamp <- function(x,...){if(length(x)==1) x else sample(x,...)}
resamp((1:10),10)
# [1] 8 2 10 6 5 4 3 7 9 1
resamp((1:10),1)
# [1] 7
resamp((1:10),1)
# [1] 4
resamp((1:10),1)
# [1] 10
resamp(10,1)
# [1] 10
resamp(10,1)
# [
Thx everybody,
finally I got it to go. I switched all three files (mysource.R,
sweavetemplate.Rnw,invokeSweave.R) to UTF-8 encoding, switched the options()
like suggested below and it just worked.
On 22.07.2010, at 16:11, David Winsemius wrote:
>
> On Jul 22, 2010, at 9:39 AM, Bunny, lautlos
On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), ": ", bar(zeta), " Boxplot from 2001 to
2009", sep = "")) but it doesn'
Hello,
Suppose one is interested in fitting a GLM with a log link to binomial data.
How does R choose starting values for the estimation procedure? Assuming I
don't supply them.
Thanks,
Tyler
__
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https://stat.ethz.ch
Hi.
This is working fine.
Thank you for your help.
Best regards,
Phil
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I see.. Thanks!
On Thu, Jul 22, 2010 at 4:39 PM, Hadley Wickham wrote:
> Did you look at the examples in sample?
>
> # sample()'s surprise -- example
> x <- 1:10
> sample(x[x > 8]) # length 2
> sample(x[x > 9]) # oops -- length 10!
> sample(x[x > 10]) # length 0
>
> ## For R >= 2.11.0
Did you look at the examples in sample?
# sample()'s surprise -- example
x <- 1:10
sample(x[x > 8]) # length 2
sample(x[x > 9]) # oops -- length 10!
sample(x[x > 10]) # length 0
## For R >= 2.11.0 only
resample <- function(x, ...) x[sample.int(length(x), ...)]
resample(x[x > 8]) #
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), ": ", bar(zeta), " Boxplot from 2001 to
2009", sep = "")) but it doesn't work. The program said they
cannot fin
On 22.07.2010 21:47, Filoche wrote:
Hi everyone.
I'm trying to display the r^2 of a linear regression on a plot using
text(...). I first build the string to display. However, I can't find why
this is not working (it display literally r^2 instead of r superscript 2).
r2string = expression(pa
Try this:
x <- 10
sample(x, 1, prob = c(rep(0, x - 1), 1))
On Thu, Jul 22, 2010 at 5:31 PM, Jon BR wrote:
> Hi All,
>I'm trying to use the "sample" function within a loop where the
> vector being sampled from (the first argument in the function) will
> vary in length and composition. When
Hi All,
I'm trying to use the "sample" function within a loop where the
vector being sampled from (the first argument in the function) will
vary in length and composition. When the vector is down in size to
containing only one element, I run into the "undesired behaviour"
acknowledged in the ?
Hi everyone.
I'm trying to display the r^2 of a linear regression on a plot using
text(...). I first build the string to display. However, I can't find why
this is not working (it display literally r^2 instead of r superscript 2).
r2string = expression(paste(r^2)," = ", r2);
Any help would be
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), ": ", bar(zeta), " Boxplot from 2001 to 2009", sep = "")) but
it doesn't work. The program said they cannot find the function "bar". Does anyone know how to do it
c
Dear all,
i think I am a step further with the issue. I am pretty sure now, that it is
not a Sweave problem. It's indeed an encoding problem with my Mac. If i source
the file without any sweave
source("myRcode.R") I already get an error message, despite the fact that the
code is correct. If
Spencer Graves structuremonitoring.com> writes:
>
> Have you tried something like the following:
>
> library(sos)
> H <- ???Hurst
> summary(H)
> H
>
>This identified 50 links in 15 packages, and displayed the
> results in a table in a web browser with links to the best match in the
Dear All,
I met problems when doing contrast and now really need some help in the
model below:
Fit=gam(y~treat+SEQUENCE+PERIOD+SEX+s(x),data=dat,
random=list(SUBJID=~1),correlation=corAR1(form=~1|SUBJID))
And error message keeps coming out when I want to compare the differences
between treatments
On Jul 22, 2010, at 2:13 PM, Gregory Gilbert wrote:
R Community,
I have a stupid little barplot I am trying to construct (Windows XP,
R.11.1,
32-bit). Whenever I run it my bars run below the horizontal axis.
Can anyone
(1) reproduce this and (2) offer a solution?
I'm not sure I should b
Hi, Bill,
Thanks. This is what I am looking for.
Jun
On Thu, Jul 22, 2010 at 1:54 PM, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen
>> Sent: Thursday, July 22, 2010 10:30 AM
>> To: R-help
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), ": ", bar(zeta), " Boxplot from 2001 to 2009", sep =
"")) but it doesn't work. The program said they cannot find the function
"bar". Does anyone know how to do it correctly? Thanks.
tin
Thanks, that seems to have cleared it up.
(I did cut and paste from the console - I didn't think most of what I tried
was very going to be very helpful and didn't want to drown people in
spurious info. Sorry for the mistakes)
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Here is a function I use to return the non-NULL elements of a list:
delete.NULLs <- function (x.list)
{
x.list[unlist(lapply(x.list, length) != 0)]
}
On Thu, Jul 22, 2010 at 2:22 PM, Ted Byers wrote:
> Here is the function that makes the data.frames in the list:
>
> funweek <- function(d
Looks like your event id is unique. If that is so, why not just do
##Not checked
events <- events[sort(events$event),]
dataF <- dataF[sort(data$event),]
if(doUpdate == 1){
if(!is.null(dataF) && nrow(dataF) > 0){
events[events$eventid %in% dataF$event, c("timestamp",
"isSynchronized","ti
Here is the function that makes the data.frames in the list:
funweek <- function(df)
if (length(df$elapsed_time) > 5) {
res = fitdist(df$elapsed_time,"exp")
year = df$sale_year[1]
sample = df$sale_week[1]
mid = df$m_id[1]
estimate = res$estimate
sd = res$sd
samplesize
R Community,
I have a stupid little barplot I am trying to construct (Windows XP, R.11.1,
32-bit). Whenever I run it my bars run below the horizontal axis. Can anyone
(1) reproduce this and (2) offer a solution? Rather simplistic code follows
(I am new to the community)> Thanks for your help!
Gre
I'm parallelizing some computation on hierarchical data, and would
find it natural to do something like this (where a call to parLapply
is embedded in outer call to parLapply):
cl <- makeCluster(rep.int('localhost', 5),
type='SOCK')
clusterExport(cl, 'cl')
parLapply(cl,
Hi,
I have a global data-frame in my R script.
At some point in my script, I want to update certain columns of this
data-frame by calling in an update function.
The function looks like this:
# get events data. This populates a global event data frame in the R-script
events <- getEvents(con, ev
Dear list,
I have a matrix, spc, that should row-wise be interpolated:
E.g.
spc <- matrix (1:1e6, ncol = 250, nrow = 4000, byrow = TRUE)
spc [1:10, 1:10]
shifts <- seq_len (nrow (spc))
wl <- seq_len (ncol (spc))
interpolate <- function (spc.row, shift, wl)
spline (wl + shift, spc.row, xout =
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jun Shen
> Sent: Thursday, July 22, 2010 10:30 AM
> To: R-help
> Subject: [R] How to fix slope and estimate intercept
>
> Dear all,
>
> Is there anyway I can fix slope in some v
On 2010-07-22 11:28, Arne Schulz wrote:
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1<- rnorm(1:25)
data2<- rnorm(1:25)
vars<- c("data1", "data2")
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quotes in the vars
saminny wrote:
>
> Is there a non animation version of newton.method
> (http://bm2.genes.nig.ac.jp/RGM2/R_current/library/animation/man/newton.method.html)
> for finding roots of a function? It should find the roots without showing
> it on the GUI.
>
Have a look at package nleqslv
/Berend
--
Try this:
apply(Vectorize(get)(ls(patt = 'data')), 2, t.test)
On Thu, Jul 22, 2010 at 2:28 PM, Arne Schulz <
arne.sch...@student.uni-kassel.de> wrote:
> Dear list,
> I'd like to do several t-test within a for loop. Example follows:
>
> data1 <- rnorm(1:25)
> data2 <- rnorm(1:25)
> vars <- c("da
On Jul 22, 2010, at 1:28 PM, Arne Schulz wrote:
Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1 <- rnorm(1:25)
data2 <- rnorm(1:25)
vars <- c("data1", "data2")
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quote
Hello, and sorry for not providing an example.
I run a regular linear regression (using lm) and use weights with it
(weights = ...).
I use "QuantPsych" package, its function lm.beta to extract
standardized regression weights from my lm regression object.
When I don't use weights, everything is fin
Dear all,
Is there anyway I can fix slope in some value and only estimate
intercept for a linear regression? I understand lm(y~1, data) will NOT
have a slope at all. This is not what I want. Thanks.
Jun
__
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Dear list,
I'd like to do several t-test within a for loop. Example follows:
data1 <- rnorm(1:25)
data2 <- rnorm(1:25)
vars <- c("data1", "data2")
for (i in vars) {
t.test(i)
}
Unfortunately, it does not work because of the quotes in the vars-vector
(running t.test(data1) by hand
On Jul 22, 2010, at 11:34 AM, Nicholas Griffin wrote:
I have been trying to get the lme4 package installed on Mac OS X...
with no
success. The Mac OS binary is not available on any CRAN, and I also
can’t
install the package from old source. Has anyone found a solution to
this
problem? I
Hi All,
I'm using R 2.11.1 on 64 bit windows XP. The little function I wrote below
I use often to import the first 1001 lines in an excel sheet to R. This
works fine on the 32 bit version of R but fails on the 64 bit [both on the
same machine, using the same function, importing the same .xls f
I have been trying to get the lme4 package installed on Mac OS X... with no
success. The Mac OS binary is not available on any CRAN, and I also can¹t
install the package from old source. Has anyone found a solution to this
problem? I am happy to use nlme for now, but I tend to prefer to do my
mi
[...]
> compo[[ which.max(sapply(compo, vcount)) ]]
That's better command for obtaining Largest Component
Sometimes, I get this error while reading pajek file. Does anybody know what
this error message means ?
g <- read.graph ("F://data/pfu_raw.net", "pajek")
Error in read.graph.pajek(file, ..
I have data as below.Please let me know how the ACF and Pacf used to
determine the order od arima model.
Is there any rules need to be followed to determine order.Please advise
> turkey.price.ts
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2001 1.58 1.75 1.63 1.45 1.56 2.07
On 2010-07-22 10:43, Wu Gong wrote:
Hi William,
I'm curious about that you used d[]<- lapply(d, factor...
Could you please tell me if there are any differences between d[] and d?
Thank you.
Why not try it both ways and inspect the result.
d[] <-
d <-
Obvious?
-Peter Ehlers
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Wu Gong
> Sent: Thursday, July 22, 2010 9:44 AM
> To: r-help@r-project.org
> Subject: Re: [R] how to force a table to be square?
>
>
> Hi William,
>
> I'm curious about that yo
Hi William,
I'm curious about that you used d[] <- lapply(d, factor...
Could you please tell me if there are any differences between d[] and d?
Thank you.
-
A R learner.
--
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I'd like to train a decision tree on a set of weighted data points. I looked
into the rpart package, which builds trees but doesn't seem to offer the
capability of weighting inputs. (There is a weights parameter, but it seems to
correspond to output classes rather than to input points).
I'm m
On 7/22/2010 5:01 AM, Allan Engelhardt wrote:
There are so many ways Here is one:
aggregate(v ~ u, data=X, function(...) length(unique(...)))
# u v
# 1 T1 2
# 2 T2 1
Hope this helps
Here is one other way, using the plyr package (which is very good for
taking a data structure (data.frame
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Liat
> Sent: Thursday, July 22, 2010 8:41 AM
> To: r-help@r-project.org
> Subject: Re: [R] how to force a table to be square?
>
>
> Thanks Peter!
> that worked, and was so easy.
Dear list,
I'm using the Sweave function in order to get some report.
Here one chunk:
<>=
report<-lapply(repor, function(x) {
(print(xtable(data.frame(x[1:2,]), align="|l|rrr|"),floating=FALSE,tabular.
environment="longtable",include.colnames=FALSE,size="\\small"))
(print(xtable(data.frame(x[3:nro
Thank you!!!
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Sent from the R help mailing list archive at Nabble.com.
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R-help@r-project.org mailing list
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Thanks Peter!
that worked, and was so easy. LOL.
I tried playing with factor and levels before I wrote here, but obviously I
didn't do it right.
Thanks again,
Liat.
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Sent from t
Hi, try ?levels
myData <- matrix(sample(c(LETTERS[1:10],NA),100,replace=T),nrow=25)
table(factor(as.vector(myData),levels=LETTERS[1:26]),useNA="ifany")
-
A R learner.
--
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Se
Hi,
I am trying to write some code that tracks changes that may have been made
after a dataframe has been edited using fix(). But if I edit only the first
cell (row 1, col 1) of the dataset below, it is like many records were edited.
What is the explanation for this error?
require(R.util
On Jul 22, 2010, at 4:00 PM, Liat wrote:
>
> Hi guys,
> I hope you can help me with this.
> Here is the problem:
> I have some data (myData) that looks similar to this:
>
> [,1] [,2] [,3] [,4]
> [1,] "A" "A" "B" "B"
> [2,] "B" "B" "B" "B"
> [3,] "C" "C" "C" "C"
>
> When I build
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