Hi John,
Aren't they in the component 'model' of the fitted object?
##
> options(contrasts = c("contr.treatment", "contr.poly"))
> house.plr <- polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing)
> head(house.plr$model)
Sat Infl Type Cont (weights)
1LowLow Tower Low
R is Not Excel, I think... But if you insist on "drawing" a table
below the graph, you may use rect() and text().
Why not add your numbers directly to your bars/lines/points/...? If
you can express the information in a table, why use a plot?
Regards,
Yihui
--
Yihui Xie <[EMAIL PROTECTED]>
Phone:
someone mentioned the RCurl package recently
On Sun, Sep 28, 2008 at 9:12 AM, June Kim <[EMAIL PROTECTED]> wrote:
> Hello,
>
> What is the best material(book, pdfs, ...) for programmers, who have
> extensive experience in other programming languages, to learn R
> programming? I think there are ma
Hello,
What is the best material(book, pdfs, ...) for programmers, who have
extensive experience in other programming languages, to learn R
programming? I think there are many materials on how to use R for
specific statistical jobs, but I haven't seen any material
particularly designed for R progr
Will this do it for you:
> set.seed(1)
> coarse <- runif(5,0,100)
> coarse <- cbind(coarse, coarse + runif(5,1,10))
> coarse <- round(coarse,3)
> fine <- runif(100,0,100)
> fine <- cbind(fine, fine + runif(100,0,1))
> result <- lapply(seq(nrow(coarse)), function(.row){
+ list(coarse[.row,], wh
Darin Brooks wrote:
Glad you were amused.
I assume that "booking this as a fortune" means that this was an idiotic way
to model the data?
Dieter was nominating this for the "fortunes" package in R. (Thanks Dieter)
MARS? Boosted Regression Trees? Any of these a better choice to extract
si
It's more a statement that it expresses a statistical perspective very
succinctly, somewhat like a Zen koan. Frank's book,"Regression
Modeling Strategies", has entire chapters on reasoned approaches to
your question. His website also has quite a bit of material free for
the taking.
--
D
I have two pairs of time intervals: coarse- and fine-grained. They're
components of their respective dataframes, looking like,
coarse:endtimestarttime
1t1_end t1_start
2 t2_end t2_start
...
fine: is the same, except that i
Glad you were amused.
I assume that "booking this as a fortune" means that this was an idiotic way
to model the data?
MARS? Boosted Regression Trees? Any of these a better choice to extract
significant predictors (from a list of about 44) for a measured dependent
variable?
-Original Messag
I think you want 'assign':
for (i in 1:3){
assign(qrnox[i], rq(nox~factor(year)+factor(state)+pcinc+I(pcinc^2)+I(pcinc^3),
tau=quant[i], data=exmp))
}
better yet, use a list:
result <- list()
for (i in 1:3){
result[[1]] <- rq()
}
result[[1]]
On Sat, Sep 27, 2008 at 4:24 PM, dimitrisk
On 27-Sep-08 21:45:23, Dieter Menne wrote:
> Frank E Harrell Jr vanderbilt.edu> writes:
>
>> Estimates from this model (and especially standard errors and
>> P-values)
>> will be invalid because they do not take into account the stepwise
>> procedure above that was used to torture the data unt
Frank E Harrell Jr vanderbilt.edu> writes:
> Estimates from this model (and especially standard errors and P-values)
> will be invalid because they do not take into account the stepwise
> procedure above that was used to torture the data until they confessed.
>
> Frank
Please book this as a f
I am sorry where is the download page or index?
Once I know that it is split out like this library(dse1) and library(dse2) work
just fine. My question now is how did you know this?
Thank you.
Kevin
Sarah Goslee <[EMAIL PROTECTED]> wrote:
> If you look at the download page, or the index,
Thanks Hadley, unfortunately doing this gives me an error:
> ggplot(polls, aes(x =Date, y = Popular_Support, colour=Party)) +
+ stat_smooth(span=0.5) +
+ geom_point(aes(shape=Source))
Error in `[.data.frame`(df, , var) : undefined columns selected
if I move it back up into ggplot, then it works f
I have two waves of a survey given to students at various middle
schools and high schools, with student id numbers for each student. I
am having difficulty reshaping the file from long to wide.
My code is below:
library(foreign)
svy <- read.spss("studsur4.SAV")
svy.wide <- reshape(svy, time
Hi,
?unique should work fine for this.
I think the problem in your implementation is that you modify x in the
loop, in particular its length may become shorter than the stopping
condition.
HTH,
baptiste
On 27 Sep 2008, at 14:30, Bastian Offermann wrote:
Hello all,
one brief question
Darin Brooks wrote:
Sorry.
Let me try again then.
I am trying to find "significant" predictors" from a list of about 44
independent variables. So I started with all 44 variables and ran
Why? What is wrong with insignificant predictors?
drop1(sep22lr, test="Chisq")... and then dropped the
Hi,
I am trying to use (i) as an index but R considers it as a function and not
as text. To be more specific I would like for example to estimate some
regressions named qrnox1, qrnox2, qrnox3,. and so on. But when I am
using qrnox(i) ot qrnox[i] it tries to find the ith element of vector qr
Hi guys,
I'm new to R and this might be a basic question. I want to have a plot with
a table right under it containing all the data in the plot. Is that possible
in R? How do to it?
Thanks!
Yuhan
--
View this message in context:
http://www.nabble.com/plotting-with-a-table-right-under-it-tp197
Hello R-helpers.
We have a question about AIC values. We ran several different GLMs (quadratic,
interactions) and our lowest AIC values were the same for several different
models. We don't know how to interpret this. Any thoughts? Thanks in advance.
Katrina Shelding
Hi,
I would like to solve a double integral of the form
\int_0^1 \int_0^1 x*y dx dy
using Gauss Quadrature.
I know that I can use R's integrate function to calculate it:
integrate(function(y) {
sapply(y, function(y) {
integrate(function(x) x*y, 0, 1)$value
})
}, 0, 1)
but I would like to use
Hello all,
one brief question
I would like to remove double/triple elements from a vector, e.g.
0 1 1 1 1 2 2 2 4 5 6 6
but keep one of these multiple ones. Should look like this finally:
0 1 2 4 5 6
My function does not work somehow
for(i in 2 : length(x)) {
If you look at the download page, or the index, or the online help
for dse, they all point out that dse is a bundle containing several
packages, none named "dse". Have you tried
library(dse1)
and
library(dse2)
If you have, and that does not work, please send the information
requested in the postin
Hello R users,
I am trying to get package 'dse' and it seems to download OK:
bundle 'dse' successfully unpacked and MD5 sums checked
But when I try to use it I get:
> help(package="dse")
Error in .find.package(pkgName, lib.loc, verbose = verbose) :
there is no package called '
Sorry, this should have the rigth subject now.
[EMAIL PROTECTED] wrote:
> I am trying to get package 'dse' and it seems to download OK:
>
> Content type 'application/zip' length 1413606 bytes (1.3 Mb)
> opened URL
> downloaded 1.3 Mb
>
> bundle 'dse' successfully unpacked and MD5 sums check
I am trying to get package 'dse' and it seems to download OK:
Content type 'application/zip' length 1413606 bytes (1.3 Mb)
opened URL
downloaded 1.3 Mb
bundle 'dse' successfully unpacked and MD5 sums checked
The downloaded packages are in
. . . \downloaded_packages
updating HTML package
It is easy enough to create the list, the real question is what do you
want in it ( provide commented, minimal, self-contained, reproducible
code). Here is the list:
> x <- vector('list', 10)
> names(x) <- paste("SS", 1:10, sep='')
> x
$SS1
NULL
$SS2
NULL
$SS3
NULL
$SS4
NULL
$SS5
NULL
$SS6
N
So, rather than the estimated between and within group variances from
a standard fixed-effects ANOVA (i.e. the Mean Sqs in the anova table),
you are looking for the estimated variance components from a model
with crossed random effects? If so, you may find the lmer function
found in package lme4 t
Is this what you want:
> l <- list(a=c(1), b=c(2,3), c=c(4,5,6))
> l
$a
[1] 1
$b
[1] 2 3
$c
[1] 4 5 6
> l$b <- c(l$b, 99)
> l
$a
[1] 1
$b
[1] 2 3 99
$c
[1] 4 5 6
>
Or is you want to dynamically specify the list element:
> x <- 'a'
> l[[x]] <- c(l[[x]], -99)
> l
$a
[1] 1 -99
$b
[1] 2
Dear list members,
The polr() function in the MASS package takes an optional weights argument
for case weights. Is there any way to retrieve the case weights from the
fitted "polr" object? Examining both the object and the code, I don't see
how this can be done, but perhaps I've missed something.
Dear R users:
Is there a way to append selectively to components of a list (if possible,
loops are to be avoided)? To illustrate the point, in the example below, I
would like to append 99 to vector b of the list l.
> l <- list(a=c(1), b=c(2,3), c=c(4,5,6))
> l
$a
[1] 1
$b
[1] 2 3
$c
[1] 4 5 6
I'm replying to this on the R-devel mailing list and the
Subject "On modes, types and R documentation"
Martin
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/po
Hi Jim:
Thanks for your suggestion from the plotrix package. That's exactly what I was
looking for, however, I can't see the dates along the X axis and neither the
temp line (I only see the thicks). Is there a way to actually plot the temp
line? I have been using the ggplot2 package and I am abl
Windows XP
R 2.7.2
I would like to create a list of varying length (length n) suitable for use as
dimnames for a matrix whose size will vary, i.e.
SS1 SS2 SS3 . . . SSn
Ansy suggestions for accomplishing this would be appreciated.
Thanks,
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and I
I have a huge data set with thousands of variable and one binary
variable. I know that most of the variables are correlated and are not
good predictors... but...
It is very hard to start modeling with such a huge dataset. What would
be your suggestion. How to make a first cut... how to eliminate m
Windows XP
R 2.7.2
I need to create a list of varying length of the form
SS1 SS2 SS3 . . . SSn
I will be using the list for the dimnames of a matrix that will have varying
dimensions. Any suggestions for a good way to accomplish this would be
appreciated.
Thanks,
John
Confidentiality Statemen
On 27-Sep-08 15:18:32, Gavin Simpson wrote:
> On Fri, 2008-09-26 at 13:17 +0100, GRANT Lewis wrote:
>> Dear All
>> I have three data sets, X1, X2 and Y. X1 is data, X2 and Y were
>> generated in (different) R programs. All three vectors have one
>> column of 60 data points.
>> I am using the code l
Yes. Also a redundent , following the first )
--- On Fri, 9/26/08, Roy Mendelssohn <[EMAIL PROTECTED]> wrote:
> From: Roy Mendelssohn <[EMAIL PROTECTED]>
> Subject: Re: [R] Bug in "is" ?
> To: "R help" <[EMAIL PROTECTED]>
> Received: Friday, September 26, 2008, 12:24 PM
> From the R-web pages
Thank you, an interesting interview and I have used the original pdf version
more than once though I tend to use it to think about how convert R to SAS
rather than SAS to R.
--- On Fri, 9/26/08, Ajay ohri <[EMAIL PROTECTED]> wrote:
> From: Ajay ohri <[EMAIL PROTECTED]>
> Subject: [R] A Book
Try this:
> BOD # comes with R
Time demand
118.3
22 10.3
33 19.0
44 16.0
55 15.6
67 19.8
> rollapply(as.zoo(as.matrix(BOD)), 3, sum)
2 6 37.6
3 9 45.3
4 12 50.6
5 16 51.4
rollapply has many more options and capabilities. See ?rollapply
On Sat, Sep 27,
Hi All,
I am trying to build a composite plot, with multiple categories, using
ggplot2.
In principle, it could be done using facetting, but I do not seem to be able
to get past the defaults, so I try building each plot separately, then
putting them all together using frameGrob and placeGrob.
For
On Fri, 2008-09-26 at 13:17 +0100, GRANT Lewis wrote:
> Dear All
> I have three data sets, X1, X2 and Y. X1 is data, X2 and Y were
> generated in (different) R programs. All three vectors have one column
> of 60 data points.
> I am using the code lm(Y~X1)$coef and lm(Y~X2)$coef.
Others have repli
Thanks Gabor,
But the function I intend to apply requires data.frame object ... not
zoo() object
have you had expirience with this kind of problems.
On Sep 27, 3:59 pm, "Gabor Grothendieck" <[EMAIL PROTECTED]>
wrote:
> rollapply in the zoo package does that. See ?rollapply and the
> three accom
Hello R list subscribers,
I am trying to use the "by" command to create line-specific variance covariance
matrices (where "x" is the original data matrix):
by(x, x$line, function(d) {
d.clean <- d[,-1]})
write.table(d.clean$line[1,1], sep = ",", file = "covariances.csv", col.names =
FALSE, row.
rollapply in the zoo package does that. See ?rollapply and the
three accompanying vignettes.
> library(zoo)
> z <- zoo(1:10)
> rollapply(z, 3, sum)
2 3 4 5 6 7 8 9
6 9 12 15 18 21 24 27
On Sat, Sep 27, 2008 at 10:45 AM, milicic.marko <[EMAIL PROTECTED]> wrote:
> Is there an implementat
Is there an implementation of moving window functionality so I can
apply any function while sliding trough window.
Thanks
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-pr
On Fri, 2008-09-26 at 06:11 -0700, glaporta wrote:
> I performed canonical correspondence analysis (cca) with the example data of
> vegan, but I'm not able to obtain a table like scores() for the constraining
> variables. I can see them in the summary() mode, but it would be great to
> have in a se
Donald Braman wrote:
Thanks, for the response!
Unfortunately, I was unclear; my problem is not that I need to know what the
percentile ranges are, but that I need to assign an appropriate percentile
range to each of the records in my dataframe. My dataframe contains
somewhere between 1000 and 9
Darin Brooks wrote:
> Sorry.
>
> Let me try again then.
>
> I am trying to find "significant" predictors" from a list of about 44
> independent variables. So I started with all 44 variables and ran
> drop1(sep22lr, test="Chisq")... and then dropped the highest p value from
> the run. Then I rer
Hi,
I would like to solve a double integral of the form
\int_0^1 \int_0^1 x*y dx dy
using Gauss Quadrature.
I know that I can use R's integrate function to calculate it:
integrate(function(y) {
sapply(y, function(y) {
integrate(function(x) x*y, 0, 1)$value
})
}, 0, 1)
but I would like to use
On Sat, Sep 27, 2008 at 1:08 AM, Tylere Couture <[EMAIL PROTECTED]> wrote:
> I have a simple plot:
>
> ggplot(polls, aes(x =Date, y = Popular_Support, colour=Party, shape=Source))
> +
> stat_smooth(span=0.5) +
> geom_point()
>
> How can I get the smooth to only render along one of the scales? ie, I
Sorry.
Let me try again then.
I am trying to find "significant" predictors" from a list of about 44
independent variables. So I started with all 44 variables and ran
drop1(sep22lr, test="Chisq")... and then dropped the highest p value from
the run. Then I reran the drop1.
Model:
MIN_Mstocked
Thanks, for the response!
Unfortunately, I was unclear; my problem is not that I need to know what the
percentile ranges are, but that I need to assign an appropriate percentile
range to each of the records in my dataframe. My dataframe contains
somewhere between 1000 and 9000 rows/records in my
mcmcsamp does not yet operate on non-normal distributions.
Hank
On Sep 27, 2008, at 9:06 AM, Parry Clarke wrote:
Hello,
I'm building a couple of mixed models using the lmer function.
The actual modelling is going well, but doing some reading on the
use of
crossed random effects and the compa
Hello,
I'm building a couple of mixed models using the lmer function.
The actual modelling is going well, but doing some reading on the use of
crossed random effects and the comparison of models with and without
random effects it is clear that I need to generate some Markov Chain Monte
Carlo sampl
Try this:
my.df$my.newvar <- quantile(my.df$my.var, probs = seq(0.01,1, 0.01))
On Sat, Sep 27, 2008 at 3:50 AM, Donald Braman <[EMAIL PROTECTED]> wrote:
> I'm wondering if there is a simple way to assign a quantile to a vector in a
> data frame, much like one could in Stata using centile. Let's
Hi
I am trying to utilize my dual core processor (and later a
High-performance clusters (HPC) ) by using the Rmpi, snow, snowfall,
... packages, but I am struggling at the beginning, i.e. to initialise
the "cluster" on my dual core computer. Whenever I try to initialize
it (via sfInit(parallel=TRU
Darin Brooks wrote:
> Good afternoon
>
> I have what I hope is a simple logistic regression issue.
>
> I started with 44 independent variables and then used the drop1,
> test="chisq" to reduce the list to 8 significant independent variables.
>
> drop1(sep22lr, test="Chisq") and wound up wit
Hi
I am using Ubuntu 8.04 64 bit, R as below, Matlab 7.6.0. I would like to
transfer mat files back and forward between R and Matlab. Whilst I have used
Matlab for years its been a long time since I have used R (hence question may
be a bit simple)
Running code
A <- c(1:10)
dim(A) <- c(2,5
C.H. wrote:
Dear R Gurus,
I have a problem related to plot.
For example, I have two variables, pre and post.
pre <- c(1,2,3,4,5)
post <- c(2,5,7,2,3)
How can I plot a line graph similar to this one?
http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=1847566&rendertype=figure&id=F1
Wou
Felipe Carrillo wrote:
Hello All:
Using the below dataset how can I make a barplot with
Date(X) and NumEggs(Y) by Site. Then plot Temp(lineplot)
It seems really simple, but I am having a hard time trying to
do it by Site. Thanks
Hi Felipe,
This might do what you want:
fdc<-r
Dear R useRs,
i try to compute the posterior mean for the parameters omega and beta
for the following
posterior density. I have simulated data where i know that the true
values of omega=12
and beta=0.01. With the function postMeanOmega and postMeanBeta i wanted
to compute
the mean values of om
On Sat, Sep 27, 2008 at 3:32 AM, Hesen Peng <[EMAIL PROTECTED]> wrote:
> Well, I finally figured out to do some algebra transformation and the
> problem was reduced to non-linear optimization on bounded areas. But
> on my way to this I ran into IMSL Fortran library function NNLPF. And
> its documen
Hi,
I've a question about the RandomForest package.
The package allows the extraction of a variable importance measure. As far as
I could see from the documentation, the computation is based on the Gini index.
Do you know if this extraction can be also based on other criteria? In
particular,
I'
dear Dr. Kubovy,
I am sorry. but the Variance table is not exactly what I want. I want
the partitioned VARIANCE for between and within the groups. the anova
()-table just gives me the SumSq and the mean Sq... I know how to run
t.test and ANOVA!
in the nlme-package there is the VarCorr functio
I have a simple plot:
ggplot(polls, aes(x =Date, y = Popular_Support, colour=Party, shape=Source))
+
stat_smooth(span=0.5) +
geom_point()
How can I get the smooth to only render along one of the scales? ie, I want
to see regressions for each colour, but not each shape.
Right now I get a trendl
Try do.call("rbind", nameofyourlist)
Nael
On Sat, Sep 27, 2008 at 8:51 AM, Matthew Pettis <[EMAIL PROTECTED]>wrote:
> Hi,
>
> I have a list output from the 'lapply' function where the value of
> each element of a list is a data frame (each data frame in the list
> has the same column types). How
67 matches
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