Hi Frank,
Thanks for the suggestion. using numericScale() does work for Dotplot,
but there're still a few issues:
1. My factor names are Plot A, PF, MSF, and YSF, so numericScale turns
that into 3, 2, 1, 4 and the x-axis is plotted 1, 2, 3, 4. Is there
any way I can retain the same order on the g
Hi,
library(gridExtra)
example(patternGrob)
provides some patterns to fill a rectangular area using Grid graphics.
It could in theory be used in lattice. I wouldn't use it either, but I
can imagine how it might be useful on very special occasions.
Best,
baptiste
On 28 July 2010 06:11, HC
ddply(z, .(groupId,location),function(d)with(d,
c(startLoc=Pos[1],endLoc=Pos[length(Pos)],
peakValue=max(sumoo),other=map[1])))
startLoc=Pos[1],endLoc=Pos[length(Pos)], peakValue=max(sumoo),numeric
value
other=map[1]===charactor value
as a result:
c(startLoc=Pos[1],endLoc=Pos[length(Pos)]
Am 28.07.2010 07:36, schrieb jd6688:
>
> DF1
> name OTHER
> ABCO
> KKKO
> QQQO
> DDDO
> PPPO
>
> DF2
> name
> ABC
> KKK
> DDD
>
> If the names in df1 mapped the names in df2, then add the mapped name to df1
> as a separate column, for instance "mappedColumn"
What do you
this is very insightful. sounds exactly like what I want to do.
thanks. Frank.
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Sent from the R help mailing list archive at Nabble.com.
good point. It seems to be important to investigate the nature of
distribution. I might be too naive to assume that a "empirical probability
distribution" would be automatically generated from a cloud of data
points.
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-gener
hi, Dennis:
points well taken. it seems to be important to investigate the nature of
distribution. I may be too naive to assume a "empirical probability
distribution" would be computed from a could of data points
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When I tried this, I'm having this error. Can somebody help me on
this. Are there any alternatives or workaround for this? I'm having
hard time to convince our admin to install X11 library and headers
since they are not included on the default OS installation.
Thanks in advance :)
> jpeg("test.jp
Hello, I too am having this problem. Some two minutes ago all was well then
all of a sudden I cannot backspace or delete or use arrows etc..
--
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Sent from the R help mailing list archive at
DF1
nameOTHER
ABC O
KKK O
QQQ O
DDD O
PPP O
DF2
name
ABC
KKK
DDD
If the names in df1 mapped the names in df2, then add the mapped name to df1
as a separate column, for instance "mappedColumn"
how to implement this? Thanks
--
View this message in context:
http
This is my first post. I am running Mac OS X version 10.6.3. I am running R
2.11.0 GUI 1.33 64 bit.
This may or may not be related to sqldf, but I experienced this problem while
attempting to use an sqldf query. The same code runs with no problem on my
Windows machine. Here is what happens
Hi,
Thanks for your responses. Although a wrist slapping for not keeping
updated with R/Fedora may be warranted, I find it quite terrifying
that by simply having a version number in a directory name R went on
to try and erase everything on the computer! Thank god I wasn't logged
in as ro
to be more clear, i'm particularly interested in trying to determine
the effect of "Class" on "Score1", while controlling for the effect
"Score2" has on "Score1" (and "Score2" is correlated with
"Class" (where "Class" is an ordered variable with exactly 4 distinct
levels)).
i can perform an ANOVA o
You are confused. You have not specified a "grouping" random effect,
so this is not a mixed effect model as it stands.
If this is a homework problem, ask a teacher or classmate for help.
Otherwise, try consulting your local statistician. You do not appear
to understand the concepts of mixed effec
Thank you for your follow up on this matter.
I did think about the "partial transparent color" option and will certainly
use it and see how it works out. For presentations and color prints, the
partial transparency is surely going to work and may even look nicer. But
for black and white printing
d <- apply(s, 2, sample, size = 1*nrow(s), replace = TRUE)
why the code above return the following error
Error: cannot allocate vector of size 218.8 Mb
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Sent from the R he
On Tue, 27 Jul 2010, Maomao Luo wrote:
> Hi,
>
> I am trying to create a spatial weight matrix based on the neighboring
> relationship among Europan countries. Can I use the map function in R to
> generate a European map and then use this map to build the w weight matrix?
> So far, I cannot creat
hi all - i'm having trouble using lme to specify a mixed effects
model.
i'm pretty sure this is quite easy for the experienced anova-er, which
i unfortunately am not.
i have a data frame with the following columns:
col 1 : "Score1" (this is a continuous numeric measure between 0 and
1)
col 2 : "Sc
hi r-help
have tried looking for a package that has analog complexing algo function.
appreciate if someone could guide me to the appropriate package.
thanks
stan
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
https:/
Hi:
I tried the following:
library(sos)
findFn('Anderson-Darling')
found 49 matches; retrieving 3 pages
2 3
It appears that packages nortest and ADGofTest might be good places to
start, but you can check out the others for yourself.
HTH,
Dennis
On Tue, Jul 27, 2010 at 6:08 PM, Roslina Zakaria
Dennis:
points well taken. It seems to be important to investigate the nature of
distribution. I might be too naive to assume a "emiprical probability
distribution" will be simply calculated from a clound of data points...
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View this message in context:
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Hi,
I have the binned data (observed and generated from model) that I would like
to
test using the anderson-darling goodness of fit test. But I'm not sure which
package in R to use.
I tried ad.test(...) but it does not recognise the test by Vito Ricci in
FITTING
DISTRIBUTIONS WITH R
> ad
Thanks to everybody for the solutions.
On Tue, Jul 27, 2010 at 3:47 AM, Dennis Murphy wrote:
> Hi:
>
> Another approach might be to use the melt() function in package reshape
> before creating the plot with xyplot, something along the lines of the
> following:
>
> library(reshape)
> mdat <- melt(
Thanks!
On Tue, Jul 27, 2010 at 4:29 PM, wrote:
> Is this the kind of thing you are looking for?
>
> > dat <- data.frame(x = 1:3, freq = 2:4)
> > dat
> x freq
> 1 12
> 2 23
> 3 34
> > newDat <- dat[rep(rownames(dat), dat$freq), ]
> > newDat
>x freq
> 1 12
> 1.1 12
> 2
In addition to the arrows function, look at the my.symbols and ms.arrows
functions in the TeachingDemos package, this is a wrapper around the arrows
function, but may be a more natural interface for what you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
If they want to generate directly from the empirical distribution, then
sampling with replacement is the best choice (others had already suggested
that). But the reference in the original post to the normal and beta
distributions suggested to me that the original poster may have wanted a smooth
On 27/07/2010 6:34 PM, vincent.deluard wrote:
Hi all,
I am dealing with very large numbers but I am only interested in their last
digits.
244^244
[1] Inf
As far as I can tell, the largest number R can take has 308 digits (1+E308).
Is there a way I could see the last digits only of 244^244?
Is this the kind of thing you are looking for?
> dat <- data.frame(x = 1:3, freq = 2:4)
> dat
x freq
1 12
2 23
3 34
> newDat <- dat[rep(rownames(dat), dat$freq), ]
> newDat
x freq
1 12
1.1 12
2 23
2.1 23
2.2 23
3 34
3.1 34
3.2 34
3.3 34
>
I'm somewhat a new user and have been trying to figure out how to repeat
rows a certain number of time based on a variable. Currently, the number of
rows is not reflective of the number of observations. To get the number of
observations (n=22 in this case), I have to multiply by the variable
NoRe
It won't happen automatically unless you make it do so yourself.
Fortunately this is not hard. See,
?arrows
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Trying To learn again
Sent: Wednesday, 28 July 2010 7:07 AM
To: r-hel
Hi,
Here are two options. I would colour the last point differently, or
make it a different shape rather than using an arrow, but its your
choice.
x <- 1:10
y <- 1:10
#With an arrow
plot(x, y)
arrows(x0 = x[length(x)], y0 = y[length(y)] - 1,
x1 = x[length(x)], y1 = y[length(y)] - .5)
#Overwr
It is probably easier to do this without eval and parse (see fortune(106)).
Something like:
val <- get(dataname)[[v1]]
hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-he
Hi I want to plot an x,y plot something like an scatter plot.
I always have the same doubt, were is the last point of my file?
Imagine it is a time series so I want the last point to indicate with an
arrow (but athomatically if posible).
Someone knows if it is posible?
Could it be posible to ad
This has been shown to yield unreliable analyses. Use the more formal
proportional odds ordinal logistic model. This is a generalization of
the Wiloxon-Mann-Whitney-Kruskal-Wallis statistic. This is
implemented in the rms package and elsewhere.
Frank E Harrell Jr Professor and Chairman
>
>
> As far as I can tell, the largest number R can take has 308 digits
> (1+E308).
>
This is not a correct statement. The magnitude of the largest 64-bit double
precision floating
point number in R is approximately 1E308. The internal storage in R (and
in most computers today)
contains only th
Look at the HSAUR package (and the book that it goes with). This may give you
(your students) enough to start on several topics. If you use these examples
then you should probably include the book as at least an optional text for the
class.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Ce
Hello,
I am trying to use copula-GARCH to model a multivariate time series. So far, I
have:
-Estimated the GARCH(1,1) model
-Obtained and standardized the residuals
-Applied the pdf (Student's t in this case) to obtain (pseudo) uniform variables
-Estimated the copula
-Obtained a random sample fr
Colleagues
I am trying to replicate some analyses performing in SAS using a "rank ANCOVA
model". I assume that this is a non-parametric ANOVA with covariates but I
have not been able to get confirmation from the people who did the original
analyses. Can anyone direct me to the comparable func
It looks to me like you keep sampling from some dataset 's' 10,000
times. Since you can sample() with replacement, I wonder if you could
just take a sample of the size you want, rather than using a loop with
sample. Perhaps along these lines:
d <- apply(s, 2, sample, size = 1*nrow(s), replac
Easiest thing is to sample with replacement from the original data.
This is the idea behind the bootstrap, which is sampling from the
empirical CDF.
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On
You can get the 'bc' package for R, or just use 'bc' itself:
bc 1.05
Copyright 1991, 1992, 1993, 1994, 1997, 1998 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
244^244
33351673199434891333790950239586798749134713899690921746779018273
I am trying to do the following to accomplish the tasks, can anybody to
simplify the solutions.
Thanks,
for (i in 1:1){
d<-apply(s,2,sample)
pos_neg_tem<-t(apply(d,1,doit))
if (i>1){
pos_neg_pool<-rbind(pos_neg_pool,pos_neg_tem)
}else{
pos_neg_pool<- pos_neg_tem
}}
--
View
Another option for fitting a smooth distribution to data (and generating future
observations from the smooth distribution) is to use the logspline package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
Hi all,
I am dealing with very large numbers but I am only interested in their last
digits.
> 244^244
[1] Inf
As far as I can tell, the largest number R can take has 308 digits (1+E308).
Is there a way I could see the last digits only of 244^244?
Many thanks for your help.
Vincent Deluard, C
On Jul 27, 2010, at 4:53 PM, Duncan Murdoch wrote:
> On 27/07/2010 4:39 PM, Jarrod Hadfield wrote:
>> Hi,
>> I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
>> tried to remove "junk files" it removed EVERYTHING in my local
>> account! (See below).
>> Can anyone tell me what hap
On 27/07/2010 4:39 PM, Jarrod Hadfield wrote:
Hi,
I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
tried to remove "junk files" it removed EVERYTHING in my local
account! (See below).
Can anyone tell me what happened, and even more importantly if I can I
restore what was lost
Write a function that incorporates "doit" and the column shuffle.
Let's call it "doitbetter"
replicate(1, doitbetter())
You'll probably want to read the help for "replicate" to make sure the
defaults are what you want.
--Gray
On Tue, Jul 27, 2010 at 4:43 PM, jd6688 wrote:
>
> myDF:
>
> d1
It all depends on what you are going with the data. First in your
scan example, I would not read in a line at a time, but probably
several thousand and then process the data. Most of your time is
probably spent in reading. I assume that you are not reading it all
in at once (but then maybe you a
Hopefully the "right now" will be a forever. The fact that trellis graphics do
not support hatching is a conscious decision or feature, not a bug or missing
feature to be corrected some day (at least I for one will be disappointed if
Deepayan (or someone else) gives in and implements it).
Have
myDF:
d1 d2 d3 d4
d5
-0.1669103510.022304377 -0.00825924 0.008330689 -0.000925938
-0.1669103510.022304377 -0.00825924 0.008330689 -0.000925938
-0.1669103510.022304377 -0.00825924 0.0
This is a frequently asked/answered question (7.21 in the FAQ). What searching
did you do and why did it not find this FAQ or previous discussion of it? How
could the documentation/search/etc. be improved so that you (and the next n
people with this question) will find the answer easier?
The
On 27/07/2010 3:20 PM, rrich...@fh-lausitz.de wrote:
I want to plot points with different colors to show different
selections of points in a 3d plot.
My problem is that if I plot a red point at a location where already a
blue point was plotted, than the point is still blue.
Is there a param
On Tue, Jul 27, 2010 at 4:17 PM, schuster wrote:
>
>
> Hello,
>
> I see some people including myself confused by the different object-oriented
> approaches in R (S3, S4, OOP, R.oo etc.).
S3 is the primary OO approach in R and if you are new you should
initially focus on that both because of its p
Hi,
I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
tried to remove "junk files" it removed EVERYTHING in my local
account! (See below).
Can anyone tell me what happened, and even more importantly if I can I
restore what was lost.
Panickingly,
Jarrod
[jar...@localhost AMan
The index indicates which samples should go into the training set.
However, you are using out of bag sampling, so it would use the whole
training set and return the OOB error (instead of the error estimates
that would be produced by resampling via the index).
Which do you want? OOB estimates or ot
I've found that opening a connection, and scanning (in a loop)
line-by-line, is far faster than either read.table or read.fwf. E.g,
here's a file (temp2) that has 1500 rows and 550K columns:
showConnections(all=TRUE)
con <- file("temp2",open='r')
system.time({
for (i in 0:(num.samp-1)){
new.gen[
Hello,
I see some people including myself confused by the different object-oriented
approaches in R (S3, S4, OOP, R.oo etc.).
Would it be ok to collect examples and solutions for the different OO-packages
in one package and add a vignette for documentation?
(assuming I find time for this t
Yes, this solved the problem. Thanks.
Jorge
On 7/27/10 3:13 PM, "Duncan Murdoch" wrote:
> On 27/07/2010 2:34 PM, Jorge A. Ahumada wrote:
>> > I am writing a function where the arguments are names of objects or
>> variable
>> > names in a data frame. To convert the strings to the objects I am u
Hello,
I'm using the package survey, version 3.22-2 in R 2.11.1, on windows
XP, but this isn't a platform issue.
I am trying to calculate some regression and probability proportional
to size (pps) estimates for a dataset I have. The data has 2 strata,
weights, and cluster IDs. However, I can't fi
I want to plot points with different colors to show different
selections of points in a 3d plot.
My problem is that if I plot a red point at a location where already a
blue point was plotted, than the point is still blue.
Is there a parameter or so which can be used to draw over a existing po
Thanks for all the help.
I had tried using the "index" in caret to try to dictate which rows of the
sample would be used in each of the tree building in RF. (e.g. use all data
from A B site for training, hold out all data from C site for testing etc)
However after running, when I cross-checked
On 27/07/2010 2:34 PM, Jorge A. Ahumada wrote:
I am writing a function where the arguments are names of objects or variable
names in a data frame. To convert the strings to the objects I am using
eval(parse(text=name)):
f.graph.two.vbs<-function(dataname,v1){
val<-paste(dataname,v1,sep
Well, the data set I am using is quite large, but RECORD is also a numeric
variable and data is a data frame.
Here is my sessionInfo():
>sessionInfo()
R version 2.11.0 (2010-04-22)
i386-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1]
Hi Jose,
What's the problem?
# some data
set.seed(123)
mydata <- data.frame(x = rnorm(10), y = rpois(10, 10))
mydata
f.graph.two.vbs<-function(dataname,v1){
val<-paste(dataname,v1,sep="$")
val<-eval(parse(text=val))
val
}
f.graph.two.vbs('mydata', 'x')
[1] -0.56047565 -0.23
Dear Jorge,
It might help if you included some sample data that replicates your
problem, or, perhaps, the results of sessionInfo(). I cannot
replicate it. Here are my results:
> data <- data.frame(A = 1:10, RECORD = 1:10)
>
> f.graph.two.vbs<-function(dataname,v1){
+ val<-paste(dataname,v1,s
I am writing a function where the arguments are names of objects or variable
names in a data frame. To convert the strings to the objects I am using
eval(parse(text=name)):
f.graph.two.vbs<-function(dataname,v1){
val<-paste(dataname,v1,sep="$")
val<-eval(parse(text=val))
val
When I ask R to compute: predict(data, int = "c"), following a linear
regression, what is it computing exactly? How are these lower and upper
"prediction" limits different than what I would get for confidence limits?
Thanks,
Matt.
[[alternative HTML version deleted]]
__
Hi,
I am running the following model:
fit1.full <- coxme(Surv(age_sym1, sym1) ~ sex + lifedxm*sex + (1|famid),
data=bip.surv)
I would like to extract the AIC from that object to calculate the AICC.
However, when I look at str(fit1.full) and summary(fit1.full) (pasted
below) I don't see anything th
Hi,
Just to extend the excellent suggestions, if you are interested in the
odds ratio, you can just use exp():
#Odds Ratio
exp(fit4$coefficients)
#Confidence interval around OR
exp(confint(fit4))
To give you an idea graphically of the log odds (or logit) look at:
p <- seq(0, 1, by = .001)
plot
Hi Jim,
Ow! Very nice job at http://mephisto.unige.ch/traminer/preview.shtml I´m
going to read more about it.
I have a lot of different steps, in a sequence. Actually, 586 different
possible steps, but I have 4269 legal cases, with a maximum of 379 steps
each one.
If you want, I can send this da
On Tue, 27 Jul 2010, Murray Jorgensen wrote:
I am introducing the scan() function to my class. Consider the following file
(Scanexamp.Rnw )
\documentclass[12pt]{article}
\begin{document}
<<>> =
height = scan()
64 62 66 65 62
69 72 72 70
part = scan(what = character(0))
"Soprano" "Soprano" "
On Tue, 27 Jul 2010, John Sorkin wrote:
Do not worry about the SE. The SE listed on the output is the SE of the log
odds. You can use the estimate (beta) and SE from the listing to compute a
confidence interval (CI)as follows:
CI exp(beta-1.96*SE) to exp(beta-1.96*SE)
The standard errors can
Hi Allan,
It helps a lot. I´ll try to read more about it.
But, as you asked me, here goes a brief explanation about the necessary
columns of the sample date paste at the end:
id_processo: identify a legal case, it is its primary key.
ordem_andamento: is the step number inside a legal case (id_pr
Do not worry about the SE. The SE listed on the output is the SE of the log
odds. You can use the estimate (beta) and SE from the listing to compute a
confidence interval (CI)as follows:
CI exp(beta-1.96*SE) to exp(beta-1.96*SE)
John
John Sorkin
Chief Biostatistics and Informatics
Univ. of Maryla
On Mon, 26 Jul 2010, Alon Friedman wrote:
Hi
I am looking for R templates to introduce the R to my students at
Seton hall university. The templates are predefined scripts in R that
will retain its primary intent when individually customized with their
own variable data or text.
Well the help
Not to beat a dead horse...
I've found that I like the useR conferences more than most statistics
conferences. This isn't due to the difference in content, but the
difference in the audience and the environment.
For example, everyone is at useR because of their appreciation of R.
At most other co
On 7/27/2010 6:00 AM, r-help-requ...@r-project.org wrote:
Date: Mon, 26 Jul 2010 11:36:29 -0700 (PDT)
From: xin wei
To:r-help@r-project.org
Subject: [R] how to generate a random data from a empirical
distribition
Message-ID:<1280169389379-2302716.p...@n4.nabble.com>
Content-Type: text/pla
Thanks for the quick reply. That makes sense, although I am still confused
why the functions prevent mincriterion from being lower than 0 as to run
cforests with "testtype" as "Teststatistic" and grow maximum depth trees you
need to put in qnorm(0) (= -inf) which is impossible? I will stick to ot
On Tue, 27 Jul 2010, Christopher David Desjardins wrote:
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file a
On Tue, 27 Jul 2010, Allan Engelhardt wrote:
Great, thanks. I couldn't quite get your syntax to work, but
Did you use R-devel? The syntax has changed ... and that's why I said
'e.g.'.
z <- packageStatus(.libPaths()[1])[[1]]
unname( z$Package[z$Status == "unavailable"] )
seems to do the
Great, thanks. I couldn't quite get your syntax to work, but
z <- packageStatus(.libPaths()[1])[[1]]
unname( z$Package[z$Status == "unavailable"] )
seems to do the trick for me.
Thanks again.
Allan
On 27/07/10 16:31, Prof Brian Ripley wrote:
If I understand you correctly, set the filter and
Dear all,
I am struggling with the calculation of standard error of the coefficient in
Binary logistic regression analysis.
I built a binary logsitic regression model as follows and got confused since
the calculation of standard error of coefficients of X1, X2 and X3 are not
the same as the Line
Hi Ana,
Does the "predict" function do what you want? Type in ?predict.lm
--Gray
On 7/27/10, Ana De Barros wrote:
> Hi,
>
> Is there any way to estimate a DEPENDENT variable through a GLM/LM model?
>
> Suppose I have the linear model: y=a0+a1*x1+a2*x2 (a0=1, a1=0.6, a2=0.8,
> x1~N(1,1), x2~N(0
On Tue, 27 Jul 2010, Matthew OKane wrote:
Hi,
Could anyone help me understand how the mincriterion threshold works in
ctree and cforest of the party package? I've seen examples which state that
to satisfy the p < 0.05 condition before splitting I should use mincriterion
= 0.95 while the docume
If I understand you correctly, set the filter and use packageStatus().
Its summary() method tells you which packages you have installed which
are 'unavailable'. E.g. my Mac (with pkgType = "source") shows in
R-devel
summary(packageStatus(.libPaths()[1]))$Libs[[1]]$unavailable
[1] "BayesX"
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
> On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
>
> > Sorry. I should have included some data. I've attached a subset of my
> > data (50/192) cases in a Rdata file and have pasted it below.
> >
> > Running anova I g
Hi,
Could anyone help me understand how the mincriterion threshold works in
ctree and cforest of the party package? I've seen examples which state that
to satisfy the p < 0.05 condition before splitting I should use mincriterion
= 0.95 while the documentation suggests I should use mincriterion =
On Tue, Jul 27, 2010 at 9:48 AM, Gabor Grothendieck
wrote:
> On Tue, Jul 27, 2010 at 7:17 AM, Allan Engelhardt wrote:
>> I only recently discovered options("available_packages_filters" = list(add =
>> TRUE, "license/FOSS")) [cf. help("available.packages", package="utils") in R
>> 2.10.0 or later]
On Tue, Jul 27, 2010 at 10:02 AM, Rnoobie wrote:
>
> Hi Ya'll
>
> I'm trying to install the zoo library... I've downloaded about 5 or 6 of
> their releases (inc. the latest one), all of them giving me the same syntax
> error when trying to install...
>
> Anyone seen this problem before? its not th
Hi Ya'll
I'm trying to install the zoo library... I've downloaded about 5 or 6 of
their releases (inc. the latest one), all of them giving me the same syntax
error when trying to install...
Anyone seen this problem before? its not the "lib" argument that is the
problem (ive been installing oth
Pablo, we've had success using
http://mephisto.unige.ch/traminer/preview.shtml to look at marketing paths.
Question would be how many distinct case step discriptions are there?
HTH, Jim
On Jul 26, 2010 9:44 AM, "Pablo Cerdeira" wrote:
Hi all,
I have no idea if this question is to easy to be an
If the x-axis variable is really a factor, xYplot will not handle it.
You probably need a dot chart instead (see Hmisc's Dotplot).
Note that it is unlikely that the confidence intervals are really
symmetric.
Frank
On Tue, 27 Jul 2010, Kang Min wrote:
Hi,
I'm trying to plot a graph with err
Hi,
On Tue, Jul 27, 2010 at 4:14 AM, Albert-Jan Roskam wrote:
> Thank you! As an R newbie I sometimes find it confusing that in R everything
> can be done in at least four different ways. My question about S4/OOP was
> motivated by the fact that those two class systems have the same designer
>
On Tue, Jul 27, 2010 at 7:17 AM, Allan Engelhardt wrote:
> I only recently discovered options("available_packages_filters" = list(add =
> TRUE, "license/FOSS")) [cf. help("available.packages", package="utils") in R
> 2.10.0 or later] which goes nicely with my options("checkPackageLicense" =
> TRUE
Hi,
I'm trying to plot a graph with error bars using xYplot in the Hmisc
package. My data looks like this.
mortstand sitetype
0.042512776 0.017854525 Plot A ST
0.010459803 0.005573305 PF ST
0.005188321 0.006842107MSFST
0.004276068 0.01
One of the nice features of R's formula syntax is that
you can create a character string containing a formula,
and pass it to the formula() function. For example:
xyplot(formula(paste(paste(paste('D',1:10,sep=''),collapse='+'),'X',sep='~')),data)
will do what you want.
On 27/07/10 13:20, Ben Bolker wrote:
Allan Engelhardt cybaea.com> writes:
I only recently discovered options("available_packages_filters" =
list(add = TRUE, "license/FOSS")) [cf. help("available.packages",
package="utils") in R 2.10.0 or later] which goes nicely with my
options("checkPack
Hi,
Is there any way to estimate a DEPENDENT variable through a GLM/LM model?
Suppose I have the linear model: y=a0+a1*x1+a2*x2 (a0=1, a1=0.6, a2=0.8,
x1~N(1,1), x2~N(0,1)).
The alphas and the auxiliary variables are given and I have to estimate y.
The point is if I estimate it, let¹s say algebr
Hi,
do.call(sum, mylist)
?do.call
baptiste
On 27 July 2010 14:36, Nicola Sturaro Sommacal
wrote:
> Hi!
>
> I have a list of 24 elements, all of the same type (dataframe, for example).
>
> I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
> mylist[[24]] to obtain the sum.
>
>
Have a look at Reduce(), e.g.,
Reduce("+", your.list)
I hope it helps.
Best,
Dimitris
On 7/27/2010 2:36 PM, Nicola Sturaro Sommacal wrote:
Hi!
I have a list of 24 elements, all of the same type (dataframe, for example).
I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
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