On Mon, Jul 30, 2012 at 10:16 PM, Bert Gunter wrote:
> David:
>
> I think one needs to carefully parse the xyplot help, where it says:
>
> "To use more than one legend, or to have arbitrary legends not
> constrained by the structure imposed by key, use the legend argument.
> "
>
> So I presume tha
On Tue, Jul 31, 2012 at 2:43 AM, Elliot Joel Bernstein
wrote:
> Is there an easy way to "thin" a lattice plot? I often create plots from
> large data sets, and use the "pdf" command to save them to a file, but the
> resulting files can be huge, because every point in the underlying dataset
> is re
On Mon, Jul 30, 2012 at 11:18 PM, jpm miao wrote:
> Hi,
>
>I let xzoo be an empty object:
>
>> xzoo<-{}
This sounds like the start of a bad idea
>
>
>and I have an existing zoo object x1zoo_f. I would like to combine
> the two to make a new zoo object, and continue doing so in a loo
Hi,
I let xzoo be an empty object:
> xzoo<-{}
and I have an existing zoo object x1zoo_f. I would like to combine
the two to make a new zoo object, and continue doing so in a loop,
which is not shown here. However, when I type
> xzoo<-cbind(xzoo, x1zoo_f)
An error message emerges
On Jul 29, 2012, at 11:26 PM, mary Joy Regidor wrote:
Greetings of Peace!
I am new in R software. I want to use the correlation dimension
corrDim in computing a map. I have installed the packages(fractal)
in R. Now my problem is this, when I tried your example in the
following site http:
On 07/29/12 08:56, arcata wrote:
> I am mapping U.S. state counties in individual states maps, e.g. A map of
> Montana, with county lines.
> How do I fill specific counties with a different color?
>
Do you mean something like:
m <- map("county", "Montana")
map("county", m$names[c(1, 4)], fill=TRU
Amazingly, RTFM. Linux typically uses "repositories" to download software from.
If you follow the instructions on CRAN specific to Ubuntu, you can add a
repository that has current versions of R. Admittedly, some of the steps are
non-obvious due to security concerns, but the instructions are pr
On 07/28/12 23:46, Thomas Steiner wrote:
> Hi,
> I'd like to have a low resolution word map in R.
> The "maps" package has this option, but if I use the argument, the map
> looses sense: Russia and Australia get empty etc
>
> library("maps")
> m=map(col="skyblue",fill=TRUE,plot=TRUE,resolution=10)
Hi,
Thank you. I've tried to find the substitute of such packages with google,
but I couldn't find any. Could you please tell me how to find it? Sorry,
I'm a beginner user of this software.
Regards,
Ario
On Tue, Jul 31, 2012 at 12:19 AM, Uwe Ligges <
lig...@statistik.tu-dortmund.de> wrote:
>
>
>
Jan,
Could you send the exact gam call and exact vis.gam call that did this
please? Also, if 'm' denotes your fitted model, what result does
'fitted(m)' give? and what is the output from print(m)?
best,
Simon
On 07/30/2012 09:19 PM, janvanhove wrote:
Hi everyone,
I ran a binomial GAM consi
I am a newbie when it comes to Ubuntu having "grown up" on Windows
using R. I installed 12.04 on a laptop and it went fine. When I went
out to CRAN and followed the directions (hopefully) to install R,
seems to have gone fine, except that when I start R I get 2.14.1:
jim@jim-winbook:~$ R
R vers
Or, assuming you only have 4 different elements :
mat<- matrix(rep(c(1,2,"A", "B"),each=10),10,10, byrow=F)
mat2 <- as.data.frame(mat)
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[2,] "1" "2" "A" "B" "1" "2" "A" "B"
Hello,
Maybe something along the lines of
J <- 10
cols <- rep(c(TRUE, TRUE, FALSE, FALSE), 3)[seq_len(J)]
for(i in which(cols)) { do something }
for(i in which(!cols)) { do something else }
Hope this helps,
Rui Barradas
Em 31-07-2012 00:18, Claudia Penaloza escreveu:
Dear All,
I would like t
Dear All,
I would like to apply two different "for loops" to each set of four columns
of a matrix (the loops here are simplifications of the actual loops I will
be running which involve multiple if/else statements).
I don't know how to "alternate" between the loops depending on which column
is "run
You might also check ?pdf on your system. On Windows the default is for
compression. Your code creates a 186K file although it is slow to load
reflecting the overhead from decompressing the file.
--
David L Carlson
Associate Professor of Anthropology
Te
Hello,
I have a problem with special characters such as "í" or "ñ" when using
hwriter. This only happens when I use windows, it works fine on mac. If I
do:
print(datosdv)
"Ciencias Sociales y Jurídicas n:74 | 33.94%"
but:
hwrite(datosdv, p, br=TRUE)
"Ciencias Sociales y Jur�dicas n:74 | 33.94
Hello,
Your str() is different from mine. It has G,Y,R, and T as factors. You can do
one thing, add stringsAsFactors=FALSE in read.table(). Even without it, I
don't have any problems as this will only affect the Summary column.
Z16<-read.table(text="
Summary G Y R T
Ac
Dear R users,
I have a hard time interpreting the covariances in the parameter estimates
output (standardized), even in the example documented (PoliticalDemocracy).
Can anyone tell me if the estimated covariances are residual covariances
(unexplained by the model), or the covariances of the obser
Hi Arun,
when i'm using the logic which you mentionI'm getting error..can you
please help me on this.
> Z16<-str(Z16)
'data.frame': 5 obs. of 5 variables:
$ Summary: chr "Accts" "AcctCov" "Cov%" "UnCov%" ...
$ G : Factor w/ 5 levels "224","39","568",..: 3 1 2 4 5
$ Y : Factor
Hello,
It's not "Colnames". It is colnames(). Also, you should use
stringsAsFactors=FALSE.
I tried to reply it to nabble. But, seems like nabble webpage is redirecting
to some porn site.
Anyway,
X<-read.table(text="
Summary G Y R
Acc 12 12 13
Hello,
Try the following.
# make up some data
dds <- 1e3
nc <- 10
xss <- data.frame(matrix(runif(nc*dds, min=-1, max=1), ncol=nc))
names(xss) <- paste0("xs", 1:10)
# two functions with descriptive names
getControlLimits <- function(x, L = 3){
mu <- mean(x)
sigma <- sd(x)
c(lcl = mu
Dear Mauro,
I believe that I've answered a version of this question three times this
month alone, so I'll be brief.
Are you aware that if you use type-III tests, even if you are careful to
employ contrasts, such as orthogonal polynomial contrasts, that are
orthogonal for different terms in the r
On Jul 30, 2012, at 2:13 PM, Elliot Joel Bernstein wrote:
Is there an easy way to "thin" a lattice plot? I often create plots
from
large data sets, and use the "pdf" command to save them to a file,
but the
resulting files can be huge, because every point in the underlying
dataset
is render
Hello,
Sorry for my earlier post, they show the problem of not providing
context. Fortunatelly, arun's post made me realize there was some data
posted somwhere after all. So, try this:
aggregate(DEPTH ~ LOG_DATE, data = dat1, quantile, probs = c(0.025, 0.975))
LOG_DATE DEPTH.2.5% DEPTH.97.5
You can't always get what you want (a data.frame with two sets of column
headings), but you do have several options. Maybe they will help you think
about what you are trying to do.
Given X and Y as Arun provided earlier, you can create Z
> Z <- rbind(X,colnames(Y),Y)
> Z
Summary G Y R
1
On 30.07.2012 15:36, Robert Baer wrote:
On Mon, 30 Jul 2012 09:58:09 +0100 (BST) (Ted Harding)
wrote:
Greetings All.
My apologies for a question whose answer is probably
readily available somewhere (for some interpetation
of "somewhere") ...
Say I have just typed (from a sheet of paper) seve
Dear Prof. John Fox,
I found the paper very useful. Thank you very much for attaching the link!
Which type of SS (II or III) do you suggest for a multivariate model with 2
unbalanced factors and 2 covariates?
I think that type III is the right one but
Mauro
2012/7/30
> Dear GMM,
>
> > --
Is there an easy way to "thin" a lattice plot? I often create plots from
large data sets, and use the "pdf" command to save them to a file, but the
resulting files can be huge, because every point in the underlying dataset
is rendered in the plot, even though it isn't possible to see that much
deta
Hello,
Try the following.
plants_herbs_input_top10 <- structure(list(total = c(28L, 17L, 11L, 6L,
6L, 5L, 4L, 3L, 3L,
2L), young = c(29L, 22L, 15L, 8L, 5L, 3L, 2L, 2L, 2L, 2L), old = c(36L,
11L, 11L, 8L, 6L, 5L, 4L, 3L, 2L, 2L)), .Names = c("total", "young",
"old"), class = "data.frame", row.
On Mon, Jul 30, 2012 at 3:17 PM, namit wrote:
> Hi Freinds,
>
> I have two data frames X,Y. I want to append both the data frames into one,
> along with the columns names from both the data frames (it should look like
> Z).
>
> X:
> SummaryG Y R
> Acc 1212
Thank you very much. I didn't know about this difference, I thought it
just behaved as in C/C++.
Thanks.
On Mon, 2012-07-30 at 14:15 -0400, jim holtman wrote:
> sprintf is working just fine. Your problem is the interpretation of
> what the results are. Displaying the object will show the escape
Hello,
You could try some non-parametric tests for comparison such as
Kolmogorov-Smirnov test (ks.test() package: truncgof)or Cramer-von Mises test
(cvmts.pval() package: CvM2SL2Test).
A.K.
- Original Message -
From: Luis Fernando García Hernández
To: r-help@r-project.org
Cc:
Sent:
Hi everyone,
I ran a binomial GAM consisting of a tensor product of two continuous
variables, a continuous parametric term and crossed random intercepts on a
data set with 13,042 rows. When trying to plot the tensor product with
vis.gam(), I get the following error message:
Error in persp.default
Hi Freinds,
I have two data frames X,Y. I want to append both the data frames into one,
along with the columns names from both the data frames (it should look like
Z).
X:
SummaryG Y R
Acc 1212 13
Bcc 1114 15
Ccc
Sorry, one error:
On Mon, Jul 30, 2012 at 1:18 PM, Bert Gunter wrote:
> ??
>
> On Mon, Jul 30, 2012 at 11:46 AM, R. Michael Weylandt
> wrote:
>> On Mon, Jul 30, 2012 at 8:49 AM, Bert Gunter wrote:
>>> Not necessarily. If the OP really meant the R list() structure, then
>>> is.element does not a
??
On Mon, Jul 30, 2012 at 11:46 AM, R. Michael Weylandt
wrote:
> On Mon, Jul 30, 2012 at 8:49 AM, Bert Gunter wrote:
>> Not necessarily. If the OP really meant the R list() structure, then
>> is.element does not apply.
>
> Perhaps...
>
> x <- list(1:5, "a", `+`, rnorm, NULL, list(letters))
Not
Hello,
You can check this link
(http://econometricsense.blogspot.com/2010/12/r-code-example-for-neural-networks.html).
A.K.
From: Rahul Bhalla
To: arun
Sent: Monday, July 30, 2012 3:51 PM
Subject: R-neuralnet code
Can you please give me a simple R-n
Hi,
If you look at the output from the given data, the lwr and upr is constant for
a particular date. So, I guess you can use that value as the percentile. I
assume that your statement of grouping all of one date together so as to
average out the observations. In that case, there will be o
Load the parallel package as David suggested, and look at
example(mclapply) -- it gives a bootstrapping example in the "don't
run" section.
Michael
On Mon, Jul 30, 2012 at 2:50 PM, Chet Seligman wrote:
> I have already consulted multicore.pdf and wondered if anyone could offer
> specific code fo
I have already consulted multicore.pdf and wondered if anyone could offer
specific code for doing loops, apply's and/or
cross-validation on a multicore machine?
On Mon, Jul 30, 2012 at 12:07 PM, David Winsemius wrote:
>
> On Jul 30, 2012, at 11:56 AM, Chet Seligman wrote:
>
> Can anyone provide
Hello,
Is this it?
function1 <- function(x) print(x)
name.function1 <- deparse(substitute(function1))
name.function1
Hope this helps,
Rui Barradas
Em 30-07-2012 11:47, Alaios escreveu:
> Dear all,
> I have a variable that I would like also to use it as a string. The reasons
> is that I want to
Sorry, I've sent this only to the op, forgetting to cc the list.
Hello,
Percentile of what? You probably have a dataset with dates and a
variable, right? What is the nature of that variable?
Post a data example:
dput(head(myData), 20) # or 30
Copy the output of that command and paste it in
Hello,
Em 30-07-2012 19:46, R. Michael Weylandt escreveu:
On Mon, Jul 30, 2012 at 8:49 AM, Bert Gunter wrote:
Not necessarily. If the OP really meant the R list() structure, then
is.element does not apply.
Perhaps...
x <- list(1:5, "a", `+`, rnorm, NULL, list(letters))
letters %in% x # Work
Stefan: Your comparison of two models, one with a nested reduced set of
parameters by fixing the others at zero, with anova() can be used to make
inference either based on a likelihood ratio form of test or the rankscore
test for a given quantile (see ?anova.rq and the vignette for literature
Hi,
Not sure what you wanted.
But, this will be just a subset of possible combinations as you described in
the post.
vector1<-sample(LETTERS[1:6])
vector2<-sample(letters[1:6])
b<-list()
for(i in 1:length(vector1)){
b[[i]]<-list()
b[[i]]<-sample(vector1)
}
d<-list()
for(j in 1:length(vector
On Jul 30, 2012, at 11:56 AM, Chet Seligman wrote:
Can anyone provide code example(s) for doing loops, apply's and/or
cross-validation on a multicore machine?
help(package=parallel)
--
David Winsemius, MD
Alameda, CA, USA
__
R-help@r-project.org
Can anyone provide code example(s) for doing loops, apply's and/or
cross-validation on a multicore machine?
Chet Seligman
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLE
On Mon, Jul 30, 2012 at 8:49 AM, Bert Gunter wrote:
> Not necessarily. If the OP really meant the R list() structure, then
> is.element does not apply.
Perhaps...
x <- list(1:5, "a", `+`, rnorm, NULL, list(letters))
letters %in% x # Works -- vectorized, mostly false, but the "a" is
there, per b
On Mon, Jul 30, 2012 at 04:11:40AM -0700, GiuseppeRicci wrote:
> Hi,
>
> thank Petr for your help.
> I have implemented you code suggestion but there is another problem.
> It seems that the code:
>
> for (i in 1:m){
> Z[i,,]=table(occ, data_matrix[,i])
> }
>
> don't charge any values in Z.
> Is
I am unable to determine what you think the problem is from your description.
A wild guess is that you just don't know what you are looking at in the output.
Do you understand the difference between the output of the "print" function and
of the "cat" function? Try
cat( sprintf("a\"bc\"d") )
an
On Mon, Jul 30, 2012 at 06:51:47AM -0700, Megh Dal wrote:
> Hi, I have following optimization problem:
>
> Min: x1 + x2 +...+ x7
> subject to:
>
> x1 + x2 >= 80
> x2 + x3 >= 65
> x3 + x4 >= 40
>
> all xi are ***positive integer***.
>
> Can somebody help me in this optimization problem?
Hi.
As
sprintf is working just fine. Your problem is the interpretation of
what the results are. Displaying the object will show the escape
characters, but if you use "cat", or output to a file, you will see
that the result is correct:
> cat( sprintf("a\nb"))
a
b>
> cat(sprintf("a\"bc\"d"))
a"bc"d>
>
Does this do what you want?
(using just three letters to keep the list of results short)
> tmp <- expand.grid(v=letters[1:3],V=LETTERS[1:3])
> tmp
v V
1 a A
2 b A
3 c A
4 a B
5 b B
6 c B
7 a C
8 b C
9 c C
>
> subset(tmp, tolower(tmp$V) != tmp$v)
v V
2 b A
3 c A
4 a B
6 c B
7 a C
8 b C
-Don
On Jul 30, 2012, at 9:29 AM, Jim Silverton wrote:
I have the following codes:
Now, suppose I have x = runif(1000,0,1, rep(1, 250), rep(0, 100) and
I want
to create a 'bin' for the 0's and the 1's and put the rest of the
values in
say about 20 bins. How can this be done?
?cut
--
David.
Hi.
I am having trouble with something that should be simple. I am unable
to get sprintf using escape sequences:
> sprintf("a\nb")
[1] "a\nb"
> sprintf("a\"bc\"d")
[1] "a\"bc\"d"
But it seems to need them any way:
> sprintf("a\"bc"d")
Error: unexpected symbol in "sprintf("a\"bc"d"
Any suggest
I was hoping to calculate the percentile for each date. So group all of one
date together, calculating the value of the 2.5 and 97.5 percentile.
--
View this message in context:
http://r.789695.n4.nabble.com/Calculating-percentiles-for-multiple-dates-tp4638183p4638378.html
Sent from the R help
Hi all,
I need to sub sample some large data sets.
The problem is using complete data samples for bats that are ± data per
1-minute time blocks using acoustic data collection and a method I
published some time back. Miller, B. W. 2001. A method for determining
relative activity of free flying
On Mon, Jul 30, 2012 at 08:40:59PM +0545, Christofer Bogaso wrote:
> Dear all, I was encountering with a typical Matching problem and was
> wondering whether R can help me to solve it directly.
>
> Let say, I have 2 vectors of equal length:
> vector1 <- LETTERS[1:6]
> vector2 <- letters[1:6]
>
>
Hello R users,
I'm working with a time-series of several years and to analyze it, I’m using
GAM smoothers from the package mgcv. I’m constructing models where
zooplankton biomass (bm) is the dependent variable and the continuous
explanatory variables are:
-time in Julian days (t), to creat a long-
1. Please read an Intro to R before posting further. In your previous
post you confused "=" with "==". These and other basic matters are
discussed there, so you can save yourself -- and this list -- a lot of
time and aggravation by familiarizing yourself with it.(Unless this
was just a typo).
2. I
Hi, I have following optimization problem:
Min: x1 + x2 +...+ x7
subject to:
x1 + x2 >= 80
x2 + x3 >= 65
x3 + x4 >= 40
all xi are ***positive integer***.
Can somebody help me in this optimization problem?
Thanks for your help
__
R-help@r-project.org
I'm attempting to install the gsl package using the following command:
install.packages("c:/users/mike/documents/R/win-library/gsl_1.9-9.tar.gz",repos=NULL,type="source")
My attempt is failing. Please help.
The message supplied is:
Installing package(s) into C:/Users/Mike/Documents/R/win-lib
Hi,
A client has asked for a series of pair-wise tests similar to Tukey's
but using medians as opposed to means. The reason is that the
distributions are highly right skewed. Are there such tests (I'm
finding nothing) and if so, is there anything in R?
Thanks,
Walt
__
Dear Prof. John Fox,
thank you very much for your suggestions.
However, I still do not know how to use the contrasts after generating them.
Once I generate the matrix with the polynomial contrasts, what are the
following steps toward the statistical test?
A whole example would be very useful.
Tha
I'm very much a novice with R, and not a statistician. But as my group
has been working with generalized additive models (semi-parametric
regression), we have followed Wood's advice about using the R anova
function to do model comparison for different regressions. I would
imagine at least some
Dear GMM,
> -Original Message-
> From: Manzoni, GianMauro [mailto:gm.manz...@auxologico.it]
> Sent: July-30-12 9:49 AM
> To: John Fox
> Cc: r-help@r-project.org; Greg Snow
> Subject: Re: [R] MANOVA polynomial contrasts
>
> Dear Prof. John Fox,
> thus all I should do to test quadratic and
Hello,
Not sure whether this is what you need.
colnames(mymatrix)<-c("function1","function2","function3")
> toString(colnames(mymatrix))
#[1] "function1, function2, function3"
A.K.
- Original Message -
From: Alaios
To: R help
Cc:
Sent: Monday, July 30, 2012 6:47 AM
Subject: [R] Con
Dear Prof. John Fox,
thus all I should do to test quadratic and cubic effects is to change the
second argument of the linearHypothesis() function, right?
So, for testing the cubic effect:
> linearHypothesis (mod, "f.C")
Is there a chapter or paragragh about contrasts in your book "An R
companion
Hi,
thank Petr for your help.
I have implemented you code suggestion but there is another problem.
It seems that the code:
for (i in 1:m){
Z[i,,]=table(occ, data_matrix[,i])
}
don't charge any values in Z.
Is there some error?
Thanks.
Giuseppe
--
View this message in context:
http://r.78969
Dear all,
I hope someone can assist with soliving my problems with the DCC function in
the BOOTRES package.
Initially did my analyses in Dendroclim, but when I tried to repeat the
analyses in dcc I encountered some problems, that I think I should report.
First I encountered that that one varia
Dear r-help members.
I would like to:
a) control the margin around my legend box. Unfortunately I did not find an
appropriate command under "?legend". The margin around the actual legend is way
too wide. There is a lot of unnecessary "empty space" on the right side.
b) increase the width of t
Hello,
I have a unconventional question arising from my current master thesis on
regression modeling. Suppose we have fitted a (linear) relationship between
a dependent variable y and an independent variable x. Now we choose two
points on the x-axis, i.e. according to percentiles x10 and x90. Thes
Dear all,
does anybody have experience with the calculation of marginal effects
(“effects”) in Mlogit (see Croissant, Package ‘mlogit’, p.8)?
1) Is there a good qualitative explanation available for the listed options
for the argument “type” (“aa”, …)? When do I have to choose aa, ar…? And
w
Hello. I hope someone can suggest a way to test log-linear models with
cells
blocked when working with survey data.
The tables I am working with come from a multi-stage stratified sample.
The
command 'svyloglin' in the package 'survey' works well when I want to
choose between models with tw
Dear all,
does anybody have experience with building logits in Mlogit?
I want to test the use of a couple of alternative specific variables with a
generic regression coefficient. However, one of them simply does not work. R
says the length of this variable is different.
Problem: If I check the l
If you type improveProb at the command line you'll see all the code, which is
quite simple. I'm glad you are not attempting to classify risk as this is
almost always futile.
Frank
Svingen, Gard Frodahl wrote
>
> Dear Sirs.
>
> I am working with the R packages Hmisc and PredictABEL to make NRI
Im not sure what SAS is doing (or if you are using it correctly). In R you
do not create marginal totals independent of the data and try to fit them to
the data. In your first example you create a matrix called raw, but you do
not use it for anything. Your loglin() call is for the all cells 16.67
On Jul 29, 2012, at 9:49 PM, Khush gg wrote:
Hi,
Can anyone tell me how to set Z-score according to my own
requirement as
the below code is taking as per the file entries. Any help would
be appreciable.
That was a bit vague and hard to parse, but I'm thinking I may need to
make allowan
FAQ 7.31
In floating point, you get about 15 digits of precision. If you look
at the value of time for today you will see:
1343668783
If you add 5 more digits after the decimal point, you will see that
there is not resolution to a microsecond level. I usually assume that
I can get millisecond
On 30.07.2012 11:42, peter dalgaard wrote:
On Jul 30, 2012, at 11:08 , Ario Ario wrote:
*Hi all,
I got problems installing RSTAR, MSVAR, and MSVECM packages. *
No such thing exists on CRAN (nor its Archive section), so little wonder.
Similarly named packages exist for the Ox matrix langua
Dear list,
I'm a bit puzzled by an ambiguity with respect to the representation of
micro-/milliseconds when using 'POSIXlt' objects.
It seems that the last digit of the 'sec' attribute sometimes seems to
differ from the digits shown when printing the 'POSIXlt' object. You'll
find a little SO
David:
I think one needs to carefully parse the xyplot help, where it says:
"To use more than one legend, or to have arbitrary legends not
constrained by the structure imposed by key, use the legend argument.
"
So I presume that this is to be interpreted as: "ONLY the legend
argument will be use
For basic hierarchical clustering use
groups <- hclust(dist(qmu))
But there are many options for function dist() and for function hclust() and
these will affect the results. For basic plotting use
plot(groups)
But as you will see, there are too many cases for the labels to be legible.
Hierarchi
On Jul 30, 2012, at 8:42 AM, Jie wrote:
Sorry to miss the OS information. I run it on Window 7 32 bit, 64
bit and
Unix (version unclear, it is a grid machine.)
Have you looked at:
?.Machine
--
David.
Best wishes,
Jie
On Mon, Jul 30, 2012 at 11:39 AM, Prof Brian Ripley
wrote:
On 30/0
This is where the sort/merge application on the mainframe has excelled
for the last 40 years. If you can not send it to a mainframe, you can
look at the SyncSort package that runs on UNIX machines.
On Mon, Jul 30, 2012 at 12:25 PM, Matthew Keller wrote:
> Hello all,
>
> I have some genetic datas
On Mon, Jul 30, 2012 at 6:00 PM, jim holtman wrote:
> try this:
>> indx <- match(x, repl.tab[, 1], nomatch = 0)
>> x[indx != 0] <- repl.tab[indx, 2]
>> x
> [1] "A" "B" "c" "D" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p"
> "q" "r" "s" "t" "u" "v" "w"
> [24] "x" "y" "z" "A" "B" "c" "D" "e" "f"
I have the following codes:
Now, suppose I have x = runif(1000,0,1, rep(1, 250), rep(0, 100) and I want
to create a 'bin' for the 0's and the 1's and put the rest of the values in
say about 20 bins. How can this be done?
Jim
On Thu, Jul 5, 2012 at 4:08 PM, Sarah Goslee wrote:
> Which gives
Hello all,
I have some genetic datasets (gzipped) that contain 6 columns and
upwards of 10s of billions of rows. The largest dataset is about 16 GB
on file, gzipped (!). I need to sort them according to columns 1, 2,
and 3. The setkey() function in the data.table package does this
quickly, but of
Hello R-helpers,
I'm trying to customize a graphic in lattice using the 'legend' argument to add
labels on my plot but in the process I'm losing the legend drawn by 'auto.key',
despite the fact that I'm actually not sticking these on the same sides of the
graphic. I worked up a quick and simple
try this:
> (x <- rep(letters,2))
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p"
"q" "r" "s" "t" "u" "v" "w"
[24] "x" "y" "z" "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m"
"n" "o" "p" "q" "r" "s" "t"
[47] "u" "v" "w" "x" "y" "z"
> values <- c("aa", "a", "b", NA, "d", "z
Dear all
I've got stuck when trying to replace values in a vector by selecting
replacements from a replacement table. I'm trying to use only base
functions. Here's a dummy example:
> (x <- rep(letters,2))
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p"
"q" "r" "s" "t" "u" "v"
[
Is this what you want:
> vector1 <- sample(LETTERS[1:6]) # randomize
> vector2 <- letters[1:6]
> # convert to lower case for matching
> vector1 <- tolower(vector1)
> vector2 <- tolower(vector2)
> # count the number of matches so order does not matter
> count <- match(vector1, vector2)
> if (length
Sorry to miss the OS information. I run it on Window 7 32 bit, 64 bit and
Unix (version unclear, it is a grid machine.)
Best wishes,
Jie
On Mon, Jul 30, 2012 at 11:39 AM, Prof Brian Ripley
wrote:
> On 30/07/2012 16:08, Jie wrote:
>
>> Dear All,
>>
>> I am curious if there is any why to use R to
On 30/07/2012 16:08, Jie wrote:
Dear All,
I am curious if there is any why to use R to know the specification of the
machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe even with
thesoftware information.
Thank you for your attention.
There are ways ... howeve
Does trace() do what you want from a .Rprofile file? It works
from the command line:
> trace(setwd,print=FALSE,quote(options(prompt=paste0(dir,"> "
Tracing function "setwd" in
package "base"
[1] "setwd"
> setwd("z:/")
z:/> getwd()
[1] "z:/"
z:/> setwd("c:/temp")
c:/temp> pi
[Env: Win XP, R 2.14.2, R 2.15.0]
I have a replacement function for setwd() in my .Rprofile which displays
the current R path in the R window
title. It no longer works in R 2.15.x, giving the error below. Worse,
the error prevents the rest of my
.Rprofile script from completing.
Is there some
Dear All,
I am curious if there is any why to use R to know the specification of the
machine that runs on,
i.e. read the cpu model, memory size, those hardware info, maybe even with
thesoftware information.
Thank you for your attention.
Best wishes,
Jie
[[alternative HTML version deleted
Dear all, I was encountering with a typical Matching problem and was
wondering whether R can help me to solve it directly.
Let say, I have 2 vectors of equal length:
vector1 <- LETTERS[1:6]
vector2 <- letters[1:6]
Now I need to match these 2 vectors with all possible ways like:
(A,B,C,D,E) & (
nevermind. all.equal() works!
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I have a data frame (trans)
123 P
143 z ...ect
I want to write a statement that says:
if ((trans[i,2]="p")&(trans[1+1,2]="z")){
but I get the warning message that states:
operations are possible only for numeric, logical or complex types
Any ideas?
Thanks
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