On 02/08/2013 10:53 PM, e-letter wrote:
Readers,
Is it possible to add a data symbol (e.g. pch='21') to an axis label?
The objective is to plot a graph with two (2) y-axes and the plotting
character for data set 1 is added to the label of y1 axis (left); plot
character for data set 2 is added to
Hi all, hope someone can help me out with this.
Background Introduction
I have a data set consisting of data collected from a questionnaire that I
wish to validate. I have chosen to use confirmatory factor analysis to
analyse this data set.
Instrument
The instrument consists of 11 subscales. Ther
Greets,
My data looks like:
> p3.18
s xbar subgroup
1 0.84 12.21
2 1.64 11.22
3 2.07 10.63
4 2.49 12.24
5 0.84 11.25
...
Using the command
> qcc(p3.18$xbar,type="xbar",sizes=5,center=mean(p3.18$xbar),std.dev=mean(p3.18$s)/0.94,title="X-bar
> C
-- Forwarded message --
From:
Date: 2013/2/8
Subject: vegdist Error en double(N * (N - 1)/2) : tama?o del vector
especificado es muy grande
To: caro.bell...@gmail.com
Message rejected by filter rule match
-- Mensaje reenviado --
From: caro bello
To: r-help@r-p
I do not promise much. But try to use XLConnect package. It requires rJava
package which requires java on your system.
Good luck
Orvalho
On Fri, Feb 8, 2013 at 8:55 PM, Edwin Isensee wrote:
> I'm using the read.xls function from gdata package to read one Excel file,
> like the example below:
>
>
dear R experts---I am trying to replicate a perl feature. I want to be
able to embed R commands inside a character string, and have the
string be printed with the command executed. my perl equivalent is
my $a=10;
my $teststring = "the expression, $a+1, is ::$a+1::, but add one
more for ::$a+2
Dear Fabrice Tourre,
Re:
> John Kane,
>
> Thanks. It makes some of sense. But it seems not exactly what I want.
> I just remember it need using mtext and adjust margin. I saw such of
> example long time ago, but I have forgotten it.
(etc...)
Maybe this is what you're after?
Fiddled a bi
Rail is an ordered factor. ?ordered for details.
Default contrasts for an ordered factor are orthogonal polynomials.
See: https://stat.ethz.ch/pipermail/r-help/2007-January/123268.html
and
http://www.ats.ucla.edu/stat/r/library/contrast_coding.htm#ORTHOGONAL
-- Bert
On Fri, Feb 8, 2013 at 1:
Hi,
I am using a data called Rail in the nlme package. The data contains two
variables: Rail and Travel.
>Rail
Grouped Data: travel ~ 1 | Rail
Rail travel
1 1 55
2 1 53
3 1 54
4 2 26
5 2 37
6 2 32
7 3 78
8 3 91
9 3 85
10
David Winsemius comcast.net> writes:
> On Feb 8, 2013, at 10:55 AM, Edwin Isensee wrote:
>
> > I'm using the read.xls function from gdata package to read one Excel file,
> > like the example below:
> >
> > library(gdata)
> > my_file <- '/Users/Desktop/Project.xlsx'
> > valores <- read.xls(my_fi
Hi,
Try:
R[,2:4]<-F[,2:4][match(R$x,F$x),]
R$p<- unique(F$p)
R
x y z w k p
1 -1 1 3 8 9 18
2 -2 2 4 9 9 18
3 -3 3 5 10 9 18
4 -1 1 3 8 9 18
5 -2 2 4 9 9 18
6 -3 3 5 10 9 18
7 -1 1 3 8 9 18
8 -2 2 4 9 9 18
9 -3 3 5 10 9 18
10 -1 1 3 8 9 18
11 -2 2 4 9 9 18
12 -3 3 5 10
On Feb 8, 2013, at 10:55 AM, Edwin Isensee wrote:
> I'm using the read.xls function from gdata package to read one Excel file,
> like the example below:
>
> library(gdata)
> my_file <- '/Users/Desktop/Project.xlsx'
> valores <- read.xls(my_file)
>
> The problem is: one of the columns at the Exc
Hi Tom,
I also teach R to newbies (materials at
http://projects.iq.harvard.edu/rtc/filter_by/r-0), and I've found
RStudio (http://rstudio.com/ide) a huge help, and much less
intimidating than the Windows GUI. File and object browsers,
integrated help window, syntax highlighting and code completion
Dear All,
I am trying to merge 2 dataframes of with different sizes.
Let's say that one dataframe R contains some raw data and another data
frame F contains the information to fix R.
For instance
F <- data.frame(cbind(x=-seq(10),
y=1:10,
z=3:12,
John Kane,
Thanks. It makes some of sense. But it seems not exactly what I want.
I just remember it need using mtext and adjust margin. I saw such of
example long time ago, but I have forgotten it.
On Fri, Feb 8, 2013 at 3:32 PM, John Kane wrote:
> Here is a rough example of what you want. You w
I know that I can get a count of histogram bins in base R with plot=FALSE.
However, I'd like to do the same thing with lattice. The problem is that I've
set up shingles, and I'd like to get the count within each bin within each
shingle. plot=FALSE doesn't seem to do it.
[[alternativ
Hello,
With the following, the first instruction will give you correlations
matrices, the second coefficients.
dat <- read.table(text = "
x y group
1 0.876751503 0.6518345 a
2 0.627067150 0.8801790 a
3 0.632465192 0.1768305 a
4 0.060359554 0.8835652 a
5 0.67586
Hello,
You're right, sorry for the misleading tip. How about seq(1, 177, 11)?
Please note that without a data example, it's not very easy to say.
Can't you post a small dataset using ?dput
dput(head(data, 20)) # paste the output of this.
Rui Barradas
Em 08-02-2013 17:10, christel lacaze e
Try:
a<- rnorm(1000,0,1)
b<- rnorm(1000,0,2)
c<- rnorm(1000,0,3)
d<- rnorm(1000,0,4)
par(mfrow=c(2,2))
hist(a,main="")
title(main="A",adj=0)
hist(b,main="")
title(main="B",adj=0)
hist(c,main="")
title(main="C",adj=0)
hist(d,main="")
title(main="D",adj=0)
cheers,
Rolf Turner
On 02/
Here is a rough example of what you want. You will need to adjust the actual
placement by hand by redefining the x & y values.
set.seed(100)
a <- rnorm(1000,0,1)
b <-rnorm(1000,0,2)
c <-rnorm(1000,0,3)
d <-rnorm(1000,0,4)
opr <- par(mfrow=c(2,2))
hist(a)
text(-3, 150, labels ="a"
Thank your reminder.
Here is the example code:
a<- rnorm(1000,0,1)
b<-rnorm(1000,0,2)
c<-rnorm(1000,0,3)
d<-rnorm(1000,0,4)
par(mfrow=c(2,2))
hist(a)
hist(b)
hist(c)
hist(d)
I want to add A, B, C and D on each top left.
On Fri, Feb 8, 2013 at 2:35 PM, John Kane wrote:
> Yes there probably is if
Yes there probably is if you would show us what you are doing.
Minimal working code helps.
https://github.com/hadley/devtools/wiki/Reproducibility
John Kane
Kingston ON Canada
> -Original Message-
> From: fabrice.c...@gmail.com
> Sent: Fri, 8 Feb 2013 14:28:17 -0500
> To: r-help@r-proje
Use POSIXct instead of POSIXlt in data frames in general, and particularly for
this task.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Dear list,
Is there a elegant way to name multiple ordered plot as A, B, C and D?
Just put A, B, C and D on the top left corner of each plot.
Usually I do it in Illustrator, but I think in R should there is a way.
Could you please provide a example?
Thank you very much in advance.
Hello,
The id.vars must "be integer (variable position) or string (variable
name)" (from the help page ?melt.data.frame)
This seems to work.
myd <- mydata
myd$date <- as.character(myd$date)
melt(myd, id.vars = c("date"))
Hope this helps,
Rui Barradas
Em 08-02-2013 18:31, John Kane escreve
Hi everyone,
I'm not exactly sure how to ask this question most clearly, but I hope that
giving the context in which it occurs for me will help: I'm trying to
compare the brain images of two patient populations; each image is composed
of voxels (the 3D analogue of pixels), and I have two images
On 02/07/2013 05:00 AM, r-help-requ...@r-project.org wrote:
I'd argue that there's an important distinction between documenting a
function (how to use it) vs. documenting an algorithm (how it works).
I think noweb can work well for describing how something works, but
it's not so good for descri
Dear Tom,
As you can imagine, I watched your video with interest and found it quite
impressive. I do have a couple of comments and a question.
The comments: You change a numeric ID variable into a factor with several
thousand levels. I think that it would be more natural and efficient to
assign t
Just posting to answer my own question, at least for the "variables
constant" error: I hadn't noticed that lda has an argument called 'tol'
that governs when variables are interpreted as constant within groups; it's
right there in the help entry for lda, so I apologize for having missed
it.As
I realize it's -12C and we're having the next best thing to a blizzard but why
can I not melt this data frame.
I am missing something terribly obvious but I just don't understand what the
error message is saying.
John Kane
Kingston ON Canada
Code and aata below
#=
I wanted to put a short example into the r-help archives how to embed
the Bera (texlive) font.
options(texlive= "/usr/local/texlive/2012/")
options(texfonts= paste0(getOption("texlive"), "/texmf-dist/fonts/"))
library(grDevices)
pdfFonts(Bera = Type1Font("Bera", paste0(getOption("texfonts"),
"afm/
I seem to be suffering from the same problem (the Friday one, not the cor one)
Have a look at
http://stats.stackexchange.com/questions/4040/r-compute-correlation-by-group
for something that looks like it will work
John Kane
Kingston ON Canada
> -Original Message-
> From: david.cars..
On 02/08/2013 05:13 AM, Pulkit Mehrotra wrote:
hello,
I was having some problems in debugging C codes assocaited with R-packages.
I don't have much experience with debugging tools used in C language. Is
there any methods similar to R-debugging tools such as browser ,debug etc
which can be used
useRs,
Using a code suggestion from the previous post in this thread (quoted
next), I was able to work through a smaller example problem and get some
results I was aiming for.
>That might not construct the bands in the blocks correctly: This might
>be better:
>
>MAT [ kronecker( matrix(1, 4,4), a
On Fri, Feb 8, 2013 at 10:14 AM, David Winsemius wrote:
>
> On Feb 8, 2013, at 8:55 AM, ilai wrote:
>
> > Like this ?
> > xyplot(4:5~4:5, groups=4:5, lex = 5 ,
> > par.settings = simpleTheme(cex=10, pch=21, lwd=5),
> > auto.key=TRUE)
> >
>
> And if I had used a more general search strategy I might
On Feb 8, 2013, at 8:55 AM, ilai wrote:
> Like this ?
> xyplot(4:5~4:5, groups=4:5, lex = 5 ,
> par.settings = simpleTheme(cex=10, pch=21, lwd=5),
> auto.key=TRUE)
>
And if I had used a more general search strategy I might have found it, too,
rather than hacking my way to a less adequate solut
that doesn't seem to be the issue i'm afraid... both j sequences produce the
same numbers:
> j<-seq(1,166,11)
> j
[1] 1 12 23 34 45 56 67 78 89 100 111 122 133 144 155 166
> j<-seq(1,176,11)
> j
[1] 1 12 23 34 45 56 67 78 89 100 111 122 133 144 155 166
> Date: Fri, 8 Feb 2
On Feb 8, 2013, at 5:46 AM, Karl Ove Hufthammer wrote:
> Dear list members,
>
> I can't figure out how get 'xyplot' or 'dotplot' in the 'lattice' package to
> respect the 'lwd' value for specifying the border with for *symbols* (for
> lines it works fine). Example:
>
> -
> # Base graphics
Like this ?
xyplot(4:5~4:5, groups=4:5, lex = 5 ,
par.settings = simpleTheme(cex=10, pch=21, lwd=5),
auto.key=TRUE)
On Fri, Feb 8, 2013 at 6:46 AM, Karl Ove Hufthammer wrote:
> Dear list members,
>
> I can't figure out how get 'xyplot' or 'dotplot' in the 'lattice' package
> to respect the 'lw
Hello,
Maybe seq(1, 16*11, 11)? (16*11 is 176, not 166)
Hope this helps,
Rui Barradas
Em 08-02-2013 16:03, christel lacaze escreveu:
hi there,
I have a dataframe in the shape vA1, vA2,..., vA11, vB1, vB2,..., VB11,...,
VP1, VP2,, VP11 (so 16 times a sequence of 11 variables)
I am
On Feb 8, 2013, at 5:13 AM, Pulkit Mehrotra wrote:
> hello,
>
> I was having some problems in debugging C codes assocaited with R-packages.
> I don't have much experience with debugging tools used in C language. Is
> there any methods similar to R-debugging tools such as browser ,debug etc
> wh
On Feb 8, 2013, at 3:53 AM, e-letter wrote:
> Readers,
>
> Is it possible to add a data symbol (e.g. pch='21') to an axis label?
> The objective is to plot a graph with two (2) y-axes and the plotting
> character for data set 1 is added to the label of y1 axis (left); plot
> character for data s
I seem to have a Friday afternoon block and can't see the easiest way of
doing this.
Given a data frame like:
dat <- data.frame(x = runif(100), y = runif(100), group = rep(letters[1:10],
each = 10))
> head(dat)
x y group
1 0.876751503 0.6518345 a
2 0.627067150 0.8801790
On Feb 7, 2013, at 8:20 AM, Giuseppe Amatulli wrote:
> Hi Rolf,
> sorry for this late answer and thanks for your kind explanation and
> relative R code. I really appreciate.
> In reality the concept that I'm trying to address is a bit more complex.
> I'm fitting a model y vs 6 predictors with MAR
hi there,
I have a dataframe in the shape vA1, vA2,..., vA11, vB1, vB2,..., VB11,...,
VP1, VP2,, VP11 (so 16 times a sequence of 11 variables)
I am trying to build a double loop so that i can apply the function (i-1)*v(i)
to the first 10 variables, then the same for the next 10 variable
On Thu, Feb 7, 2013 at 9:58 PM, James Jong wrote:
> What are the *RDS counterparts? What is the difference?
>
apropos("RDS")
? readRDS
? saveRDS
They allow serialization of an individual object and restoring it --
the advantage is that you can restore to a different name instead of
clobbering an
Ah, I didn't know this! Thanks Ben.
On Fri, Feb 8, 2013 at 10:18 AM, Ben Tupper wrote:
> Hi,
>
> On Feb 8, 2013, at 9:40 AM, Brian Smith wrote:
>
> > Hi,
> >
> > I am trying to append tables on file with this sample code:
> >
> >for(i in 1:2){
> >mat <- data.frame(sample(1:30
Hi,
library(lubridate)
ymd_hms(testtime)
1 parsed with %Y-%m-%d %I:%M:%S %p
#[1] "2013-01-01 13:00:01 UTC"
A.K.
- Original Message -
From: e-letter
To: r-help@r-project.org
Cc:
Sent: Friday, February 8, 2013 4:44 AM
Subject: [R] convert 12 time stamp to 24 hour
Readers,
For a 12 h
Thanks, fixed upstream.
David
On 2013-02-08 07:20, Achim Zeileis wrote:
On Fri, 8 Feb 2013, Richard M. Heiberger wrote:
require(vcd)
mosaic(matrix(1:6, 2, 3), gp=gpar(fill=c("red","gray","blue")),
main="three
colors in each row")
mosaic(matrix(1:3, 1, 3), gp=gpar(fill=c("red","gray","blue")),
I want to make a script in R (I use RStudio if it matters) in which the user
chooses the initial data file interactively using tk_choose.files
the code goes like this:
fname1<-tk_choose.files(default = "", caption = "Choose file",multi = FALSE,
filters = NULL, index = 1)
however, it pops out new
Hi,
On Feb 8, 2013, at 9:40 AM, Brian Smith wrote:
> Hi,
>
> I am trying to append tables on file with this sample code:
>
>for(i in 1:2){
>mat <- data.frame(sample(1:30,9),3,3)
>colnames(mat) <- letters[1:3]
>ifelse(i ==
> 1,write.table(mat,paste('te
Thanks Louis! That seems to work!
On Fri, Feb 8, 2013 at 10:06 AM, Louis Aslett wrote:
> I believe your problem stems from using ifelse() actually ... it
> requires the statements which it runs to return a value with the same
> shape as the test, which write.table() isn't doing.
>
> Just change i
I believe your problem stems from using ifelse() actually ... it
requires the statements which it runs to return a value with the same
shape as the test, which write.table() isn't doing.
Just change it to a regular if with an else and you'll be fine:
for(i in 1:2){
mat <- data.frame(sample(1:30
Hi,
I am trying to append tables on file with this sample code:
for(i in 1:2){
mat <- data.frame(sample(1:30,9),3,3)
colnames(mat) <- letters[1:3]
ifelse(i ==
1,write.table(mat,paste('test.txt',sep=''),row.names=F),
write.table(mat,paste('test.txt',sep
Dear Janesh,
Re:
> Dear Franklin Bretschneider,
>
> Thank you so much for your reply and explanation about the filter using the
> stats and signal package.
>
> I decided to opt the filter method in signal package. I have a simple
> question about the cut off frequency here.
>
> I have 30
Dear Elisha,
It's impossible to tell from the information that you provide what the
source of the problem is. The gold standard for a problem report is
"commented, minimal, self-contained, reproducible code" (quoting from the
text at the bottom of all r-help postings) that would allow me to track
Hi,
On Feb 8, 2013, at 2:49 AM, Mat wrote:
> Hello together,
>
> i have a problem with an export.
> I want to export a data.frame, only if there ist data in it.
> My data.frame (CO.PHB) ist empty but with this code, the .txt-file gets
> exported every time, with no content.
>
> {
> pfad<-paste(
You can do this directly with the survfit function (survival verstion 2.37 or
greater)
If status is a factor variable with levels of censored, relapse and death
fit <- survfit(Surv(time, status) ~ sex, data=mydata)
plot(fit, xlab=...
The primary advantage of this over the cuminc package
Dear list members,
I can't figure out how get 'xyplot' or 'dotplot' in the 'lattice'
package to respect the 'lwd' value for specifying the border with for
*symbols* (for lines it works fine). Example:
-
# Base graphics works fine (gives a 'fat’ circle)
plot(5, cex=10, pch=21, lwd=10)
# B
Hi
Does not work in some locale environment
> strptime(testtime, "%Y-%m-%d %I:%M:%S %p")
[1] NA
> Sys.getlocale("LC_TIME")
[1] "Czech_Czech Republic.1250"
> Sys.setlocale("LC_TIME", "C")
[1] "C"
> strptime(testtime, "%Y-%m-%d %I:%M:%S %p")
[1] "2013-01-01 13:00:01"
Regards
Petr
> -Orig
hello,
I was having some problems in debugging C codes assocaited with R-packages.
I don't have much experience with debugging tools used in C language. Is
there any methods similar to R-debugging tools such as browser ,debug etc
which can be used to debug such C files.
I was using the foreign p
New versions 2.4 of our distr-family of packages are submitted to
CRAN and will soon be available on all mirrors
(i.e.; startupmsg, SweaveListingUtils, distr, distrEx, distrDoc,
distrEllipse, distrMod, distrSim, distrTEst, distrTeach)
In addition, there is a new package distrRmetrics with distr s
Readers,
Is it possible to add a data symbol (e.g. pch='21') to an axis label?
The objective is to plot a graph with two (2) y-axes and the plotting
character for data set 1 is added to the label of y1 axis (left); plot
character for data set 2 is added to label of y2 (right) axis.
Thanks.
--
r2
Hello,
Try this:
> testtime<-("2013-01-01 01:00:01 PM")
> strptime(testtime, "%Y-%m-%d %I:%M:%S %p")
[1] "2013-01-01 13:00:01"
HTH,
Pascal
Le 08/02/2013 18:44, e-letter a écrit :
Readers,
For a 12 hour time stamp:
testtime<-("2013-01-01 01:00:01 PM")
testtime
[1] "2013-01-01 01:00:01 PM"
Hello,
I don't know whether it matches what you are looking for. Here are some
example using different spacing:
> par(las=1, mfrow=c(2,2))
> plot(1,1,ylab=expression(paste('rate (',degree~C^-1,')')),
main='Spacing with ~')
> plot(1,1,ylab=expression(paste('rate (',degree,phantom(0),C^-1,')'))
Readers,
For a 12 hour time stamp:
> testtime<-("2013-01-01 01:00:01 PM")
> testtime
[1] "2013-01-01 01:00:01 PM"
> testtime24hour<-strftime(testtime,'%H:%M:%S')
> testtime24hour
[1] "01:00:01"
how to convert to 24 hour format so that the object 'testtime24hour' is:
'13:00:01'
Thanks.
--
r215
Hello,
There is also that previous discussion:
http://r.789695.n4.nabble.com/peak-detection-td3086410.html
Pascal
Le 07/02/2013 03:22, Johannes Graumann a écrit :
Grrr ... new trial with code here: http://pastebin.com/RjHNNG9J
Maybe the amount of inline-code prevented posting?
Hello,
I am
Hi
You must use indexing based on loops
sim(ss[i],cc[j],oo1[k],oo2[l])
an of course assign result to some object.
result[...] <- sim(ss[i],cc[j],oo1[k],oo2[l])
where ... means some suitable index either chosen from loop index or
constructed during loop execution.
Maybe it is time to go through
Hi
I am not sure if it suits your needs but I use this function for identifying
peaks in different spectral data.
peaks <- function (series, span = 3, ties.method = "first")
{
if ((span <- as.integer(span))%%2 != 1)
stop("'span' must be odd")
z <- embed(series, span)
s <- s
Hello together,
i have a problem with an export.
I want to export a data.frame, only if there ist data in it.
My data.frame (CO.PHB) ist empty but with this code, the .txt-file gets
exported every time, with no content.
{
pfad<-paste(paste(paste("C:/Test/","Reminder",sep=""),
"Orders",sep="/"),"t
70 matches
Mail list logo
