On Jan 21, 2012; 7:39am stephen sefick wrote:
> I plot the combined value with plot(four.comb, type="G"). What do the
> colors mean? I have both grey
> and black bars.
The help file for fourthcorner plainly tells you (sub Details). You can also
work out the meaning by looking at the summary s
Hello,
Maybe it's no longer needed but, another way would be the function below.
It's more complicated because you don't need to know that there are only for
categories.
Only the original form and the wanted output.
fun <- function(x, var.to.transform){
f <- function(x, nm){
if(is.n
I can't replicate the problem on my system (OS X) and I don't know
enough about the Gtk + tcltk frameworks to help diagnose it outside of
R -- it might be worth working up a minimal (non-)working example
along the lines of:
michaelweylandt$ R --vanilla
library(playwith)
playwith(plot(1:10))
and
I have used the fourthcorner function as suggest by dray and legendre
(model 2 and 4 then combine). I plot the combined value with
plot(four.comb, type="G"). What do the colors mean? I have both grey
and black bars.
many thanks,
Stephen
--
Stephen Sefick
***
The Institute for Social Research (ISR) and its Statistical
Consulting Service (SCS) at York University are pleased to
announce our Summer Program In Data Analysis (SPIDA) for
2012. The Program runs from May 24th to June 1st, 2012.This
year’s Program focuses on the theory and practice of linear
mo
Even if you're not doing medical research, I like a lot about
Spiegelhalter's book:
http://www.amazon.com/Bayesian-Approaches-Health-Care-Evaluation-Statistics/dp/0471499757/ref=sr_1_1?ie=UTF8&qid=1327112075&sr=8-1
For interacting with R and JAGS/BUGS my two favorite books that cover theory
are C
Thanks Michael and Brian
Thanks for your time.
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I'm not sure if this is appropriate. If the sum of your variables is
always the same constant, then you might try a Ternary plot (
http://en.wikipedia.org/wiki/Ternary_plot ).
The "vcd" package can make ternary plots.
On 01/20/2012 02:56 PM, Roary wrote:
I have two observed categorical variab
Why not use the pre-compiled RPM's from EPEL (
http://fedoraproject.org/wiki/EPEL )?
Version 2.14 of R is still in the testing folder here:
http://download.fedora.redhat.com/pub/epel/testing/5/
Jason
On 01/20/2012 02:34 PM, Erik Wright wrote:
Hello,
I am trying to upgrade to the latest R r
As Sarah said, you have a path problem.
Are you saying that RawData is a sub-folder (sub-directory) of
SampleProject?
And you are running the script with the working directory set to
SampleProject?
[check using getwd() as Sarah suggested]
If so, it looks like it would work if you use './RawData/F
That's exactly what I wanted. Thanks!
Helios
>>> El día 21/01/2012 a las 0:33, David Winsemius
escribió:
> On Jan 20, 2012, at 6:26 PM, Helios de Rosario wrote:
>
>> Hi, a quick question:
>>
>> Is there a way to retrieve the default value of a function argument
-
>> if it exists?
>> (I know I
On Jan 20, 2012, at 6:26 PM, Helios de Rosario wrote:
Hi, a quick question:
Is there a way to retrieve the default value of a function argument -
if it exists?
(I know I can see it if I type the function name, but I would like get
the value programaticaly.)
?formals
--
David Winsemius,
Hi, a quick question:
Is there a way to retrieve the default value of a function argument -
if it exists?
(I know I can see it if I type the function name, but I would like get
the value programaticaly.)
Thanks,
--
Helios de Rosario Martínez
Researcher
INSTITUTO DE BIOMECÁNICA DE VALENCI
You're running into time zone issues. If you're in a time zone that
recognizes daylight savings time for part of the year, the difference
in the value you calculate and the expected answer will vary.
Check your current system timezone:
Sys.timezone()
If it doesn't return "UTC" (aka GMT), you're
On 1/20/2012 9:12 AM, cameron wrote:
Can anyone please help me with this?
I have a list of business dates. What I want is to have last day of last
month and paste them on next month.
What i haveWhat i want
5725 2011-09-22
5726 2011-09-23
5727 2011-09-26
5728 2011-09-27
5
Hi Fred,
It seems you don't have an rbind() problem, but a path problem. The
error you're getting means R can't find your file.
When specifying a relative path, as you do with "../RawData/File1.csv"
it's relative to your current working directory, not necessarily the
directory where your script i
Hello there,
Much thanks in advance for any help. I have a few questions:
1) Why do I keep getting the following error:
File1 <- read.csv("../RawData/File1.csv",as.is=TRUE,row.names=1)
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
c
Hi Luke,
Thank you for the answer.
On 20/01/2012, at 6:11 , Luke Miller wrote:
> 666.1751 sure seems like it should return 2010-10-29 04:12:09 based on
> your example.
>
> 666.1751 days from 2009-01-01 is 2010-10-29 + some hours/min/seconds.
>
> 0.1751 days * 24 hrs/day = 4.2024 (i.e. 4:00AM + s
Thanks for your reply. The advantage of nonparametric statistics is
that you do not need to specify the distribution because the estimator
(e.g. kernel estimator) converges to the true density as the number of
observations converges to infinite. The mode in this case is the
maximum of this e
I have two observed categorical variables X1 and X2, with X3=X1+X2, and a
continuous response Y. I can interpolate the surface and construct an
ordinary 2D square contour plot (with X1,X2 axes and X3 on the diagonal).
However, I would like to change the orientation of the plot so that the
axes fit
to use ggplot:
dat<-data.frame(num=1:3,usage=c(4,2,5),cap=c(10,20,10),diff=c(6,18,5))
dat.melt<-melt(dat,id.var=c('num','cap'))
ggplot(dat.melt)+geom_bar(aes(x=num,y=value,fill=variable),stat='identity')
On Fri, Jan 20, 2012 at 12:30 PM, Jean V Adams wrote:
> Bart6114 wrote on 01/20/2012 08:
On Fri, Jan 20, 2012 at 03:14:21PM -0500, Sam Steingold wrote:
> On Fri, Jan 20, 2012 at 14:05, Sarah Goslee wrote:
> >> then I need to convert each 6/8 character string into an integer base 36
> >> or 64 (depending on the field) - how?
> >
> > base 36?
>
> 10 decimal digits + 26 english characte
Here part of it. This is the conversion of base 36 to numeric that is
case insensitive. This makes use of mapping the alphabetics to
characters that start just after '9' and then doing the conversion.
You can extend it to base 64 using the same approach.
> base36ToInteger <- function (Str)
+ {
You might look at John Kruschke's book, Doing Bayesian Data Analysis (AP),
which starts with basics and goes from there. It also relies on R and Bugs.
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the p
Bart6114 wrote on 01/20/2012 08:54:39 AM:
> Hey,
>
> I want to create a stacked barchart in R for the following dataset
> (http://pastebin.com/pyHUNgr2):
>
> # usage capacity diff
> 1 4 10 6
> 2 2 20 18
> 3 5 10 5
>
> The stacked barchart should, in one plot sho
On Fri, Jan 20, 2012 at 14:05, Sarah Goslee wrote:
> Reproducible example, please. This doesn't make a whole lot of sense
> otherwise.
here is the string:
"1288915200|070400905a0A118"
I want the following data extracted from it:
1. the decimal number before "|": 1288915200
2. the str
On Fri, Jan 20, 2012 at 14:05, Sarah Goslee wrote:
>> then I need to convert each 6/8 character string into an integer base 36
>> or 64 (depending on the field) - how?
>
> base 36?
10 decimal digits + 26 english characters = 36.
ThusThisLongWordWithLettersAndDigitsFrom0to9isAnIntegerBase36
(case
You realize, I trust, that "the mode" of a continuous distribution has
no meaning without prior specification of the distribution, which, for
a fitted density estimate would mean specification of the fitting
parameters (bandwidth, etc.) at a minimum.
So perhaps you need to rethink what you are try
Hello,
I am trying to upgrade to the latest R release on a machine running Red Hat
el5. Previously I was successful at building R 2.11, but now I am having
troubles with R 2.14.
Configure goes fine, but then make throws a lot of errors (output below). Any
idea what I am doing wrong this time
I believe you are looking for analysis of covariance.
I recommend the ancova function in the HH package.
## install.packages("HH") ## if needed
library(HH)
tmp <- ancova(y ~ x + group, data=mydata)
tmp
## You suggested that you want the superpose panel. You can get
## that on a single page with
On 20.01.2012 16:38, Sam Chand wrote:
Hello,
I have 2 variables - x and y, that belong to separate groups.
I want to plot all the x and y together, but show separate abline for each
group. It can be done in ggplot2, but is there a simpler way to draw
ablines by group?
e.g.,
mydata<- data.fram
Try this:
date <-
c("9/22/2011","9/23/2011","9/26/2011","9/27/2011","9/28/2011","9/29/2011","9/30/2011","10/17/2011",
"10/18/2011","10/19/2011","10/20/2011","10/21/2011","10/24/2011","10/25/2011","10/26/2011","11/17/2011","11/18/2011","11/21/2011","11/22/2011","11/23/2011","11/25/2011","11/28/201
Hello,
I have 2 variables - x and y, that belong to separate groups.
I want to plot all the x and y together, but show separate abline for each
group. It can be done in ggplot2, but is there a simpler way to draw
ablines by group?
e.g.,
mydata <- data.frame(x = 1:20+rnorm(20, -3, 1), y = seq(1,20,
Dear R-helpers,
I am trying to fit my data to a 4-parameter lognormal distribution (aka
Johnson Sb dist) with fitdist function from the library(fitdistrplus). So
far, I have learnt that with "mle" method it's not always possible to
estimate the gamma and delta parameters even if the bounding estim
Hello,
I have 2 variables - x and y, that belong to separate groups.
I want to plot all the x and y together, but show separate abline for each
group. It can be done in ggplot2, but is there a simpler way to draw
ablines by group?
e.g.,
mydata <- data.frame(x = 1:20+rnorm(20, -3, 1), y = seq(1,20,
Hi,
Would like to analyze this survey:
https://docs.google.com/spreadsheet/ccc?key=0Al8vE0D1FPpldEVKWEpZalVELXRhdXo1RU5pTXJWUlE
I know how to read in the data like this:
require(RCurl)
myCsv <-
getURL("https://docs.google.com/spreadsheet/pub?hl=en_US&hl=en_US&key=0Al8vE0D1FPpldEVKWEpZalVELXRhdXo1
Can anyone please help me with this?
I have a list of business dates. What I want is to have last day of last
month and paste them on next month.
What i haveWhat i want
5725 2011-09-22
5726 2011-09-23
5727 2011-09-26
5728 2011-09-27
5729 2011-09-28
5730 2011-09-29
5731 201
Hi all,
I am trying to estimate the mode of a 4-dimensional nonparametric density
estimator (any) using a sample of size n=10,000. I have tried using the
package 'ks' and 'np' but they are extremely slow; this is related to the
estimation of the bandwidth matrix. I also checked the package
Sam:
On Fri, Jan 20, 2012 at 10:52 AM, Sam Steingold wrote:
> Hi,
> I have a data frame with one column containing string of the form
> "ABC...|XYZ..."
> where ABC etc are fields of 6 alphanumeric characters each
> and XYZ etc are fields of 8 alphanumeric characters each;
> "|" is a mandatory se
Reproducible example, please. This doesn't make a whole lot of sense
otherwise.
On Fri, Jan 20, 2012 at 1:52 PM, Sam Steingold wrote:
> Hi,
> I have a data frame with one column containing string of the form
> "ABC...|XYZ..."
> where ABC etc are fields of 6 alphanumeric characters each
> and XYZ
Hi,
I have a data frame with one column containing string of the form
"ABC...|XYZ..."
where ABC etc are fields of 6 alphanumeric characters each
and XYZ etc are fields of 8 alphanumeric characters each;
"|" is a mandatory separator;
I do not know in advance how many fields of each kind will each r
Not sure if I understand the question: If you have more data the grid
produced by image() or contour() will be finer anyway...
Perhaps we just need an example what you are actually asking for.
Uwe Ligges
On 20.01.2012 13:28, Roary wrote:
Hi All,
I have 3 variables which present a perfect lin
On 20.01.2012 09:01, Ashy wrote:
Could anyone please tell how to pass parameters of form to server in rsp?
Could you please ask questions that are semantically understandable and
additionally read the posting guide?
Thanks,
Uwe Ligges
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View this message in context:
http://r.789695.n
So you are still unable to cite the previous part of the thread?
On 19.01.2012 00:13, Jonas Stein wrote:
Jonas, I've just seen your function 'sistring' code and it's different from
the code in
Thanks a lot for reporting this bug. It is fixed now in the git
repository.
I added some examples,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Sula2011
> Sent: Friday, January 20, 2012 7:00 AM
> To: r-help@r-project.org
> Subject: [R] Incorrect DateTime using ISOdatetime in R
>
> Dear list,
>
> I need to transform the D
Thanks for your reply Michael,
gwindow() command does work. and here is my sessionInfo
> gwindow()
guiWidget of type: gWindowRGtk for toolkit: guiWidgetsToolkitRGtk2
*(opens a window)*
> sessionInfo()
R version 2.14.1 (2011-12-22)
Platform: i486-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.
If you use apply the result will be a matrix,
not a data.frame. You could use a for loop
for(j in seq_len(ncol(x))) {
x[,j] <- scale(x[,j])
}
or the odd looking
x[] <- lapply(x, scale)
to scale all the columns and keep x a data.frame.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap ti
Hi Martin,
On Fri, Jan 20, 2012 at 12:04 PM, Martin Batholdy
wrote:
> Hi,
>
>
> I am currently trying to z-transform (that is subtracting the mean and divide
> by the standard deviation) multiple columns of a data.frame at the same time.
>
>
> My first approach was:
>
> x <- data.frame(c(0:10),
666.1751 sure seems like it should return 2010-10-29 04:12:09 based on
your example.
666.1751 days from 2009-01-01 is 2010-10-29 + some hours/min/seconds.
0.1751 days * 24 hrs/day = 4.2024 (i.e. 4:00AM + some minutes).
0.2024 hours * 60 min/hr = 12.144 (i.e. 12 minutes + some seconds).
0.144 mi
great, thank you!
On 20.01.2012, at 18:10, R. Michael Weylandt wrote:
> ? scale
> apply(x, 2, scale)
>
> Michael
>
> On Fri, Jan 20, 2012 at 12:04 PM, Martin Batholdy
> wrote:
>> Hi,
>>
>>
>> I am currently trying to z-transform (that is subtracting the mean and
>> divide by the standard
? scale
apply(x, 2, scale)
Michael
On Fri, Jan 20, 2012 at 12:04 PM, Martin Batholdy
wrote:
> Hi,
>
>
> I am currently trying to z-transform (that is subtracting the mean and divide
> by the standard deviation) multiple columns of a data.frame at the same time.
>
>
> My first approach was:
>
>
On Fri, 2012-01-20 at 18:04 +0100, Martin Batholdy wrote:
> Hi,
>
>
> I am currently trying to z-transform (that is subtracting the mean and divide
> by the standard deviation) multiple columns of a data.frame at the same time.
>
>
> My first approach was:
>
> x <- data.frame(c(0:10), c(10:20
On Fri, Jan 20, 2012 at 9:04 AM, Martin Batholdy
wrote:
> Hi,
>
>
> I am currently trying to z-transform (that is subtracting the mean and divide
> by the standard deviation) multiple columns of a data.frame at the same time.
>
>
> My first approach was:
>
> x <- data.frame(c(0:10), c(10:20))
> (
Hi, I have data of the form:
tx y trip
t1x1+e y1+eA
t2x2+e y2+eA
t3x3+e y3+eB
t4x4+e y4+eB
t5x5+e y5+eB
... ... ... ...
where t is time and
Hi,
I am currently trying to z-transform (that is subtracting the mean and divide
by the standard deviation) multiple columns of a data.frame at the same time.
My first approach was:
x <- data.frame(c(0:10), c(10:20))
(x - colMeans(x)) / apply(x, 2, sd)
This is obviously not working.
Is th
On Jan 20, 2012, at 10:46 AM, Christof Kluß wrote:
Hi
I like to use "axis.POSIXct" to plot days from 2006 till 2008. But I
only have datas for the summer months. Is it possible to get two axis
breaks, to have not so long distances without points?
There are worked examples of broken axes in p
It may be possible, but perhaps not a good idea. A better approach would be to
use multiple graphs (panels in lattice or facets in ggplot2).
Provide a reproducible example and you might get a more concrete example.
---
Jeff N
?gc
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (S
TOC_NI<-read.csv2("C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv",
sep=";", dec=",", encoding="UTF-8")
circ<-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,]
plot(NI~TOC,data=TOC_NI,col="blue", pch=16, xlim=c(0,450))
abline(lm(NI~TOC,data=TOC_NI),col = "red",lwd=3)
points(NI~TOC,data=
Hi all,
Lets say I have a huge list which is indexed in the following format:
mylist[[i]][[j]][[k]]
where the size is 100 x 100 x 10
If I want to set
mylist[[2]][[3]]=NULL
How do I free the memory used by that sub-list?
Thanks a lot!
[[alternative HTML version deleted]]
___
Hi
I like to use "axis.POSIXct" to plot days from 2006 till 2008. But I
only have datas for the summer months. Is it possible to get two axis
breaks, to have not so long distances without points?
thx
Christof
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https:
Dear list,
I need to transform the DateTime of my GPS data from:
"666.1751" into "/mm/dd hh:mm:ss"
I have the following code:
d$Date <- ISOdatetime(2009, 1, 1, 0, 0, 0, tz = "GMT")+d$Date*(24*3600)
This gives me: 2010-10-29 04:12:09, which is wrong. It should be 2010-10-29
06:12:09
Ano
Hey,
I want to create a stacked barchart in R for the following dataset
(http://pastebin.com/pyHUNgr2):
# usage capacitydiff
1 4 10 6
2 2 20 18
3 5 10 5
The stacked barchart should, in one plot show each l
Dear Petr and Justin,
my problem ist, that I only want to have the 4 highest values for "Ni" as a
red point or with a red circle. The other points should not be modificated.
In your proposals always all points get a red circle or a red point not only
the 4 highest Ni values!
I hope you could unders
Hi Michael,
I think you're right, I should be looking for "predict" instead of
"forecast". I'm still fairly new to R so often don't know what to look
for. As a simplified example (let's neglect the fourier terms):
fit = auto.arima(data)
but now I have data.latest, so I want to use the ARIMA te
Hi. I have a question about the taylor.diagram() in plotrix package. How can
I control the label "correlation"? In the embedded figure you can see the
label "correlation" is too close to the ticks. How can I move it and make it
larger?
Another problem is the labels "0.95" and "0.99" are too close
First time ever that I try to call subroutines in a Win DLL using R.
Have done this before using VBA and Python.
The C code's function argument list contains only double pointers
(double *x). The function is declared void, the output value is one of
the arguments.
C calling sequence is used.
On Thu, 2012-01-19 at 19:23 -0500, C W wrote:
> Thanks, Rich, I will look at the book.
>
> I agree, there are many nice packages, but what if the package changes in a
> few years? I would have no idea what is going on! I've heard
> from predecessor in the industry who emphasize the learning, not
If you are going to resurrect old threads, it's kind to at least read
them. Use the chron package as suggested by Gabor.
Michael
On Fri, Jan 20, 2012 at 7:54 AM, uday wrote:
> Hi,
> How I should convert Fractional days in Year Month Days Hour Minute and Day
> format ?
>
>
>
>
> --
> View this me
Hi
Yes with layout she can put several plots on one page, but if I consider
20 plots per page and one minute per page inspection, 54000 plots is still
more than 2000 pages and about 40 hours, which I do not consider as a well
spent time.
Regards
Petr
>
> Use layout() to put multiple plots on
Hello again,
No I managed to do everything correctly...
the code now looks like:
var1 <-seq(1,5)
var2 <-c("A","B","C","D","E")
var3 <-c("00","01-1;02-3;04-1","01-2;02-1","01-0;04-2",NA)
x <- data.frame(var1,var2,var3)
#create new columns and prefill with 0
x$var3_01 <- 0
x$var3_02 <- 0
x$var3_
Hello,
I found in the forum two threads about point biserial correlation. One of
them (1) mentioned "a point-biserial correlation is just a Pearson
correlation where one of the variables is dichotomous. Thus, the command
is just the normal cor function". The other (2) mentioned "Professor Fox
Hi,
How I should convert Fractional days in Year Month Days Hour Minute and Day
format ?
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Sent from the R help mailing list archive at Nabble.com.
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R
Hi All,
I have 3 variables which present a perfect linear dependency such that the
third is the sum of the first two. I have an ordinary 2D contour plot on a
square grid with the first two variables forming the axes and the third
naturally being the diagonals. From an interpretive point of view it
Use layout() to put multiple plots on a single page. (Note that with >50k plots
you'll want layout() and multiple pages.)
Michael
On Jan 20, 2012, at 6:46 AM, Petr PIKAL wrote:
>>
>> Thank you very much,
>>
>> but in *.pdf I can see 1 plot, may I ask you another question?
>> How can see mo
You also might look at grepl() if you have time: it allows regular expressions
and will be a little (a lot?) more flexible in how you define a match if you
want to ignore things like capitalization.
(mnemonic: the L in grepl indicates its like grep but returns logicals instead
of positions)
M
Hello all,
I think I am now on the way to correctly split the vector as I want it
using for loops.
I got now to a point where I got stuckedSo maybe someone can help
me out...
Remember the result I am looking for should look like (for
the input vector I want to split see below: var3)
var1 va
Hi,
thank you very much... %in% is the operator I was looking for.
cheers,
johannes
Original-Nachricht
> Datum: Fri, 20 Jan 2012 13:01:54 +0100
> Von: Rainer M Krug
> An: Johannes Radinger
> CC: R-help@r-project.org
> Betreff: Re: [R] test if text is part of vector
> -BE
Dear all,
I would like to know how to print the variance-covariance matrix used in the
function anova.rq () when investigating whether the coefficients of a quantile
regression model is the same for a range of quantiles. To be more precise, when
I use the function summary.rq(, cov=TRUE) conside
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 20/01/12 12:50, Johannes Radinger wrote:
> Hello,
>
> this is a very simple question: How can I find out if a word is
> part of a list of words
>
> like: a <- "word1" b <- "word4"
>
> vector <- c("word1","word2","word3")
>
> I tried it with matc
Hi
> Hello,
>
> this is a very simple question:
> How can I find out if a word is part of a list of words
>
> like:
> a <- "word1"
> b <- "word4"
>
> vector <- c("word1","word2","word3")
>
> I tried it with match(a,vector)
> but this gives the position of the word.
>
Perhaps
a %in% vector
Hello,
this is a very simple question:
How can I find out if a word is part of a list of words
like:
a <- "word1"
b <- "word4"
vector <- c("word1","word2","word3")
I tried it with match(a,vector)
but this gives the position of the word.
I am not sure if and how that can be
done with a logical
>
> Thank you very much,
>
> but in *.pdf I can see 1 plot, may I ask you another question?
> How can see more than one in each page?
You just press page down or up for moving through pdf document.
But seriously, try it yourself
lll <- split(rnorm(100) , rep(1:10, each=10))
pdf("test.pdf")
f
Hi
You shall at least read what others wrote you about your code.
here is a quote from what I wrote you yesterday and what stays valid for
today too.
"Also be aware that all levels of a factor are preserved in a subset
unless you specifically strip the unused levels. Therefore there is a
mism
Huh
If you spend only 10 seconds inspecting one plot you will need about 150
hours for that task. I would recommend to reconsider this issue for your
own sanity.
Anyway you can save them either to separate files or in multi page PDF
document although I do not know if there is some limit in pdf
This works but I do not know if there is a better way
tmp = df[df$myvalue<2000,]
ind = match(tmp$myvalue, df$myvalue)
res = df$DateTime[ind]
solution = list(ind[1], res[1])
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Hi All,
I want small modification in apperance of legend. I want seperate legend for
each graph representing the lines present in that graph only (not all the lines
in all graphs) .
Can you please help?
Thank you
Regards
Devarayalu
Orange1 <- structure(list(REFID = c(7, 7, 7, 7, 7, 7, 7, 7, 8,
Thank you Jorge and Florent for your responses.
Now, I 'd like to get the date *(and its index) *where myvalue < 2000 for
the first time.
I expect for a result like (index, date) = (3, 2012-01-07 )
This way does not work:
ind = match(df$myvalue <2000, df$myvalue)
res = df$DateTime[ind]
--
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Dear All
I have 54000 plots in R,
How can I observe them?
If Iâ have to save them one-by-one?
Soheila
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PLEASE do read the
Dear all,
I would like to ask you if you know there is a symlet package that can be used
to create those and apply them in a DWT .
B.R
Alex
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Could anyone please tell how to pass parameters of form to server in rsp?
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Thank You!
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This is discussed in the help of seq(). See the details section of ?sec
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderl
Hcan you give session info (after loading playwith)? I'm able
to get that code to work...also -- can you get the basic RGtk
functions (like gwindow() ) to work?
Michael
On Thu, Jan 19, 2012 at 5:28 PM, Farhat Maha wrote:
> Hello, I managed to install playwith package and all its prerequi
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