Vishal vishalps at gmail.com writes:
I have a matrix(3000x2) of numbers and I have plotted a scatterplot
(defined in the ``car'' library.)
scatterplot(r$V1,r$V2,ellipse=TRUE)
The ellipse plotted is an error ellipse.
I want the find the major(semi), minor(semi) minor axis length of
On Wed, 2 Sep 2009 17:12:18 +1200 Worik R wor...@gmail.com wrote:
WR I have a data frame, df that I want to extract some rows from
What you need is subset, see
?subset
Example:
df-data.frame(TTE=(0:10)/100,SOME=rnorm(11))
df
# subset with all columns
subset(df,TTE0.02)
# the same as:
Duncan Murdoch murd...@stats.uwo.ca
on Tue, 01 Sep 2009 08:02:57 -0400 writes:
On 9/1/2009 6:37 AM, (Ted Harding) wrote:
On 01-Sep-09 10:25:53, Duncan Murdoch wrote:
Jim Lemon wrote:
Duncan Murdoch wrote:
On 8/31/2009 11:50 AM, Mark Knecht wrote:
On Mon, Aug 31,
PaCo == p connolly p_conno...@slingshot.co.nz
on Wed, 02 Sep 2009 12:19:31 +1200 writes:
PaCo On Mon, 31-Aug-2009 at 08:25PM +1000, Jim Lemon wrote:
PaCo [...]
PaCo | Hi Liviu,
PaCo | I was going to steer clear of this one, as my favorite editor
(NEdit)
PaCo | has
Hi
that is a question which comes almost so often as why R does not think
that my numbers are equal. So even I, non statistician, can deduct that
hist with probability =T can have any y axis range but the sum below curve
has to be below 1.
Regards
Petr
r-help-boun...@r-project.org napsal dne
Jonathan Greenberg:
Quick informal poll: what is everyone's favorite text editor for working
with R? I'd like to hear from people who are using editors that have
some level of direct R interface (e.g. Tinn-R, Komodo+SciViews). Thanks!
I use RKward. It’s a KDE app, based on the same editor
As suggested in the article R News 4/1, I used
as.Date(as.character(Phenology_VE$Date), %Y-%m-%d), however this function
returns me only NA values
as.Date(as.character(Phenology_VE$Date), %Y-%m-%d)
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA
[26] NA NA NA NA
Thanks. I have no idea how I did not try that. Sigh!
all good now!
Worik
On Wed, Sep 2, 2009 at 7:04 PM, Stefan Grosse singularit...@gmx.net wrote:
On Wed, 2 Sep 2009 17:12:18 +1200 Worik R wor...@gmail.com wrote:
WR I have a data frame, df that I want to extract some rows from
What you
R ships with the R Data Import/Export manual which is a good start if
your questions is as unspecific. Please also note the very last sentence
of the R-help message: PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html;
Best,
Uwe Ligges
joel ulises sevilla wrote:
See, e.g.,
cloud() in package lattice
plot3d() in package rgl
scatterplot3d() in package scatterplot3d
Uwe Ligges
ivo...@gmail.com wrote:
dear R experts: I am trying to plot an empirical likelihood function in 3d.
The values are not over a regular grid---I just searched the likelihood
Here's one way (assuming your data frame is named dat):
with(dat,
data.frame(a,t(sapply(a,function(x){
apply(dat[a - x = -5 a - x =
0,c('b','c')],2,sum)}
- Phil Spector
Dear Gavin, Simon,
this is the result of str:
str(dist_scot24_vector_with_climate)
'data.frame': 2265025 obs. of 14 variables:
$ X : int 1 2 3 4 5 6 7 8 9 10 ...
$ tetrad_i: Factor w/ 1505 levels HP61A,HP61I,..: 1505 1504 1503 1502
1501 1500 1499 1498 1497 1496 ...
$ tetrad_j:
Hi,
I want to save the image produced by heatmap.2 in a tiff file. To do so, I call
tiff(filename ='test.tiff', res = 300)
heatmap.2(c2, col = greenred(200),labCol = , labRow = ,breaks =
seq(-3,3,0.03), trace = none, dendrogram = none, keysize = 2, scale =
none)
But I get the following error
I have a data that looks like this:
http://dpaste.com/88561/plain/
And I intend to create multiple density curve into one plot, where each curve
correspond to the unique ID.
I tried to use sm package, with this code, but without success.
__BEGIN__
library(sm)
dat - read.table(mydat.txt);
plotfn
On Wed, 2009-09-02 at 09:26 +0100, Corrado wrote:
Dear Gavin, Simon,
this is the result of str:
str(dist_scot24_vector_with_climate)
'data.frame': 2265025 obs. of 14 variables:
$ X : int 1 2 3 4 5 6 7 8 9 10 ...
$ tetrad_i: Factor w/ 1505 levels HP61A,HP61I,..: 1505 1504
Héctor Villalobos wrote:
Hi,
This also happens in 2.9.1 and 2.9.2. My personal solution is to look
under ' C:\Program Files\R\R-2.9.0\library' for a bizarre-named
directory (starts with an 'f' and has numbers and letters mixed) ,
which contains the package directory (in this case
Dear R-help fellows
good afternoon.
I am struggling in the attempt to impose some graphical conditions (changing
point symbols, colors, etc) to biplot function (I am using it to visualize the
results of princomp) but I can't apparently manage to change anything but the
axis... and I have
Have a look at the ggplot2 package.
library(ggplot2)
dat - read.table(mydat.txt)
ggplot(dat, aes(x = V1, colour = factor(V2))) + geom_density()
#or a few alternatives
ggplot(dat, aes(x = V1, fill = factor(V2))) + geom_density(alpha = 0.2)
ggplot(dat, aes(x = V1)) + geom_density() +
Hello!
I estimate vector error correction model (vecm) model. I have only one
cointegratio relationship. I write :
joh.vecm.rls - cajorls(joh.vecm, r=1)
The output estimation is :
Call:
lm(formula = substitute(form1), data = data.mat)
Coefficients:
up.d expl.d
I'm afraid that mgcv:gam can't cope with this size of data set with this
complexity of model. The model matrix alone for your first model would
require around 3 terabytes of storage. For the simplest additive model the
model matrix is `only' 1.6 Gb, but that's before you do anything with it
Hi
I had similar problem some time ago. I was advised to use eqscplot but it
did not suite my purpose so I used a little twist of biplot. (I added a
result)
fit-princomp(some.data, cor=T)
biplot(fit, xlabs=rep(, no.of.poins)) #biplot without points
#some scaling of plot parameters
Taking the example from ?biplot.princomp with reproducible code (hint,
hint):
biplot(princomp(USArrests))
biplot(princomp(USArrests), col=c(2,3), cex=c(1/2, 2))
clearly changes the color and font size on my system. For changing the
point symbols (which are text by default), try setting xlabs
Try this also:
xtabs(value ~ site + parameter, data = DF)
On Tue, Sep 1, 2009 at 8:59 PM, Krista Chin 0574...@acadiau.ca wrote:
Hi,
The lab in which I send my samples return the results in a format that
is difficult for me to run my analysis. The lab outputs the results
where each
Hi, You can make a biplot on Your own, it is not so hard. And in this
case You can change parameters for every low level function as You
wish.
PC - prcomp (iris[,1:4])
lambda - PC$sdev * sqrt(nrow(PC$x))
plot (t(t(PC$x)/lambda),pch=16,col=as.numeric(iris[,5]))
par (new=T)
Rot -
Hi Thierry,
Thanks for the reply. I tried this:
dat - read.table(http://dpaste.com/88561/plain/;)
But I got such error:
ggplot(dat, aes(x = V1, colour = factor(V2))) + geom_density()
Error in density.default(data$x, adjust = adjust, kernel = kernel,
weight = data$weight, :
need at least
Hi Thierry,
I am sorry for coming back to you.
Maybe I misunderstood you, but I got this:
colnames(dat)
[1] V1 V2
ggplot(dat, aes(x = V1, colour = factor(V2))) + geom_density()
Error in data.frame(..., check.names = FALSE) :
arguments imply differing number of rows: 1, 200
ggplot(dat,
My orginal code should work with those colnames. Note that you must not use
variable.name but just variable.name.
I guess that somethings wrong with your dataframe. What does str(dat) and
summary(dat) gives?
Thierry
str(dat)
'data.frame': 200 obs. of 2 variables:
$ V1: num 0.98 0.19 1.09 0.21 0.26 0.98 0.31 0.88 0.23 0.2 ...
$ V2: int 1 0 1 0 0 1 0 1 0 0 ...
summary(dat)
V1 V2
Min. :0. Min. :0.000
1st Qu.:0.1600 1st Qu.:0.000
Median :0.2950 Median :0.000
Mean
Hello,
I have a number of files output1.dat, output2.dat, ... , output20.dat,
each of which monitors several variables over a fixed number of
timepoints. From this I want to create a data frame which contains the
mean value between all files, for each timepoint and each variable.
The
This code work without errors for me.
library(ggplot2)
dat - read.table(http://dpaste.com/88561/plain/;)
ggplot(dat, aes(x = V1, colour = factor(V2))) + geom_density()
ggplot(dat, aes(x = V1, fill = factor(V2))) + geom_density(alpha = 0.5)
Have a look at table(dat$V2). Some classes have only 1
Manca Marco (PATH m.manca at path.unimaas.nl writes:
.
I am struggling in the attempt to impose some graphical conditions (changing
point symbols, colors, etc)
to biplot function (I am using it to visualize the results of princomp) but I
can't apparently manage to
change anything but the
Why not create a single data frame with all the results and then you
can easily use tapply/sapply/aggregate/reshape package
# combine into a single DF
combinedDF - do.call(rbind, MyList)
On Wed, Sep 2, 2009 at 8:55 AM, Ed Longe.l...@ucl.ac.uk wrote:
Hello,
I have a number of files
Hello,
i have this dataset http://www.umass.edu/statdata/statdata/data/pharynx.txt.
the variables GRADE, T_STAGE anda N_STAGE are qualitative or quantitative
variables???
i only have this simple doubt...!
another example: why in the dataset ovarian (library survival) the variable
ecog.ps:
A reproducible example would help. What is Phenology_VE$Date? This works
as.Date(2009-09-01, %Y-%m-%d)
[1] 2009-09-01
Is this the date you wanted:
as.Date(39936, origin='1900-2-1')
[1] 2009-06-05
On Wed, Sep 2, 2009 at 2:09 AM, swertiev_coudr...@voila.fr wrote:
As suggested in the
I'm using the latest version: 0.8.3 on R 2.9.2
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
Which version of ggplot2 you use?
Mine is:
Version: 0.7
Date: 2008-10-03
Somehow it can't recognize facet_wrap()
ggplot(subset(dat, V2 = 2), aes(x = V1)) + geom_density() + facet_wrap(~V2)
Error: could not find function facet_wrap
- G.V.
On Wed, Sep 2, 2009 at 10:01 PM,
Here is a solution assuming that all files have the same structure and a
variable TimePoint which contains the time info.
CombinedData - do.call(rbind, lapply(seq_len(20), function(i){
fileName - paste(output, i, .dat, sep=)
read.table(fileName, header=TRUE)
}))
library(plyr)
On Tue, Sep 1, 2009 at 9:06 AM, Mark Knechtmarkkne...@gmail.com wrote:
Hi,
I'm not understanding how the width height parameters are
supposed to work. When I execute the following 4 commands:
X11()
X11(width=20, height=20)
X11(width=20, height=10)
X11(width=40, height=40)
I get the
Though, from my limited understanding, the 'apply' family of functions
are actually just loops. Please correct me if I'm wrong. So, while
more readable (which is important), they're not necessarily more
efficient than explicit 'for' loops.
allie
On 9/2/2009 3:13 AM, Phil Spector wrote:
Here's
Thanks a million Thierry. I solved the problem
with new installation. I apologize for troubling you
in this trivial matter.
- G.V.
On Wed, Sep 2, 2009 at 10:26 PM, ONKELINX,
Thierrythierry.onkel...@inbo.be wrote:
I'm using the latest version: 0.8.3 on R 2.9.2
Would like some tips on how to avoid loops as I know they are slow in R
They are not slow. They are slower than vectorised equivalents, but
not slower than apply and friends.
Hadley
--
http://had.co.nz/
__
R-help@r-project.org mailing list
If you can do it- try a for loop and another solution to prove this to
yourself. A for loop can get a little unwieldy for a novice like me
to understand the code, but doable. The simpler the better, but they
are not terribly slow. I have run into a couple of situations where a
vectorized
Currently, I am doing it this way.
x - mtcars$mpg
h-hist(x, breaks=10, col=red, xlab=Miles Per Gallon,
main=Histogram with Normal Curve)
xfit-seq(min(x),max(x),length=40)
yfit-dnorm(xfit,mean=mean(x),sd=sd(x))
yfit - yfit*diff(h$mids[1:2])*length(x)
lines(xfit, yfit, col=blue, lwd=2)
But
I've been trying to compile 2.9.2 under solaris 10 the last couple days without
success. configure runs fine and I'm using GNU Make. We're trying to build
under 64 bit (and I'm wondering if this might be part of our problem). Here is
the output from the make...
ld: warning: file
That is not very complex with densities instead of counts.
library(ggplot2)
ggplot(mtcars, aes(x = mpg)) +
geom_histogram(aes(y = ..density..), fill = red) +
stat_function(
fun = dnorm,
args = with(mtcars, c(mean = mean(mpg), sd = sd(mpg)))
Another advantage of the apply family of functions is that
they determine the size and type of their output in an
efficient way, which is sometimes tricky when you write
the loop yourself.
- Phil Spector
Given:
mitest - matrix(1:16,ncol=4)
dimnames(mitest)[[1]] - c(a,b,c,d)
dimnames(mitest)[[2]] - c(a,b,c,d)
mitest
a b c d
a 1 5 9 13
b 2 6 10 14
c 3 7 11 15
d 4 8 12 16
I can do:
mitest[cbind(c(1,2,3),c(2,3,2))]
[1] 5 10 7
but using the names does not work:
Hello,
I have been having difficulty getting boxplot to give the output I want -
probably a result of the way I have been handling the data.
The data is arranged in columns: each date has two sets of data. The number
of data points varies with the date, so each column is of different length.
Hi,
Thank you for your responses.
Héctor, I had also discovered the solution you suggested. However, when I
install files from a list (such as pckg.list), failure in the installation of
one package in the list results in none of the packages later in the list being
installed. After moving
Jari, thanks for the quick answer.
sqrt(eigen(cov.trob(mydataforellipse)$cov)$values)
what will this return?
For my data, I get:
sqrt(eigen(cov.trob(r)$cov)$values)
[1] 1.857733e-05 4.953181e-06
Is this Left hand value the major or the semi major length?
I also try to plot a circuit
Loris Bennett loris.benn...@fu-berlin.de writes:
I get the same problem using R version 2.9.2.
I would be very grateful if anyone could shed some light on this
issue.
Regards
Loris
loris.benn...@fu-berlin.de (Loris Bennett) writes:
Hi,
I am getting the following error
#
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one minor point: The new variables
are added to the data in reverse order.
For example:
x - data.frame(a = 1:3, b = 4:6)
y - within(x, {
c = a^2
d =
I forgot to mention that the ellipse has a rotation. It's horizontal
axis is not parallel to the x-axis.
Just repeating, when I try to draw a circuit keeping the center same
as the ellipse, and radius equal to the first value returned by
sqrt(eigen(cov.trob(mydataforellipse)$cov)$values)
taking
I too am looking to do the same thing. Anyone have any insight as to this can
be done?
Thanks,
Jason
Chris Stubben wrote:
I wrote a package which includes a number of genome sequencing project
statistics on the web like http://www.ncbi.nlm.nih.gov/genomes/lproks.cgi.
I included some
Crudely but I think it works
x - data.frame(aa - mtcars$mpg)
b - ggplot(x, aes(aa)) + geom_histogram(aes(y=..density..)) +
stat_function(fun=dnorm, args=list(mean=mean(x$aa), sd=sd(x$aa)))
b
--- On Wed, 9/2/09, Gundala Viswanath gunda...@gmail.com wrote:
From: Gundala Viswanath
Hi,
See: ?system.file
-steve
On Sep 2, 2009, at 11:39 AM, jbryer wrote:
I too am looking to do the same thing. Anyone have any insight as to
this can
be done?
Thanks,
Jason
Chris Stubben wrote:
I wrote a package which includes a number of genome sequencing
project
statistics on
On Tue, 1 Sep 2009, dolar wrote:
Would like some tips on how to avoid loops as I know they are slow in R
If I understand your criterion (and calling your data.frame 'dat'):
criterion - as.matrix(dist(dat$a)) = 5 outer(dat$a,dat$a,=)
criterion %*% as.matrix(dat[, c(b,c)])
b c
1 5
Hello dear R community.
I just started playing with the snowfall package (a wrapper for the snow
package), and found it very convenient.
(See also this great website:
http://www.imbi.uni-freiburg.de/parallel/ )
I was wondering if it is possible to connect snowfall with the foreach
package (since
Hi Tal,
On Sep 2, 2009, at 11:52 AM, Tal Galili wrote:
Hello dear R community.
I just started playing with the snowfall package (a wrapper for the
snow
package), and found it very convenient.
(See also this great website:
http://www.imbi.uni-freiburg.de/parallel/ )
I was wondering if it
Hi Steve,
Thanks for asking - I forgot to mention I use windows XP.
And the only support foreach has for windows (as far as I read from the
manuals so far) - is for the snow package.
Cheers,
Tal
On Wed, Sep 2, 2009 at 6:57 PM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
Hi Tal,
Dear R Users,
I am using the LLTM and the LRSM functions in the eRm package to do repeated
measurements where there are 2 measurement points for a list of 10 items. I
am trying to get ability estimates but am having trouble. I don't think
that it is appropriate to use the pmat function since
Alexander Shenkin wrote:
Though, from my limited understanding, the 'apply' family of functions
are actually just loops. Please correct me if I'm wrong. So, while
more readable (which is important), they're not necessarily more
efficient than explicit 'for' loops.
Hi Allie -- This uses an
Dear R-users,
I am trying to use the grid.text and expression functions to display
several character strings and plotmath text on a viewport. Some strings
can include a variable portion (PI.limits in the following example),
which I thought could be implemented by combining the bquote and the
R Import and Export manual on the R site is a start.
Also what kind of data are you trying to read?
given a simple csv file called sss.csv like this
a, b, c
1,2,3
4,5,6
on your C drive
you can read it into R with
read.csv(C:/sss.csv)
See ?read.table for more information
--- On Tue,
Dear list,
I've written a function that plots subjects. Something like:
myplot - function(subject) { plot(subject) }
Subjects are vectors, e.g. ...
s1 - c(200,200,190,180)
... and plotting them works fine, e.g. ...
myplot(s1)
Now I want to have s1 etc appear in the plot title, but I don't
Hi miksanta.
They seem to be ordered variables (with specific notation for NA).
Some are qualitative and some quantitative
On Wed, Sep 2, 2009 at 4:19 PM, miksa...@gmail.com wrote:
Hello,
i have this dataset
http://www.umass.edu/statdata/statdata/data/pharynx.txt.
the variables GRADE,
Hi,
Try this,
library(grid)
value - c(0.1)
lab - c(test,
expression(bquote(paste(.(value[1]*100), and percentiles1,
sep=))),
bquote(expression(.(value[1]*100)* and percentiles2)),
bquote(paste(.(value[1]*100), and percentiles3, sep=)) )
grid.newpage()
Not sure I completely understand what you want. You might try looking at
the irt.ability() function in the MiscPscyho package for ability
estimates.
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Tessari
Sent: Wednesday,
Hi,
Try this,
myplot - function(subject) { plot(subject,
main=deparse(substitute(subject))) }
s1 - c(200,200,190,180)
myplot(s1)
see ?deparse
HTH,
baptiste
2009/9/2 Marianne Promberger mprom...@psych.upenn.edu
Dear list,
I've written a function that plots subjects. Something like:
Try this:
mapply(function(x, y)mitest[x, y], c(a, b, c), c(b, c, b))
or
diag(`[`(mitest, i = c(a, b, c), j = c(b, c, b)))
On Wed, Sep 2, 2009 at 6:57 AM, Agustin Lobo aloboa...@gmail.com wrote:
Given:
mitest - matrix(1:16,ncol=4)
dimnames(mitest)[[1]] - c(a,b,c,d)
Martin,
Thanks for showing the timing tests. It is important
to see how the time (and memory usage) grows with
the size of the problem, where size may be the number
of rows or length of the lag.
Here is another function to toss in the hat. It uses no
loops and does all the sum by diff'ing
Your data ranges from 0.67 to 1.21, just to simplify things let's assume the
that the histogram will go from the pretty numbers of 0.65 to 1.25 for a total
width of 0.6. Now consider the simplest histogram consisting of 1 single bar
going from 0.65 to 1.25 (very uninteresting histogram, but
Hi there!!!
I am having trouble with *paste* function. I dont know how to proceed. I
tried many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is *244901_at* )
Then I am using the paste function
paste(set
Try this:
transform(transform(x, c = a ^ 2, d = b ^2), e = c + d)
On Wed, Sep 2, 2009 at 10:27 AM, RINNER Heinrich
heinrich.rin...@tirol.gv.at wrote:
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one
I write about R every weekday at the Revolutions blog:
http://blog.revolution-computing.com
In case you missed them, here are some articles from last month of
particular interest to R users.
http://bit.ly/11YkB0 listed seven reasons of an anthropology professor
for using R.
http://bit.ly/9sbno
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one cell in the row contains a NA.
Presently, I am running a loop that is supposed to extract the rows
that are full of numbers into a second data frame and skips the rows
Hi there!!
I am having trouble with *paste* function. I dont know how to proceed. I
tried many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is *244901_at* )
Then I am using the paste function
paste(set g=,
?is.na
?sum
?if
?else
I think that is how I would approach it, but I could be far off.
On Wed, Sep 2, 2009 at 12:09 PM, Payam
Minoofarpayam.minoo...@meissner.com wrote:
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one
Dear Payam,
Here is a suggestion:
index - apply(yourdata, 1, function(x) any( is.na(x) ) )
yourdata[ !index, ]
Above creates an index (TRUE) when any row of the data contains a missing
value. Then it filters up (extract) the rows that have complete
observations.
See ?any, ?is.na, ?! and ?apply
Hi Baptiste,
Thank for the help. One thing though that I found while transposing your
syntax to my problem: lab must be of class expression for your syntax to
work. For instance, if one replaces the second elements of the lab
variables by a simple integer, lab is not more of class expression,
Try this:
DF[complete.cases(DF),]
On Wed, Sep 2, 2009 at 2:09 PM, Payam Minoofar
payam.minoo...@meissner.comwrote:
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one cell in the row contains a NA.
Presently, I am
Fahim -
Apparently dataline is a factor, so you'd need to use
paste('set g=',as.character(f1))
- Phil Spector
Statistical Computing Facility
Department of
On 9/2/2009 9:27 AM, RINNER Heinrich wrote:
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one minor point: The new variables
are added to the data in reverse order.
For example:
x - data.frame(a = 1:3, b
Payam -
Take a look at na.omit . This is exactly what it
was written for.
- Phil
On Wed, 2 Sep 2009, Payam Minoofar wrote:
Hello everyone,
I am trying to prune a data frame for partial least squares analysis.
I need to delete an entire row if one cell
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Payam Minoofar
Sent: Wednesday, September 02, 2009 10:09 AM
To: r-help@r-project.org
Subject: [R] pruning data
Hello everyone,
I am trying to prune a data frame for partial
Heinrich,
You could create your own function mywithin()
by inserting a couple of rev()'s in within.data.frame().
In within.data.frame(), replace the two commented lines
with those immediately following:
mywithin -
function (data, expr, ...)
{
parent - parent.frame()
#e -
Hello,
I'm using RJDBC/DBI to query a MS SQL DB that looks like:
YMDHMs Num1stsAboveBaseline
2009-05-18 00:001
2009-05-18 00:012
2009-05-18 00:022
...
The first column will get converted to POSIXct.
However, dbGetQuery converts the first column to a
Hi all,
Is there a way in R to plot points using symbols as defined in another
vector without adding a separate points line for each symbol? For
example, I have the results from an ordination with ~ 35 points and an
associated vector that corresponds to different symbols in pch for the
35
I think
install.packages(akima)
library(akima)
example(interp)
would be most useful.
--
View this message in context:
http://www.nabble.com/%22simple%22-3-dimensional-plots--tp25247706p25262822.html
Sent from the R help mailing list archive at Nabble.com.
On Wed, Sep 2, 2009 at 9:27 AM, RINNER
Heinrichheinrich.rin...@tirol.gv.at wrote:
Dear R community,
I am using function 'within' in R.2.9.1 to add variables to an existing
data.frame. This works wonderful, except for one minor point: The new
variables are added to the data in reverse order.
Good point, I failed to spot this kink. You might be interested in a recent
discussion on r-help and r-devel,
http://markmail.org/message/4hvdmwqjyqwprbwf
Best,
baptiste
2009/9/2 Sebastien Bihorel sebastien.biho...@cognigencorp.com
Hi Baptiste,
Thank for the help. One thing though that I
I did something very similar in ggplot2 about a year ago. Use the
unique sampling location from the species count data (rownames) as a
factor column, and then use that in a geom_color or geom_shape add on
to qplot
untested:
library ggplot2
a - metaMDS(foo, k=3)
b - rownames(foo)
d -
On 9/2/2009 1:31 PM, Peter Ehlers wrote:
Heinrich,
You could create your own function mywithin()
by inserting a couple of rev()'s in within.data.frame().
In within.data.frame(), replace the two commented lines
with those immediately following:
mywithin -
function (data, expr, ...)
{
On 9/2/2009 1:28 PM, Phil Spector wrote:
Fahim -
Apparently dataline is a factor, so you'd need to use
paste('set g=',as.character(f1))
That shouldn't be necessary:
f1 - factor(abc)
paste('set g=', f1)
[1] set g= abc
I think we need reproducible code to diagnose this one.
On Wed, 02-Sep-2009 at 07:02AM -0700, Mark Knecht wrote:
| On Tue, Sep 1, 2009 at 9:06 AM, Mark Knechtmarkkne...@gmail.com wrote:
| Hi,
| I'm not understanding how the width height parameters are
| supposed to work. When I execute the following 4 commands:
|
| X11()
| X11(width=20,
I am having trouble with paste function. I dont know how to proceed. I tried
many options but i failed miserably.
I am using a variable f1 to assign a string as below:
f1=dataLine[locAffyProbeID];
( the value of f1 is 244901_at )
Then I am using the paste function
paste(set @g1=, f1);
Thanks Erik and Henrique,
That's what I was after.
Jonas
On Tue, Sep 1, 2009 at 8:08 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
sapply(vec.names, get)
But for this example, you don't need for, try:
dat - 1
On Tue, Sep 1, 2009 at 2:52 PM, jonas garcia
Hi R-users,
I have a problem for updating the estimates of correlation coefficient in
simulation loop.
I want to get the matrix of correlation coefficients (matrix, name: est) from
geese by using loop(500 times) .
I used following code to update,
nsim-500
est-matrix(ncol=2, nrow=nsim)
for(i in
Hello, I'm new to R and I'm building a system that uses a cluster of
computers to parallelize computations. To handle cluster setup I'm using the
R package Snowfall, but knowledge of Snowfall may not be necessary to answer
my question. I'm using a socket-based approach to cluster setup, so all I
William, thank you so much! /tmp has noexec in the fstab and doesn't allow any
file
to be executable. I temporary changed those settings and I was able to
install rJava and RJDBC.
Perhaps the installation routine shouldn't use /tmp and use /var/tmp instead.
Many thanks,
Matt
-Original
1 - 100 of 136 matches
Mail list logo
