Hi all,
Is there a way to write my own error criteria function in the
cross-validation part of the Boosting algorithm?
I am talking about the GBM package. Of course, if you know other good
Boosting implementation in R, please give me some pointers! I am
looking for a good classifier (not neccessa
I'm guessing that your 'data' and 'data1'
are just vectors so your 'rbind' command
returns a 2 by 3 matrix.
Jim showed you already that:
rbind(as.matrix(data), as.matrix(data1))
will probably get you what you are looking
for.
However, I'm suspicious that just:
c(data, data1)
will serve you j
Chris Friedl gmail.com> writes:
> I have two questions about the built-in function step. Ultimately I want to
> apply a lm fitting and subsequent step procedure to thousands of data sets
> groups by a factor defined as a unique ID.
>
> Q1. The code below creates a data.frame comprising three mar
Now that I've actually read the question,
I'm in a better position to answer it.
I have no idea how you are getting the
results that you show, but you can use
'rownames' to set whatever row names you
like. As in:
rownames(result) <- 1:6
Pat
Patrick Burns wrote:
I'm guessing that your 'data'
Dear R-users,
For reporting purpose (using Sweave and LaTeX), I am creating complex tables
with the cat function such as
> x<-c("A", "B", "C", NA)
> cat(x, '\n')
A B C NA
For convenience, I would like to change all my NA value to something else
like '.' (as in SAS for example). Is there a globa
mau...@alice.it wrote:
I was suggested to install two tar-red and gzip-ped packages that are not part
of CRAN or BioConductors yet.
I read the R manual about Administration and could only find a good description of how to install packages
not canonically included in CRAN repository, on UNIX s
Hi all,
I am creating dendrograms using agnes and was wondering if it is
possible to add color to the leaves (and just the leaves).
For example, in the documentation, they have an example using the
"votes.repub" data set. If I wanted to make the word "Washington" green
(and only Washington
This does exactly what is needed.
Thanks guys!
-Original Message-
From: markle...@verizon.net [mailto:markle...@verizon.net]
Sent: Thursday, June 18, 2009 5:30 PM
To: rene.schoenem...@tu-berlin.de
Subject: Re: [R] filtering number of values in a data frame
do.call(rbind,lapply(un
Hi, everybody
OK, I got it working with "recursive". Don't know why this argument
slipped my mind, as I use filter() so often!
Now it is 44 times faster, which is good enough for me. :-)
Thank you, Gabor and Jim.
Best,
Sergey
On Fri, Jun 19, 2009 at 15:23, jim holtman wrote:
> check out 'filter
Martin Maechler wrote:
Hallo Sebastian,
"SP" == Sebastian Pölsterl
on Sun, 14 Jun 2009 14:04:52 +0200 writes:
SP> Hello Martin,
SP> I plotting the silhouette of a clustering and storing it as png. When I
SP> try to store the image as png the bars are missing. The bars are
I'm stopped at a browser in a loop where the following objects look
like this:
Browse[1]> jk
[1] 1
Browse[1]> leg.ab[jk]
[1] "Snails Rep1"
Browse[1]> top.k
[1] "LT95=7.5; LT99=8.8"
I can join them and a few other characters together like this easily
enough:
Browse[1]> paste(jk, ": ", leg.ab[jk],
I want to create a number of vectors like :
vec1 <- rnorm(1)
vec2 <- rnorm(2)
vec3 <- rnorm(3)
and so on...
Here I tried following :
for (i in 1:10) paste("vec", i, sep="") <- rnorm(i)
However obviously that is not working. Here vectors I need to be seperated
i.e I do not want to cre
megh:
> I want to create a number of vectors like :
>
> vec1 <- rnorm(1)
> vec2 <- rnorm(2)
> vec3 <- rnorm(3)
>
> and so on...
Maybe try the assign() function. Something like:
for (i in 1:10) assign ( paste ( "vec" , i , sep = "" ) , rnorm(i) )
Kind regards, Nikos
I have attached 2 files sample.csv and sample.r to illustrate the problem.
Thank you for considering.
--
View this message in context:
http://www.nabble.com/Customize-axis-labels-in-xyplot-tp24126788p24143742.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi,
I'm running the following code to produce lattice plots of microfibril
angle versus ring number in Scots pine. There are 12 trees and 5 sample
positions ("Position") in each tree:
xyplot(MFA ~ RN | Tree, data = MFA.data,
groups = Position, subscripts=TRUE,
auto.key=list
Dear R Users,
I'm finding that when I execute the following bit of code, that the new line
argument is actually displayed as text in the graphics device. How do I avoid
this happening?
mtext(side=2, line=5.5, expression(paste("Monthly Summed Runoff (mm/month)",
"/n", "and Summed Monthly Preci
I'm trying to determine if a set of data is normal from a qq plot but seem to
be having a bit of difficulty.
I have a file of the following form
9 36
3 37
6 38
7 39
.
where the left column is the frequency of the number in the right column.
I've found the prob
Steve Murray wrote:
Dear R Users,
I'm finding that when I execute the following bit of code, that the new line
argument is actually displayed as text in the graphics device. How do I avoid
this happening?
mtext(side=2, line=5.5, expression(paste("Monthly Summed Runoff (mm/month)", "/n", "an
Hi
jim holtman napsal dne 19.06.2009 15:06:55:
> I have wondered about this way of testing for equality:
>
> > x <- c(1,0,3,0)
> > x[1] * length(x) == sum(x)
> [1] TRUE
> > x <- rep(1,4)
> > x[1] * length(x) == sum(x)
> [1] TRUE
> This would seem to indicate that both vectors contain the same
Thanks for the response, however, whilst this eliminates the 'new line'
character from appearing, it doesn't actually cause a new line to be printed!
Instead, I have the last few characters of the first character string
overlapping with the first few characters of the next.
How best should I c
Consider the use of a 'list' so you don't clutter up the global enviroment
with a lot of objects that you might forget about:
> vec <- lapply(1:10, rnorm)
> vec
[[1]]
[1] -0.6264538
[[2]]
[1] 0.1836433 -0.8356286
[[3]]
[1] 1.5952808 0.3295078 -0.8204684
[[4]]
[1] 0.4874291 0.7383247 0.575781
> "TobiasV" == Tobias Verbeke
> on Sun, 21 Jun 2009 08:25:07 +0200 writes:
TobiasV> Hi Ken,
>> I have been using R for a while. Recently, I have begun converting my
>> package into S4 classes. I was previously using Rdoc for documentation.
>> Now, I am looking to us
On Mon, Jun 22, 2009 at 7:12 AM, Martin
Maechler wrote:
>> "TobiasV" == Tobias Verbeke
>> on Sun, 21 Jun 2009 08:25:07 +0200 writes:
>
> TobiasV> Hi Ken,
> >> I have been using R for a while. Recently, I have begun converting my
> >> package into S4 classes. I was previously
Greetings,
I usually read this mailing list through google groups
(http://groups.google.com/group/r-help-archive), but when I opened the
webpage this morning it said: "The group named r-help-archive has been
removed because it violated Google's Terms Of Service."
Is there an alternative website w
On 6/22/2009 7:58 AM, Steve Murray wrote:
Thanks for the response, however, whilst this eliminates the 'new line'
character from appearing, it doesn't actually cause a new line to be printed!
Instead, I have the last few characters of the first character string
overlapping with the first few c
On 6/22/2009 7:23 AM, Tony Breyal wrote:
Greetings,
I usually read this mailing list through google groups
(http://groups.google.com/group/r-help-archive), but when I opened the
webpage this morning it said: "The group named r-help-archive has been
removed because it violated Google's Terms Of S
Dear David,
Try this:
x <- c("A", "B", "C", NA)
x[is.na(x)] <- "."
x
HTH,
Jorge
On Mon, Jun 22, 2009 at 2:11 AM, David_D wrote:
>
> Dear R-users,
>
> For reporting purpose (using Sweave and LaTeX), I am creating complex
> tables
> with the cat function such as
>
> > x<-c("A", "B", "C", NA)
>
Dear list,
>From an earlier post I got the impression that one could promote
warnings from a glm to errors (presumably by putting
options(warn=1)?), then try() would flag them as errors. I’ve spent
half the day trying to do this, but no luck. Do you have an explicit
solution?
My problems is that
Auty, Dave forestry.gsi.gov.uk> writes:
>
> I'm running the following code to produce lattice plots of microfibril
> angle versus ring number in Scots pine. There are 12 trees and 5 sample
> positions ("Position") in each tree:
>
> xyplot(MFA ~ RN | Tree, data = MFA.data,
>
>groups =
On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates wrote:
> On Mon, Jun 22, 2009 at 7:12 AM, Martin
> Maechler wrote:
>>> "TobiasV" == Tobias Verbeke
>>> on Sun, 21 Jun 2009 08:25:07 +0200 writes:
>>
>> TobiasV> Hi Ken,
>> >> I have been using R for a while. Recently, I have begun c
Fredrik Nilsson-5 wrote:
>
>>From an earlier post I got the impression that one could promote
> warnings from a glm to errors (presumably by putting
> options(warn=1)?), then try() would flag them as errors. I’ve spent
> half the day trying to do this, but no luck. Do you have an explicit
> sol
Dear list,
I have been struggling to find how I would go about changing the
bakground colors of groups in a lattice barchart in a way so that the
auto.key generated also does the right thing and pick it up for the
key.
I have used the "col" argument (which I guess is sent to par()) in a
way so tha
Look at show.settings() and str(trellis.par.get()). This will show you
what the default settings are. The group colors are set by the
superpose.* elements (e.g. superpose.line is for group lines). To set
them, I usually create a list and pass it to par.settings. For
example,
my.theme <- list(super
Does anybody know of a function that implements the derivative (gradient) of
the multivariate normal density with respect to the *parameters*?
It’s easy enough to implement myself, but I’d like to avoid reinventing the
wheel (with some bugs) if possible. Here’s a simple example of the result
I’
Hi all, Eigen vectors obtained from the function eigen() are ortho-normal? I
see the documentation however there is no formal mention on that. If no,
then is there any direct function to do the same?
--
View this message in context:
http://www.nabble.com/Eigen-value-calculation-tp24147807p241478
What error is it giving? Please include the exact error.
What happens if you do this:
if (file.exists(findings)) cat('File',findings,'exists\n') else
cat('File',findings,not found\n')
Your description suggests that you are using 'if' expression inside a
loop. If that is the case, try
i
Thanks again for a very useful comment. That seems to have separated the text
and put it onto separate lines.
However, whilst this results in the text being centralised in relation to the
axis, it means that the lower line is left-justified in relation to the upper
line, rather than being cent
Hi Karl,
I am not aware of any. May I ask what your purpose is? You don't really
need this if you are going to use it in optimization, since most optimizers
use a simple finite-difference approximation if you don't provide the
gradient. Using the numerical approximation from "numDeriv" will be
On Jun 22, 2009, at 4:08 AM, Dieter Menne wrote:
Chris Friedl gmail.com> writes:
I have two questions about the built-in function step. Ultimately I
want to
apply a lm fitting and subsequent step procedure to thousands of
data sets
groups by a factor defined as a unique ID.
Q1. The code
On 6/22/2009 10:30 AM, Steve Murray wrote:
Thanks again for a very useful comment. That seems to have separated the text
and put it onto separate lines.
However, whilst this results in the text being centralised in relation to the
axis, it means that the lower line is left-justified in relatio
Hello R Users,
I have a question regarding fitting a model with GAM{mgcv}. I have data
from several predictor (X) variables I wish to use to develop a model to
predict one Y variable. I am working with ecological data, so have data
collected many times (about 20) over the course of two years. P
Hi,
I have a simple question, suppose I have the date "05/16/2008", what would
be the command to get the month, day and year?
Thanks,
-Jack
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Couple of choices depending on what you want to do with the data:
> x <- as.POSIXct("05/16/2008", format="%m/%d/%Y") # if you want to use the
date
> format(x, "%d") # day
[1] "16"
> format(x, "%m") # month
[1] "05"
> format(x, "%Y") # year
[1] "2008"
> # or
> y <- strsplit("05/16/2008", '/') # t
Dear All,
I used an AR(1) model to explain the process of the stationary residual and
have used an 'ar' command in R. From the results, i tried to extract
the standard error and p-value for the estimated parameter, but unfortunately,
i never find any way to extract it from the output.
What
Resending, as am not sure about the original "To:" address. Sorry for
any redundancy.
- Cliff
-- Forwarded message --
From: Clifford Long
Date: Mon, Jun 22, 2009 at 11:04 AM
Subject: question about using _apply and/or aggregate functions
To: r-h...@lists.r-project.org
Hi R-lis
On Jun 22, 2009, at 12:04 PM, Clifford Long wrote:
Hi R-list,
I'll apologize in advance for (1) the wordiness of my note (not sure
how to avoid it) and (2) any deficiencies on my part that lead to my
difficulties.
I have an application with several stages that is meant to simulate
and explore
On 6/22/2009 9:23 AM, Tobias Verbeke wrote:
On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates wrote:
On Mon, Jun 22, 2009 at 7:12 AM, Martin
Maechler wrote:
"TobiasV" == Tobias Verbeke
on Sun, 21 Jun 2009 08:25:07 +0200 writes:
TobiasV> Hi Ken,
>> I have been using R for a while. Rec
Jorge,
Thanks a lot for your reply. It's indeed working in this case. However, with
data frames mixing numeric and character is not working anymore. Morever I
am working with many variables and I don't want to modify them. What I would
really appreciate is a global option (in the Rprofile?) that
Ravi Varadhan skreiv:
I am not aware of any. May I ask what your purpose is? You don't really
need this if you are going to use it in optimization, since most optimizers
use a simple finite-difference approximation if you don't provide the
gradient. Using the numerical approximation from "numD
I have been trying to draw histogram for my manscript and found some strange
things that I could not figure out why.
Using the same code listed below I have successfully draw histograms for a
few figures with fraction labeled on Y axis less than 1 (acturally between 0
to 0.1). But one dataset gi
> "DM" == Duncan Murdoch
> on Mon, 22 Jun 2009 09:41:12 -0400 writes:
DM> On 6/22/2009 9:23 AM, Tobias Verbeke wrote:
>> On Mon, Jun 22, 2009 at 2:18 PM, Douglas Bates
wrote:
>>> On Mon, Jun 22, 2009 at 7:12 AM, Martin
>>> Maechler wrote:
> "TobiasV" == To
There are many ways to do this in R. See R News 4/1 for info on dates.
Here is one method:
> library(chron)
> mdy <- month.day.year("05/16/2008")
> mdy
$month
[1] 5
$day
[1] 16
$year
[1] 2008
> mdy$month
[1] 5
On Mon, Jun 22, 2009 at 11:36 AM, Jack Luo wrote:
> Hi,
>
> I have a simple questi
Karl,
You may want to look at the paper by Dwyer on "Some applications of matrix
derivatives in multivariate analysis" (JASA 1967), especially the Table 2 on
p. 617.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Profe
Dear R users,
I am having problems in trying to run R2jags on a Linux Cluster. When I
tried to run the model given in the R2jags manual I get the following error
error message:
Error in FUN(X[[3L]], ...) : object 'J' not found
Can any one help me with this problem?
Looking to hear from you all.
Hello, I have this generalized linear formula:
log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
where the the x[i] and the n[i] are known.
Is there a way to program the GLM procedure to input the formula above and
get the beta[i] estimates? If not the GLM is there another procedure to do
that? The aim
When freq=FALSE then the y axis is not the proportion in each group (what I am
assuming you mean by fraction), but rather is scaled so that the total area of
the histogram is 1 (making comparing to theoretical densities easier). If all
the data values are between 0 and 1, then the height of at
Hi David,
It works for me when handling data frames mixing characters and numeric by
using either print() or a function called "foo":
# Some data
x <- c("A", "B", "C", 3.5, 1, 0, NA, "a character",NA, "another character")
mydata <- data.frame(x = x, y = sample(x))
mydata
# Option 1: print()ing
pr
On Jun 22, 2009, at 1:21 PM, Jorge Ivan Velez wrote:
Hi David,
It works for me when handling data frames mixing characters and
numeric by
using either print() or a function called "foo":
# Some data
x <- c("A", "B", "C", 3.5, 1, 0, NA, "a character",NA, "another
character")
mydata <- dat
On Jun 22, 2009, at 12:35 PM, francogrex wrote:
Hello, I have this generalized linear formula:
log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
where the the x[i] and the n[i] are known.
Is there a way to program the GLM procedure to input the formula
above and
get the beta[i] estimates? If not t
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of SAS recoding:
IF TYPE='TRUCK' and count=12 THEN
Dear R-helpers:
May I ask a question related to storing a number of lmer model fit into a
list.
Basically, I have a for-loop (see towards the bottom of this email)
in the loop, I am very sure that the i-th model fit (i.e.,fit_i) is
successfully generated and the character string (i.e., tmp_i) is c
On Jun 22, 2009, at 2:27 PM, Mark Na wrote:
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of
On 6/22/2009 2:27 PM, Mark Na wrote:
> Dear R-helpers,
>
> I am helping a SAS user run some analyses in R that she cannot do in
> SAS and she is complaining about R's peculiar (to her!) way of
> recoding variables. In particular, she is wondering if there is an R
> package that allows this kind of
Mark Na wrote:
Dear R-helpers,
I am helping a SAS user run some analyses in R that she cannot do in
SAS and she is complaining about R's peculiar (to her!) way of
recoding variables. In particular, she is wondering if there is an R
package that allows this kind of SAS recoding:
IF TYPE='TRUCK'
Le lundi 22 juin 2009 à 09:35 -0700, francogrex a écrit :
> Hello, I have this generalized linear formula:
> log(x[i]/n[i])=log(sum(x)/sum(n)) + beta[i]
I do not understand the problem as stated. if x[i] and n[i] are known,
and unless sum(n)=0, your dataset reduces to a set of nrow(dataset)
indepe
Gmane has already been mentioned, and you might also want to consider
Nabble -- judging from my referrer logs many people use it to read
r-help. If you use Gmail, you might also want to consider subscribing
to the list and using a simple filter. Details and links at
blog.revolution-computing.com, h
Hi, I need help to perform a Shapiro.test on a data frame, I know that
this test works only with vector but I guess there most be a way to
permor it on a data frame instead of vactor by vector (i.e. I've got 40
variables to analyze and its kinda annoying to do it one by one)
Thanks to anyone that
Hey,
A basic question. Is there anyone that knows a good text on using the
R packages map and maptools for R? I have read the references for both
but having trouble getting started.
What I want to do is to load a map (.shp file) and display statistics
on this maps using colours.
Thanks in advanc
Emmanuel Charpentier wrote:
>
> I do not understand the problem as stated. if x[i] and n[i] are known,
> and unless sum(n)=0, your dataset reduces to a set of nrow(dataset)
> independent linear equations with nrow(dataset) unknowns (the beta[i]),
> whose solution is trivially beta[i]=log(x[i]/n[
It is not the case as you described. In any case, the total area should be 1
and labeled fraction on y axis should be far less than 1, since I have more
than 1 data points. I also test differerent bin size by change the
break.
I draw the graph using only 1 group, the same result was obtaine
Try this:
x <- data.frame(A = runif(10), B = rnorm(10))
lapply(x, shapiro.test)
On Mon, Jun 22, 2009 at 3:15 PM, Gonzalo Quiroga
wrote:
> Hi, I need help to perform a Shapiro.test on a data frame, I know that
> this test works only with vector but I guess there most be a way to
> permor it on a
(a) Your code is unnecessarily convoluted.
(b) The example of things *not* working is not reproducible. (Read
the posting guide!!!)
(c) Nonetheless the phenomenon you describe is weird/interesting.
On my system, the following runs without error:
fit.list <- NULL
a <- factor(rep(1:10,each=
On Monday 22 June 2009, Gonzalo Quiroga wrote:
> Hi, I need help to perform a Shapiro.test on a data frame, I know that
> this test works only with vector but I guess there most be a way to
> permor it on a data frame instead of vactor by vector (i.e. I've got 40
> variables to analyze and its kind
On Mon, 22 Jun 2009, charles78 wrote:
It is not the case as you described. In any case, the total area should be 1
and labeled fraction on y axis should be far less than 1, since I have more
than 1 data points. I also test differerent bin size by change the
break.
Try reading Greg's res
Hello list,
I'm trying to fit a model like beta[trt]/(1+alpha*x) where the data
include some grouping factor. The problem is that the estimate for alpha
is undefined for some of the treatments - any value greater than 20 is
equally good and a step function would suffice. Ignoring the grouping
stru
Hi John,
The SIDS examples in the spatial graphics gallery may be helpful:
http://r-spatial.sourceforge.net/gallery/
For a general reference, see Applied Spatial Data Analysis with R.
Note that the book's webpage includes figures with code
http://www.asdar-book.org/
There's also the spatial Ta
Hi R-list,
I'll apologize in advance for (1) the wordiness of my note (not sure
how to avoid it) and (2) any deficiencies on my part that lead to my
difficulties.
I have an application with several stages that is meant to simulate
and explore different scenarios with respect to product sales (in
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify "list=()" in the call to aggregate. I made the correction,
and now get the following:
> slope.mult = simarray[,1]
> adj.slope.value = simarray[,2]
> adj.slope.level = simarray[,2]
> qc.run.violation = sim
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
> datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function:
> datac$Mean2<-apply(datac,2,mean)
so that I
thanks Deepayan, that works great!
2009/6/19 Deepayan Sarkar :
> On 6/18/09, Katharina May wrote:
>> Hi there,
>>
>> sorry for troubling everybody once again, I've got a problem rewriting
>> Sarkar's function for
>> rewriting the tick locations in a logaritmic way (s.
>> http://lmdvr.r-forge
Tena koe Mark
I think you might want
apply(datac, 1, mean)
i.e., apply the function to the first dimension (rows) rather than the
second (columns).
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ma
Mark Na wrote:
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function:
datac$Mean2<-apply(datac,2,m
Hi all,
I am thinking of extending a package by directly adding stuff to its
C++ code. And then I have to recompile the package.
Do I have to download the whole R source repository, in order to do
the recompilation? What is the minimal setup requirement for such a
recompilation?
I am using MSVC
I have used all 3 packages for decision trees (SAS/EM, CART and R). As
another user on the list commented, the algorithms CART uses are
proprietary. I also know that since the algorithms are proprietary, the
decision tree that you get from SAS is based on a "slightly different"
algorithm so as to n
I am trying to run jags.model and need the data read in a suitable
format. The package (rjags) documentation describes "read.data", while
the "classic-bugs" examples use "read.jagsdata". I am running R (R
2.8.1) on a PowerBook G4 (Mac OS X 10.5.7), and my installation
recognizes neither read
Hi,
I have a list made up of character strings with each item a different
length (each item is between 1and 6 strings long). Is there a way to
convert a "ragged" list to a matrix such that each item is its own row?
Here is a simple example:
a=list();
a[[1]] = c("a", "b", "c");
a[[2]] = c
try this:
> a
[[1]]
[1] "a" "b" "c"
[[2]]
[1] "d" "e"
[[3]]
[1] "f" "g" "h" "i"
> # find max row length
> rowMax <- max(sapply(a, length))
> # now create output matrix by lengthening rows
> do.call(rbind, lapply(a, function(x){
+ length(x) <- rowMax
+ x
+ }))
[,1] [,2] [,3] [,4]
[1,]
Try this:
matrix(unlist(lapply(a, '[', 1:max(sapply(a, length, ncol = 4, byrow =
TRUE)
or
do.call(rbind, lapply(a, '[', 1:max(sapply(a, length
On Mon, Jun 22, 2009 at 8:15 PM, Kenneth Takagi wrote:
> Hi,
>
>
>
> I have a list made up of character strings with each item a different
>
Thanks for all the feedback; I found a previous post covering this
question:
https://stat.ethz.ch/pipermail/r-help/2009-February/189232.html
Problem solved!
Kenneth Takagi
kat...@psu.edu
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Monda
Try this:
> unname(t(do.call(cbind, lapply(a, ts
[,1] [,2] [,3] [,4]
[1,] "a" "b" "c" NA
[2,] "d" "e" NA NA
[3,] "f" "g" "h" "i"
On Mon, Jun 22, 2009 at 7:15 PM, Kenneth Takagi wrote:
> Hi,
>
>
>
> I have a list made up of character strings with each item a different
> length
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify "list=()" in the call to aggregate. I made the correction,
and now get the following:
slope.mult = simarray[,1]
adj.slope.value = simarray[,2]
adj.sl
On Jun 22, 2009, at 6:19 PM, Mark Na wrote:
Hi R-helpers,
I have been struggling with calculating row and column statistics,
e.g. standard deviation.
I know that
datac$Mean<-rowMeans(datac,na.rm=TRUE)
will give me row means.
I have tried to replicate those row means with the apply function
Greetings,
I have two files which contain responses to a series of multiple
choice questions. One
file contains responses before an "intervention" and the other
contains the responses afterward.
There were three possible responses to each question: D, F, T (for
Don't Know, False, and Tru
On Jun 22, 2009, at 7:55 PM, David Winsemius wrote:
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but
forgot
to specify "list=()" in the call to aggregate. I made the
correction,
and now get the following:
slo
Try this:
table(factor(pre, levels = c("D", "F", "T")),
factor(post, levels = c("D", "F", "T")))
On Mon, Jun 22, 2009 at 8:56 PM, Jeffrey Edgington wrote:
> Greetings,
>
> I have two files which contain responses to a series of multiple choice
> questions. One
> file contains responses
On Jun 22, 2009, at 7:56 PM, Jeffrey Edgington wrote:
Greetings,
I have two files which contain responses to a series of multiple
choice questions. One
file contains responses before an "intervention" and the other
contains the responses afterward.
There were three possible responses to
On 22/06/2009 6:52 PM, Michael wrote:
Hi all,
I am thinking of extending a package by directly adding stuff to its
C++ code. And then I have to recompile the package.
Do I have to download the whole R source repository, in order to do
the recompilation? What is the minimal setup requirement for
Hi,
I have a data frame with two columns: dt and tf. The dt column is
datetime and the tf column is a temperature.
dt tf
1 2009-06-20 00:53:00 73
2 2009-06-20 01:08:00 73
3 2009-06-20 01:44:00 72
4 2009-06-20 01:53:00 71
5 2009-06-20 02:07:00 72
...
I need a
Hi,
I'm a beginner using R and I'm modeling a time series with ARIMA.
I'm looking for a way to determine the p-values of the coefficients of my model.
Does ARIMA function return these values? or is there a way to determine them
easily?
Thanks for your answer
Myriam
Convert dt to POSIXct class:
wtd$dt <- as.POSIXct(wtd$dt)
subset(wtd, format(dt, '%M') == 53)
On Mon, Jun 22, 2009 at 9:58 PM, Keith Jones wrote:
> Hi,
>
> I have a data frame with two columns: dt and tf. The dt column is datetime
> and the tf column is a temperature.
>
>
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