On May 22, 2013, at 01:54 , David Winsemius wrote:
>> {
>> test0 <- transform(test, (get(var.names[i])) = 0)
>
> There is no `get<-` function. You need to use assign.
Also note that transform's "left hand sides" are really names of function
arguments and therefore syntactically cannot be expre
Thanks.
As to the data " warpbreaks", if I want to analysis the impact of
tension(L,M,H) on breaks, should I order the tension or not?
Many thanks.
At 2013-05-21 20:55:18,"David Winsemius" wrote:
>
>On May 20, 2013, at 10:35 PM, meng wrote:
>
>> Hi all:
>> If the explainary variab
Dear Rxperts,
Sorry about that..forgot to update the numeric part of the multipanel group
indicator...
Below is the updated code... in addition to getting rid of the curly
braces, is there a better way to control the position of
panel.text flexibly instead of hardcoding.
Thanks,
santosh
q <- dat
On Sat, May 18, 2013 at 7:05 AM, Stephen Milborrow wrote:
> Paul Johnson wrote:
>>
>> m1 <- lm(log(y) ~ log(x), data = dat)
>>
>> termplot shows log(y) on the vertical. What if I want y on the vertical?
>>
>
> plotmo in the plotmo package has an inverse.func argument,
> so something like the fo
Thanks a lot for all your help.
best wishes,
Mitra
On 21 May 2013 21:04, arun wrote:
> You could also use:
>
> sapply(seq_len(ncol(mdat)),function(i) mdat[,i]*100/x[i])
> # [,1] [,2] [,3] [,4] [,5]
> #[1,] 7.142857 13.3 30 36.36364 27.8
> #[2,] 78.571429 80.00
I want to make an index using IRT, I use eRm package.
I have a data frame (24 items, 4 categories, 427 cases) that works if I use
PCM (partial credit model), but, when I run RSM (rating scale model) this
appear:
Error in solve.default(parest$hessian) :
Lapack routine dgesv: system is exactly sin
HI,
I am not sure about what you expect as output.
dat1<- read.table(text="
Offense Play
Y A
N B
Y A
Y C
N B
N C
",sep="",header=TRUE,stringsAsFactors=FALSE)
with(dat1,tapply(Play,list(Offense),table))
#$N
#
#B C
#2 1
#
#$Y
#
#A C
#2 1
#or
with(dat1
Hi,
Try this:
lst1<-lapply(1:5,function(i) {pdf(paste0(i,".pdf"));
hist(rnorm(100),main=paste0("Histogram_",i));dev.off()}) #you can change the
numbers
A.K.
>I'm trying to generate a pdf called 1.pdf, 2.pdf, 3.pdf etc and it isn't
>working. My code is:
>x <- 0
>for(i in 1:1000){
>x <- x +
I recommend that you not plan on waiting for the hash package to be redesigned
to meet your expectations. Also, your response to discovering this feature of
the hash package seems illogical.
>From a computer science perspective, the hash mechanism is an implementation
>trick that is intended to
Hi,
May be this helps:
dat1<-read.table(text="
Date Var day
1/1/2013 1 Tue
1/2/2013 2 Wed
1/3/2013 3 Thu
1/4/2013 4 Fri
1/5/2013 5 Sat
1/6/2013 6 Sun
1/7/2013 7 Mon
1/8/2013 8 Tue
1/9/2013 9 Wed
1/10/2013 10 Thu
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$days<-as.num
Dear Rxperts,
Ok The "curly braces" as we talked before,...
They appear if the "group" argument of "xyplot" function is entered as a
numeric value; and don't when the values are letters.
I just figured how to hide the strip borders...and also control the ticks
in different axes...
Any suggesti
On May 21, 2013, at 12:40 PM, Roni Kobrosly wrote:
> Hello, I'm having a problem using the "transform" and "get" functions. I'm
> probably making a dumb mistake, and I need help!
>
> I start by making a small simulated dataset. I save the names of the
> variables in "var.names." Without getting
it works! Thanks!
On Tue, May 21, 2013 at 1:24 PM, Sarah Goslee wrote:
> You can use rep() to create the Date column, and data.frame() to combine
> it.
>
> For your simple example,
>
> newdata <- data.frame(dat, Date=rep(1:3, each=2))
>
> On Tue, May 21, 2013 at 4:16 PM, Ye Lin wrote:
> > I wan
rep(c(1:30),each=1440) works!
On Tue, May 21, 2013 at 1:25 PM, Don McKenzie wrote:
> Do you want each number in "Date" to be repeated once (as in your example)
> or appear 48 times (so that you start over every 1440 rows, as in your
> request).
>
> For the former,
>
> rep(c(1:30),each=48)
>
> f
Hey I have a dataset like this:
Date Var day 1/1/2013 1 Tue 1/2/2013 2 Wed 1/3/2013 3 Thu 1/4/2013 4
Fri 1/5/2013 5 Sat 1/6/2013 6 Sun 1/7/2013 7 Mon 1/8/2013 8 Tue
1/9/2013 9 Wed 1/10/2013 10 Thu
And I want to plot Var~day
Here is the code I use:
plot(Dataset$Var~Dataset$day,xlab='Day
hi
I've been using this code to set a reference level for uni and
multivariate analysis
rmix$interviewmethodcode<-relevel(mix$interviewmethodcode,ref="SAQ")
summary(rma.1<-rma(yi,vi,mods=~interviewmethodcode,data=rmix,method="SJ",knha=F,weighted=F,intercept=T))
giving
Hello, I'm having a problem using the "transform" and "get" functions. I'm
probably making a dumb mistake, and I need help!
I start by making a small simulated dataset. I save the names of the
variables in "var.names." Without getting into the details of it, I have to
create a custom function to p
I am new to mapping with R, and I would like to use the point.in.polygon
function from the sp package, but I am unsure of how to get my data in the
correct format for the function. The generic form of the function is as
follows:
point.in.polygon(point.x, point.y, pol.x, pol.y, mode.checked=FALSE)
Dear Rxperts,
Using the same example above, is there a way to remove the borders of
multi-panel strips and control the display of the borders of each panel..
for example, I would like to keep only side 1 & 2 of a panel...
Thanks,
Santosh
On Wed, May 1, 2013 at 11:11 PM, Santosh wrote:
> Th
On 21/05/2013 21:24, Bert Gunter wrote:
At the risk of misunderstanding... (inline)
On Tue, May 21, 2013 at 12:17 PM, Milan Bouchet-Valat wrote:
Le mardi 21 mai 2013 à 08:17 -0700, Jeff Newmiller a écrit :
That is like complaining that your hammer does not fit these
newfangled Philips screws.
Am Dienstag, den 21.05.2013, 16:17 +0100 schrieb Prof Brian Ripley:
> On 21/05/2013 16:11, Dimitri Liakhovitski wrote:
> > Sorry, I am using package "vars"
>
> So you need to report the bug in that package to its maintainer.
> ?logLik says
>
> Value:
>
> Returns an object of class ‘logLi
May be this helps:
dat<- read.table(text="
ID Var
1 1
2 4
3 6
4 7
5 7
6 8
7 9
",sep="",header=TRUE)
dat$Date<-cumsum(seq_len(nrow(dat))%%2)
dat
# ID Var Date
#1 1 1 1
#2 2 4 1
#3 3 6 2
#4 4 7 2
#5 5 7
Duh -- I mean "for the latter"
On 21-May-13, at 1:25 PM, Don McKenzie wrote:
Do you want each number in "Date" to be repeated once (as in your
example) or appear 48 times (so that you start over every 1440
rows, as in your request).
For the former,
rep(c(1:30),each=48)
fills the first 14
Do you want each number in "Date" to be repeated once (as in your
example) or appear 48 times (so that you start over every 1440 rows,
as in your request).
For the former,
rep(c(1:30),each=48)
fills the first 1440 rows.
On 21-May-13, at 1:16 PM, Ye Lin wrote:
I want to add identifier col
At the risk of misunderstanding... (inline)
On Tue, May 21, 2013 at 12:17 PM, Milan Bouchet-Valat wrote:
> Le mardi 21 mai 2013 à 08:17 -0700, Jeff Newmiller a écrit :
>> That is like complaining that your hammer does not fit these
>> newfangled Philips screws.
>>
>> These are different tools. Do
You can use rep() to create the Date column, and data.frame() to combine it.
For your simple example,
newdata <- data.frame(dat, Date=rep(1:3, each=2))
On Tue, May 21, 2013 at 4:16 PM, Ye Lin wrote:
> I want to add identifier column (Date) to a time series data frame. I want
> to name the "Date
I want to add identifier column (Date) to a time series data frame. I want
to name the "Date" column be from 1 to 30 every 1440 rows.
Say I have a data like this (I simply my actual data here):
$dat
ID Var
1 1
2 4
3 6
4 7
5 7
6 8
How can
Le mardi 21 mai 2013 à 08:17 -0700, Jeff Newmiller a écrit :
> That is like complaining that your hammer does not fit these
> newfangled Philips screws.
>
> These are different tools. Do not expect them to interoperate.
I understand that Lattice and ggplot2 do not use settings from par().
I'm fine
Hi,
library(stringr)
b[str_detect(colnames(b),"^y")]
# y y.1 y.2
#1 0.0 0.00 0.00
#2 19.55811 17.023812 15.354880
#3 10.74991 9.024250 8.177128
#4 5.91924 4.789331 4.367188
#or
b[,!is.na(match(gsub("\\..*","",names(b)),"y"))]
# y y.1 y.2
Hi,
Try:
set.seed(28)
dat1<-
as.data.frame(matrix(c(rep(c(5,10,15,20,25),each=3),sample(1:20,15,replace=TRUE),sample(15:35,15,replace=TRUE),sample(20:40,15,replace=TRUE)),ncol=4))
names(dat1)[1]<- "concentration"
lapply(seq_len(ncol(dat1[,-1]))+1,function(i) {x1<-
cbind(dat1[,1],dat1[,i]);colnam
that is exactly what I wanted! Thank you Sarah!
Andras
--- On Tue, 5/21/13, Sarah Goslee wrote:
> From: Sarah Goslee
> Subject: Re: [R] help with data.frame
> To: "Andras Farkas"
> Cc: r-help@r-project.org
> Date: Tuesday, May 21, 2013, 2:07 PM
> So if I understand you correctly, and
> I may
I don't think the problem is with the plotting, I think the problem is
with your loop. R isn't selecting random numbers, it's doing exactly
what you told, but what you told it doesn't make sense. Hint: use the
same loop but simply print out i at each iteration.
Here's a better way to do it, assumi
So if I understand you correctly, and I may not, you want to extract
the columns from a dataframe that start with y?
Using your reproducible example (thanks!):
> b[, grepl("^y", colnames(b))]
y y.1 y.2
1 0.0 0.00 0.00
2 19.55811 17.023812 15.354880
3 10.74991
Dear All
I have the following code for list "a":
a <-list(structure(c(0, 4, 8, 12, 0, 19.5581076131386, 10.7499105081144,
5.91923975728553, 0, 4.08916328337685, 2.26872955281708, 1.24929641535359
), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("time", "y", "b"
)), istate = c(2L, 107L, 250L, NA, 5L
Greetings,
I cannot find solution for this problem (I was searching on web, but without
success):
I want to plot dose-response models for one concentration and many responses
(lets say 200) and I don´t want to do it manually.
So I use loop for this:
for (i in mydata[,2:201]){#f
Thanks a million Uwe.
I used the following from a forum to do it and it worked out:
mylevels <-c(-150,-100,-50,-20,-10,-5,-2,-1,1,2,5,10,20,50,100,150)
cols <- jet.colors(length(mylevels) - 1)
customAxis <- function() {
n <- length(mylevels)
y <- seq(min(mylevels), max(mylevels), length.out=n
Here is a reproducible example:
library(vars)
data("Canada")
mymodel <- VAR(Canada, p = 2, type = "const")
myLL<-logLik(mymodel)
AIC(myLL)
Why does logLik(mymodel) say that df=NULL?
Might this be the reason for AIC(myLL) being numeric(0)?
Dimitri
On Tue, May 21, 2013 at 11:17 AM, Prof Brian Ri
Jannis,
Not strange. Try with numbers that can be exactly represented in binary
(that's what computers use), e.g.,
a<-seq(5,70,by=5)
b<-15
a==b
or fractions,
a<-seq(1/32,14/32,by=1/32)
b<-3/32
a==b
Most decimal fractions cannot be represented exactly although a bit of
magic will display t
On 21/05/2013 16:11, Dimitri Liakhovitski wrote:
Sorry, I am using package "vars"
So you need to report the bug in that package to its maintainer.
?logLik says
Value:
Returns an object of class ‘logLik’. This is a number with at
least one attribute, ‘"df"’ (*d*egrees of *f*reedom
That is like complaining that your hammer does not fit these newfangled Philips
screws.
These are different tools. Do not expect them to interoperate.
---
Jeff NewmillerThe . . Go Li
Sorry, I am using package "vars"
On Tue, May 21, 2013 at 11:09 AM, Prof Brian Ripley
wrote:
> On 21/05/2013 16:00, Dimitri Liakhovitski wrote:
>
>> Hello!
>>
>> I am using package "VAR".
>>
>
> What is that? There is no such package on CRAN nor BioC.
>
>
> I've fitted my model:
>> mymodel<-VAR
On 21/05/2013 16:00, Dimitri Liakhovitski wrote:
Hello!
I am using package "VAR".
What is that? There is no such package on CRAN nor BioC.
I've fitted my model:
mymodel<-VAR(mydata,myp,type="const")
I can extract the Log Liklihood for THE WHOLE MODEL:
logLik(mymodel)
How could I calculat
Hello!
I am using package "VAR".
I've fitted my model:
mymodel<-VAR(mydata,myp,type="const")
I can extract the Log Liklihood for THE WHOLE MODEL:
logLik(mymodel)
How could I calculate (other than manually) the corresponding Akaike
Information Criterion (AIC)?
I tried AIC - but it does not take
On 05/18/2013 05:00 AM, r-help-requ...@r-project.org wrote:
hi all
this command used tt function for all variables.
How can i define a different function for each variable?
> exCox.all<-coxph(Surv(SURVT,STATUS) ~
RX+LOGWBC+SEX+tt(RX)+tt(LOGWBC)+tt(SEX),
data=rem.data,tt=function(x,t,...) log
Thank you, that really worked. Actually received an even shorter version:
rowSums((t(apply(D > 0, 1, cumsum)) <= 3) * D)
2013/5/21 Xiao He
> Oops, a couple of missing brackets in the previous reply:
>
>
> foo<-function(x){
> temp=x[x>0]
> if(length(temp)>=3) sum(temp[1:3])
> else sum(te
Please see my answers below.
Best,
Wolfgang
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Moon Qiang
> Sent: Thursday, May 16, 2013 19:12
> To: r-help
> Subject: [R] using metafor for meta-analysis of before-after studies
>
Le mardi 21 mai 2013 à 23:30 +1000, Duncan Mackay a écrit :
> Hi
>
> See par.settings in xyplot
>
> Things are also controlled by
> trellis.par.get()
> to see values
> trellis.par.set()
>
> eg
> xyplot(~Freq|Year, data = sheep2,
> groups = farm,
> par.settings = list(st
That's a good one, Using cumsum + rowsum would definitely be faster,
On Tue, May 21, 2013 at 6:40 AM, José Verhoeven wrote:
> Thank you, that really worked. Actually received an even shorter version:
>
>
> rowSums((t(apply(D > 0, 1, cumsum)) <= 3) * D)
>
>
>
> 2013/5/21 Xiao He
>
>> Oops, a co
Hi,
You could try:
set.seed(24)
mat1<- matrix(sample(0:5,10*20,replace=TRUE),ncol=20)
mat2<- mat1
mat2[2,1:19]<-0
sapply(split(mat2,row(mat2)),function(x) sum(x[x!=0][1:3],na.rm=TRUE))
# 1 2 3 4 5 6 7 8 9 10
# 5 2 12 8 5 14 6 10 10 8
mat3<- mat2
mat3[2,]<- 0
sapply(split(mat3,row(ma
Hi Oihane
lmFit is from the limma package, and the limma package is part of the
Bioconductor project; the Bioconductor mailing list
http://bioconductor.org/help/mailing-list/
(follow the 'Post' link) may be a more appropriate place to post this question
(after searching the archive) if the
Hi
See par.settings in xyplot
Things are also controlled by
trellis.par.get()
to see values
trellis.par.set()
eg
xyplot(~Freq|Year, data = sheep2,
groups = farm,
par.settings = list(strip.background = list(col = "transparent"),
axis.text =
On 21-05-2013, at 13:16, shyam basnet wrote:
> Dear R-Users,
>
> While reading a GDX file from GAMS-software, the R-program does not read the
> string (text) observations. Instead, it assigns some numerical values to each
> text. Do you have some idea on how to read string observations?
>
You could also use:
sapply(seq_len(ncol(mdat)),function(i) mdat[,i]*100/x[i])
# [,1] [,2] [,3] [,4] [,5]
#[1,] 7.142857 13.3 30 36.36364 27.8
#[2,] 78.571429 80.0 130 90.90909 55.6
#[3,] 14.285714 20.0 40 45.45455 33.3
#[4,] 71.428571 60.0
You are absolutely right.
I am storing POSIXct objects into a hash (from the hash package). However, if I
try to get them out as a vector using the values() function, they are
unclassed. And that breaks my (highly vectorized) code. Take a look at this:
> h = hash()
> h[["a"]] = Sys.time()
> st
On May 20, 2013, at 10:35 PM, meng wrote:
> Hi all:
> If the explainary variables are ordinal,the result of regression is different
> from
> "unordered variables".But I can't understand the result of regression from
> "ordered
> variable".
>
> The data is warpbreaks,which belongs to R.
>
> If
Hello,
That's FAQ 7.31.
Note that the increment, 0.05, is not a power of 2 so the third value of
the sequence is not exactly equal to 0.15.
Hope this helps,
Rui Barradas
Em 20-05-2013 13:36, Jannis escreveu:
Dear R users,
I ran into the strange situation where a number does not seem to e
Hi. Check R FAQ, section 7.31 Why doesn't R think these numbers are equal?
http://cran.r-project.org/doc/FAQ/R-FAQ.html
all.equal(a,b)
Andrija
On Mon, May 20, 2013 at 2:36 PM, Jannis wrote:
> Dear R users,
>
>
>
> I ran into the strange situation where a number does not seem to equal its
> v
Hello,
I am using the package arules to generate association rules. I would like
to restrict the rules so that in the left-hand side there's only one
particular element, let's call it "potatoe".
If I do this:
rules <- apriori(dtm.mat, parameter = list(sup = 0.4, conf =
0.9,target="rules"), appea
Dear R users
I have the matrix of the centres of some clusters, e.g. 20 clusters each
with 100 dimentions, so this matrix contains 20 rows * 100 columns numeric
values.
I have collected new data (each with 100 numeric values) and would like to
keep the above 20 centres fixed/'unmoved' whilst jus
Hi! I'm sorry for bothering you. I'm a new R-user and I'm having some problems
while doing a microarray analysis. I'm comparing the whole genome array of a
Salmonella serovar to another 25, and my goal is to determine which genes are
differentially expressed. I'm using limma package and running
Dear R users,
I ran into the strange situation where a number does not seem to equal
its value. Try this:
a <- seq(0.05,0.7,0.05)[3]
b <- 0.15
a == b
Should this not be TRUE? a-b yields a very small number (and not 0) so
this most probably is a numerical error, but why does seq create
Dear R-Users,
While reading a GDX file from GAMS-software, the R-program does not read the
string (text) observations. Instead, it assigns some numerical values to each
text. Do you have some idea on how to read string observations?
Example in GDX file:
fid out year value
1_2_3 RI
Hi!
When inserting R plots into a document using odfWeave, I fought for a
while to get Lattice plots use the same text size as base plots. I
eventually discovered that specifying a point size via e.g.
svg(pointsize=10) has no effect on Lattice plots. One needs to adjust
the size manually via:
trel
Oops, a couple of missing brackets in the previous reply:
foo<-function(x){
temp=x[x>0]
if(length(temp)>=3) sum(temp[1:3])
else sum(temp)
} #
Does this work? Probably not the fastest, but I think it does the job.
foo<-function(x){
temp=x[x>0]
if(length(temp)>=3) sum(temp[1:3])
else sum(temp)
set.seed(2)
mat<-matrix(sample(0:4, 25, replace=T, prob=c(1/2,rep(1/8,4)), ncol=5)
mat
# [,1] [,2] [,3] [,4] [,5]
#[1,]012
Hi Riu,
Very helpful. Thanks a million!
Andrew
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 20 May 2013 23:49
To: Lorentz, Andrew
Cc: r-help@r-project.org
Subject: Re: [R] Gamma curve fit to data with specific bins
Hello,
You are fitting a vector other tha
Hi there,
I've got this matrix D with, say 10 rows and 20 columns. For each row I want
to sum the first 3 non zero elements and put them in a vector z.
So if the first row D[1,] is
0 3 5 0 8 9 3 2 4 0
then I want z
z<-D[1,2]+D[1,3]+D[1,5]
But if there are less than 3 non zero elements, those sho
Thanks for your reply Pascal.
I am presently using it with sweep. But here in the question I just gave
one simple example. In reality I need several functions to run. Thus I
was wondering, if without sweep, I can use loop. Also want to learn how to
do this using loop.
Any help will be really grea
Hi
maybe someone could help
Program:
NS=10
ONS=100
. some declarations
## f. START
. functions definitions
## f. STOP
for (osym in 1:
Hello,
If you *really* need a loop:
Data <- array(NA, dim(mdat))
for(i in 1:ncol(mdat)){
Data[,i] <- mdat[,i]*100/x[i]
}
But I would consider first the thread cited by Berend.
Regards,
Pascal
On 05/21/2013 06:19 PM, Suparna Mitra wrote:
Thanks for your reply Pascal.
I am presently using
Hi All,
I am looking for a package that performs Repeated k-fold cross-validation
with Stepwise Regression.
I would greatly appreciate if someone could share with us a package(s) that
include this type of analysis.
Thank you very much in advance.
Chris
[[alternative HTML version delete
On 21-05-2013, at 09:16, Suparna Mitra wrote:
> Hello R Experts,
> I need a bit of help in using loop.
> I have a matrix onto which I need to use several functions.
>
> In a simplified form suppose my matrix is
>> mdat
> [,1] [,2] [,3] [,4] [,5]
> [1,]12345
> [2,] 11
Hi,
?sweep
mdat <-
matrix(c(1,11,2,10,5,2,12,3,9,6,3,13,4,8,7,4,10,5,9,8,5,10,6,10,4),5,5)
x <- c(14,15,10,11,18)
sweep(mdat*100, 2, x, FUN='/')
[,1] [,2] [,3] [,4] [,5]
[1,] 7.142857 13.3 30 36.36364 27.8
[2,] 78.571429 80.0 130 90.90909 55.6
[3,] 14
Hello R Experts,
I need a bit of help in using loop.
I have a matrix onto which I need to use several functions.
In a simplified form suppose my matrix is
> mdat
[,1] [,2] [,3] [,4] [,5]
[1,]12345
[2,] 11 12 13 10 10
[3,]23456
[4,] 109
I have the same problem you mention here. As there is no answer, did you
find the way to obtain the optimal number of groups? Thank you!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
WIth x, y are vectors, the grid is plotted by surface3d. Thanks for your
answer.
My code is here;
#SurfacePlot is plotted by above surface data
Surface <- read.csv("D:/R/Surface.csv", header=F)
Surface <- as.matrix(Surface)
x <- Surface[2:17,1]
x <- as.numeric(x) # x is vector 1x16
y <- Surface[
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