Dear all. Apologies if I am asking a stupid question, but I have been unable to
find a solution so far.
I would like to run a logistic regression in which individual data points are
assigned different weights (related to my confidence in their validity). These
individual observations are binar
HI users,
Is it possible to check the significance of peaks in a power
spectrum in R
Thanks and regards
nuncio
--
Nuncio.M
Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
ph off 91 832 2525636
ph: cell 91 9890357423
[[altern
On Oct 25, 2012, at 10:32 PM, Santini Silvana wrote:
> Dear R users,
> I have used the following function (in blue)
No, we do not do "in blue" here. This is a monochrome mailing list.
> aiming to find the linear regression between MOE and XLA and nesting my data
> by Species. I have obtained t
On Oct 25, 2012, at 8:49 PM, Rlotus wrote:
> Plz help me ;(( I lost 7 hours for thinking and doing this code ;( I need to
> create points.20 points(dots). All these dots have to be in one graph...
> when i run this codeit gives me only one dote.
You got twenty dots. It's just that th
On Oct 26, 2012, at 03:17 , phantastic wrote:
> How do I generate two bell curves, both with the same mean but different
> standard deviations? I used to have a script for it but I lost it.
>
> Thanks.
curve(, from=..., to=...)
curve(, add=TRUE)
If you plot the tallest one first, you w
On Oct 25, 2012, at 5:14 PM, Rlotus wrote:
> plot(y,x, ylab="sample mean",xlab="sample size")
>
> for (i in seq(1:20))
>> {rsidpVector= rsidp(1)
>> x<-mean(rsidpVector)
>> y<-length(number)
>> print (rsidpVector)
>> print(x)
>
> sorry, but how i can use points function in the loop?
Tested c
Dear R users,
I have used the following function (in blue) aiming to find the linear
regression between MOE and XLA and nesting my data by Species. I have obtained
the following results (in green).
model4<-lme(MOE~XLA, random = ~ XLA|Species, method="ML")summary(model4)
Linear mixed-effects model
One dote..and its a last dote of the loop ;(
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Dear R users,
I need to run 1000 simulations to find maximum likelihood estimates. I
print my output as a vector. However, it is taking too long. I am running 50
simulations at a time and it is taking me 30 minutes. Once I tried to run
200 simulations at once, after 2 hours I stopped it and saw t
Plz help me ;(( I lost 7 hours for thinking and doing this code ;( I need to
create points.20 points(dots). All these dots have to be in one graph...
when i run this codeit gives me only one dote.
number<- (0,2,3,4,5,6,8)
for (i in seq(1:20))
{rsidpVector=rsidp(i)
mean(rsidpVector)
pl
How do I generate two bell curves, both with the same mean but different
standard deviations? I used to have a script for it but I lost it.
Thanks.
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Hey David,
my answers are delayed here, although I am not using my gmail email
address:)
Yep thats right, those bands of zeros are one of the most important values
to define one group, and have a nice distance from the rest of the groups
:). I cannot really get rid of those, I bet it would not h
Hey Bert,
thanks for your fast reply. Yes, based on svd it is singular. The "no way"
statement was because of the source of the dataset. I would not expect that.
I never used the stats Maha dist calc, but after giving it a shot, not a
surprise still singular.
Any idea how to manipulate the data
plot(y,x, ylab="sample mean",xlab="sample size")
for (i in seq(1:20))
> {rsidpVector= rsidp(1)
> x<-mean(rsidpVector)
> y<-length(number)
>print (rsidpVector)
> print(x)
sorry, but how i can use points function in the loop?
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If they have the same number of rows, you can use cbind() to create one object
to write out.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
A general question that I have been pursuing for some time but have set
aside. When finishing some analysis, I can have multiple matrices that
have specific column names. Ideally, I would like to combine these
separate matrices for a final output as a csv file.
A generic example:
Matrix 1
var1A
On Fri, Oct 26, 2012 at 12:14 PM, langvince wrote:
> Whatever I do, however I subset it I get the "system is computationally
> singular: reciprocal condition number" error.
> I know what it means and I know what should be the problem, but there is no
> way this is a singular matrix.
>
> I have up
You can also just replicate() as follows:
# x is the vector
# s is the size of the sample
# B is the number of samples
# ... arguments passed to sample()
f <- function(x, s, B, ...) replicate(B, sample(x, s, ...))
f(x, 3, 10, TRUE)
f(x, 3, 10, FALSE)
HTH,
Jorge.-
On Fri, Oct 26, 2012 at 12:05
Hello,
The other options is to use the sample() function.
test2 <- matrix (rep(sample(number1, size = 5), times=3), nrow=3)
Pradip Muhuri
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Rui Barradas [ruipbarra...@sapo.pt
On Oct 25, 2012, at 4:41 PM, Bert Gunter wrote:
> 1. I don't know what StatMatch is. Try using stats::mahalanobis.
>
> 2. It's the covariance matrix that is **numerically** singular and
> can't be inverted. Why do you claim that there's "no way" this could
> be true when there are hundreds of va
On Oct 25, 2012, at 2:45 PM, Johnson, Franklin Theodore wrote:
> Hello R-help,
>
>
>
> I am using R version 2.15.1.
>
> I upgraded from R version 2.13 a few months back.
>
>
>
> Previously, I was able to plot error bars on an xy scatter plot using the
> errbar function:
>
> errbar(RAEthy
Hi Arun,
Thank you ! Very much appreciated.
[Only added t() at the end to preserve the original orientation.]
Also, many thanks to Rui Barradas.
Cheers,
Ved
On Thu, Oct 25, 2012 at 12:06 PM, arun wrote:
>
>
> Hi Ved,
>
> Sorr
On Oct 25, 2012, at 2:24 PM, Rlotus wrote:
> I have his code. I wanna draw the graph according to x and y. But at the end
> when I run the code it gives me only one dot...it is strange cuz from
> the loop it has to give 20 dots. Help me plz to draw a graph(((
>
> for (i in seq(1:20))
> {rsid
1. I don't know what StatMatch is. Try using stats::mahalanobis.
2. It's the covariance matrix that is **numerically** singular and
can't be inverted. Why do you claim that there's "no way" this could
be true when there are hundreds of variables (= dimensions).
3. Try calculating the svd of your
Hello,
You don't need the loop, the sample() argument 'size' is there for that.
See 'sample.
number <- c(0,1,3,4,5,6,8)
rsidp <- function(n) sample(number, n, replace = TRUE)
rsidp(5)
Hope this helps,
Rui Barradas
Em 25-10-2012 20:24, Rlotus escreveu:
I wanna generate random numbers from a
Hi folks,
I know, this is a fairly common question and I am really disappointed that I
could not find a solution.
I am trying to calculate Mahanalobis distances in a data frame, where I have
several hundreds groups and several hundreds of variables.
Whatever I do, however I subset it I get the "s
Hello R-help,
I am using R version 2.15.1.
I upgraded from R version 2.13 a few months back.
Previously, I was able to plot error bars on an xy scatter plot using the
errbar function:
errbar(RAEthylene$TIME,RAEthylene$AVE,RAEthylene$AVE+RAEthylene$STD,RAEthylene$AVE-RAEthylene$STD,add
= T
I have code in R. I need also demo code of my output and demo code of code. I
dont know what is demo. An d how to create it...can you tell how to do that
plz. thank you.
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I have his code. I wanna draw the graph according to x and y. But at the end
when I run the code it gives me only one dot...it is strange cuz from
the loop it has to give 20 dots. Help me plz to draw a graph(((
for (i in seq(1:20))
{rsidpVector= rsidp(1)
x<-mean(rsidpVector)
y<-length(number
thank u so much! i got it.
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Hi,
Try this:
number1<-c(0,1,3,4,5,6,8)
rsidp<-function(x){
y<-sample(x,5,replace=TRUE)
y
}
rsidp(number1)
#[1] 3 0 6 8 4
rsidp(number1)
#[1] 1 8 8 6 4
rsidp(number1)
#[1] 8 3 6 6 6
A.K.
- Original Message -
From: Rlotus
To: r-help@r-project.org
Cc:
Sent: Thursday,
You are very lucky. This question was answered last week in
https://stat.ethz.ch/pipermail/r-help/2012-October/326597.html
For the future, please read the posting guide referenced below and include
an example of your
data and the result of the calculation you would like to see.
Please use R-help
hello,
I want to make a two dimensional matrix from the data table.That data
table has three column 1st is userid,2nditemid and 3rd is rating
corresponding to the itemid given by userid.I want to make a m*n matrix in
which 'm' is userid and 'n' to be itemid .That 2D matrix should b filled
with
Hi Rui,
I thought the OP was looking for something like this: May be I am wrong.
dat2<-dat1[order(dat1$Trip_id,dat1$Vessel,dat1$CommonName,dat1$Length,dat1$Count),]
dat3<-dat2
dat3$Prop<-unlist(tapply(dat3$Count,list(dat3$Trip_id,dat3$CommonName),function(x)
x/sum(x)))
head(dat3)
# Trip_id
n <- 10; # Sample size
fitglm <- function(sigma,tau){
x <- rnorm(n,0,sigma)
intercept <- 0
beta <- 0
ystar <- intercept+beta*x
z <- rbinom(n,1,plogis(ystar))
xerr <- x + rnorm(n,0,tau)
model<-glm(z ~ xerr, family=binomial(logit))
int<-coef(model)[1]
slope<-coef(
HI,
In my previous solution, the order got messed up. I should have ordered the
columns.
Try this:
dat1<-read.table(text="
Trip_id Vessel CommonName Length Count
1 230 Sunlight ShadAmerican 19 1
2 230 Sunlight ShadAmerican 20 1
3 230
Hi Jean - Thank you for your help. The code worked great.
- Original Message -
From: "Jean V Adams [via R]"
To: "Sally_roman"
Sent: Thursday, October 25, 2012 2:48:45 PM
Subject: Re: trying ti use a function in aggregate
Sally,
It's great that you provided data and code. To make
Hi,
Did you mean this?
group1<-c(40,50,"60",70)
#or
group2<-c(50,"var1","var2",60)
In either of the above cases, when you check
str(group1) # all are converted to character.
# chr [1:4] "40" "50" "60" "70"
str(group2)
# chr [1:4] "50" "var1" "var2" "60"
Suppose, I am comparing the test1 datase
I wanna generate random numbers from a vector...
for example number<-c(0,1,3,4,5,6,8)
so
rsidp<-function(x){
i=0
for (i in seq(1:x))
{y<-sample(number,x, replace=T)}
return(y)
}
so all random numbers have to be from vector "number";
so if I type rsidp(5). i
David Winsemius wrote:
>
> On Oct 24, 2012, at 2:14 AM, Johannes Graumann wrote:
>
>> Hello,
>>
>> testclass <- setRefClass(
>> "testclass",
>> fields = list(testfield = "logical"),
>> methods = list(validate=function(){testfield<<-TRUE}))
>>
>>> test <- testclass$new()
>>> test$testfield
>>
You might have better luck posting on stats.stackexchange.com, a
statistical help list.
-- Bert
On Thu, Oct 25, 2012 at 1:08 PM, Keith Weintraub wrote:
> Folks,
> I am working on a credit card defaults and transition probabilities. For
> example a single credit card account could be in a number
> n <- 10; # Sample size
>
> fitglm <- function(sigma,tau){
+ x <- rnorm(n,0,sigma)
+ intercept <- 0
+ beta <- 0
+ ystar <- intercept+beta*x
+ z <- rbinom(n,1,plogis(ystar))
+ xerr <- x + rnorm(n,0,tau)
+ model<-glm(z ~ xerr, family=binomial(logit))
+ int<-coef(model)[1]
+ slope<-coef(model)[2]
+ p
Mohamed Radhouane Aniba gmail.com> writes:
>
> Which one ? :)
>
https://stat.ethz.ch/mailman/listinfo/r-help
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/
Folks,
I am working on a credit card defaults and transition probabilities. For
example a single credit card account could be in a number of states:
up-to-date, 30, 60, 90 days in arrears or in default.
* Are there packages in R that do estimation of the transition probabilities
given historica
Your code is still not runnable.
It gives the error message
Error in fitglm(0.05, 1) : could not find function "prediction"
Berend
On 25-10-2012, at 21:55, Adel Powell wrote:
> I think I have corrected it. Can you tell me are my spec and sens values
> correct
> n <- 1000; # Sample size
>
>
On 25-10-2012, at 21:28, Adel Powell wrote:
> I am running my code in a loop and it does not work but when I run it
> outside the loop I get the values I want.
>
> n <- 1000; # Sample size
>
> fitglm <- function(sigma,tau){
>x <- rnorm(n,0,sigma)
>intercept <- 0
>beta <- 0
>ysta
I am running my code in a loop and it does not work but when I run it
outside the loop I get the values I want.
n <- 1000; # Sample size
fitglm <- function(sigma,tau){
x <- rnorm(n,0,sigma)
intercept <- 0
beta <- 0
ystar <- intercept+beta*x
z <- rbinom(n,1,plogis(ystar))
x
Hi,
Try this:
dat1$Percent<-unlist(tapply(dat1$Count,list(dat1$Trip_id,dat1$CommonName),function(x)
x/sum(x)))
head(dat1)
# Trip_id Vessel CommonName Length Count Percent
#1 230 Sunlight ShadAmerican 19 1 0.01408451
#2 230 Sunlight ShadAmerican 20 1 0.01408451
#3
Sally,
It's great that you provided data and code. To make it even more
user-friendly for R-help readers, supply your data as Rcode, using (for
example) the dput() function.
The reason you were getting all 1s with your code, is that you had told it
to aggregate by trip, LENGTH, and species.
On Oct 25, 2012, at 7:30 PM, Mohamed Radhouane Aniba wrote:
> Hello folks,
>
> I am currently receiving a lot of emails from the list which proves that this
> is a very important place to get good feedbacks and tips and that the
> community is here to help .. Excellent thing.
> I am not thou
Or using ddply from plyr,
library(plyr)
myframe <- data.frame (ID=c("Ernie", "Ernie", "Ernie", "Bert", "Bert",
"Bert"), Timestamp=c("24.09.2012 09:00", "24.09.2012 10:00", "24.09.2012
11:00"), Hunger=c(1,1,1,2,2,1) )
myframe
myframestime <- as.POSIXct (strptime(as.character(myframe$Timestamp),
"%
Which one ? :)
On Oct 25, 2012, at 2:34 PM, "R. Michael Weylandt "
wrote:
>
>
> On Oct 25, 2012, at 7:30 PM, Mohamed Radhouane Aniba
> wrote:
>
>> Hello folks,
>>
>> I am currently receiving a lot of emails from the list which proves that
>> this is a very important place to get good fe
Hi,
May be this helps:
dat1<-read.table(text="
Trip_id Vessel CommonName Length Count
1 230 Sunlight ShadAmerican 19 1
2 230 Sunlight ShadAmerican 20 1
3 230 Sunlight ShadAmerican 21 1
4 230 Sunlight ShadAm
Hello,
Try the following.
(I've changed your function a bit. And named the data.frame 'dat', not
'data', which is an R function.)
myfun <- function (x) ifelse(sum(x) == 0, 0, x/sum(x))
aggregate(Count ~ Trip_id + Length + CommonName, data = dat, myfun)
The output shows that each and every gr
Hello folks,
I am currently receiving a lot of emails from the list which proves that this
is a very important place to get good feedbacks and tips and that the community
is here to help .. Excellent thing.
I am not though able to login to my subscriber space to change the email
reception into
On Oct 25, 2012, at 10:56 AM, jim holtman wrote:
> Here is a function I use to get the size of objects:
>
> Here is an example output:
>
>> my.ls()
> Size Mode
> allStores 7,303,224 list
> convertedStores 0 NULL
> f.createCluster40,508 function
Here is a function I use to get the size of objects:
Here is an example output:
> my.ls()
Size Mode
allStores 7,303,224 list
convertedStores 0 NULL
f.createCluster40,508 function
x 41,672 list
**Total 7,385,404 -
> > I want to add an additional column at the right and get in
> each row a value which shows the mean of "hunger" of the last two hours.
Isn't that just a moving average?
if so, see
http://stackoverflow.com/questions/743812/calculating-moving-average-in-r,
which mentions zoo for RollingMeans,
Marte Lilleeng gmail.com> writes:
>
[snip]
> I have a simple model that i would like to plot with 95% CIs.
> It is like follows:
> m1<-lmer(Richness~Grazing+I(Grazing^2)+(1|Plot),family=poisson)
>
> By using the effects package I get two plots, one for the linear term
> and one for the squ
Hi,
For the sake of completeness, I thought I might follow up my email
from earlier in the month with what I found out by digging into the
code. The answer to my earlier question perhaps should have been
obvious to me from the "Usage" description from the documentation.
In the three examples I o
Hello,
Or
ct <- cut(myframe2$myframestime, breaks = "2 hour")
ave(myframe2$Hunger, ct)
And assign the output of 'ave' to a new column.
Hope this helps,
Rui Barradas
Em 25-10-2012 15:50, arun escreveu:
Hi,
May be this helps:
new1<-with(myframe2,aggregate(cbind(Hunger,myframestime),by=list(
On Oct 25, 2012, at 6:00 AM, S Ellison wrote:
>
>
>> You typed a ` without closing it:
>> barplot(xtab(`profits`,data=Forbes2000))
>>
>> anyway: pushing the escape button should also return you to
>> the R-prompt (at least on a Windows platform)
Also on Macs.
>
> But why is the OP using a
On Oct 25, 2012, at 8:00 AM, Dheeraj Pandey wrote:
> How Can I run all these codes in VBA using RExcel
> library(rugarch)
I predict there are a very small number (possibly zero) of persons reading this
list that are both Excel coders and users of that package. You might consider
posting it on
It seems you don't quite understand how foreach works. foreach (..)
%dopar% { ... } takes the last value from each of the second {...}
evaluations and feeds them to the .combine function (in your case
rbind()). Since your last call in the %dopar% {...} block is assign(),
you are not getting anythin
Hi,
May be this helps:
new1<-with(myframe2,aggregate(cbind(Hunger,myframestime),by=list(ID=ID),
function(x) mean(tail(x,2[,1:2]
merge(myframe2,new1,by="ID",all=TRUE)
# ID Hunger.x myframestime Hunger.y
#1 Bert 2 2012-09-24 09:00:00 1.5
#2 Bert 2 2012-09-24 10:
Hello,
I wrote a request yesterday on how to select a random observation that met
certain criteria. I think mostly I have it figured out, but I can't figure
out how to append the observations. The output just has one observation in
it.
Any help would be appreciated,
thanks.
Christy
setwd ("C:
How Can I run all these codes in VBA using RExcel
library(rugarch)
spec=ugarchspec(variance.model=list(model="sGARCH",garchOrder=c(1,1)),
mean.model=list(armaOrder=c(1,1), arfima=FALSE), distribution.model="std")
fit=ugarchfit(data=d, spec=spec)
z=sigma(fit)
Dheeraj
[[alternative HT
As David said, you sorted by Date.
But sorting by rownames is not really the point. The point is that
rownames are not line numbers.
The rownames were assigned when the data frame was created, and then
preserved when you sorted. Sometimes, rownames contain meaningful
information that is associat
Hi -I am using R v 2.13.0. I am trying to use the aggregate function to
calculate the percent at length for each Trip_id and CommonName. Here is a
small subset of the data.
Trip_id Vessel CommonName Length Count
1 230SunlightShad,American 19 1
2 2
So easy solution..., thank you very much :)
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Hi
I want to implement Egarch (1,1) with t distribution model using RExcel and VBA.
May I know the syntax.
Following is the code that I 'm using.
rinterface.RRun
"spec=ugarchspec(variance.model=list(model=(eGARCH),garchOrder=c(1,1)),
mean.model=list(armaOrder=c(1,1), arfima=FALSE), distribution.
Hello,
I have a data frame somewhat like that:
myframe <- data.frame (ID=c("Ernie", "Ernie", "Ernie", "Bert", "Bert",
"Bert"), Timestamp=c("24.09.2012 09:00", "24.09.2012 10:00", "24.09.2012
11:00"), Hunger=c(1,1,1,2,2,1) )
myframestime <- as.POSIXct (strptime(as.character(myframe$Timestamp),
"%
Hi everyone!
I have a simple model that i would like to plot with 95% CIs.
It is like follows:
m1<-lmer(Richness~Grazing+I(Grazing^2)+(1|Plot),family=poisson)
By using the effects package I get two plots, one for the linear term
and one for the squared term.
Q1: Can I get all in one? I.e. with one
Please use this model to send your data and code
> mydata <- data.frame(x, y, fx, fy)
> head(mydata) ## just the first 6 rows are displayed on screen
x y fx fy
1 4 3 37 40
2 4 3 21 20
3 0 0 65 50
4 12 0 91 49
5 1 5 22 20
6 2 3 6 10
> dump("mydata", "")
mydata <-
structure(list(x = c(4, 4
With effort I re-inserted the line feeds in your code that the
pdf file deleted.
There is an error in your code
Error in 0:(x[t == i][l]) (from #9) : NA/NaN argument
>
> x[t == i]
[1] 8 2 9
> l
[1] 4
>
On Thu, Oct 25, 2012 at 7:45 AM, Loukia Spineli
wrote:
> x and y are the frequencies of missi
Thank you very much Berend!:)
On Thu, Oct 25, 2012 at 4:37 PM, Berend Hasselman wrote:
> Loukia,
>
> Please send your replies to the R-help list. I am CC'ing this to the
> R-help list.
>
> On 25-10-2012, at 15:30, Loukia Spineli wrote:
>
> > Dear Berend,
> >
> > I am not familiar at all with as
Loukia,
Please send your replies to the R-help list. I am CC'ing this to the R-help
list.
On 25-10-2012, at 15:30, Loukia Spineli wrote:
> Dear Berend,
>
> I am not familiar at all with ascii text. If you could suggest me a good link
> to learn about it, then it would be really helpful!
If t
You were asked to provide R code in ascii text.
Also use dput to get the objects into a mail.
Nobody is going to make the effort to copy from your pdf. I certainly will not.
Berend
On 25-10-2012, at 13:45, Loukia Spineli wrote:
> x and y are the frequencies of missing participants in the inter
> You typed a ` without closing it:
> barplot(xtab(`profits`,data=Forbes2000))
>
> anyway: pushing the escape button should also return you to
> the R-prompt (at least on a Windows platform)
But why is the OP using a backtick at all? It's not necessary in this instance
(and in fact it's ver
Please use dput while supplying data.
Best Regards,
Bhupendrasinh Thakre
Sent from my iPhone
On Oct 25, 2012, at 5:31 AM, Soheila Khodakarim wrote:
> var1
>
> var2
>
> var3
>
> var4
>
> var5
>
> var6
>
> var7
>
> var8
>
> var9
>
> var10
>
> gold
>
> 2
>
> 3
>
> 1
>
> 2
>
> 4
>
This could well be out of date because I have not paid any attention to the
official POLR code in a year or more. Attached is fixed-polr.R.
cheers,
Tim
On Wed, Oct 24, 2012 at 10:38 PM, ahs [via R] <
ml-node+s789695n4647311...@n4.nabble.com> wrote:
> Great!
> You can skip my question about s0 th
... and note that the "{" ,"}" are unnecessary:
> plot(1, main = bquote(f^-1*(x)))
-- Bert
On Thu, Oct 25, 2012 at 4:32 AM, Rui Barradas wrote:
> Hello,
>
> Use an asterisk to put a space between the exponent an (x).
>
> plot(1, main = expression(f^{-1}*(x)))
>
> Hope this helps,
>
> Rui Barr
Hi,
You can also use lapply() or sapply() without the get() function.
list1<-list(x=x,y=y)
sapply(list1,object.size)
# x y
#80040 840
do.call(rbind,lapply(list1,object.size))
# [,1]
#x 80040
#y 840
A.K.
- Original Message -
From: Jorge I Velez
To: Purna chander
No, you didn't miss anything in the survival package. I've never found post-hoc tests
interesting so have little motivation to add such (and a very long "to do" list of things
I would like to add).
If you simply must have them, why not do all pairwise tests?
chisq <- matrix(0., 4,4)
for (i in
x and y are the frequencies of missing participants in the intervention and
the control treatment respectively! vector t contains the code of the
interventions (we have 11 interventions). I re-attach the PDF with some
small modifications. I am trying to create a list, where each list element
is a v
HI Petr,
Thanks for sharing the function. True, very efficient than cut.
dat1$cat<-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T)
^^^
# May be the interval is 0:6.
dat1$cat1<-findInterval(dat1$V1, 1:6, rightmost.closed = T, al
Hello,
Use an asterisk to put a space between the exponent an (x).
plot(1, main = expression(f^{-1}*(x)))
Hope this helps,
Rui Barradas
Em 25-10-2012 12:21, stat.kk escreveu:
I would like to have an expression f^(-1)(x) in a legend of plot. For this I
used expression(f^{-1}(x)), but variabl
Hi Stuart,
So, I guess my result (below) serves the purpose!
A.K.
- Original Message -
From: Stuart Leask
To: arun
Cc:
Sent: Thursday, October 25, 2012 3:13 AM
Subject: RE: [r] How to pick colums from a ragged array?
Even confusing myself now, serves me right for replying late at n
Dear Berend,
Many thanks for the advice. I'll try to work with these packages and I hope
to find a solution.
again, many thanks
Pina.
--
View this message in context:
http://r.789695.n4.nabble.com/equation-solver-tp4647287p4647388.html
Sent from the R help mailing list archive at Nabble.com.
Hi,
I am trying to parallel computing with foreach function, but not able to
get the result. I know that in parallel processing, all result is collected
in list format, but I am not able to get input there.
Any help is really appreciated.
esf.m <-foreach (i = 1:n.s, .combine=rbind) %dopar% {
Hi
can you please tell me how to quit R script & return to R prompt.
As i tried following but still cannot able to return on to R prompt..
R> barplot(Forbes2000$profits)
R> barplot(xtab(`profits,data=Forbes2000))
+ barplot(xtab(~profits,data=Forbes2000))
+ )
+ Q()
+ ?barplots
+
+
+
+
+ barplo
I would like to have an expression f^(-1)(x) in a legend of plot. For this I
used expression(f^{-1}(x)), but variable is always in exponent. How could I
change it in order to (x) be in a line, not in exponent?
Thank you for your responses!
--
View this message in context:
http://r.789695.n4.na
you can see it goes wrong at:
> barplot(xtab(`profits,data=Forbes2000))
You typed a ` without closing it:
barplot(xtab(`profits`,data=Forbes2000))
anyway: pushing the escape button should also return you to the R-prompt (at
least on a Windows platform)
--
View this message in context:
ht
Sorry, forgot to cc to rhelp
Petr
> -Original Message-
> From: PIKAL Petr
> Sent: Thursday, October 25, 2012 11:19 AM
> To: 'Stuart Leask'; arun (smartpink...@yahoo.com)
> Subject: RE: [r] How to pick colums from a ragged array?
>
> Hi
>
> If I understand correctly you now want only to
On 10/25/2012 05:06 AM, bwone wrote:
I am using the package plotrix radial.plot(). Yes, radial.plot() has a line
type argument, lty, but that is for the polygons or the radial lines, not
the radii or axes of the radial plot.unless I am doing something wrong.
Hi bwone,
No, there is no way to
var1
var2
var3
var4
var5
var6
var7
var8
var9
var10
gold
2
3
1
2
4
0
1
4
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2
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2
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2
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0
0
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0
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0
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3
Yes, but you'll need to learn vector (multivariate) time series methods. See,
perhaps firstly, B Pfaff's book and corresponding R packages. It's dense, but
not too long and will get you going the right way. Terms like VAR and VECM will
help guide your googling
Michael
On Oct 25, 2012, at 8:34
Thank you Dr. Pikal for this alternative.
Best,
Jorge.-
Sent from my phone. Please excuse my brevity and misspelling.
On Oct 25, 2012, at 8:29 PM, PIKAL Petr wrote:
> Hi
>
> Maybe also findInterval can be used
>
> dat1$cat<-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T)
>
> It
Dear All,
My main objective was to compute the distance of 10 vectors from a
set having 900 other vectors. I've a file named "seq_vec" containing
10 records and 256 columns.
While computing, the memory was not sufficient and resulted in error
"cannot allocate vector of size 152.1Mb"
So I
Hi
Maybe also findInterval can be used
dat1$cat<-findInterval(dat1$V1, 1:6, rightmost.closed = T, all.inside = T)
It is said to be more efficient than cut.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of aru
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