Hello, R experts
I am a new user of rgl package.
I want axis label to be parallel to the axis.
However, I cannot find any options in package manual.
Could you show me how to how to make label be parallel to crresponding axis.
Especially, Z axis label, in case that it is long, protrudes out of the p
hi, i am a newbie to R
using rocr package to find out accuracy of a model using rpart
> perf.acc <- max(cust.rp.perf.acc at y.values[[1]])
Error: unexpected symbol in ;perf.acc plot(cust.rp.perf.acc)
>plot(cust.rp.perf.acc)
how to interpret this graph on x-axis cutoff values and on y
-values-accur
Hi Qamar,
I guess you are looking for row means vs. days plot.
If that is the case, try this:
datQ<-cbind(c(6,5,6,7,4,3,5),c(2,4,35,32,4,6,6),c(2,4,2,3,423,4,5),c(2,3,13,5,3,5,3))
datQmean<- apply(datQ,1,mean)
[1] 3.00 4.00 14.00 11.75 108.50 4.50 4.75
datQmean<-data.frame(datQmean)
d
On Jun 1, 2012, at 4:09 PM, jcrosbie wrote:
I can not get the point function to work with in my code.
The code:
AggOfTempMatrix$CumSumPercentSize<-c(
0.05265450, 0.05738490, 0.05865403, 0.05911553, 0.05957703,
0.06003854,
0.06058368,
0.06098750, 0.06147208, 0.06187589, 0.06291427, 0.0633
hi peter,
another question for you, if you are willing. well actually, both this
question and the question i just asked Bert are for anybody willing to
answer! I very much appreciate your opinions!
My dependent variable is an annual total. However most (tho not all) of the
250+ variables are at a
Bert, if you have caught your breath yet (i would insert a smilely here, but
i hate smileys), can i ask you another question?
I see your affiliation has the word Genotech in it. Of all the things i have
been reading about, it seemed my best hope could be in following tools
statistical/data dredgi
Hello! I'm collecting data on a refrigerator that I'm using to cure
meat. Specifically I am collection humidity and temperature readings.
The temperature readings look sinusoidal (due to the refrigerator
turning on and off).
I'd like to calculate the frequency and period of the wave so that I ca
Dear R users,
I have weekly data in the following manner
[,1][,2][,3][,4]
6 2 2 2
5 4 4 3
6 35 2 13
7 32 3 5
4 4 423 3
3 6 4 5
5 6 5 3
I drew curve of each colum
Could somebody look at this? I'm new to ggplot, so any help would be greatly
appreciated. I provided an example to facilitate. Thanks!
dadrivr wrote
>
> I'm fitting a lme growth curve model with two predictors and their
> interaction as predictors. The multilevel model is nested so that level
Hello,
I am looking to scrape this Webpage:
http://toast.gasunie.de/gud/search.aspx?soid=GUD&lang=de
The page uses the method "POST", it contains various HTML Forms, mostly
lists and a couple of radio buttons. After submit, I should get forwarded to
a new page. Which selections are being made in
I can not get the point function to work with in my code.
The code:
AggOfTempMatrix$CumSumPercentSize<-c(
0.05265450, 0.05738490, 0.05865403, 0.05911553, 0.05957703, 0.06003854,
0.06058368,
0.06098750, 0.06147208, 0.06187589, 0.06291427, 0.06331808, 0.06354884,
0.06408533,
0.06489296, 0.07
Is there away of putting an excel style gradient background? I want to have
dark blue in the middle and shad to white on the top and bottom.
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Hi,
Try this:
DF1<-DF[with(DF,rev(order(Company,Salary))),]
DF2<-within(DF1,rank<-ave(Salary,Company,FUN=function(x)rev(order(x
> DF2
Company Person Salary rank
15 Toyota O 68278.808 1
5 Toyota E 57487.220 2
11 Toyota K 56090.075 3
12 Toyota L 52602.77
Thank you, works fine.
Regards,
Phil
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Hi,
I have encounter a problem that my data a_data.frame is of data.frame class
with columns: sequenceID, eventID, items.
when I convert it to transaction class:
as(a_data.frame, "transactions")
All columns are grouped together under items, which is like this:
{sequenceID, eventID, items}
inst
Sorry, I meant to put that it didn't work for "GM", not "Toyota".
> DF[DF$Company=="GM",]
Company PersonSalary rank
1 GM A 7067.9054
3 GM C 31627.1718
5 GM E 66200.5087
7 GM G 91287.5923
10 GM J 33239.4676
13 GM
Ah. I'm very new to this and I definitely wasn't clear enough about what I'm
trying to do... sorry.
I am extracting information (in this case, DNA sequences to be exported to a
fasta file) from just two columns of a table that has a lot of extraneous...
stuff. I want a prompt that asks the user
Before using ddply, try adding an id variable to uniquely identify each
record (this is a good data integrity practice anyway). Then you can simply
create the new data frame by using all the ids that aren't in your
'To_remove' subset.
Here's the code for your example:
library(plyr)
library(outlie
Many, many blank lines deleted.
On Jun 1, 2012, at 2:53 AM, Akkara, Antony (GE Energy, Non-GE) wrote:
Hi iLai,
What you showed below, almost same like I am also expecting.
There is two matrix,
1) 1st - matrix contain values like this,
ABC
(Dear farmedgirl,
I agree with Bert's assessment: you're just plain out of luck
until you either get more (a lot more) data or reduce the number of
candidate predictors on some scientifically reasonable basis.
But here are a couple of additional comments:
1. with regsubsets()'s default of nvmax=
On Fri, Jun 1, 2012 at 5:23 PM, David Winsemius wrote:
>
> On Jun 1, 2012, at 2:37 PM, Sarah Goslee wrote:
>
>> There are several ways. The easiest to understand is probably using
>> if() statements: see ?if for help and examples.
>>
>> Sarah
>>
> I would have thought ifelse() to be the necessary
On Jun 1, 2012, at 2:37 PM, Sarah Goslee wrote:
There are several ways. The easiest to understand is probably using
if() statements: see ?if for help and examples.
Sarah
I would have thought ifelse() to be the necessary function, but for
such simple cases I find boolean math to be clearer.
Try using the rank function instead of the order function.
rank(x) is order(order(x)) if there are no ties.
Since you want reverse ranks do either rank(-x) or length(x)+1-rank(x).
E.g.,
> DF <- within(DF, rank2 <- ave(Salary, Company, FUN=function(x)rank(-x)))
> DF[DF$Company=="Toyota",]
C
Just to throw out an alternative using the ?reshape function:
# See the example using 'df3' on the help page
DF.Long <- reshape(score, direction = "long", idvar = "subject",
varying = 2:8, sep = "")
> head(DF.Long, 24)
subject timetest
ab.1 ab1 0.17687
cl.1
On Jun 1, 2012, at 12:27 PM, pigpigmeow wrote:
Dear all,
I have a lot of problems on R-programming.
for example
my csv. file is ..
That certainly does not look like any csv file I have ever seen.
Date wrfRH wrfsolar wrfwindspeed wrfrain wrftd wrfta
21/10/2010 92.97 22.11 53
Hello,
Try
score2$subject <- rep(score$subject, 7)
Hope this helps,
Rui Barradas
Em 01-06-2012 20:47, Jason Love escreveu:
Hello R users,
I'd like to ask a question about how to add a new column. So, below is my
situation.
In order to perform the repeated ANOVA, I first imported the followi
Hello R users,
I'd like to ask a question about how to add a new column. So, below is my
situation.
In order to perform the repeated ANOVA, I first imported the following
table.
score=read.csv("patients_tests.csv");
subject test1 test2 test3 test4test5test6
test7
1
Hi Sarah,
That I was making things too complicated doesn't surprise me. A skilled
programmer makes everything look easy I think. And someone who is still
learning does just the opposite.
Am going to spend some time now looking through your tweaks.
Thank you very much for your help.
Paul
Hi Paul,
I think you're making it far too complicated. With some minor tweaking
to your function, I can easily process the entire data frame you
originally presented.
nearTerms <- function(rawtext, target, before, after){
Text <- unlist(strsplit(rawtext, " "))
Target <- grep(target, Text)
Hello,
I am trying to obtain the partial r-square values (r^2 or R2) for
individual predictors of an outcome variable in multiple linear
regression. I am using the 'lm' function to calculate the beta
coefficients, however, I would like to know the individual %
contributions of several indepenent v
Hello Florian,
The best fit only depends on the relative statistical errors. The
estimated parameter error is a purely statistical error and can be
estimated from the sample. Systematic parameter errors are not
estimated. Yes, parameter errors grow with measurement errors.
Take care
Yes, that did it. Thanks.
*Ben Caldwell*
On Fri, Jun 1, 2012 at 11:41 AM, Sarah Goslee wrote:
> Hi,
>
> Look at this line:
>
> cross.val.error<-append(cross.val.error.temp,value.temp)
>
> at each iteration you're overwriting cross.val.error with
> cross.val.error.temp and value.temp
>
> You pr
Hello Bert and Sarah,
Thank you for your replies. Helped me understand how people might perceive my
question and why they might not respond.
Spent some time learning about R's debugging tools this morning. Began to
realize why my function didn't work. My second argument was the name of a
vari
Hi,
not sure this is the right mailing list, but anyway - I maintain the
WGCNA package, and was just alerted by a Mac user that it is not
available. Looking at the error log at
http://www.r-project.org/nosvn/R.check/r-release-macosx-ix86/WGCNA-00check.html
reveals
checking package dependencies
Hi,
Look at this line:
cross.val.error<-append(cross.val.error.temp,value.temp)
at each iteration you're overwriting cross.val.error with
cross.val.error.temp and value.temp
You probably actually need something like
cross.val.error.temp <- append(cross.val.error.temp,value.temp)
print(cross.va
Hello all,
Let me first say that this isn't a question about outliers. I am using
the outlier function from the outliers package but I am using it only
because it is a convenient wrapper to determine values that have the
largest difference between itself and the sample mean. Where I am
running int
There are several ways. The easiest to understand is probably using
if() statements: see ?if for help and examples.
Sarah
On Fri, Jun 1, 2012 at 11:34 AM, jfca283 wrote:
> Hi
> I need to do something very simple. I have 2 variables, Y and M. I need to
> multiply Y by 1 if M=1, by 2 if M=3 and by
Hi,
On Fri, Jun 1, 2012 at 12:27 PM, pigpigmeow wrote:
> Dear all,
> I have a lot of problems on R-programming.
> for example
> my csv. file is ..
> Date wrfRH wrfsolar wrfwindspeed wrfrain wrftd wrfta
> 21/10/2010 92.97 22.11 53.27 0 1546.337861 61.00852664
> 22/10/2010 87.35 21
Hi,
On Fri, Jun 1, 2012 at 12:05 PM, pigpigmeow wrote:
> Thank, it OK now
> but I don't understand what is the meaning of y <- y[!is.na(y[5]),]
> Thank in advance!
This is very basic R syntax. You should definitely read the
Introduction to R, and also
?"["
?"!"
?is.na
Sarah
--
Sarah Goslee
ht
Hello all,
*
*
I'm having some difficulty, and I think the problem is with how I'm using
append() nested inside a for loop. The data are:
y,x
237537.61,873
5007.148438,227
17705.77306,400
12396.64369,427
228703.4021,1173
350181.9752,1538
59967.79376,630
140322.7774,710
42650.07251,630
5382.858702,
Frank -- where are you?!
(To the OP: Your post leaves me simply breathless. You are embarked on
a fool's errand. Filoche's "help" will continue you down that path.
IMHO only of course.
Bottom line: You CANNOT do what you wish to do. Or to quote John Tukey:
"The combination of some data and an ac
Hello,
Try
colSums(apply(...etc...))
Hope this helps,
Rui Barradas
Em 01-06-2012 18:37, Filoche escreveu:
Hi everyone.
I would like to count number of row higher than 0 for each column in a
dataframe.
For instance, I was using something like:
YY = replicate(5, rnorm(10))
apply(YY,2,">",0)
Great thanks.
If anyone else have an idea of statistics that can represent different
aspects of time-series please let me know.
--
Hello,
I am trying to collect several global measures or statistics for
time-series as well as packages of R that can compute them. I have found
several of them in p
Hi,
Couldn't find any problems if I understand what you mean.
A.K.
> DF <-within(DF,rank<-ave(Salary,Company,FUN=function(x)rev(order(x
> DF
Company Person Salary rank
1 GM A 7067.905 4
2 Ford B 9946.616 3
3 GM C 31627.171 8
4 Toyota D
thanks Phil, I will give it a try!
-Kim
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Hi.
I would take a look to the /forward.se/l function in the /packfor/ package.
http://r-forge.r-project.org/R/?group_id=195
Good luck,
Phil
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Hi
I need to do something very simple. I have 2 variables, Y and M. I need to
multiply Y by 1 if M=1, by 2 if M=3 and by 3.6678 if M=9. How do i make it?
Thanks for your time and interest
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Hi everyone.
I would like to count number of row higher than 0 for each column in a
dataframe.
For instance, I was using something like:
YY = replicate(5, rnorm(10))
apply(YY,2,">",0)
This give me a boolean matrix, but how to have the count of row with numbers
> 0 by column?
Thank in advance
I noticed that nls treats weights as relative and that the absolute size of the
weights w in
the following script has therefore no influence on the errors of the parameters
reported in the summary
a<-1
b<-3
x<--100:100
y<-a*x+b
yeps<-y+rnorm(length(x),sd=1)
w<-rep(1,length(x))
plot(x,yeps)
lin
Hello Martin,
I am not sure that PAM works with pre-defined medoids. The avg.width is
always the same and when i check the medoids that were used for the
calculation they are the same as before, i.e. before defining my own
medoids.
Please advise
BAHKO
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Hi
i need to create a model from 250 + variables with high collinearity, and
only 17 data points (p = 250, n = 750). I would prefer to use Cp, AIC,
and/or BIC to narrow down the number of variables, and then use VIF to
choose a model without collinearity (if possible). I realize that having a
huge
Hello James,
Thanks for the pointer, but Rredland is not maintained any more.
Take care
Oliver
On Thu, May 31, 2012 at 8:21 PM, J Toll wrote:
> On Thu, May 31, 2012 at 6:40 AM, Oliver Ruebenacker wrote:
>> Hello,
>>
>> Is there a convenient way to import RDF/OWL data into
Dear all,
As a novice user of R I ran into a problem that's quite hard for me to
resolve. I have a database containing data of a clinical trial in which
patients are included that survived or died:
x <- matrix(data=c(1:5,0, "1/1/2012 00:00:00",0,0,"1/7/2012 00:00:00"),
nrow=5, ncol=2, dimnames= l
Thank, it OK now
but I don't understand what is the meaning of y <- y[!is.na(y[5]),]
Thank in advance!
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Dear all,
I have a lot of problems on R-programming.
for example
my csv. file is ..
Date wrfRH wrfsolar wrfwindspeed wrfrain wrftd wrfta
21/10/2010 92.97 22.11 53.27 0 1546.337861 61.00852664
22/10/2010 87.35 21.99 40.89 0 1300.408288 62.85352227
23/10/2010 88.38 21.71 28.04 0.01 1
So that we could explore the structure of a HDF5 file?
Thanks a lot!
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.or
Thanks - figured it out, I rebooted the physical server and
everything is back to normal. Guess would be state is held
somewhere that a restart of the R server didn't clear, probably
/tmp, I have the server on a Linux system.
Jim
On 6/1/12 12:38 PM, "Jeff Newmiller" wrote:
>Ask the RStudio deve
The reviewers are right.
But this is not an R question at all. Post on a statistics list like
stats.stackexchange.com.
Much better yet, consult a local statistician, as it is abundantly
clear that you have insufficient statistical background to properly
analyze your data.
Cheers,
Bert
On Fri, J
Hello,
Try
library(zoo) # needed for as.yearmon
x <- c(201009, 201010, 201011, 201101, 201102)
(y <- as.Date(paste(x, "01", sep=""), format="%Y%m%d"))
y - y[1]
12*(as.yearmon(y) - as.yearmon(y[1]))
Note that to have a time difference between dates we need days (paste).
Hope this helps,
Rui B
Hi all:
There was a concern raised by reviewers of a manuscript of mine over the
proper execution of a Pearson's correlation. In brief, this was undertaken
in order to determine the relationship between the extent of wheel running
(y axis) and ethanol intake (x axis) across three, separate 10 day
Whoops, forgot one line of code; below.
On 2012-06-01 09:37, Peter Ehlers wrote:
On 2012-06-01 07:09, Apoorva Gupta wrote:
a<- data.frame(name=c(rep("a",5), rep("b",5)), year=c(1989:1993, 1989:1993),
var=c(1:10))
str(a)
b<- pdata.frame(a, index=c("name","year"))
str(b)
Now, I want to convert
Ask the RStudio developers.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Play
On 2012-06-01 07:09, Apoorva Gupta wrote:
a<- data.frame(name=c(rep("a",5), rep("b",5)), year=c(1989:1993, 1989:1993),
var=c(1:10))
str(a)
b<- pdata.frame(a, index=c("name","year"))
str(b)
Now, I want to convert b into a data frame and have a structure
similar to a. How do I do that?
I assume
Hi all,
My console is no longer responding to commands, I'm using the web-based console
running
off of a server. I have tried to interrupt R, I have deleted the data and
profile files in the
user directory, and restarted the server, relogged in, flushed the cache on the
browser,
but the console
Antony,
I am now utterly confused. The conditions involve column names of X1. In
your first post I assumed you just meant "check for each row of x" not
columns. After Arun replied I realized that may have been a wrong
assumption, but I just don't understand how you get, for example the last
TRUE (
You need the package compiled for your platform. You may also need to have a
version of R that is compatible with that package, depending on how that
package was constructed.
You can either obtain the source for the package and install RTools and compile
it yourself, or convince whoever develop
Hi Tammy,
I suspect that the first 4 digits compose the year, so with
Year_Month<-c(2010*100+9:11,2011*100+1:2)
#check composition
Year_Month
ym<-(Year_Month%/%100*12+Year_Month%%100)
ym-ym[1]
should give the difference in months.
Differences in Days are a bit trickier, since not all months hav
Hopefully this is an easy problem...
I'm trying to add a partitioned rank column to a data frame where the
rank is calculated separately across a partition by categories, the
way you could easily do in SQL. I found this solution in the archives
that looked like it might work:
http://tolstoy.newc
Yes, you need to modify both the R and the underlying C code. It's the the
source package on CRAN (the .tar.gz file).
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Josh Browning
Sent: Friday, June 01, 2012 10:48 AM
To: r
Hi All,
I'm using R's randomForest package (and it's quite awesome!) but I'd
really like to do some stratified sampling with a regression problem.
However, it appears that the package was designed to only accommodate
stratified sampling for classification purposes (see
https://stat.ethz.ch/pipe
> a <- data.frame(name=c(rep("a",5), rep("b",5)), year=c(1989:1993, 1989:1993),
> var=c(1:10))
> str(a)
> b <- pdata.frame(a, index=c("name","year"))
> str(b)
Now, I want to convert b into a data frame and have a structure
similar to a. How do I do that?
--
Apoorva Gupta
Consultant
National Insti
Hello,
I 'd like to use some functions in myLib. So I do:
library(myLib)
Then I get this message:
Error: package 'myLib' is not installed for 'arch=i386'
> sessionInfo()
R version 2.13.2 (2011-09-30)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=French_France.1252 LC_CTYPE=Fr
One of the most important concepts is most certainly Stationarity (see “unit
root test").
the most common r-package will be: tseries.
see:
Brockwell/Davis (2006): Time Series: Theory and Models.
Brockwell/Davis (2002): Introduction to Time Series and Forecasting.
Cowpertwait/Metcalfe (2009): Intr
>
> Hello,
>
> I am trying to collect several global measures or statistics for
> time-series as well as packages of R that can compute them. I have found
> several of them in papers and books, but the literature is so big i am sure
> i am missing several of them.
>
> skewness
> kurtosis
> min
> ma
how about just removing those network related package (including CRAN) from
your copy of R?
R can be used portably, as long as you have the package you need installed
already within your R.
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Dear All,
I hava maps in sp fomat and I like to use different map projection.
SHouls I somehow convert sp to maps and make projection and ploting there?
Or, is ther any tools to make new map project in sp?
Cheers,
Jari H
Department of Public Health
Faculty of Medicine
University of Helsinki
t
Hi,
> one <- c("ciao","zio","caio","bello")
> two<-c("ciao","zio")
setdiff(one,two)
[1] "caio" "bello"
A.K.
- Original Message -
From: Cren
To: r-help@r-project.org
Cc:
Sent: Friday, June 1, 2012 6:13 AM
Subject: Re: [R] Subtracting test string from vectors
Rui Barradas wrote
>
>
Hello,
It works with me.
> one <- c("ciao","zio","caio","bello")
> two <- c("caio","zio")
> setdiff(one, two)
[1] "ciao" "bello"
Are you sure of the values in your 'one' and 'two'?
Rui Barradas
Cren wrote
>
>
> Rui Barradas wrote
>>
>> Hello,
>>
>> ?setdiff
>> setdiff(one, two)
>>
> Than
Dear R-listers,
I am giving part of my R code :
###
n=15
m=1
library("partitions")
library("gregmisc")
library("combinat")
x = t(restrictedparts(n-m,m))
l = length(x[,1])
for(u in 1:l){
A= unique(matrix( unlist(permn(x[u,])), ncol=m,
Hi Peter,
Thanks for the help. It is exactly what I wanted.
Sorry, I forgot to post the library name.
A.K.
- Original Message -
From: Peter Ehlers
To: arun
Cc: R help
Sent: Friday, June 1, 2012 7:32 AM
Subject: Re: [R] Fit lines in intxplot
On 2012-05-31 21:41, arun wrote:
> Dear
To put double quotes in a string in VBA one has to use double double quotes
str = "a string with a ""quoted"" word"
On Jun 1, 2012, at 12:26 AM, Duncan Murdoch wrote:
> On 12-05-31 4:40 PM, Bert Jacobs wrote:
>> Hi,
>> I'm trying to run on Windows 7 a scriptfile with Rscript.exe from within
>>
Dear Vito,
Thanks for the quick reply.
In the case of splitting up the data, would it be advisable to
independently fit a GLM on the subsets of the data, or use one single
GLM for comparison in the 'segmented' function?
Kind regards,
Peter
On Fri, Jun 1, 2012 at 2:21 PM, Vito Muggeo (UniPa)
w
yes it handles completely on one side.
your sample included that case.
Sent from my iPhone
On Jun 1, 2012, at 2:15, Roberto Brunelli wrote:
> Thanks to both of you,
>
> close to the solution, but I think that your proposals do not address
> the case when a bar is completely positive (or negati
dear Peter,
Currently segmented handles multiple breakpoints for several variables
(the limit discussed in the msg 2006 has been fixed..).
However you are looking for a somewhat complicated model where the
breakpoint of the relationship 's.size' and 'R.AUC' depends on another
covariate 'bedek
On 2012-05-31 21:41, arun wrote:
Dear R help,
I tried to add linear fit lines in intxplot by adding "index.cond = function(x,y)
coef(lm(y ~ x))[1]" inside the intxplot(). It didn't help. I would appreciate any
help.
My dataset and codes are pasted below.
Thanks,
A.K.
datGreen<- read.table
Hi,
You assumed
scale=229678.21 and shape=0.41
and the procedure estimated
\hat{scale}=427196.6 and \hat{shape}=0.2092887
I think these estimates are natural, since
1) you use a simulated generated data with n=100, and 2) the true value your
scale parameter is very large.
Hope this helps
Be
HI, R-Users:
I got a questions. have been struggling so long time
I have this data:
> m1$Year_Month
201009 201010 201011 201101 201102
> min(m1$Year_Month)
201009
I want to calculate the following two answers, how do I program it?
> difference in Month?
[1] 0 1 2 4 5
>differenc
On Fri, 1 Jun 2012, Mabille, Geraldine wrote:
Thanks! I had seen this section but hadn't understood that the dashed
line represented the mean. Does that mean in my case (two tests hardly
significant P=0.03 and one NS) that I should reject the hypothesis of
different levels of survival probabil
Thanks! I had seen this section but hadn't understood that the dashed line
represented the mean.
Does that mean in my case (two tests hardly significant P=0.03 and one NS) that
I should reject the hypothesis of different levels of survival probability for
different groups of females??
-Ori
On Fri, 1 Jun 2012, Mabille, Geraldine wrote:
Hi and thanks again for your answer. I have just a last question regarding the
choice of the functional...if you have time to help on that again.
I have tried running the sctest using the functionals you recommended
sctest(gmass2,functional=meanL2
Dear Researchers,
I am looking for a library or a function to calculate SMAPE with same
doneminator to avoid a problem to the presence of 0 values in the series.
I find this variotion of SMAPE called mSMAPE
mSMAPE page 13
http://www.stat.iastate.edu/preprint/articles/2004-10.pdf
yesterday nigt
On Friday, June 1, 2012, Roberto Brunelli wrote:
> Thanks to both of you,
>
> close to the solution, but I think that your proposals do not address
> the case when a bar is completely positive (or negative): I do not
> think that the histogram can not be extended easily to this case ...
Why not?
Hi and thanks again for your answer. I have just a last question regarding the
choice of the functional...if you have time to help on that again.
I have tried running the sctest using the functionals you recommended
sctest(gmass2,functional=meanL2BB)
sctest(gmass2,functional=rangeBB)
Rui Barradas wrote
>
> Hello,
>
> ?setdiff
> setdiff(one, two)
>
Thank you for your help, Rui.
But
*> setdiff(one,two)
[1] "ciao"*
Where's "bello"?
--
View this message in context:
http://r.789695.n4.nabble.com/Subtracting-test-string-from-vectors-tp4632049p4632053.html
Sent from the R hel
Hello,
?setdiff
setdiff(one, two)
Hope this helps,
Rui Barradas
Cren wrote
>
> Hi all,
>
> let I have two text string:
>
> *one <- c("ciao","zio","caio","bello")
> two <- c("caio","zio")*
>
> I would like to obtain a new text string which is* one - two* like this
> one:
>
> [1] "ciao" "bel
Hi all,
let I have two text string:
*one <- c("ciao","zio","caio","bello")
two <- c("caio","zio")*
I would like to obtain a new text string which is* one - two* like this one:
[1] "ciao" "bello"
because "caio" and "zio" elements have been subtracted from *one*.
What's the most efficient way t
[1]transfert [2]transfert
[3]fichier [4]fichier
[5]sécurisé [6]sécurisé
Bonjour,
Vous êtes plus de 200 000 utilisateurs à échanger vos fichiers
professionnels en tout
Hi,
I have a problem in fitting GPD distribution.
i generate random numbers from gpd distribution from specific parameters
using pot packege then i used fitgpd function to estimete the parameters.The
estimated parameters are not matched with the given parameters i.e.from
which i generate random nu
Yes this is what i wanted.
thanks indeed!!!
On Fri, 1 Jun 2012 07:40:49 +
Ben Bolker wrote:
QAMAR MUHAMMAD UZAIR polito.it> writes:
Dear R users,
I have data in the following manner
[,1][,2][,3][,4]
6 2 2 2
5 4 4 3
[snip]
I want to a
On 12-06-01 5:05 AM, Vito Muggeo (UniPa) wrote:
dear all,
I do not if it is a nonsense question..
Is it possible in the R session to get the name of the current .Rdata
file that I ran?
I mean: suppose I double click the file myfile.Rdata. ls() returns the
names of the objects in the current wor
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