Dear All,
I have .csv file it looks like this :
rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t'
, header=FALSE)
> head(rawdata,n=5)
TenantDomain Owner Current State
1\\N ROOTadmin Running
2\\N ROOTadmi
Hi Richard,
> ## for an array with
> ## dim(a) == c(3,4,2)
> ## a[i,j,k] means select the element in position
> ##i + (j-1)*3 + (k-1)*3*4
My understanding;
e.g.
1)
dim(a) == c(3,4,2)
3 + (4-1)*3 + (2-1)*3*4
3+9+12=24
2)
## dim(a) == c(1,2,1)
1 + (2-1)*3 + (1-1)*3*4
1+3+0=4
3)
## dim(a)
Dear R users
Topic: Linear effect model fitting using the nlme package (recomended by
Pinheiro et al. 2008 for unbalanced data set).
The R help provides much info about the controversy to use the anova(lme.model)
function to present numerator df and F values. Additionally different p-values
ca
Dear All
I have some problem with calculate probability.
Assume I have data with normal distribution with mean = 5 sd = 2.
I want to approximate probability = 2.4.
I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE), correct
or not.
Many Thanks
Jumlong
__
Try this:
gsub("N", "BlankSpace", rawdata$Tenant)
On Sat, Nov 6, 2010 at 8:15 AM, Mohan L wrote:
> Dear All,
>
> I have .csv file it looks like this :
> rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t'
> , header=FALSE)
>
> > head(rawdata,n=5)
> TenantDoma
On 06-Nov-10 11:16:28, Jumlong Vongprasert wrote:
> Dear All
> I have some problem with calculate probability.
> Assume I have data with normal distribution with mean = 5 sd = 2.
> I want to approximate probability = 2.4.
> I used pnorm(2.4, 5, 2) - pnorm(2.4, 5, 2, lower.tail = FLASE),
> correct o
Hi there, can anyone tell me how to extract to values of a particular slot for
some S4 object? Let take following example:
> library(fOptions)
> val <-GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08, b =
> 0.08, sigma = 0.30)
> val
Title:
Black Scholes Option Valuation
Call:
Try this: v...@price
On Sat, Nov 6, 2010 at 11:41 AM, Megh Dal wrote:
> Hi there, can anyone tell me how to extract to values of a particular slot
> for some S4 object? Let take following example:
>
> > library(fOptions)
> > val <-GBSOption(TypeFlag = "c", S = 60, X = 65, Time = 1/4, r = 0.08,
Hi list,
I just got stuck with this one:
In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F and
NA). Then getting some nice plots using
ggplot(data, aes(age[na.exclude(gender)])) +
geom_histogram( binwidth = 3, aes(y = ..density.. ), fill = "lightblue" )
+
facet_grid( gende
Create a subset of your data that excludes the NAs before you feed it to ggplot.
"Ottar Kvindesland" wrote:
>Hi list,
>
>I just got stuck with this one:
>
>In Data I have the sets age (numbers 1 to 99 and NA) and gender (M, F
>and
>NA). Then getting some nice plots using
>
>ggplot(data, aes(age[
Hi Joshua,
Thanks for your advice.
1)
Re your advice:-[quote]
> a3d
, , 1 <--- this is the first position of the third dimension
[,1] [,2] [,3] [,4] <--- positions 1, 2, 3, 4 of the second dimension
[1,]147 10
[2,]258 11
[3,]369 12
^ the first dim
Sounds to me like you should really be seeking help from your local
statistician, not this list. What you request probably cannot be done.
What is wrong with what you get from lme, whose results seem fairly
clear whether the P values are accurate or not?
Cheers,
Bert
On Sat, Nov 6, 2010 at 4
Hi folks,
Debian 600 64-bit
Is rcom for Linux available?
rcom
rcom: R COM Client Interface and internal COM Server
http://cran.r-project.org/web/packages/rcom/index.html
If YES please advise where to get it.
TIA
B.R.
Stephen L
__
R-help@r-proje
isn't COM a Windows-only technology?
Shige
On Sat, Nov 6, 2010 at 12:12 PM, Stephen Liu wrote:
> Hi folks,
>
> Debian 600 64-bit
>
> Is rcom for Linux available?
>
> rcom
> rcom: R COM Client Interface and internal COM Server
> http://cran.r-project.org/web/packages/rcom/index.html
>
> If YES pl
Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1. Any idea?
Robert
___
Robert Ruser gmail.com> writes:
>
> Hello R Users,
> I'm wondering if there exists any elegant and simple way to do the
> following: I have a data.frame X fills in numbers. I have a vector y with
> numbers as well. Every value in X that is equal to any values in y should
> be replaced by e.g. 1.
On 11/06/2010 11:36 AM, Robert Ruser wrote:
Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1.
This is a "tableplot", available on R-Forge at
https://r-forge.r-project.org/projects/tableplot/
install.packages("tableplot", repos="http://R-Forge.R-project.org";)
will install, as long as you are using R 2.12.x; otherwise, you'll
have to download the source package and install from source.
-
Thank you vary much Ben and Erik.
It's exactly what I want. Below is my a little modified example.
set.seed(12345)
X = sample(c(40:60),40,replace=TRUE)
x = matrix(X,nc=5)
y = c(40,43,55,60)
x[x %in% y] <- -1
__
R-help@r-project.org mailing list
https://
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Stephen Liu
> Sent: Saturday, November 06, 2010 7:38 AM
> To: Joshua Wiley
> Cc: r-help@r-project.org
> Subject: Re: [R] About 5.1 Arrays
>
> Hi Joshua,
>
> Thanks for your advic
> Date: Sat, 6 Nov 2010 07:45:26 -0700
> From: gunter.ber...@gene.com
> To: sibylle.stoec...@gmx.ch
> CC: r-help@r-project.org
> Subject: Re: [R] anova(lme.model)
>
> Sounds to me like you should really be seeking help from your local
> statistician, not this list. What you request probably cannot
Can anyone show me how to test for significant simple slopes of a 3-way
interaction, with covariates.
my equation
tmod<-(glm(PCL~ rank.f + gender.f + MONTHS + CEXPOSE.M + bf.m +
MONTHS*CEXPOSE.M*bf.m,
data=mhatv, family=gaussian ,na.action=na.omit))
Thank you
Mike
[[alternative HTML vers
Hello everybody,
I have usually solved this problem by repeating lines of codes instead of a
loop, but it's such a waste of time, I thought I should really learn how to
do it with loops:
What I want to do:
Say, I have several data files that differ only in a number, e.g. data
points (or vector
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Tuatara
> Sent: Saturday, November 06, 2010 9:22 AM
> To: r-help@r-project.org
> Subject: [R] Using changing names in loop in R
>
>
> Hello everybody,
>
> I have usually solved
Hello,
I am trying to alter the way in which lattice functions (specifically xyplot)
print the axis labels when one uses the 'scales' parameter.
I can obtain the effect I want by using
scales=list(y=list(log=10, labels=expression(yvalues)))
where yvalues are the values that would have been prin
On Sat, Nov 6, 2010 at 3:02 PM, Marc Paterno wrote:
> Hello,
>
> I am trying to alter the way in which lattice functions (specifically xyplot)
> print the axis labels when one uses the 'scales' parameter.
> I can obtain the effect I want by using
> scales=list(y=list(log=10, labels=expression(yv
On Sat, Nov 6, 2010 at 5:22 PM, Tuatara wrote:
>
> Hello everybody,
>
> I have usually solved this problem by repeating lines of codes instead of a
> loop, but it's such a waste of time, I thought I should really learn how to
> do it with loops:
>
Would the following construct help?
> for(i in 1:
Mike Marchywka hotmail.com> writes:
>
>
> > Date: Sat, 6 Nov 2010 07:45:26 -0700
> > From: gunter.berton gene.com
> > To: sibylle.stoeckli gmx.ch
> > CC: r-help r-project.org
> > Subject: Re: [R] anova(lme.model)
> >
> > Sounds to me like you should really be seeking help from your local
> >
Hi,
I'm trying to write a general-purpose "lexicon" class and associated methods
for storing and accessing information about large numbers of specific words
(e.g., their frequencies in different genres). Crucial to making such a class
practically useful is to get hashing working correctly so t
Hi All,
I am confused with SMATR's test for common slope. My null hypothesis here is
that all slopes are parallel (common slopes?), right?
So if I get a p value < 0.05 means that we can have confidence to reject it?
That slopes are different?
Or the other way around? it means that we have statisti
Hi,
I've been trying to use optim to minimise least squares for a
function, and then get a guess at the error using the hessian matrix
(calculated from numDeriv::hessian, which I read in some other r-help
post was meant to be more accurate than the hessian given in optim).
To get the standard erro
Hi, all:
I want the user to give a function g as the parameter of our function f. In
function f, we use user's function g to compute something. Since our
function f is implemented in C++, how do I further pass this function to C++
code?
Thank you very much!
best
Liping LIU
[[alternativ
blackscorpio wrote:
>
> Dear community,
>
> I am currently trying to fit an ordinal logistic regression model with the
> polr function. I often get the same error message :
>
> "attempt to find suitable starting values failed", for example with :
> ...
> Does anyone have a clue ?
>
Yes. The
I wish to generate 100 by 1 vector of x1 and x2 both are uniform distributed
with covariance matrix \Sigma.
Thanks,
Michael
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
The start value generation code in polr is also known to fail quite
frequently. For example, against the Iris data as recently posted to this
list by blackscorpio ( Sep 6, 2010).
> polr(Species~Sepal_Length+Sepal_Width+Petal_Length+Petal_Width,data=iris)
Error in polr(Species ~ Sepal_Length + Se
Note that that the enhancements in my original post solve the unresolved
problem of Chaehyung Ahn (22 Mar 2005) whose data I reproduce:
y,x,lx
0,3.2e-02,-1.49485
0,3.2e-02,-1.49485
0,1.0e-01,-1.0
0,1.0e-01,-1.0
0,3.2e-01,-0.49485
0,3.2e-01,-0.49485
1,1.0e+00,0.0
0,1.0e+00,0.0
1,3.
I would make make an environemnt called wfreqsEnv
whose entry names are your words and whose entry
values are the information about the words. I find
it convenient to use [[ to make it appear to be
a list (instead of using exists(), assign(), and get()).
E.g., the following enters the 100,000 word
Hi,
If you have data that is similar enough to warrant only changing the
extension (i.e., 1, 2, etc.) and that you (at least at times) wish to
perform operations on together, it is time to start thinking of a more
flexible framework. Fortunately, such a framework already exists in
lists. Lists l
Eugenio -
> I am confused with SMATR's test for common slope. My null hypothesis here is
> that all slopes are parallel (common slopes?), right?
> So if I get a p value < 0.05 means that we can have confidence to reject it?
> That slopes are different?
> Or the other way around? it means that we
On 06-Nov-10 21:54:41, michael wrote:
> I wish to generate 100 by 1 vector of x1 and x2 both are uniform
> distributed with covariance matrix \Sigma.
>
> Thanks,
> Michael
First, some comments.
1. I don't think you mean a "100 by 1 vector of x1 and x2" since
you have two variables. "100 by 2"
some of this can be automated using the CRAN package
hash.
Kjetil
On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap wrote:
> I would make make an environemnt called wfreqsEnv
> whose entry names are your words and whose entry
> values are the information about the words. I find
> it convenient to
great thanks a lot!
On Sat, Nov 6, 2010 at 3:54 PM, Kevin Middleton wrote:
>
> Eugenio -
>
> > I am confused with SMATR's test for common slope. My null hypothesis here
> is
> > that all slopes are parallel (common slopes?), right?
> > So if I get a p value < 0.05 means that we can have confiden
Ted,
Thanks for your help, it is right on the money!
for your comments:
1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
2. The correlation is the only free parameter.
Michael
On Sat, Nov 6, 2010 at 7:07 PM, Ted Harding wrote:
> On 06-Nov-10 21:54:41, michael wrote:
> > I wish
OK, any reason why ggplot2 does not allow filtering of NA?
ottar
On 6 November 2010 15:23, Jeff Newmiller wrote:
> Create a subset of your data that excludes the NAs before you feed it to
> ggplot.
>
> "Ottar Kvindesland" wrote:
>
> >Hi list,
> >
> >I just got stuck with this one:
> >
> >In D
On Sat, Nov 6, 2010 at 4:43 PM, Ottar Kvindesland
wrote:
> OK, any reason why ggplot2 does not allow filtering of NA?
It is not so much that ggplot2 does not allow the filtering of NA
values, it is that you need to use data from the dataset you
specified. By subsetting in aes() rather than in da
A more detailed example:
Say I would like to read in data files that are set-up identically and have
identical (but somewhat) different text names (see below):
data_1 <- read.csv("data1.txt")
data_2 <- read.csv("data2.txt")
data_3 <- read.csv("data3.txt")
How do I automate this process?
(I ass
Dear Michael,
On Sat, Nov 6, 2010 at 7:27 PM, michael wrote:
> Ted,
>
> Thanks for your help, it is right on the money!
>
> for your comments:
> 1. Yes I mean 100 by 2, each variable x1, x2 is 100 by 1.
> 2. The correlation is the only free parameter.
>
> Michael
>
>
I like Ted's solution.
On Sat, Nov 6, 2010 at 3:13 PM, Tuatara wrote:
>
> A more detailed example:
>
> Say I would like to read in data files that are set-up identically and have
> identical (but somewhat) different text names (see below):
>
> data_1 <- read.csv("data1.txt")
> data_2 <- read.csv("data2.txt")
> data_3 <-
Jay,
Yes I'm looking for unif(0,1) and your method works just fine. I
suppose your method should work for dimensions greater than 2, am I right?
Michael
On Sat, Nov 6, 2010 at 8:05 PM, G. Jay Kerns wrote:
> Dear Michael,
>
> On Sat, Nov 6, 2010 at 7:27 PM, michael wrote:
> > Ted,
> >
>
On Sat, Nov 6, 2010 at 8:22 PM, michael wrote:
> Jay,
>
> Yes I'm looking for unif(0,1) and your method works just fine. I
> suppose your method should work for dimensions greater than 2, am I right?
>
> Michael
>
Yes, but it gets that much more tricky to specify the covariance
matrix. Two
- Original Message
From: Shige Song
To: Stephen Liu
Cc: r-help@r-project.org
Sent: Sun, November 7, 2010 12:17:42 AM
Subject: Re: [R] Where to get rcom for Linux
> isn't COM a Windows-only technology?
Hi Shige
Thanks.
I see. I was surprised for unable to find it after having turned
Dear R users, simple figure:
postscript(file="~/Desktop/figure.ps", horizontal=T, width=20, height=10)
par(mfcol=c(2,5))
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
plot(rnorm(100), type='l')
##-
plot(rnorm(100), type='l')
plot(
Hi Robert,
You need to add paper = "special" to the postscript() call.
postscript(file="~/Desktop/figure.ps", horizontal=TRUE,
width=20, height=10, paper = "special")
plot()
...
plot()
dev.off()
Otherwise it is reset because you are specifying a size outside of
what can fit on the
Hi,
I've got a problem that sounds a lot like this,
http://r.789695.n4.nabble.com/Re-R-R-2-12-0-hangs-while-loading-RGtk2-on-FreeBSD-td3005929.html
under windoze 7.
but it seems to hang with this stack trace,
#0 0x77830190 in ntdll!LdrFindResource_U ()
from /cygdrive/c/Windows/system32/n
On Nov 6, 2010, at 9:06 AM, Michael Wood wrote:
Can anyone show me how to test for significant simple slopes of a 3-
way
interaction, with covariates.
You might start by defining what you mean by "simple slopes" when
discussing a model with a three way interaction. You might also
includ
Hi Daniel,
> I am correcting you. :-) You are using dim() incorrectly, and not accessing
>the array correctly.
>
> In all of your examples you should be using dim(3,4,2). Then you need to
>specify the indexes
>
> of the array element you want to look at. So, to use your example
Thanks
Wow, that is perfect: the hash package is exactly what I needed. Thank you!
Roger
On Nov 6, 2010, at 4:09 PM, Kjetil Halvorsen wrote:
> some of this can be automated using the CRAN package
> hash.
>
> Kjetil
>
> On Sat, Nov 6, 2010 at 10:43 PM, William Dunlap wrote:
>> I would make make an e
Hi,
I have this symmetric matrix, at least I think so.
col1 col2 col3
[1,] 0.20 0.05 0.06
[2,] 0.05 0.10 0.03
[3,] 0.06 0.03 0.08
or
structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08
), .Dim = c(3L, 3L), .Dimnames = list(NULL, c("var1", "var2",
"var3")))
But isSymmetric() d
Hello R-helpers
I hope that my subject line is not detering anyone from helping me out:)
I have been stuck of a few hours now, and I don't seem to pinpoint
where the problem is.
I have a data.frame which is structured as follow:
str(hDatPretty)
'data.frame': 1665 obs. of 8 variables:
$ time
I am guessing this is a very simple question, but this is only my second day
with R so it is all still a bit imposing.
I am trying to run an autocorrelation.
I imported a CSV file, which has one column labeled "logistic".
I ran the command:
ACF(data$logistic,maxLag=10)
However, I received the
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