On May 8, 2012, at 08:34 , array chip wrote:
> Thank you Peter, so if I observe a significant coefficient, that significance
> still holds because the standard error of the coefficient has taken the
> residual error (which is large because large R square) into account, am I
> correct?
In esse
On May 8, 2012, at 05:10 , array chip wrote:
> Hi, what does a low R-square value from an ANCOVA model mean? For example, if
> the R square from the model is about 0.2, does this mean the results should
> NOT be trusted? I checked the residuals of the model, it looked fine...
It just means tha
Unfortunately, there's a no-homework policy on this list so we can't
really give you much help.
Googling "simulate blackjack r-project" does turn up some interesting
leads though.
Best,
Michael
On Mon, May 7, 2012 at 5:37 PM, Mark G wrote:
> Hello,
>
> I am trying to do a project on R-Code invo
I know this is not a revolution support forum, but as anyone noticed the
following?
I have a foreach loop to generate random samples. If I run the exact code
below in normal r (2.14.1) it works as expected, but if I run it from
revolution 4.2.0 each loop returns the same numbers.
The only way I c
Hello,
Try
plot(1)
text(1, 1.4, expression(bar(X)[cv] == 552.01))
[subscript] and ==
Hope this helps,
Rui Barradas
Dan Abner wrote
>
> Hello everyone,
>
> I am trying to add the following text (in proper notation) to a
> graphic using expression().
>
> X-bar (with a subscript of cv) =
Hello,
I am trying to do a project on R-Code involving black jack simulations for a
class of mine and I need help starting it. I have tried so many different
options and am getting no where and am looking for some help. If anyone can
help me out it would be amazing. thank you
Gabriel
Dear List,
Here is an example of survival data in counting process format
(detailed record of each day)
> data[data$Id == 11,]
# extracted one person's record
Id Event Fup Start Stop sex Drug1
601 11 0 6 01 0 0
602 11 0 6 12 0 0
603 11 0 6 2
Hi, what does a low R-square value from an ANCOVA model mean? For example, if
the R square from the model is about 0.2, does this mean the results should NOT
be trusted? I checked the residuals of the model, it looked fine...
Thanks for any suggestion.
John
[[alternative HTML version
You might have been thinking of Hadley's webpage for reshape2:
http://had.co.nz/reshape/ or the JSS article:
http://www.jstatsoft.org/v21/i12
Hope this helps,
Michael
On Mon, May 7, 2012 at 8:54 PM, Rich Shepard wrote:
> On Mon, 7 May 2012, R. Michael Weylandt wrote:
>
>> What makes you think re
contentsData() will report the names of the "sheets" in an Excel file,
or the names of the datasets in a SAS transport file, or the names
of the tables in a datbase. importData has a pageNumber argument
to let you import from a specific Excel sheet.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap ti
On Mon, 7 May 2012, R. Michael Weylandt wrote:
What makes you think reshape2 has a vignette? I don't see one on CRAN:
http://cran.r-project.org/web/packages/reshape2/index.html
Michael,
I don't find one there, either. But, I thought that last autumn I read one
or two about the package. It w
Hello experts!!
I apologize for posting SPlus related query here..badly in need of relevant
info..
I usually use R (and your advice/tips) for my daily work. Was wondering if
there is an equivalent of "sheetCount" of the package "gdata" avaailable in
Splus 8.2? I would like to get the total number
If all you want are binary 0/1, then look at the 'bit' package which
will let you create a vector of bits. Even for your matrix, you will
need almost 1GB of memory to store a copy.
On Mon, May 7, 2012 at 5:23 PM, Lucas wrote:
> Dear R people.
> I惴 facing a big problem.
> I need to create a matri
What makes you think reshape2 has a vignette? I don't see one on CRAN:
http://cran.r-project.org/web/packages/reshape2/index.html
Michael
On Mon, May 7, 2012 at 6:19 PM, Rich Shepard wrote:
> Running 2.15.0 on Slackware-13.1 with all packages in /usr/lib/R/library.
> Package reshape2 is in the
On May 7, 2012, at 5:16 PM, Suhaila Haji Mohd Hussin wrote:
Hello.
Now the median is solved but then I'm still figuring out how to put
the updated column back to replace the original column of the whole
data. I'll show you what I meant:
Continuing from the previous commands you guys helped
On May 7, 2012, at 6:15 PM, stat curio wrote:
Hello,
I have a data frame which looks like
head(d)
a1 a2n j col
1 1 1 88341002 11 #E7E7E7
2 1 2 25094882 11 #E7E7E7
3 1 3 16916246 11 #E7E7E7
4 1 4 14289229 11 #E7E7E7
5 1 5 11945929 11 #E7E7E7
6 1 6 8401235 11 #E7E7E
The following works for me:
plot(1:10)
text(4,8,expression(bar(X)[cv]==552.01))
cheers,
Rolf Turner
On 08/05/12 09:39, Dan Abner wrote:
Hello everyone,
I am trying to add the following text (in proper notation) to a
graphic using expression().
X-bar (with a subscript of
Running 2.15.0 on Slackware-13.1 with all packages in /usr/lib/R/library.
Package reshape2 is in the library but there are no vignettes seen by R. If
this is the result of something I did incorrectly, please point out how to
install the vignettes with the package.
I can get help for melt and
Hello,
I have a data frame which looks like
> head(d)
a1 a2n j col
1 1 1 88341002 11 #E7E7E7
2 1 2 25094882 11 #E7E7E7
3 1 3 16916246 11 #E7E7E7
4 1 4 14289229 11 #E7E7E7
5 1 5 11945929 11 #E7E7E7
6 1 6 8401235 11 #E7E7E7
The values in 'j' run from 1 to 11. I would li
The code below should have been
plot(0, main = expression(bar(X)[cv]))
Apologies for the noise.
Jorge.-
On Mon, May 7, 2012 at 5:44 PM, Jorge I Velez <> wrote:
> Hi Dan,
>
> If I understood correctly, the following will do:
>
> plot(0, main = expression(bar(X)^{cv}))
>
> HTH,
> Jorge.-
>
>
>
>
Hi Dan,
If I understood correctly, the following will do:
plot(0, main = expression(bar(X)^{cv}))
HTH,
Jorge.-
On Mon, May 7, 2012 at 5:39 PM, Dan Abner <> wrote:
> Hello everyone,
>
> I am trying to add the following text (in proper notation) to a
> graphic using expression().
>
> X-bar (wit
Hello everyone,
I am trying to add the following text (in proper notation) to a
graphic using expression().
X-bar (with a subscript of cv) = XX.
Note: Ideally "cv" would be a subscript, but it doesn't have to be.
I have the following code:
> text(625,.012,expression(bar(X)cv = 552.01))
Error
See ?sparseMatrix from package Matrix.
Eloi
On 12-05-07 02:23 PM, Lucas wrote:
> Dear R people.
> I´m facing a big problem.
> I need to create a matrix with 10.000 columns and 750.000 rows.
> matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
> as you can see, the data frame has huge
Dear R people.
I´m facing a big problem.
I need to create a matrix with 10.000 columns and 750.000 rows.
matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
as you can see, the data frame has huge dimesions. I was able to find out
about thr "L" in data, this way I´m telling that my data
Hello.
Now the median is solved but then I'm still figuring out how to put the updated
column back to replace the original column of the whole data. I'll show you
what I meant:
Continuing from the previous commands you guys helped out I continued as
followed:
Original Data: http://i1165.
8th Annual International R User Conference
useR! 2012, Nashville, Tennessee USA
===> Regular registration rates and most blocks of discounted
===> hotel rooms are available only 5 more days
Registration Deadlines:
Early Registration: Passed
Regular Registration: Mar 1- May 12
Late Registration:
On May 7, 2012, at 19:39 , Bert Gunter wrote:
> 1. As this is a statistical, rather than an R issue, you would do
> better posting on a statistical help site like stats.stackexchange.com
> (although some generous soul here may respond).
>
> 2. You would also probably do better consulting with a
Hello,
I am using SSfol in nlme to fit some data for the change of N concentration
(N) in plant tissue over time (gdd). The model works nicely for 2 out of 3
treatments, so I would really like to use it, but it consistently has a bad
fit for my third treatment. I am pasting the figure for the t
Hello,
meredith wrote
>
> Afternoon-
> I am trying to subtract a matrix, basically a vector of 28 values, each
> by the same number to account for differences in regression fitting. I am
> taking the 1x28 and minus it by the mean value of the matrix, each number.
> The result I receive is a 1X
R has functions for computing kappa, fleiss's kappa, etc., but can it compute
Gwet's AC1?
Thanks,
Matt.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read t
I'm sorry but where is 28 coming from? It looks to me like you have a vector
of length 34 and the result is of length 34. Oh, I see, the 28 and 29 are the
indices for the first number on the line not the total length.
John Kane
Kingston ON Canada
> -Original Message-
> From: mmbal..
I'm sorry but where is 28 coming from? It looks to me like you have a vector
of length 34 and the result is of length 34. Oh, I see, the 28 and 29 are the
indices for the first number on the line not the total length.
John Kane
Kingston ON Canada
> -Original Message-
> From: mmbal..
*
Disregard message please
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values, each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each number.
The result I receive is a 1X29 matri
On May 7, 2012, at 3:26 PM, meredith wrote:
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values,
each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each
number.
The result I
Afternoon-
I am trying to subtract a matrix, basically a vector of 28 values, each by
the same number to account for differences in regression fitting. I am
taking the 1x28 and minus it by the mean value of the matrix, each number.
The result I receive is a 1X29 matrix. Does anyone know why the r
On 07/05/2012 3:05 PM, Suhaila Haji Mohd Hussin wrote:
I might be silly but if I was going to type in dput() then how should I send
the data over here?
Cut and paste. For example, if I have a dataframe named x and type
dput(x), I see
structure(list(a = 1:10), .Names = "a", row.names = c(N
Thank you so much!
Suhaila.
> Date: Mon, 7 May 2012 15:08:47 -0400
> Subject: Re: [R] Problem with Median
> From: jholt...@gmail.com
> To: bell_beaut...@hotmail.com
>
> Your problem is that a.AC is a dataframe:
>
>
> > x <- read.table(text = " a b c class
> + 1 12 0 90
Hi Lincoln,
Some thoughts:
1) Did you intend to use "cohort" as a factor and not as a numeric? (at
least that is what it looks like in your output)
2) Is there a strong correlation between "cohort" and the
other explanatory variables you are trying in your model?
Contact
Detail
I might be silly but if I was going to type in dput() then how should I send
the data over here?
Instead, I've just uploaded the image online, you can access it via the link
below.
http://i1165.photobucket.com/albums/q585/halfpirate/data.jpg
> Date: Mon, 7 May 2012 14:55:24 -0400
> Subject: R
Please use dput() to give us your data (eg dput(data) ) rather than
simply pasting it in.
Sarah
On Mon, May 7, 2012 at 2:52 PM, Suhaila Haji Mohd Hussin
wrote:
>
> Hello.
> I'm trying to compute median for a filtered column based on other column but
> there was something wrong. I'll show how I
Hello.
I'm trying to compute median for a filtered column based on other column but
there was something wrong. I'll show how I did step by step.
Here's the data:
a b c class
1 12 0 90 A-B2 3 9711 A-B3 78 NA123 A-C4
NA NA12A-C5
Thank Sarah!
> Date: Mon, 7 May 2012 14:06:31 -0400
> Subject: Re: [R] Problem in executing R-script
> From: sarah.gos...@gmail.com
> To: bell_beaut...@hotmail.com
> CC: r-help@r-project.org
>
> Hi Suhaila,
>
> You don't need to make a function: your script should just contain:
> med <- read.cs
Perhaps I haven't explained it that well as I would have liked to.
To me this was an R issue because I didn't understand why the binomial GLM
is getting these results and I believed this was something due to the way I
am implementing it in R, not to the binomial GLM itself.
If I was wrong and this
On 07-05-2012, at 19:41, Suhaila Haji Mohd Hussin wrote:
>
> Hello. I'm a newbie here.
> In my script (I name it readData.R), I wrote the followings:
> readData <-function(){med = read.csv("medicalData.csv");}
> Then I tested the script by 'Source R Code' then on the command I typed
> '
Hello,
I have an experimental design where I would like to use separate ranking
events to predict an independent ranking event. I have been using function
clmm in the ordinal library but now realize that I am violating one of the
assumptions. I have included a subset of the data.
I am looking a
Chris,
I think that you really need to quantify what you mean by correlation.
Things to consider would depend on what the matrices represent--are they
the estimates of the same set of N geographic points, are they traces of
the same line, are they traces of the same polygon outline? If either
Hi Suhaila,
You don't need to make a function: your script should just contain:
med <- read.csv("medicalData.csv")
If you do want to make a function, then you need to assign the
resulting value to something, eg:
med <- readData()
but there's no reason to do that. Values that are assigned within
R is a functional language so, by default, assignments (and other
things) within function scope doesn't have global effects. This is
generally considered a _very good thing_ in language design. You'd
perhaps prefer something like:
readData <- function() {
read.csv("medialData.csv")
}
med <- r
Hello. I'm a newbie here.
In my script (I name it readData.R), I wrote the followings:
readData <-function(){ med = read.csv("medicalData.csv");}
Then I tested the script by 'Source R Code' then on the command I typed
'readData()' then I typed 'med' to check if the variable contains the medical
1. As this is a statistical, rather than an R issue, you would do
better posting on a statistical help site like stats.stackexchange.com
(although some generous soul here may respond).
2. You would also probably do better consulting with a local
statistical resource if available, as it is difficul
Hi all,
I can't find the error in the binomial GLM I have done. I want to use that
because there are more than one explanatory variables (all categorical) and
a binary response variable.
This is how my data set looks like:
> str(data)
'data.frame': 1004 obs. of 5 variables:
$ site : int 0 0
Hi All,
Sorry for posting the same question again. I was not sure if the message
was sent initially since it was my first post the forum.
Can the MNP package available in R be used to analyze panel data as well?
*i.e., *if there are 3 observed discrete choices for three time periods for
the same
Hey all who have responded to this post. I am a newbie to ANOVA analysis in
R, and let me tell you- resources for us learners are scant, horrible,
unclear, imprecise.. in other words.. the worst ever. So advice like "go
look it up" in your "classical" textbook or on google is not helpful at all.
I
Hello
I need to use a coxme model with my data (survival analysis with
right-censoring and hierarchical nesting), but I cant find a way to get
predicted values from a new data table (or even from the original one).
Has anyone had this problem before? I cant find anything about that
anywhere.
Tha
Thank you Arun!
Collin.
On Mon, 7 May 2012, arun wrote:
> Hi Collin,
Look in the package 'pwr' for 'pwr.r.test'.
A.K.
- Original Message -
From: Collin Lynch
To: r-help@r-project.org
Cc:
Sent: Monday, May 7, 2012 1:44 AM
Subject: [R] Statistical power of correlations.
My apo
Great thanks Peter!
Collin.
On Mon, 7 May 2012, peter dalgaard wrote:
>
> On May 7, 2012, at 07:44 , Collin Lynch wrote:
>
> > My apologies for the statistical naivete of my question but...
> >
> > Is there an established method or calulating the statistical power of a
> > correlation te
Thank you for replies. I sort out the problem by defining the reassign
matrix.
Best Wishes,
efulas
--
View this message in context:
http://r.789695.n4.nabble.com/Repeating-tp4614371p4615072.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi David,
I am sorry for cross posting. I will keep this in mind next time.
Regarding details, my doubt is regarding the documentation in the
package where, it speaks about applying the function to held-out words
but the input parameters don't take new datasets, rather they use the
output generated
Hello,
YN wrote
>
> Hi all,
>
> One of my variables looks like this:
>
> .7_-.3_-.2_.9
>
> And this is a character variable. I made this by combining four different
> number like .7, -.3, -.2, and .9 using paste function.
> Now, I want to go back to original format from this one combined char
Oops,
Sorry, wrong number.
x <- 20101020
as.Date(as.character(x), format="%Y%m%d")
as.POSIXct(as.character(x), format="%Y%m%d")
Rui Barradas
--
View this message in context:
http://r.789695.n4.nabble.com/Dates-in-R-tp4614266p4614756.html
Sent from the R help mailing list archive at Nabble.co
Hello,
efulas wrote
>
> By the way, my "for" function is below, I can't find the mistake
>
>
> rand.max.t<- function(n){
> f<-rep(NA,n)
>
> for (i in 1:n) {
> reassign[i]<-matrix(c(sample(id),data1),203,3)
> new.data<-reassign[,1]
> random.cas=reassign[new.data==0,2:3]
> random.con=reassign[
Hello,
Try
x <- 20102010
as.Date(as.character(x), format="%Y%d%m")
[1] "2010-10-20"
as.POSIXct(as.character(x), format="%Y%d%m")
[1] "2010-10-20 BST"
Note that you must pass x as a character vector.
If not, the date functions will see it as the number of days since an origin
such as 1970-01-01
On Mon, May 7, 2012 at 3:21 PM, Apoorva Gupta wrote:
> Dear R users,
> I am working with panel data and I want the difference of a variable with
> its t+1 value.
>
> Could you tell me if such a function exists in the plm package?
>
Perhaps diff() or lag(). See the plm vignette.
Liviu
_
I tried the subset argument as Michael suggested, which led to the
same results. The results between meta-regression and subgroup
analyses were only slightly different as Wolfgang had suggested. I
also believe that these minor differences must be arising from the use
of random effects.
Thank you v
Hi Bryan,
On 07/05/2012 15:33, "Bryan Hanson" wrote:
>I don't know the answer, Jessica gave some insight.
>
>I avoid the biplot at all costs, because IMHO it violates one of the
>tenets of good graphic design: It has two entirely different scales on
>axes. These are maximally confusing to the
Hi Jessica,
Yes, that does help. It confirms my digging around in the prcomp object.
I was plotting $x, but wasn't sure whether this was appropriate. Mainly
because the data ranges are different in $x than when plotted by biplot()
- as I mentioned my reply to Bryan. Do you know if this difference
Hi Jessica,
THanks for pointing that out. The scaling in biplot() doesn't seem to make
sense to me, however. The default value for scale=1 therefore lambda ^
(1-scale) -> lambda ^ 0 which is 1 regardless of what lambda is. Which
can't be right?
Anyway, I won't worry about it anymore as you and B
In RStudio select the lines to be commented (or uncommented) and press
Ctrl+/ or select comment/uncomment on the Edit menu tab
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Me
Hi Bryan,
Many thanks for the replies.
The data is gene expression data for 36 samples over 11k genes.
I see that I can plot PC1 vs PC2 by using $x, but compared to biplot() I
can see that the range of values are different. For example, if I use
plot() the PC1 scale ranges from -150 to 150 wher
On May 7, 2012, at 6:28 AM, efulas wrote:
By the way, my "for" function is below, I can't find the mistake
rand.max.t<- function(n){
f<-rep(NA,n)
for (i in 1:n) {
reassign[i]<-matrix(c(sample(id),data1),203,3)
new.data<-reassign[,1]
random.cas=reassign[new.data==0,2:3]
random.con=reassign[ne
I don't see anything that looks like it should throw an error, but I
haven't tested your code without "data1" and "id" -- you might look at
? replicate() though -- it's designed for these sorts of things. E.g.,
replicate(100, mean(rexp(50)))
gets me a hundred draws of the mean of 50 random expone
I don't know the answer, Jessica gave some insight.
I avoid the biplot at all costs, because IMHO it violates one of the tenets of
good graphic design: It has two entirely different scales on axes. These are
maximally confusing to the end-user. So I never use it.
If it is gene expression dat
Thanks a lot for pointing me to that!
Best,
Simone
2012/5/7 David Winsemius :
>
> On May 7, 2012, at 7:15 AM, Simone Gabbriellini wrote:
>
>> Hello List,
>>
>> I have some plots with the wireframe() function, and I'd like to
>> display them in a single jpeg file. I know that par(mfrow=c(x,y)) wil
On May 7, 2012, at 7:15 AM, Simone Gabbriellini wrote:
Hello List,
I have some plots with the wireframe() function, and I'd like to
display them in a single jpeg file. I know that par(mfrow=c(x,y)) will
divide my display window in x rows and y columns, and although this
works with plot(), it l
And i always forget the question..
I haven't understood biplots a 100%, but from what i gleaned this scaling is
done so it looks better/is easier to read, while the scaling retains certain
properties of the biplot (something about projecting).
If you want to use the data for anything else, i wo
On May 7, 2012, at 6:20 AM, Beatriz De Francisco wrote:
I making an xyplot and the y label is too long and needs to be in
two rows, but when I brake it there is a huge gap between the last
text string and the expression, and I can't get rid of it. Any ideas?
My first idea would be that y
Biplot, depending on what parameters you give it, scales the data in a certain
way.
See
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/biplot.princomp.html
scale
The variables are scaled by lambda ^ scale and the observations are scaled by
lambda ^ (1-scale) where lambda are the sin
Hi YN,
I use strsplit for this:
x <- ".7_-.3_-.2_.9"
> strsplit(x, split = "_")
[[1]]
[1] ".7" "-.3" "-.2" ".9"
> strsplit(x, split = "_")[[1]][3]
[1] "-.2"
Best,
Ista
On Mon, May 7, 2012 at 9:54 AM, YN Kim wrote:
> Hi all,
>
> One of my variables looks like this:
>
> .7_-.3_-.2_.9
>
> And
Apologies for cross-posting
Members of this mailing list may be interested in the following book:
Zero Inflated Models and Generalized Linear Mixed Models with R.
Zuur, Saveliev, Ieno. (2012)
This book is only available from:
http://www.highstat.com/book4.htm
Kind regards,
Alain Zuur
--
D
Hi all,
One of my variables looks like this:
.7_-.3_-.2_.9
And this is a character variable. I made this by combining four different
number like .7, -.3, -.2, and .9 using paste function.
Now, I want to go back to original format from this one combined character
variable. For instance, I want to
Hello,
I have been using 'lda' package for topic modeling and wish to predict
new topics using a fitted model by giving new dataset as input. While
doing this, I came across the function predictive.distribution(). But
I am not able to apply it on a new dataset as the input parameter asks
for docume
Christian, is that 36 samples x 11K variables? Sounds like it. Is this
spectroscopic data?
In any case, the scores are in the list element $x as follows:
answer <- prcomp(your matrix)
answer$x contains the scores, so if you want to plot the 1st 2 pcs, you could do
plot(answer$x[,1], answer$x
To add: If thats not it, maybe you could be a bit more specific about what you
consider the "result", and how you want it visualized.
Am 07.05.2012 um 15:24 schrieb Jessica Streicher:
> That depends on what you want to plot there. Basically, you could just use
> plot() with pcaResult$x. You mig
On 07.05.2012 10:24, BrittD wrote:
Hi everyone,
I have a file in which the dates are subscribed as for instance: 20101020.
This is 20th Octobre 2010.
strptime("20101020", format="%Y%m%d")
seems to work for me...
UWe Ligges
My problem is that R won't except this as a date, since there is
That depends on what you want to plot there. Basically, you could just use
plot() with pcaResult$x. You might need to define which PCs you want to plot
there though.
pcaResult<-prcomp(iris[,1:4])
plot(pcaResult$x) # gives the first 2 PCs
plot(pcaResult$x[,2:3]) #gives the second vs the 3rd PC
o
Dear R users,
I am working with panel data and I want the difference of a variable with
its t+1 value.
For example, I have a data frame as below.
> a <- data.frame(c(rep(2,5), rep(3,5)), c(2005:2009, 2004:2008),
c(NA,10,34,23,12, 23,45, NA, 45, NA))
> colnames(a) <- c("firm","year","var")
I want
Hi Collin,
Look in the package 'pwr' for 'pwr.r.test'.
A.K.
- Original Message -
From: Collin Lynch
To: r-help@r-project.org
Cc:
Sent: Monday, May 7, 2012 1:44 AM
Subject: [R] Statistical power of correlations.
My apologies for the statistical naivete of my question but...
Is there a
Thank you both so much for your help! I ended up using
bquote(expression(...)), and it's working perfectly!
On Sat, May 5, 2012 at 1:05 PM, David Winsemius wrote:
>
> On May 5, 2012, at 2:42 PM, Josh Browning wrote:
>
> Hello useRs!
>>
>> So, I have a random question. I'm trying to build a cha
By the way, my "for" function is below, I can't find the mistake
rand.max.t<- function(n){
f<-rep(NA,n)
for (i in 1:n) {
reassign[i]<-matrix(c(sample(id),data1),203,3)
new.data<-reassign[,1]
random.cas=reassign[new.data==0,2:3]
random.con=reassign[new.data==1,2:3]
f<- list(x=random.cas[,1],y=ra
I have a decent sized matrix (36 x 11,000) that I have preformed a PCA on
with prcomp(), but due to the large number of variables I can't plot the
result with biplot(). How else can I plot the PCA output?
I tried posting this before, but got no responses so I'm trying again.
Surely this is a commo
Hi there,
I am trying to interface c++ code in R and make a package. With R CMD SHLIB
the dll was created, but when I try R CMD check, I am getting 'undefined
reference to..' linkage error messages.
The relevant c++ source from conf-infomap.cpp:
#include "conf-infomap.h"
#include "R.h" // R fu
Dear All
I would really appreciate some help with a script which a colleague wrote for
me, but I am having problems running (and have not been able to contact my
colleague).
The script is designed to compare the area of suitable habitat in binary
projections of a large number of species c
Dear All,
I have a codes which calculates the result of Ripley's K function of my
data. I want to repeat this process 999 times. However, i am getting an
error when i use the "for i in" function. Is there any way to repeat this
analysis 999 times. Here are the codes i used ;
data4 <- matrix(c(s
Hi everyone,
I have a file in which the dates are subscribed as for instance: 20101020.
This is 20th Octobre 2010.
My problem is that R won't except this as a date, since there is no sign to
seperate the Year, Month and Day
and that it will only see it as an origin, which it is not.
Does anyone k
Thanks. It does work now.
I also get another problem when I use naivebayes and prediction in myapplication
There was anerror message:
Error in table(predict(nb.obj, test.data[, subset, drop = FALSE]), test.data[, :
all arguments must have the same length
The length of test.data[,subset
And if the y-axis starts with 0: Instead of ylim=c(1,2), you take ylim=c(0,2).
Thus barplot(x[,1], ylim=c(0,2)) Hopefully it is now correct. Kind
regards,Esra Atescelik
> Date: Mon, 7 May 2012 09:55:06 +0200
> From: stude...@gmail.com
> To: r-help@r-project.org
> Subject: [R] y-axis-problem (b
Thanks Tal for answering,
Anyway I still have no idea on why the binomial GLM is missing the
relationship between the response variable and the explanatory variable
"cohort".
Is there anyone who might help me to understand this?
--
View this message in context:
http://r.789695.n4.nabble.com/Bi
using reshape:
library(reshape)
m <- melt(my.df, id.var="pathway", na.rm=T)
cast(m, pathway~variable, sum, fill=NA)
Jan
On 05/07/2012 12:30 PM, Karl Brand wrote:
Dimitris, Petra,
Thank you! aggregate() is my lesson for today, not melt() | cast()
Really appreciate the super fast help,
Karl
Hello List,
I have some plots with the wireframe() function, and I'd like to
display them in a single jpeg file. I know that par(mfrow=c(x,y)) will
divide my display window in x rows and y columns, and although this
works with plot(), it looks like it's not working with wireframe.
here's my code:
Hello,
I'm using fArma package to estimate the value of Hurst exponent using R/S
method. However, for a certain set of data I get H ~ 1.8. How do I
interpret this?
Following are the output that I get for this set:
> mean(data[,2])
[1] 400.5433
> sd(data[,2])
[1] 1139.786
>
> rsFit(data[,2], leve
1 - 100 of 118 matches
Mail list logo
