hello mike,
it would be helpful to know what you have done, what system you are working
on etc.
"JGR for Windows and Mac OS X consists of two parts: JGR R package which is
available from CRAN and JGR launcher. The latter is a small program that
starts JGR from its R package. It can be also used
Hi,
I need to permute the rows of a matrix, where each row is independently
rearranged. A simple solution is this:
shuffled <- datamatrix <- matrix(1:24, ncol = 4)
for (i in 1:nrow(datamatrix)) { shuffled[i, ] <- sample(datamatrix[i, ]) }
> datamatrix
[,1] [,2] [,3] [,4]
[1,]17 1
Dear all,
I have
-one matrix sr ([200,200]) which is the original file
-25 matrixes estimatedsr ([200,200]) which are an estimation of the sr.
As I am plotting these matrixes I want to have ONE common way that the cells
are colored. Eg. Values from 10-20--> red color, 20-30-->blue color (or
som
Using complex names, like res[, 3+i] or res$var, in the formula for a model is
a very bad idea, especially if eventually you want eventualluy to predict to
new data. (In fact it won't work, so that makes is very bad indeed.) So do
not use '$' or '[..]' terms in model formulae - this is going t
On Feb 10, 2011, at 12:24 AM, Ramya wrote:
hi there,
I need to subtract each value in the dataframe from the mean of the
column
and divide by the
standard deviation of the column.
dim(a)
[1] 2201152
data2 <- sapply(seq(from = 2, by = 1, length = 50), function(e){
rbind((a[[e]] - me
hi there,
I need to subtract each value in the dataframe from the mean of the column
and divide by the
standard deviation of the column.
> dim(a)
[1] 2201152
data2 <- sapply(seq(from = 2, by = 1, length = 50), function(e){
rbind((a[[e]] - mean(a[,e]))/sd(a[,e]))
})
Does this look right
Refresh list was used, but a restart of the client was necessary for
updates to become active.
Please advise as to appropriate list. Many thanks!
On Wednesday, February 9, 2011, David Winsemius wrote:
>
> On Feb 9, 2011, at 3:29 PM, matthew.schmidt wrote:
>
>
>
> I'm a newb with R and am trying
Hi,
How can I unsubscribe from the mailing list?
Best,
Julie mith
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org
Hi Gabor
Thank you for your reply
I tried it
> dbWriteTable(conn, "table6",mylargedata,identifier.quote ="\"")
Error in .sql.qescape(names(value), TRUE, conn@identifier.quote) :
The JDBC connection doesn't support quoted identifiers, but table/column name
contains characters that must be qu
I'm using stepAIC to fit a model. Then I'm trying to use that model to
predict future happenings.
My first few variables are labeled as their column. (Is this a problem?)
The dataframe that I use to build the model is the same as the data I'm
using to predict with.
Here is a portion of what is
Very well, Bill. all = TRUE seemed to fix the problem.
On Wed, Feb 9, 2011 at 7:54 PM, wrote:
> You want advice?
>
> 1. Write sentences that contain a subject and where appropriate, an object
> as well. This makes your email just that bit more polite. This list is not
> a paid service.
>
> 2.
Hi,
On Wed, Feb 9, 2011 at 11:18 PM, Benjamin Caldwell
wrote:
> So I needed to merge 17 .csv files, and did so by brute force, but I might
> need to do so again. Anyone have suggestions for a for loop that might do
> the below for me (where a:r are separate .csv files)
>
> ab<-merge(a,b,all=TRUE)
you might try merge_recurse(list(DF1,DF2,DF3,DF4,DF17)) in the reshape
package
--
View this message in context:
http://r.789695.n4.nabble.com/for-loop-to-merge-csvs-tp3298549p3298565.html
Sent from the R help mailing list archive at Nabble.com.
__
So I needed to merge 17 .csv files, and did so by brute force, but I might
need to do so again. Anyone have suggestions for a for loop that might do
the below for me (where a:r are separate .csv files)
ab<-merge(a,b,all=TRUE)
cd<-merge(c,d,all=TRUE)
ef<-merge(e,f,all=TRUE)
gh<-merge(g,h,all=TRUE)
On Wed, Feb 9, 2011 at 11:05 PM, Armand Pirvu (gmail)
wrote:
> Hi Gabor
>
> Thank you for your reply
>
> I tried it
>
>> dbWriteTable(conn, "table6",mylargedata,identifier.quote ="\"")
> Error in .sql.qescape(names(value), TRUE, conn@identifier.quote) :
> The JDBC connection doesn't support quot
You want advice?
1. Write sentences that contain a subject and where appropriate, an object as
well. This makes your email just that bit more polite. This list is not a
paid service.
2. The "sheets" may have variables in common, but do they have the same name in
both, and the same class, and
On Wed, Feb 9, 2011 at 9:06 PM, Armand Pirvu (gmail)
wrote:
> Dear all
>
>
> Backend is Ingres DBMS
> I use RJDBC with Ingres JDBC driver
>
> I have this csv file
>
> "","Strategy","par1","m.1997.09.01"
> "1","ALF",2,0.1244
>
>
> which I try to load it RJDBC
>
>
> require (RJDBC)
> drv <-
> JDBC(
The function factor.scores does not "inherit" anything. It is a generic
function that provieds methods for a number of classes, including those you
mention. (The terminology is important if you are to understand what is going
on here):
> library(ltm)
Loading required package: MASS
Loading re
On 02/09/2011 09:21 PM, Benjamin Caldwell wrote:
Am trying to merge about 15 .csv tables - tried a test run but came up with
0 rows (no data for each variable/column head)
CAHSEE.EA.feb.2009<-read.csv("2009 CAHSEE EA feb 2009.csv", header=TRUE)
CAHSEE.IM.MATH.2009<-read.csv("2009 CAHSEE Impact
Am trying to merge about 15 .csv tables - tried a test run but came up with
0 rows (no data for each variable/column head)
> CAHSEE.EA.feb.2009<-read.csv("2009 CAHSEE EA feb 2009.csv", header=TRUE)
> CAHSEE.IM.MATH.2009<-read.csv("2009 CAHSEE Impact Math.csv", header=TRUE)
> testmerge<-merge(CAHSE
On Feb 9, 2011, at 3:29 PM, matthew.schmidt wrote:
I'm a newb with R and am trying to move from proprietary stats tools
to open
ones. Hopefully this post might help others doing the same.
I downloaded the R package for Mac
Posted to the wrong list.
and hacked together a couple scripts,
The function factor.scores is used with package ltm and others to estimate IRT
type scores for various models. It inherits objects of class grm, gpcm and a
few others. What I would like to do is to use the factor.scores function, but
feed it my own item parameters (from a bifactor model where
Have
> actualsdf
ID Name datadate val
1 23 Acme Corp1 23
2 23 Acme Corp2 43
3 23 Acme Corp3 54
4 23 Acme Corp4 65
5 23 Acme Corp5 23
6 23 Acme Corp6 43
7 23 Acme Corp7 NA
8 23 Acme Corp8 43
9 23 Acme Corp
Dear all
Backend is Ingres DBMS
I use RJDBC with Ingres JDBC driver
I have this csv file
"","Strategy","par1","m.1997.09.01"
"1","ALF",2,0.1244
which I try to load it RJDBC
require (RJDBC)
drv <-
JDBC("com.ingres.jdbc.IngresDriver","/home/ingres/ingresv1/ingres/lib/iijdbc.jar")
conn <- dbC
If I have understood correctly, then as an example for 4 columns how about
d=data.frame(C1=c(1,1,1,1,1),C2=c(2,2,2,2,2),C3=c(3,3,3,3,3),C4=c(4,4,4,4,4))
combs=combn(paste("C",seq_len(4),sep=""),2,simplify=FALSE)
one = sapply(combs,function(x) d[x[1]])
two = sapply(combs,function(x) d[x[2
On Wed, Feb 9, 2011 at 6:12 PM, Nikhil Joshi wrote:
> I am trying to read an xlsx spreadsheet (1506 rows, 501columns) all
> populated but getting the following error:
> Please advise as to how to get around this issue.
>
>> res <- read.xlsx("c:\\BSE_v2.xlsx",1)
> Error in .jcall("RJavaTools", "Lja
Has anyone heard whether Revolution Analytics is going to release this
capability to the R community?
http://www.businesswire.com/news/home/20110201005852/en/Revolution-Analytics-Unlocks-SAS-Data
Dan
Daniel Nordlund
Bothell, WA USA
__
R-help@r-projec
Read the help page for ifelse again. It return a value; you do not
assign values with the the function. You would use it something like
this:
cbitotal <- ifelse(cond1
, result1
, ifelse(cond2
, result2
On 11-02-09 3:47 PM, eck 1 wrote:
Hello R-Help users,
I have a data frame named fd, a sample of which looks like this:
cbi_A
cbi_B
cbi_B1
cbi_B2
cbi_C
cbi_D
cbi_E
cbi_F
2.183451
1.047546
NA
NA
NA
NA
0.428528
NA
0.795837
0.510152
0.510152
NA
NA
NA
NA
NA
0.795837
I am trying to read an xlsx spreadsheet (1506 rows, 501columns) all
populated but getting the following error:
Please advise as to how to get around this issue.
> res <- read.xlsx("c:\\BSE_v2.xlsx",1)
Error in .jcall("RJavaTools", "Ljava/lang/Object;", "invokeMethod", cl, :
java.lang.OutOfMemor
[New to the community; still in early part of R's learning curve.]
Several months ago, I was requested to generate some graphs on a
periodic basis. Accordingly, I managed to figure out a way to do so,
using a combination of Perl and R (in a FreeBSD environment).
While I've needed to adjust a few
ClassS *<-* setRefClass*(*'ClassS',
fields = c*(*
'field.s'
*)*,
methods = list*(*
initialize = *function**()* *{*
*return**(*initFields*(*field.s = 0*))*
*}*
*)*
*)*
ClassM *<-* setRefClass*(*'ClassM',
fields = c*(*
'field.m'
*)*,
methods = list*(*
initialize = *function**()*
Hi,
im searching for a package or function to transform a data frame.
Initially the columns represents the ranks and the rows the test
persons, which sorts different samples by ranks. Unfortunately sometimes
one samples is not differentiable from another one, so they share one rank.
So, the orig
Thanks,
Nopes.. Does not seem to work.
> lapply(z, as.numeric)
[[1]]
[1] 1 1
[[2]]
[1] 2 2
[[3]]
[1] 3 3
[[4]]
[1] 4 4
> typeof(z[[1]][1])
[1] "character"
Shank
From: Henrique Dallazuanna [mailto:www...@gmail.com]
Sent: Wednesday, February 09, 2011 12:34 PM
To: Khanvilkar, Shashank
Cc: r-help@
I'm a newb with R and am trying to move from proprietary stats tools to open
ones. Hopefully this post might help others doing the same.
I downloaded the R package for Mac and hacked together a couple scripts, but
couldn't seem to get the right output. After installing the psych package, I
tried
Sorry.. my bad.. I should have done
z = lapply(z, as.numeric)
That works.
Shank
From: Khanvilkar, Shashank
Sent: Wednesday, February 09, 2011 1:33 PM
Cc: r-help@r-project.org
Subject: RE: [R] Question on sapply
Thanks,
Nopes.. Does not seem to work.
> lapply(z, as.numeric)
[[1]]
[1] 1 1
[[2]]
[1
Hello R-Help users,
I have a data frame named fd, a sample of which looks like this:
cbi_A
cbi_B
cbi_B1
cbi_B2
cbi_C
cbi_D
cbi_E
cbi_F
2.183451
1.047546
NA
NA
NA
NA
0.428528
NA
0.795837
0.510152
0.510152
NA
NA
NA
NA
NA
0.795837
1.149577
0.843485
1.122334
NA
Gabor,
Looking at your 2nd response, your suggestion is similar to that of
Derek Ogle. It was my misinterpretation from your first response that
elicited my original rebuttal. I apologize.
Thanks,
Tony
On 2/9/2011 2:47 PM, Gabor Grothendieck wrote:
On Wed, Feb 9, 2011 at 1:02 PM, Anthony La
Hi Bert,
Thanks for your reply. If I understand correctly, prop.test() is not suitable
to
my situation. The input to prop.test() is 2 numbers for each group (# of
success
and # of trials, for example, groups 1 has 5 success out of 10 trials; group 2
has 3 success out of 7 trials; etc. prop.te
Xiaokuan -
Maybe this will get you started:
mypanel = function(x,y,...){
agg<-aggregate(y,list(x),function(x)c(mean(x,na.rm=TRUE),sd(x,na.rm=TRUE)))
mns = agg[,2][,1]
sds = agg[,2][,2]
for(i in 1:nrow(agg))llines(c(i-.1,i+.1),rep(mns[i],2),lwd=3)
for(i in 1:nrow(agg)){llines(c(i
We spotted that there were problems, and have realized that the as.Date.numeric
was a left-over form earlier; it will be removed in the next version of the Epi
package.
Thanks for the help.
br.
Bendix
> -Original Message-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent:
Hi, I have a dataset that has 2 groups of samples. For each sample, then
response measured is the number of success (no.success) obatined with the
number
of trials (no.trials). So a porportion of success (prpop.success) can be
computed as no.success/no.trials. Now the objective is to test if th
Christina Hemphill hsph.harvard.edu> writes:
>
> I'm attempting to run an LME model with random intercepts for location and
have received an error message.
> The model and error message are given below:
>
> > mixed.distance <- lme (fixed= log(min5inpnc) ~ i93centerline,
random=~1|location, dat
To answer part 2: You should read up on statistical distributions and
when a sample size is (or isn't) large enough to produce reliable
statistical parameters such as mean or variance. I suspect David was
implying that your yardstick, based on studentized residual, removes
valid samples.
For your number 2, look at the outliers data set in the TeachingDemos package
and run the 1st set of examples, yes it uses a different rule than you use, but
still a common one. Think about what is happening in the example, doesn't that
make you a little nervous about methods that automatically
Alberto,
you simply need a font that has those characters. For example on 10.6 this
works:
plot.new(); plot.window(0:1,0:1); text(0.5,0.5,"精市",cex=10,family="Hei")
You may want to check your system for fonts that you can use (those that have
chinese characters when you click on them in Font Bo
On 9 February 2011 18:33, Philipp Grueber wrote:
> thanks for the quick answer & sorry for the mistake. Please find a corrected
> version of the code below. Unfortunately, the model still does not work –
> due to an error I believed to have overcome: “In log(2 * pi * sig2[i]) :
> NaNs produced”
Y
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Tim Howard
> Sent: Wednesday, February 09, 2011 12:44 PM
> To: r-help@r-project.org
> Subject: [R] assign factor levels based on list
>
> All,
>
> Given a data frame and a list
On Feb 9, 2011, at 4:18 PM, David Winsemius wrote:
On Feb 9, 2011, at 3:44 PM, Tim Howard wrote:
All,
Given a data frame and a list containing factor definitions for
certain columns, how can I apply those definitions from the list,
rather than doing it the standard way, as noted below.
Hi,
I have a data frame like this:
ScoreDoseSex
2.81Dose1M
1.81Dose1M
1.22Dose1M
0.81Dose1M
0.49Dose1M
0.22Dose1M
0.00Dose1M
-0.19Dose1M
-0.17Dose1F
-0.32Dose1F
-0.46Dose1F
-0.58Dose1F
-0.70Dose1F
-0.81Dose1F
-0.91Dose1F
-1.00Dose1F
-1.77Dose2M
-1.85Dose2M
-1.93Dose2M
-2.00Dose2M
-2.07Dose2M
-2.14
On Feb 9, 2011, at 3:44 PM, Tim Howard wrote:
All,
Given a data frame and a list containing factor definitions for
certain columns, how can I apply those definitions from the list,
rather than doing it the standard way, as noted below. I'm lost in
the world of do.call, assign, paste, and
On Feb 9, 2011, at 3:50 PM, Erin Hodgess wrote:
Dear R People:
What is the difference between setIs and setAs, please?
Consider the semantic difference btwn coercion and membership.
Thanks,
Sincerely,
Erin
--
David Winsemius, MD
West Hartford, CT
_
Dear R People:
What is the difference between setIs and setAs, please?
Thanks,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
R
On Wed, Feb 9, 2011 at 1:02 PM, Anthony Lawrence Nguy-Robertson
wrote:
> Thank you R-forum for you generous help.
>
> Gabor Grothendieck, I am not sure if anova in the form that you suggested is
> the most appropriate (This is probably more of a statistics related, rather
> than R related at this
All,
Given a data frame and a list containing factor definitions for certain
columns, how can I apply those definitions from the list, rather than doing it
the standard way, as noted below. I'm lost in the world of do.call, assign,
paste, and can't find my way through. For example:
#set up d
Try this:
lapply(z, as.numeric)
On Wed, Feb 9, 2011 at 5:16 PM, Khanvilkar, Shashank
wrote:
> Hello All,
> Thanks in advance for all help
> I have the following vector of strings that I want to split..
>
> > y = c("1/1","2/2", "3/3", "4/4")
> > y
> [1] "1/1" "2/2" "3/3" "4/4"
> > z = strsplit(y,
Hello All,
Thanks in advance for all help
I have the following vector of strings that I want to split..
> y = c("1/1","2/2", "3/3", "4/4")
> y
[1] "1/1" "2/2" "3/3" "4/4"
> z = strsplit(y,"/")
> z
[[1]]
[1] "1" "1"
[[2]]
[1] "2" "2"
[[3]]
[1] "3" "3"
[[4]]
[1] "4" "4"
Now how do I convert all
I'm attempting to run an LME model with random intercepts for location and have
received an error message. The model and error message are given below:
> mixed.distance <- lme (fixed= log(min5inpnc) ~ i93centerline,
> random=~1|location, data = nosmokecomp)
Error in y[revOrder] - Fitted : non-
On Wed, Feb 9, 2011 at 11:20 AM, Erik Iverson wrote:
>
>
> Mark Knecht wrote:
>>
>> Title asks it all.
>>
>> Thanks in advance,
>> Mark
>>
>> a = 1:5
>> b1 = 2:6
>> Z = data.frame(a,b1)
>> Z
>>
>> Z$b1
>>
>> count = 1
>>
>> MyName = paste("b",count,sep="")
>> MyName
>>
>> Z$MyName
>
> Z[[MyName]]
On Feb 9, 2011, at 2:20 PM, Erik Iverson wrote:
Mark Knecht wrote:
Title asks it all.
Thanks in advance,
Mark
a = 1:5
b1 = 2:6
Z = data.frame(a,b1)
Z
Z$b1
count = 1
MyName = paste("b",count,sep="")
MyName
Z$MyName
Z[[MyName]]
Also:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-
If you insist ...
1. You are reinventing wheels (poorly).
RSiteSearch("outlier tests",restr="fun")
##RsiteSearch is a handy interface to search facilities on CRAN.
# Go to the site directly for more. Or use Google or other search engines.
will show you that a R package, outlier, already exists t
Hi Anoop,
Not sure whether it's too late for you, but is this along the lines of
what you want?:
x<-0:1000
y<-sin(x/500)+rnorm(1001)/10
z<-cos(x/500)+rnorm(1001)/100
plot(x,y,col=rgb(1,(z-min(z))/(max(z)-min(z)),0))
Or possibly like:
plot(x,y,col=rainbow(256)[255*(z-min(z))/(max(z)-min(z))])
M
Mark Knecht wrote:
Title asks it all.
Thanks in advance,
Mark
a = 1:5
b1 = 2:6
Z = data.frame(a,b1)
Z
Z$b1
count = 1
MyName = paste("b",count,sep="")
MyName
Z$MyName
Z[[MyName]]
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Try:
for(i in angus) {
cut <- StoreData[[paste("P", i)]]
table(StoreData$CompanyID, !is.na(cut))
}
On Wed, Feb 9, 2011 at 1:02 PM, Rita Carreira wrote:
>
>
> Hello!
> I would like to do some tables for several variables and I would like to
> write a loop that does the table for each variable
On Feb 9, 2011, at 1:25 PM, kirtau wrote:
I have two questions,
1) if the solutions is only three or four lines of code is there
anyway you
can share those lines instead of stating that the solution is easy and
providing no code. I prefer not to use an R-Package but have a "raw
function".
Title asks it all.
Thanks in advance,
Mark
a = 1:5
b1 = 2:6
Z = data.frame(a,b1)
Z
Z$b1
count = 1
MyName = paste("b",count,sep="")
MyName
Z$MyName
N = as.name(MyName)
Z$N
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Folks,
I need to create matched test and / control pairs from a (dis)similarity
matrix. I am not sure if the existing matching softwares (e.g., optmatch,
matchit ) will allow me to input my own dissimilarity matrix.
It seems that these algorithms are focusing more on observational studies
with t
Hi Stella,
in your coding 'cut' is a string, not a data object.
something like
cut <- paste("P",i, sep="")
table(StoreData$CompanyID, !is.na(StoreData[,cut]))
should work.
hth.
Am 09.02.2011 19:02, schrieb Rita Carreira:
>
>
> Hello!
> I would like to do some tables for several variables an
On 09/02/2011 11:11 AM, koooee wrote:
is this the expected behavior of as.character ?
resultset is a data.frame from a sqlQuery() using RODBC
> a = as.character(as.vector(resultset[1]))
> a
[1] "c(-1, 1, 2, 3, 4, 5, 6, 7, 8, 9)"
I would expect the statement above to return similar to the res
We've scheduled R courses for February-April 2011 in Washington DC,
San Francisco, Seattle, New York City and other USA cities
http://www.xlsolutions-corp.com/Rcourses
February-April 2011
*** R/S: Programming Essentials
*** R Fundamentals and Programming Techniques
*** Advanced Statistical Model
Thank you R-forum for you generous help.
Gabor Grothendieck, I am not sure if anova in the form that you
suggested is the most appropriate (This is probably more of a statistics
related, rather than R related at this point). The way I understand
anova is that you are testing the variance betwe
> Date: Wed, 9 Feb 2011 12:18:57 -0500
> From: yanliu...@gmail.com
> To: r-help@r-project.org
> Subject: [R] heatmap-how to change the order of the rows (genes)
>
> Hi,
>
> I have a question about the heatmap dendrogram in R. I loaded my data
> matrix in
koooee wrote:
is this the expected behavior of as.character ?
resultset is a data.frame from a sqlQuery() using RODBC
a = as.character(as.vector(resultset[1]))
a
[1] "c(-1, 1, 2, 3, 4, 5, 6, 7, 8, 9)"
I would expect the statement above to return similar to the result below, am
I missing so
I have two questions,
1) if the solutions is only three or four lines of code is there anyway you
can share those lines instead of stating that the solution is easy and
providing no code. I prefer not to use an R-Package but have a "raw
function".
2) Can you explain why you feel that this is "
I am not sure about this... I think they have logFC larger than 2... you are
simply seeing them in scientific notation.
Why don't you try setting the option scipen (penalty for scientific notation)
"on the high side"?
Something like:
> options(scipen = 50)
should be more than enough...
An
I have two questions,
1) if the solutions is only three or four lines of code is there anyway you
can share those lines, without disrespecting me further
2) Can you explain why you feel that this is "statistical malpractice"
-
- AK
--
View this message in context:
http://r.789695.n4.nabbl
The knee jerk thought I had was to express the correlation matrix as a generic
Choleski decomposition, then randomly populate the triangular decomposed
matrix. When you remultiply, you can simply rescale to 1s on the diagonals.
Then rmnorm as usual.
In R, see ?chol
If you want to get fancy,
Hello!
I would like to do some tables for several variables and I would like to write
a loop that does the table for each variable. I should also point out that my
data set has several missing observations and sometimes the observations that
are missing are not the same for all my variables.
ab = paste(a,b,sep=";~;~;~")
flag = length(ab)==length(unique(ab))
This should work unless you use 3 consecutive winking elephants in other places
in your program.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nipesh Bajaj
Sen
is this the expected behavior of as.character ?
resultset is a data.frame from a sqlQuery() using RODBC
> a = as.character(as.vector(resultset[1]))
> a
[1] "c(-1, 1, 2, 3, 4, 5, 6, 7, 8, 9)"
I would expect the statement above to return similar to the result below, am
I missing something?
> as.
I'm going to work with R and .NET in a few weeks,
I read that I need R version 11 to be able to call R from .NET,
is it true, or is it possible with version 2.12.1?
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Hi Jim,
Thanks a lot for the response. I finally decided to do it with ggplot2
library, qplot and layers. It works pretty well, although I did not know how
to get the time working... so thanks!
Anoop
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The generic is what the user calls, the method is what the generic calls.
For example, the "summary" function is a generic function, when you issue a
command like summary(x) then the generic function looks at what type of object
x is and calls the appropriate method which might be the summary.lm
Sounds like a job for "trellising" using ?xyplot or "faceting" using
ggplot in the gglplot2 package (two terms for the same idea, but
different API's for implementation). I suggest you check the R Graph
Gallery at:
http://addictedtor.free.fr/graphiques/
for ideas. I'm sure you'll find both desig
Hi list.
I'm trying to plot a graph "by" factors. Exactly, the x axis are my depths
(as.factor), my left y axis are my transects (also as.factor) and I want to
plot the mean and standard deviation (three samples per factor combination)
of my SW (numeric) variable. The second y axis (at the right)
On Wed, Feb 09, 2011 at 07:39:55AM -0800, PtitBleu wrote:
>
> Thanks for all your answers. I didn't know this limit.
>
> I finally found another way to go around this problem: luckily the 4 first
> figures are always the same for all these numbers.
> I created a new column with as.numeric(substr(
*** APOLOGIZES FOR THOSE READING THE LIST THROUGH NABBLE THIS WAS ALREADY
POSTED THERE BUT NOT FORWARDED TO THE LIST FOR SOME UNKNOWN REASON ***
I have a dataset that looks like:
$ V1: factor with 4 levels
$ V2: factor with 4 levels
$ V3: factor with 2 levels
$ V4: num (summing up to 100 wit
Hi Arne,
thanks for the quick answer & sorry for the mistake. Please find a corrected
version of the code below. Unfortunately, the model still does not work –
due to an error I believed to have overcome: “In log(2 * pi * sig2[i]) :
NaNs produced”
So my questions remain the same: Why doesn't thi
> Date: Wed, 9 Feb 2011 18:23:21 +0100
> From: savi...@praha1.ff.cuni.cz
> To: r-help@r-project.org
> Subject: Re: [R] precision of gamma function
>
> On Wed, Feb 09, 2011 at 05:23:27PM +0200, Chuse chuse wrote:
> > Dear R users,
> >
> > I have to calcula
Hi:
On Wed, Feb 9, 2011 at 6:29 AM, Claudio Carneiro <
claudio_carneiro...@hotmail.com> wrote:
>
> Hi, all
>
> I'm trying to solve an optimization problem with two
> variables, in wich Iuse the following functions(they all are working
> perfectly):
>
> fxx1=function(x){
> calc=((2.5/3)*((x/3)^(2.
Hi,
On Wed, Feb 9, 2011 at 12:18 PM, yan liu wrote:
> Hi,
>
> I have a question about the heatmap dendrogram in R. I loaded my data
> matrix in command heatmap(), and it gave me a heatmap accordingly, and the
> rows (genes) also were clustered accordingly. But now I don't want the
> genes clust
On Wed, Feb 09, 2011 at 05:23:27PM +0200, Chuse chuse wrote:
> Dear R users,
>
> I have to calculate gamma functions for negative numbers beyond -171.4.
> e.x. gamma(-500.4)
> I got following:
>
> > gamma(-170.4)
> [1] -5.824625e-308
> > gamma(-171.4)
> [1] 0
> Warning message:
> underflow occurr
Hi,
I have a question about the heatmap dendrogram in R. I loaded my data
matrix in command heatmap(), and it gave me a heatmap accordingly, and the
rows (genes) also were clustered accordingly. But now I don't want the
genes clustered in that way, I have a new order of these genes, and want the
Your basic problem seems to be that the function RxMixed does not
return a value of the same length as its input parameter. I just used
the 'debug' library and here is what I got:
> mtrace(RxMixed)
> integrate(RxMixed,0,10)
D(3)>
[1] 0.8826402
D(3)> x
[1] 5. 0.13046736 9.86953264 0.674
Hi:
Here's one approach to the problem you posed (don't know if it is what you
want for the problem you intend you use it on...)
df1 <- read.table(textConnection("
id sexage area
01 male adult LP
01 male adult LP
01 male
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of PtitBleu
> Sent: Wednesday, February 09, 2011 4:15 AM
> To: r-help@r-project.org
> Subject: [R] Problem with long number (from character to numeric class)
>
>
> Hello,
>
> I hav
Hi Rick,
you can generate an arbitrary matrix X and calculate X'X, which is at
least positive semi-definite (and definite, when X has full rank)
X<-matrix(rnorm(16),4)
covmat<-t(X)%*%X
#check
!any(eigen(covmat)$value<0)
#corresponding correlation matrix
cov2cor(covmat)
Random multivariate can b
Hi all,
I have a hypothesis test of proportion question without using R. I hope you
could help to answer my question.
Here is the statement.
*
Preliminary data from a feasibility study showed a potential failure level
of 1%. A sample size was calculated to detect a potential
Hi David,
I'm CC-ing R-help inorder to finish this one off ;-)
On Wed, Feb 9, 2011 at 10:59 AM, Robinson, David G wrote:
[snip]
> One of you comments pointed me in the right direction and I found the
> problem. I simply commented out the line " if (j%%100==0) { ...print(N)}"
> and the original p
Hi BioC user,
I have a problem extracting the gene set I would like to work with.
Here is I work with my data:
normData <- read.delim("normalizedData.txt",sep ="\t")
# two class unpaired comparison
# y must take values 1,2
classes <- c(-1,-2,1,2)
#prepere the data for the samr analysis
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