Dear R user
how to solve the error of singular matrix in composition package
thanks in advance
ros
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PLEASE do read the posting
It seems that tcl/tk doesn't work fine on R-2.11 version.
Anyone can run this code to confirm ?
[CODE]require(tcltk)
tt <- tktoplevel()
topMenu <- tkmenu(tt)
tkconfigure(tt,menu=topMenu)
fileMenu <- tkmenu(topMenu,tearoff=FALSE)
tkadd(fileMenu,"command",label="Quit",command=function() tkdestroy(t
I didn't see you got an answer posted to this question:
You can't modify a pdata.frame object. Your transforms turn it back to a
normal data frame and diff and lag won't work as expected.
Try:
Grunfeld.p <- pdata.frame(Grunfeld,c("firm","year"))
tmp <- transform(Grunfeld.p, d.value = diff(Grun
Hi Experts,
I am new to R, using following sample code for capping outliers using
percentile information. Working on large data (3 observations and 150
variables), loop I am using in the below mentioned code for detecting
outliers and capping to upper /lower percentile value is taking much ti
Hi
I have a character class and i need to convert into dataframe
data=("0","0","0","0")
I want a dataframe with each one should under a separate column
Please help me
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Hi, r-users
I got a problem when I try to call a *gls* function in loop structure.
The gls function seems not able to recognize the parameters that I pass
into the loop function!
(But, if I use lm function, it works.)
The code looks like this:
=
gls
Dear Alessio,
A few remarks.
- R-sig-mixed models is a better list for this kind of questions
- use the glmer() function if you want logistic or poisson regression
- the error you are getting is an indication that the model is too complex for
the data
- watch for colinearity in the covariates
B
On Mon, Nov 21, 2011 at 1:28 PM, Giovanni Azua wrote:
>
> On Nov 21, 2011, at 8:31 PM, Bert Gunter wrote:
>> we disagree is that I think data analysts with limited statistical
>> backgrounds should consult with local statisticians instead of trying
>> to muddle through on their own thru lists like
Dear useRs & experRts,
I have the feeling that the 'name' argument to the attach function is
ignored when 'what' is a file name. Here is an example:
> save(letters, file="letters.RData")
> letters.env <- attach("letters.RData", name="letters")
> search()
> letters.env
The name on the search path
On Tue, 22 Nov 2011, habasque wrote:
It seems that tcl/tk doesn't work fine on R-2.11 version.
As the posting guide says, there is no such version, and this list is
only for current versions of R: you were asked to upgrade *before*
posting.
But it does work correctly in R 2.11.0 on my Linu
http://weka.sourceforge.net/doc.packages/rotationForest/weka/classifiers/meta/RotationForest.html
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Sent from the R help mailing list archive at Nabble.com.
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Essentially, this is a side-by-side forest plot, where the plot on the left is
for sensitivity and the plot on the right is for specificity. For 2x2 table
data from diagnostic studies, it is easy to calculate the sensitivity and
specificity values (and corresponding sampling variances) by hand.
I believe the command you are looking for is as.data.frame(), though you are
probably going to need as.double() rather soon as well.
Do note that data frames can, and often do, have character elements.
Best,
Michael
On Nov 22, 2011, at 1:21 AM, arunkumar wrote:
> Hi
>
> I have a charac
You can easily vectorize this code using pmin/pmax.
Sent from my iPad
On Nov 22, 2011, at 1:06, Aher wrote:
> Hi Experts,
>
> I am new to R, using following sample code for capping outliers using
> percentile information. Working on large data (3 observations and 150
> variables), loop I
I am using the "fitdistrplus" package in R and would like to do a goodness of
fit test. But there does not seem to be any option to do that. Any ideas on how
I can do that?
Thanks & Regards,
Indrajit
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R-
Hello Manuel.
Yes please, would you send me the data to reproduce the example? Else I
cannot tell, although this error is typical of undefined logs (zero or
negative argument).
Two general observations, for now:
- inserting special characters like '*' in variable names is looking for
trouble
- a
Being new to R myself, I always get trapped by factors. Taking the data you
have provided, this worked for my understanding of your intention:
> x <- rep( "0", 4 )
> x
[1] "0" "0" "0" "0"
> df <- data.frame( matrix( x, 1 ), stringsAsFactors = FALSE )
> df
X1 X2 X3 X4
1 0 0 0 0
> is.characte
Here is the solution using pmin/pmax for 10,000 rows.
> min_pctle_cut <- 0.01
> max_pctle_cut <- 0.99
> library(outliers)
>
> n <- 1
> x1 <- runif(n)
> x2 <- runif(n)
> x3 <- x1 + x2 + runif(n)/10
> x4 <- x1 + x2 + x3 + runif(n)/10
> x5 <- factor(sample(c('a','b','c'),n,replace=TRUE))
> x6 <-
On Nov 22, 2011, at 10:35 AM, Joshua Wiley wrote:
> It is true the way you use general lists is not our business, but the
> R-help list is a community and there are community rules. One of
I meant that my use of the lists is not of __his__ business I wasn't referring
to you nor other people in
Hi List,
Working on the large data frame (number of records=35000 and number of
variables=160).
Using redun() for dropping variables before using into model.
V <- redun(~., data = data.frame, r2 = 0.8)
It takes enormously high time for execution, is there anything wrong in the
script?
Suggest an
Specifying nk=0 to force all effects to be linear will speed things up.
Frank
aajit75 wrote
>
> Hi List,
>
> Working on the large data frame (number of records=35000 and number of
> variables=160).
> Using redun() for dropping variables before using into model.
>
> V <- redun(~., data = data.fr
Hi everybody,
I'm trying to select a model with the function step. It is a mixed
generalized linear model fitted by the function glmmML. I have one random
variable (id), one response variable (var) and many independent variables
(x1, x2, x3..). I obtain the following error:
Error in if (all(is.f
Dear all
I am trying to make a graphic with the "vioplot" package. I use the following
code:
library(vioplot)
x1 <- GSMrxDL
x2 <- WIFI
x3 <- UMT
vioplot(x1, x2, x3, ylim=c(0, 10), names=c("GSMrxDL", "WIFI", "UMT"),
col="gold")
title("NIS Strahlung", xlab="Sender", ylab="V/m")
Now I want to sca
Thanks all for suggestions.
I now have a nice plot showing the temperature of 6 different sites, each
site distinguished by different coloured points, using nested ifelse. My
apologies I thought I could change the type to "l" and the same arguments
would be applied to line graph, with 6 different
Hi. I have some problems trying to make cluster via "snow" package and
haven't found a solution for my problem in archives and relative topics.
I installed ssh server using cygwin and set a password-less SSH Login.
In R session after starting ssh service:
> system("ssh 10.10.5.15 date")
Tue
Dear all,
I have a problem with calculation of SE of parameters with a fisher
matrix. the fisher matrix is calculated from the hessian matrix which
is an output from optim(). However, can I calculate it this way if the
optimisation criterion was to minimise RMSE? Maybe it works only with
a log-lik
On 11/21/2011 12:43 PM, jazevedo wrote:
Hello, all,
I'm a new R user (new to any programming language in general, really), so I
apologize if this is easy/has already been answered (I've attempted
searching online but did not understand the pages I found).
My data is stored in text files with th
On 22.11.2011 12:37, french-connect...@gmx.net wrote:
Dear all
I am trying to make a graphic with the "vioplot" package. I use the following
code:
library(vioplot)
x1<- GSMrxDL
x2<- WIFI
x3<- UMT
vioplot(x1, x2, x3, ylim=c(0, 10), names=c("GSMrxDL", "WIFI", "UMT"),
col="gold")
title("NIS St
On 22.11.2011 12:37, french-connect...@gmx.net wrote:
Dear all
I am trying to make a graphic with the "vioplot" package. I use the following
code:
library(vioplot)
x1<- GSMrxDL
x2<- WIFI
x3<- UMT
vioplot(x1, x2, x3, ylim=c(0, 10), names=c("GSMrxDL", "WIFI", "UMT"),
col="gold")
title("NIS St
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html and provide commented,
minimal, self-contained, reproducible code!
On 22.11.2011 05:13, nur mohd wrote:
Dear R user
There are thousands on this list, none of them able to help you.
how to solve the error of singul
On 21.11.2011 21:51, yesitsjess wrote:
So basically I have made a HTML file with a table in it.
Column 3 contains a GenBank number and is always proceeded by "=GenBank">".
I want to read the file and return the number which comes directly after
this (the contents of column 3).
Ideally I woul
On Tue, Nov 22, 2011 at 2:09 PM, Giovanni Azua wrote:
> Mr. Gunter did not read/understand my problem, and there were no useful tips
> but only ad hominem attacks. By your side-taking I suspect you are in the
> same "party club" if you want to defend him maybe you should start by "tying
> bette
Hello All:
I am using the leaps package on scale and centered data for an
exhaustive search. There are Cp values of -Inf being returned for all
models. I was going to look at the source before contacting the list,
but it has been a while since I have looked under the hood. There are
.rdb a
On Nov 22, 2011, at 3:52 PM, Liviu Andronic wrote:
> On Tue, Nov 22, 2011 at 2:09 PM, Giovanni Azua wrote:
>> Mr. Gunter did not read/understand my problem, and there were no useful tips
>> but only ad hominem attacks. By your side-taking I suspect you are in the
>> same "party club" if you wa
On Fri, Nov 18, 2011 at 11:22 AM, Ashim Kapoor wrote:
> Dear all,
>
> I want to draw ticks on the 3rd and 4th row of a lattice. How do I do this
> ? In my search of the help, I discovered a parameter alternating,which kind
> of says where the ticks will be but does not suffice for me.
You need to
http://jennys.cz/modules/mod_wdbanners/yes.php?html143
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-con
There's also the lines() command which takes a col argument if you want to do
multiple lines (I usually wind up wrapping it in a for loop though there might
be something smarter)
ggplot2 is great, though the learning curve is a little rough: you can get good
help here but if you go down that pa
Dear Researchers,
someone know the right syntax to chose a 5% signiï¬cance level (threshold)
for the inclusion of the model variables in a multiple (linear) regression
in backward way?
I set the formula in this way, but I don't know to choose the 5%
significance?
lmodelV <-
step(lm(formula=MyF
On 22/11/2011 16:30, gianni lavaredo wrote:
Dear Researchers,
someone know the right syntax to chose a 5% signiï¬cance level (threshold)
for the inclusion of the model variables in a multiple (linear) regression
in backward way?
I set the formula in this way, but I don't know to choose the 5%
Hi,
I tried to run the VARMA model in the dse package. I specified a model:
> arma
A(L) =
1+0.244L10+0.05L1
0-0.325L11-0.234L1
B(L) =
1-0.277L10+0.211L1
0-0.206L11+0.238L1
and have a TSdata object:
> dfdata
output data:
Series 1 Series 2
1 "difex2" "difem2"
but I get this wa
Hi,
Is there an easy way to remove dataframe rows without duplicated values of
a specified column ('id')? e.g.,
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4), value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
2 1 6 4
3 1 7 3
4 2 4 3
5
Hi.
I need to compile an R package under Windows, to get a zip file.
I can't used the web services, because it is avalaible only for the current
version of R while I need of a package compiled with R 2.13.1.
The package contain C code that requires the GSL C library.
In your experience, what is
This is ugly, but it gets what you want.
dat[which(dat[,1] %in% unique((dat[duplicated(dat[,1], fromLast = T),
1]))),]
AC Del Re wrote
>
> Hi,
>
> Is there an easy way to remove dataframe rows without duplicated values of
> a specified column ('id')? e.g.,
>
> dat <- data.frame(id = c(1
On 22.11.2011 18:57, Nicola Sturaro Sommacal wrote:
Hi.
I need to compile an R package under Windows, to get a zip file.
I can't used the web services, because it is avalaible only for the current
version of R while I need of a package compiled with R 2.13.1.
The package contain C code that
Dear R People:
Hope you're having a nice day.
Here is a character vector:
> yz
[1] "pexp(3.2,rate=1)"
> str(yz)
chr "pexp(3.2,rate=1)"
>
And I would like to evaluate that vector.
I tried:
> eval(as.expression(yz))
[1] "pexp(3.2,rate=1)"
>
But that doesn't work.
Any suggestions would be most
Good morning Erin,
eval(parse(text = "pexp(3.2,rate=1)"))
seems to work
But the general rule applies:
library(fortunes)
fortune("parse()")
Best,
Michael
On Tue, Nov 22, 2011 at 1:23 PM, Erin Hodgess wrote:
> Dear R People:
>
> Hope you're having a nice day.
>
> Here is a character vector:
>
> On 22/11/11 13:04, Andy Bunn wrote:
> > Apologies for thickness - I'm sure that this operates as documented
> and with good reason. However...
> >
> > My understanding of arima.sim() is obviously imperfect. In the
> example below I assume that x1 and x2 are similar white noise processes
> with a
On 22.11.2011 19:23, Erin Hodgess wrote:
Dear R People:
Hope you're having a nice day.
Here is a character vector:
yz
[1] "pexp(3.2,rate=1)"
str(yz)
chr "pexp(3.2,rate=1)"
And I would like to evaluate that vector.
I tried:
eval(as.expression(yz))
[1] "pexp(3.2,rate=1)"
But th
Hi:
Here's one way:
do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
id value value2
1.1 1 5 1
1.2 1 6 4
1.3 1 7 3
3.5 3 5 4
3.6 3 4 3
HTH,
Dennis
On Tue, Nov 22, 2011 at 9:43 AM, AC Del Re wrote:
> Hi,
>
> Is there an easy
Sorry, you need this first:
L <- split(dat, dat$id)
do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
D.
On Tue, Nov 22, 2011 at 10:38 AM, Dennis Murphy wrote:
> Hi:
>
> Here's one way:
>
> do.call(rbind, lapply(L, function(d) if(nrow(d) > 1) return(d)))
> id value value2
> 1.
one approach is the following:
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4),
value2 = c(1,4,3,3,4,3))
ind <- ave(dat$id, dat$id, FUN = length) > 1
dat[ind, ]
I hope it helps.
Best,
Dimitris
On 11/22/2011 6:43 PM, AC Del Re wrote:
Hi,
Is there an easy way to remove dat
On Nov 22, 2011, at 12:43 PM, AC Del Re wrote:
Hi,
Is there an easy way to remove dataframe rows without duplicated
values of
a specified column ('id')? e.g.,
dat <- data.frame(id = c(1,1,1,2,3,3), value = c(5,6,7,4,5,4),
value2 =
c(1,4,3,3,4,3))
dat
id value value2
1 1 5 1
On Wed, Nov 23, 2011 at 3:55 AM, Stephen Sefick wrote:
> Hello All:
>
> I am using the leaps package on scale and centered data for an exhaustive
> search. There are Cp values of -Inf being returned for all models. I was
> going to look at the source before contacting the list, but it has been a
Dear all, I was looking for the C program found within approxfun() function. I
already have a list of available C programs which are being used with R here,
https://svn.r-project.org/R/trunk/src/main/. However this list does not contain
above C function. Can somebody help me on where to find tha
I have used LME to fit a mixed effects model on my data. The data has
274 subjects with 1 to 6 observations per subject. Time is not linearly
associated with the outcome, so I used ns to fit a natural cubic spline
with 3 auto knots. Subject and the natural cubic time of spline are both
treated as r
Hi,
I just start to use R today! I am reading the R Help on read.csv and the
description for header says "header is set to TRUE if and only if the first
row contains one fewer field than the number of columns". Why is that? My
data has the same number of fields in the first row as the number of
co
Hey All,
So - I promise to write a blog post on this topic and post it somewhere on
the internet once I get to the bottom of this. Basically, the set-up to the
problem is like this:
1. I have a data frame with dim (2547290, 4)
2. I need to make SQL like lookups on the dataframe. I have been u
Hi all
I was wondering if it is possible to get rid of the horizontal strips and
produce each barchart with a left y axes and lower x axes only. Also can you
specify an exact size of graph ie 88mm wide with a font size of 'x'.
library(lattice)
library(latticeExtra)
n=as.factor(c(1:5,1:5))
Breed=a
Dear All,
in some functions of my package, I use the Matrix S4 class, as defined
in the Matrix package.
I don't want to depend on Matrix, however, because my package is
perfectly fine without Matrix, most of the functionality does not need
Matrix. Matrix is so included in the 'Suggests' line.
I
How about calling Matrix's namespace directly?
Matrix:::rowSums()
Michael
On Tue, Nov 22, 2011 at 3:16 PM, Gábor Csárdi wrote:
> Dear All,
>
> in some functions of my package, I use the Matrix S4 class, as defined
> in the Matrix package.
>
> I don't want to depend on Matrix, however, because m
Hi Gábor,
You could import rowSums. This will not fully attach Matrix. I am
not sure there is a really good solution for what you want to do. To
fully use and validate your package, Matrix appears to be required.
This is different from simply, for example, enhancing the Matrix
package.
You cou
take a look at using the 'data.table' package. Here are some times to
do the lookup using dataframes, matrices and data.tables: data.tables
give the answer is less than 0.1 seconds.
> str(x.df)
'data.frame': 250 obs. of 4 variables:
$ x : Factor w/ 455063 levels "","AAAB",..: 200683
I think, here is the solution. If NA is included in read.table list the row
becomes a factor:
$ Q21: Factor w/ 3 levels " 1"," 2"," NA": 1 2 3 2 2. This will not work
with rowSums.
If I put the missing value as a blank, then it is still read as NA but the
whole row is considered as an integer an
Hi everyone,
I am trying to get a point of intersection between a
polyline and a straight line ….. and get the x and y coordinates of this point.
For exemplification consider this:
set.seed(123)
k1 <-rnorm(100, mean=1.77, sd=3.33)
k1 <- sort(k1)
q1 <- rnorm(100, mean=2.37, sd=0
Hi dear all,
I am wondering if there is a function existing in R that did the quadratic
bezier curve interpolation? I hope to generate a bezier curve based on
three sets of points: two end of the line and a control point.
Thanks in advance.
Tengfei
--
Tengfei Yin
MCDB PhD student
1620 Howe Hal
If it's a one off, the identify() function might be of help -- if you need
something algorithmic it's harder due to floating point stuff and sampling
frequencies. Let me know if that's the case.
Michael
On Nov 22, 2011, at 3:40 PM, Monica Pisica wrote:
>
>
>
> Hi everyone,
>
>
>
> I a
require(Hmisc)
?bezier
?drawPlot
Frank
Tengfei Yin wrote
>
> Hi dear all,
>
> I am wondering if there is a function existing in R that did the quadratic
> bezier curve interpolation? I hope to generate a bezier curve based on
> three sets of points: two end of the line and a control point.
>
>
I was wondering what the best approach is for missing data in a time series.
I give an example using xts but I would like to know what seems to be the
"best" method. Say I have
library(xts)
xts.ts <- xts(1:4,as.Date(c("1970-01-01", "1970-1-3", "1980-10-10",
"2007-8-19")), frequency=52)
I w
Thanks, I have tried that, it does not work, because rowSums() calls
callGeneric():
> Matrix:::rowSums(W)
Error in callGeneric() :
'callGeneric' must be called from a generic function or method
G.
On Tue, Nov 22, 2011 at 3:20 PM, R. Michael Weylandt
wrote:
> How about calling Matrix's namespa
Couldn't you use seq.Date() to set up the time index and then just fill as
appropriate?
Alternatively, to.weekly if you are starting with a daily series.
Michael
On Nov 22, 2011, at 4:00 PM, "Kevin Burton" wrote:
> I was wondering what the best approach is for missing data in a time series.
Hi Josh,
On Tue, Nov 22, 2011 at 3:31 PM, Joshua Wiley wrote:
> Hi Gábor,
>
> You could import rowSums. This will not fully attach Matrix. I am
> not sure there is a really good solution for what you want to do. To
> fully use and validate your package, Matrix appears to be required.
> This is
Hi Prof. Frank Harrell,
The bezier function in Hmisc package is exactly what I am looking for.
Thanks a lot!
Tengfei
On Tue, Nov 22, 2011 at 2:55 PM, Frank Harrell wrote:
> require(Hmisc)
> ?bezier
> ?drawPlot
>
> Frank
>
> Tengfei Yin wrote
> >
> > Hi dear all,
> >
> > I am wondering if
On 11/22/2011 01:16 PM, Gábor Csárdi wrote:
Hi Josh,
On Tue, Nov 22, 2011 at 3:31 PM, Joshua Wiley wrote:
Hi Gábor,
You could import rowSums. This will not fully attach Matrix. I am
not sure there is a really good solution for what you want to do. To
fully use and validate your package, Ma
Hi,
No it is not one off, the situation is even more complicated i will have a
series of straight lines like the red one parallel with each other that
intersect the black polyline and i need to get all the points (x, y).
Meanwhile i was thinking if it will not be easier if somehow i can r
I also wondered why it is important to not mention
Matrix in the DEPSCRIPTION file's Depends or Imports
lines, even though some functions in the package
require it. If this is a hard requirement you could
split your package into two packages, pkgBasic and
pkgEnhanced. pkgBasic would not not depen
Hello,
Is there a way to save plots in vector format like SVG or smth else?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commen
a b
A 1 2
B 3 4
C 5 6
The assumption above is that the rownames don't have a header and
the columns do. Therefore the default is header=TRUE.
r a b
A 1 2
B 3 4
C 5 6
In the second example, the first column is called "r" and it is not
clear whether that is a column with a variable
Dear R users,
do you know an easy way (other than star plot) of making several points
laying one over another visible? Is it any simple way of increasing such
"multipoint" symbols - or shifting their positions randomly to make several
points in one place visible?
Cheers,
sz.
--
Szymon Drobniak
I believe the Cairo package provides SVG faculties.
Michael
On Nov 22, 2011, at 4:48 PM, Antonio Rodriges wrote:
> Hello,
>
> Is there a way to save plots in vector format like SVG or smth else?
>
> __
> R-help@r-project.org mailing list
> https://
On Nov 22, 2011, at 3:40 PM, Monica Pisica wrote:
(edited out excessive white space)
I am trying to get a point of intersection between a
polyline and a straight line ….. and get the x and y coordinates of
this point.
For exemplification consider this:
set.seed(123)
k1 <-rnorm(100, mean=1.7
On Nov 22, 2011, at 4:58 PM, Szymek Drobniak wrote:
Dear R users,
do you know an easy way (other than star plot) of making several
points
laying one over another visible? Is it any simple way of increasing
such
"multipoint" symbols - or shifting their positions randomly to make
several
Thank you for the suggestions.
The only problems I see with 'to.weekly' is converting from the OHLC format
and realizing that the date is the last day of the week rather than the
first day of the week. Very minor compared to doing the whole thing myself.
-Original Message-
From: R. Michae
On 11/18/2011 1:22 AM, ren...@vannieuwkoop.ch wrote:
\documentclass[11pt,a4paper]{article}
\usepackage{Sweave}
\begin{document}
<<>>=
x = runif(100, 1, 10)
y = 2 + 3 * x + rnorm(100)
@
<>=
library(xtable)
print(xtable(summary(lm(y~x)),
align="r",
caption="Summary statistics for
Also with to.weekly there seems to be a problem with when the week starts.
For example:
>xts.ts <- xts(1:4, c(as.Date("2011-01-01"), as.Date("2011-01-10"),
as.Date("2011-10-09"), as.Date("2011-10-10")), frequency=52)
> to.weekly(xts.ts)
xts.ts.Open xts.ts.High xts.ts.Low xts.ts.Close
20
On 22-Nov-11 21:25:56, Monica Pisica wrote:
> Hi everyone,
>
> I am trying to get a point of intersection between a
> polyline and a straight line
.. and get the x and y
> coordinates of this point.
> For exemplification consider this:
>
> set.seed(123)
> k1 <-rnorm(100, mean=1.77, sd=3.33)
> k1
Hi,
I am relatively new to R and Bioconductor and am trying to filter the
topTable that I generated of differentially expressed genes from my
normlized eset file comprised of ~ 40 HG-133A Affy microarrays . I would
like to see if particular probesets are represented in this list.
Alternatively I
Wow, these specs are fantastic:
> user system elapsed
>0.330.000.39
>
I wonder how much of that is because of the capacity of the box that you are
running R on. Can you post pertinent specs? This suggest to me that
hardware upgrades (RAM specifically) may also be in order.
Inv
I'm working in a Gen/Marker-Phenotype association study in wheat and I'm
using a Mixed Model Approach to estimate the effect of the markers. My model
has the atribute measured as y (response variable), the markers and the
blocks (of a complete random block design) as fixed and the genotypes and
the
So, here is the result time from using the datatable package:
> user system elapsed
> 0.800 0.012 1.847
>
Here are the methods that I am using:
ush <- data.table(read.csv(...))
setkey(ush, product_id)
s1 <- subset(ush, product_id == product.id)
Seems like a minor improvement but not
Hello there,
when using the function numSummary in Rcmdr and selecting more than one
variable (without grouping), the grand mean across all variables is returned
for each variable instead of the mean of each single variable. However, this
happens only for the mean, and not for sd, quantiles and
On Tue, Nov 22, 2011 at 4:27 PM, Martin Morgan wrote:
[...]
> No need to Depend:. Use
>
> Imports: Matrix
>
> plus in the NAMESPACE file
>
> importFrom(Matrix, rowSums)
>
> Why do you not want to do this? Matrix is available for everyone, Imports:
> doesn't influence the package search path. Ther
Update from email outside of this thread:
Justin Haynes writes:
> matrices will help, but two quick solutions:
>
> if you are looking for single items to go in the some_value space, use ==
> instead of %in% and you'll notice speedups. The second more involved
> option is to take a look at the
Answer to my own question:
ush <- data.table(read.csv(...))
setkey(ush, product_id)
s1 <- ush[J[product.id]]
> user system elapsed
> 0.000 0.000 0.003
>
It seems like that's the method to use! Amazing.
--
View this message in context:
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Success with the lines command and col argument! I have some nice point and
line plots.
Thanks so much for you help. Ongoing project - I will probably be back!
Sarah
--
View this message in context:
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S
Dear Boris,
This bug is traceable to a change in the mean() function in R and is fixed in
the current version (1.7-3) of the Rcmdr package on CRAN.
Best,
John
John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMa
On 23/11/11 07:31, R. Michael Weylandt wrote:
Good morning Erin,
eval(parse(text = "pexp(3.2,rate=1)"))
seems to work
But the general rule applies:
library(fortunes)
fortune("parse()")
The fortune notwithstanding I find this *specific* use of parse() to be
very, uh, useful! :-)
cheers,
Dear all,
I'm working on some data from an experiment on the breeding behavior of
birds. In short, I have been measuring how the time spent on performing a
certain task (variable 'mean_on_active') differs over time (variable 'day',
2 levels) across three experimental categories (variable 'treat';
If "Suggests" doesn't work for you, perhaps you need to put more effort into
reinventing the wheel, and depend less on other packages.
---
Jeff NewmillerThe . . Go Live...
DCN:
Void of any other suggestions this approach makes sense but for my case I
think I need to use zoo objects rather than xts. If I sequence the data
generally I don't know if there will be 365 days in the year or 366. So I
have to sequence the dates as:
seq(from=as.Date("2011-01-01"), to=as.Date("201
On 11/22/2011 03:06 PM, Gábor Csárdi wrote:
On Tue, Nov 22, 2011 at 4:27 PM, Martin Morgan wrote:
[...]
No need to Depend:. Use
Imports: Matrix
plus in the NAMESPACE file
importFrom(Matrix, rowSums)
Why do you not want to do this? Matrix is available for everyone, Imports:
doesn't influen
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