I am not sure whether you are working under windows. Hope the following
message helps.
Using the RMySQL package under Windows
http://www.stat.berkeley.edu/users/spector/s133/RMySQL_windows.html
On 2010-8-27 5:03, quant wrote:
I installed MySQL 5.0.67 and R. I installed RMySQL and added env var
On Thu, 26 Aug 2010, Jorge Ivan Velez wrote:
Hi pdb,
Take a look at
http://r.789695.n4.nabble.com/test-if-a-package-is-installed-td1750671.html#a1750674
You are citing yourself with a poor solution (calling
installed.packages()) -- so please don't.
The help page for installed.packages in R
require() does what you want. Run "?require" for details.
require() returns 'FALSE' and gives a warning (rather than an error as
'library()' does by default) if the package does not exist.
'require' returns (invisibly) a logical indicating whether the required
package is available. (You can c
On Fri, 27 Aug 2010, Ruben Garcia Berasategui wrote:
Dear list members,
I'm trying to use JAGS 2.1.0 from within R but every time I try to
load the rjags package I receive the following message:
Which version of 'the rjags package' and where did you get it from?
I have 2.1.0-6, installed as
On 26/08/10 19:48, David Winsemius wrote:
On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote:
I need the parameters estimated for a non-linear equation, an example
of the
data is below.
# rm(list=ls()) I really wish people would add comments to destructive
pieces of code.
Time<-c( 0, 0,
Hi Greg,
Test01 was created with;
> Test01=read.table(file=file.choose(), header=TRUE)
the file selected was a .txt file.
DayMonthMo.NumberDay_of_yearDraft_No.
1Jan11305
2Jan12159
3Jan13251
4Jan14215
..
1Fe
When you run any graphics command (layout in this case) and there is not a
current graphics device (more technically only the null device) then a default
graphics device is opened, that is what you are seeing. What you need to do
instead is open the device yourself before calling layout. Which
Ben Bolker gmail.com> writes:
>
> Ruben Garcia Berasategui binus.ac.id> writes:
>
> >
> > Dear list members,
> > I'm trying to use JAGS 2.1.0 from within R ...
>
Sorry, didn't see the subject line ("R 2.11.0") but still
worth posting sessionInfo()
___
Ruben Garcia Berasategui binus.ac.id> writes:
>
> Dear list members,
> I'm trying to use JAGS 2.1.0 from within R ...
Just a guess: are you using an up-to-date version of R?
I think you might be picking up an older version of rjags
which in turn is looking for an older version of JAGS ...
Hi Greg,
Thanks for your advice.
I'm not prepared altering the shape of the graphs to be plotted. What I'm
trying to do is to pop up a rectangle layout window with following command.
The command;
layout(matrix(1:2, nrow=1))
pop up a square window. What I need is a rectangular window for the
Dear list members,
I'm trying to use JAGS 2.1.0 from within R but every time I try to load the
rjags package I receive the following message:
Loading required package: coda
Loading required package: lattice
Error : .onLoad failed in loadNamespace() for 'rjags', details:
call: dyn.load(file, DL
On Aug 26, 2010, at 9:07 PM, pdb wrote:
Hi,
I am writing a function that requires a specific package to be
installed.
Is there a way of checking if the package is installed and returning
a TRUE
/ FALSE result so my function can return an appropriate error
message and
exit the functio
Hi pdb,
Take a look at
http://r.789695.n4.nabble.com/test-if-a-package-is-installed-td1750671.html#a1750674
HTH,
Jorge
On Thu, Aug 26, 2010 at 9:07 PM, pdb <> wrote:
>
> Hi,
>
> I am writing a function that requires a specific package to be installed.
>
> Is there a way of checking if the pac
Hi,
I am writing a function that requires a specific package to be installed.
Is there a way of checking if the package is installed and returning a TRUE
/ FALSE result so my function can return an appropriate error message and
exit the function gracefully rather than just bombing out?
I'm thi
There is a graphical parameter that controls whether a plot is square or takes
up the maximum amount of room (rectangle), see ?par and look at the entry for
pty.
It is possible that you set pty='s' or it may be that the plot method sets it,
without us knowing what type of object Date and Test0
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of chipmaney
> Sent: Thursday, August 26, 2010 3:00 PM
> To: r-help@r-project.org
> Subject: [R] Non-standard sorts on vectors
>
>
> I have a dataset I need to sort:
>
> test.df
Hi all
I was playing with termplot(), and came across what appears to be an
inconsistency.
It would appreciate if someone could enlighten me:
> # First, generate some data:
> y <- rnorm(100)
> x <- runif(length(y),1,2)
> # Now find the log of x:
> logx <- log(x)
>
> # Now fit two models that a
A factor with 5000 levels looks like it may be a numeric variable that was
accidently coded as a factor (functions like read.table will do this if there
is a non numeric character in with the numbers).
If you really have a 5000 level factor, which levels can be discarded or
combined is a questi
I agree. I typically do not use non-linear functions, so am seeing the
"art" in describing functions of non-linear plots. One last thing. I tried
to use a self-starting Weibull function with the posted data and received
the following error.
model<-nls(Level~ SSweibull(Time,Asym,Drop,lrc,pwr))
E
On Aug 26, 2010, at 5:20 PM, Marlin Keith Cox wrote:
> The background you requested are energetic level (joules) in a group
> of starved fish over a time period of 45 days. Weekly, fish (n=5)
> were removed killed and measured for energy. This was done at three
> temperatures. I am compa
On Aug 26, 2010, at 6:10 PM, Jorge Ivan Velez wrote:
Hi Chipper,
Try
test.df[unlist(sort.v),]
That does work, and I suspect it is because data.frame converts
character vectors to factors by default. So this also works:
test.df[factor(c("Lake","Shoreline","Floodplain")), ]
But why? I mu
Hi Christian!
Sorry for being late.
David Winsemius wrote:
On Aug 18, 2010, at 2:58 PM, christiaan pauw wrote:
Hi Ricardo and everybody
In this old post to rhelp you say that the problem was solved but not
what
the sollution was. I have the same problem now. I want to read a Excel
file
fr
Here's one possibility:
sort.v<-c(Lake=1,Shoreline=2,Floodplain=3)
test.df[order(sort.v[as.character(test.df$Zone)]),]
Zone Cover
2 Lake60
3 Shoreline70
1 Floodplain50
- Phil Spector
Sta
Hi Chipper,
Try
test.df[unlist(sort.v),]
HTH,
Jorge
On Thu, Aug 26, 2010 at 5:59 PM, chipmaney <> wrote:
>
> I have a dataset I need to sort:
>
> test.df<-data.frame(Zone=c("Floodplain", "Lake",
> "Shoreline"),Cover=c(50,60,70))
>
> However, I don't want it sorted ascending/descending but in
I have a dataset I need to sort:
test.df<-data.frame(Zone=c("Floodplain", "Lake",
"Shoreline"),Cover=c(50,60,70))
However, I don't want it sorted ascending/descending but in an order that I
define via a vector:
sort.v<-data.frame(c("Lake","Shoreline","Floodplain"))
I realize I could probabl
Thank you. Could you lastly help me with this error. I was trying to use a
self starting function (Weibull).
model<-nls(Level~ SSweibull(Time,Asym,Drop,lrc,pwr))
Error in qr.default(.swts * attr(rhs, "gradient")) :
NA/NaN/Inf in foreign function call (arg 1)
On Thu, Aug 26, 2010 at 1:42 PM, D
My opinions only below; consume at your own risk.
On Thu, Aug 26, 2010 at 2:20 PM, Marlin Keith Cox wrote:
> The background you requested are energetic level (joules) in a group of
> starved fish over a time period of 45 days. Weekly, fish (n=5) were removed
> killed and measured for energy. Th
OK, I think that I figured out what is going on. You have some of your x
values that are very close to each other in value, but not exactly the same.
If we look at how many unique x values you have we get:
> length(unique(approx.data$x))
[1] 901
But inside the approxfun function the tapply fu
The background you requested are energetic level (joules) in a group of
starved fish over a time period of 45 days. Weekly, fish (n=5) were removed
killed and measured for energy. This was done at three temperatures. I am
comparing the rates at which the fish consume stored body energy at each o
Dear Anderson,
please see
http://lists.nongnu.org/archive/html/igraph-help/2007-07/msg00010.html
for the solution of the 'postscript font not included' problem.
Best,
Gabor
On Wed, Aug 25, 2010 at 11:50 PM, anderson nuel wrote:
> Dear r-help,
>
> I took your advice into consideration and i trie
On 26/08/10 18.42, Martin Morgan wrote:
On 8/26/2010 8:43 AM, Niels Richard Hansen wrote:
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
function(x,y,...) call
I installed MySQL 5.0.67 and R. I installed RMySQL and added env variable
MYSQL_HOME. But R still does not want to load the library. It says
Error : .onLoad failed in loadNamespace() for 'RMySQL', details:
call: NULL
error: MYSQL_HOME was set but does not point to a directory
Error: package/n
A ddply solution is
dat.out <- ddply(dat, .(time), transform, slope = scale(slope))
but this is not faster than the loop, and slower than the ave() solution:
> system.time(
+ for (i in 1:3) {
+mat <- dat[dat$time==i, ]
+outi <- data.frame(mat$time, mat$id, slope=scale(mat$slope))
+if
On Thu, Aug 26, 2010 at 4:33 PM, Bos, Roger wrote:
> I created a small example to show something that I do a lot of. "scale"
> data by month and return a data.frame with the output. "id" represents
> repeated observations over "time" and I want to scale the "slope"
> variable. The "out" variabl
On Aug 26, 2010, at 3:40 PM, Marc Schwartz wrote:
> On Aug 26, 2010, at 3:33 PM, Bos, Roger wrote:
>
>> I created a small example to show something that I do a lot of. "scale"
>> data by month and return a data.frame with the output. "id" represents
>> repeated observations over "time" and I wa
It looks like you have found a bug, I can confirm that with your data on my
computer that I am getting nonsense results for some cases. I even found that
when calling the function on element 164 of the input vector that I don't even
get consistent results, I ran it several times and many times
On Aug 26, 2010, at 3:33 PM, Bos, Roger wrote:
> I created a small example to show something that I do a lot of. "scale"
> data by month and return a data.frame with the output. "id" represents
> repeated observations over "time" and I want to scale the "slope"
> variable. The "out" variable sh
Thx... The following seems to work. However I´m sure there is a more elegant
solution to it
tre<-dat
b<-length(dat)
attr(dat,"label.table")->a
for (i in 1:b){
if(!is.null(a[[i]]) &
length(levels(as.factor(dat[,i])))==length(a[[i]]))
{
tre[,i]<-factor(dat[,i
I created a small example to show something that I do a lot of. "scale"
data by month and return a data.frame with the output. "id" represents
repeated observations over "time" and I want to scale the "slope"
variable. The "out" variable shows the output I want. My for..loop
does the job but is
On Thu, Aug 26, 2010 at 2:31 PM, Bond, Stephen wrote:
> Please correct the following
>
>> sqldf("update esc left join forwagg on esc.ym=forwagg.Date set
>> esc.ri2=forwagg.N1 where esc.age=12","select * from main.esc")
> Error in sqliteExecStatement(con, statement, bind.data) :
> RS-DBI driver:
On Aug 26, 2010, at 1:48 PM, David Winsemius wrote:
On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote:
I need the parameters estimated for a non-linear equation, an
example of the
data is below.
# rm(list=ls())I really wish people would add comments to
destructive pieces of code.
Exact, "efficiency", I didn't see that. Thank you very much.
2010/8/26 Gavin Simpson :
> On Thu, 2010-08-26 at 10:42 +0200, David Hajage wrote:
>> Hello useRs and guRus,
>>
>> I was trying to add the support of str() in the ascii package, and I
>> realized that str() does not have a print method.
No need to do that. They have some instances that run 64-bit ubuntu.
If I remember correctly we had to install 64-bit R from the debian
packages on the ubuntu instance.
On Wed, Aug 25, 2010 at 6:12 PM, noclue_ wrote:
>
>
>>> You have a 64 bit Linux? If so...
>
>>>Dowload the sources
>
> Do you m
I have a dataset of multiple variables and a response. For example,
> str(x)
'data.frame': 3557238 obs. of 44 variables:
$ response : Factor w/ 2 levels
$ var2: Factor w/5000 levels
If var2 for example is a factor with 5000 levels, what is the best
approach to determine which of these level
Hello all-
Just wondering if anyone has heard of a package performing latent factor
analysis, similar to what is done with LatentGold. I know that PoLCA does
latent class analysis, but I don't think it can accomodate latent factor
models.
Thanks,
David Joubert
Dept Criminology
University of
On Aug 26, 2010, at 2:35 PM, Le Wang wrote:
Thanks, David.
I did try to use predict() to obtain the graph,
"Try"? How? Code?
( predict() is used to obtain numbers, not to do graphing. )
but it somehow looks
different from the one generated by "plot" command.
I have no idea what you actu
Thanks, David.
I did try to use predict() to obtain the graph, but it somehow looks
different from the one generated by "plot" command. So, I was
wondering if there is any way that I can get the one generated by
"plot" so that I can compare. Thank you.
Le
On Wed, Aug 25, 2010 at 11:15 PM, David
Please correct the following
> sqldf("update esc left join forwagg on esc.ym=forwagg.Date set
> esc.ri2=forwagg.N1 where esc.age=12","select * from main.esc")
Error in sqliteExecStatement(con, statement, bind.data) :
RS-DBI driver: (error in statement: near "left": syntax error)
Thanks.
Step
On Aug 26, 2010, at 1:35 PM, Marlin Keith Cox wrote:
I need the parameters estimated for a non-linear equation, an
example of the
data is below.
# rm(list=ls())I really wish people would add comments to
destructive pieces of code.
Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3,
Look for a vectorized solution such as ?table or ?aggregate to obtain a list
of combination counts, followed by logical indexing or ?subset to get a set of
valid combinations, and then use ?sample to get your random selections of
locations/surveyors and then process only those combinations from
I need the parameters estimated for a non-linear equation, an example of the
data is below.
rm(list=ls())
Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8)
Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40, 41, 50,
47,
44, 44, 42,
On Thu, Aug 26, 2010 at 1:18 PM, stephenb wrote:
> is it possible to open a channel to a data frame in the default environment?
> there are cases when using a sql update statement is the simplest
> alternative, so instead of dumping the df and then updating and then
> reimporting it I would like t
Dear All,
I'm looking to perform an ANOVA between two nested panel fixed effects models.
I tried with anova, as well as with waldtest. anova tells me there is no method
available for plm objects, while waldtest tells me my models are not nested. I
think they are instead. The difference between
Greetings Gabor,
is it possible to open a channel to a data frame in the default environment?
there are cases when using a sql update statement is the simplest
alternative, so instead of dumping the df and then updating and then
reimporting it I would like to update the df directly in R.
Thank y
Did anyone manage to have a look at this? I have now tried foreach with doMC
backen and multicore, but still no luck. If someone can help that would be
much appreciated!
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-clusterCall-Error-in-checkForRemoteErrors-lapply-c
On Aug 26, 2010, at 11:01 AM, Dirk Eddelbuettel wrote:
>
> On 26 August 2010 at 11:28, R P Herrold wrote:
> | On Thu, 26 Aug 2010, Gavin Simpson wrote:
> |
> | > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
> | >> How reliable is R-Forge? http://r-forge.r-project.org/
> | >>
> | >> It is
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Shvorob
> Sent: Thursday, August 26, 2010 3:16 AM
> To: r-help@r-project.org
> Subject: [R] Quick GREP challeng
I'm sure this has appeared before on this list, but the biggest
help to me has been: "Gelman and Hill", Data Analysis and
Regression Using Multilevel/Hierarchical Models, which contains
clear explanations and the R code to go along. Also check out
http://lme4.r-forge.r-project.org/book/
Looking
On 8/26/2010 8:43 AM, Niels Richard Hansen wrote:
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
function(x,y,...) callGeneric(x,un
On Thu, 26 Aug 2010, Dirk Eddelbuettel wrote:
On 26 August 2010 at 11:28, R P Herrold wrote:
| Is anyone aware of explanations, other than a release process
| that does not require unique versioning of differing content?
| [it seems pretty basic to me that a 'receiver' of new content
| could
try
lapply(df, class)
Steve E
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer
wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
On Thu, Aug 26, 2010 at 4:44 PM, Marc Schwartz wrote:
>> sapply(iris, class)
> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
> "numeric" "numeric" "numeric" "numeric" "factor"
Note that comparing the result of class(foo) is a bad way of telling
if something is
Certainly. The link at the bottom of each and every email to this list has
the web address at which you may do so.
For your convenience:
https://stat.ethz.ch/mailman/listinfo/r-help
Sarah
On Thu, Aug 26, 2010 at 9:05 AM, Sonia Spirling
wrote:
> Hello,
>
> I was wondering if I could be taken off
???
You were provided exactly what you requested. I think you either need
to read up on what random effects models mean or more clearly
communicate what YOU mean. (It's unclear to me, anyway).
--
Bert Gunter
Genentech Nonclinical Statistics
On Thu, Aug 26, 2010 at 8:57 AM, Samantha Patrick
wrot
Ben Holt bio.ku.dk> writes:
> I have data similar to this:
>
> Location Surveyor Result
> A1 83
> A2 76
> A3 45
> B1 71
> B4 67
> C2 23
> C5 12
> D3 34
> E4
Hello,
I was wondering if I could be taken off the list to get emails?
Thanks
Sonia
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://
Thanks, but my problem is that I have my formulas in functions.
How can I do that for functions or e.g. for the body(Df) below?
And: How can I refer to the first solution of expression?
Thanks a lot!
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesend
Many thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/Quick-GREP-challenge-tp2339486p2339818.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo
On Thu, Aug 26, 2010 at 11:31 AM, Daniel Brewer wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
> apply(df,2,class)
> but all th
On 26 August 2010 at 11:28, R P Herrold wrote:
| On Thu, 26 Aug 2010, Gavin Simpson wrote:
|
| > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
| >> How reliable is R-Forge? http://r-forge.r-project.org/
| >>
| >> It is down now (for me). Reporting "R-Forge Could Not Connect to Database:
"
Hi
Thanks for your help - however if I have a model of:
mod1<-lmer(B~ A+C+(A|bird), family=quasibinomial)
coef then gives me an individual slope for factors A and C. However the random
effect is only nested within factor - so I am only trying to allow the slope to
vary in relation to effect
Hi,
Gavin Simpson wrote:
> What do you want to do with the dissimilarities?
Clustering.
Gavin Simpson wrote:
> If clustering, try the clara() function
I can't because the variables are mixed (numeric and categorical) and (I
suppose..) I should use "gower" distance in advance, shouldn't I?
On Thu, 2010-08-26 at 07:35 -0700, abanero wrote:
> Hi,
>
> I'm trying to apply the function daisy() to a data.frame 1x10 but I have
> not enough space (error message: cannot allocate vector of length
> 1476173280).
>
> I didn't imagine I was not able to work with a matrix of just 1
> obs
On Aug 26, 2010, at 10:36 AM, David Winsemius wrote:
>
> On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
>
>> Hello,
>>
>> Is there a simple way to get the class type for each column of a
>> data.frame? I am in the situation where I would like to get all the
>> columns of a data.frame that
Joris,
I looked at the problem. Here is a minimal example reproducing
the error
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
func
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
> apply(df,2,class)
> but all the
On Thu, 2010-08-26 at 10:42 +0200, David Hajage wrote:
> Hello useRs and guRus,
>
> I was trying to add the support of str() in the ascii package, and I
> realized that str() does not have a print method. It uses cat() from inside
> the function and returns nothing.
>
> Since it is not usual in R
Since a data.frame is a list, you should use lapply() or sapply():
> df <- data.frame(a=1:5, b=LETTERS[1:5])
> lapply(df, class)
$a
[1] "integer"
$b
[1] "factor"
> sapply(df, class)
a b
"integer" "factor"
HTH,
Ivan
Le 8/26/2010 17:31, Daniel Brewer a écrit :
> Hello,
>
> I
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
but all the columns come ba
That's because apply works on arrays/matrices, not data.frames.
It therefore coerces your data.frame to a matrix of type
character, since you have factors, thus the result.
You want sapply or lapply, since a data.frame is actually a
list.
sapply(df, class)
and then to get what you want:
df[sa
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote:
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
lapply(df, class)
but a
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
but all the columns come back as class "character".
Thanks
Dan
--
On Thu, 26 Aug 2010, Gavin Simpson wrote:
On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote:
How reliable is R-Forge? http://r-forge.r-project.org/
It is down now (for me). Reporting "R-Forge Could Not Connect to Database: "
late to chime in, so had tossed the first piece. As this
relate
A big thanks David! Eliminating the env = parent.frame() seems to fix
everything. I hate it when I'm that close... I included that because from
?as.formula it seemed to be the default, but on re-read, I guess it doesn't
really say that. In addition, I had included it even though it was the
defau
coef(mod1)$bird will give you a matrix with two columns. The first
column is the intercept for each bird and the second column is the
slope for each bird.
ranef(mod1) will also give you a matrix of two columns. These
represent the random effects. That is, how much the intercept (or
slope) is shift
Hi!
I didn't see the toy data, my bad.
I guess David's solution fixed the problem.
Ivan
Le 8/26/2010 16:10, Bryan Hanson a écrit :
> Hi Ivan, there is toy data given in the original post.
>
> The object passed to aov is a matrix, it is called scores, and it is passed
> via the formula that is
Dear Anne,
> -Original Message-
> From: Anne Mimet [mailto:ami...@mnhn.fr]
> Sent: August-26-10 7:45 AM
> To: John Fox
> Cc: r-help@r-project.org
> Subject: RE: [R] SEM : Warning : Could not compute QR decomposition of
> Hessian
>
> Dear John,
>
> Thank a lot for your answer. Indeed, i m
On Aug 26, 2010, at 10:44 AM, Stephen Liu wrote:
Hi folks,
Following command prints 2 graphs side-by-side:-
layout(matrix(1:2, nrow=1))
?layout
#Read the details for the widths and heights arguments more carefully
and (as it suggests) see the examples.
plot(Date,Input_No.)
plot(Test01$
Hi folks,
Following command prints 2 graphs side-by-side:-
layout(matrix(1:2, nrow=1))
plot(Date,Input_No.)
plot(Test01$Date, Test01$Input_No.)
However each is a square graph I need a rectangular layout. Pls advise how to
make it. TIA
B.R.
satimis
_
Hi,
I'm trying to apply the function daisy() to a data.frame 1x10 but I have
not enough space (error message: cannot allocate vector of length
1476173280).
I didn't imagine I was not able to work with a matrix of just 1
observations... I have setted in Rgui --max-mem-size=2G (I'm not abl
On Aug 26, 2010, at 8:47 AM, Bryan Hanson wrote:
Hello Again Gurus and Lurkers:
I’m trying to build a very user-friendly function which does aov
without
having the user type in a formula (which would be tedious in this
case).
The idea is to take the response from a PCA score matrix, and th
On Thu, Aug 26, 2010 at 9:45 AM, Immanuel Seeger wrote:
> Thanks, but my problem is that I have my formulas in functions.
> How can I do that for functions or e.g. for the body(Df) below?
Try this:
> library(Ryacas)
> Solve(deparse(body(f)), "x")
[1] "Starting Yacas!"
expression(list(x == -((10
Hi,
First, some toy data would have helped a lot, and that can explain why
you got so few answers...
Second, I have maybe some thoughts about it, not sure if this can help.
The error tells you that the data passed to aov is not a matrix. In your
function, you don't specify explicitly the dat
On Aug 26, 2010, at 8:15 AM, Gabor Grothendieck wrote:
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
grep("f[0-9]+=", "f1=5,f22=3,", value = T)
[1] "f1=5,f22=3,"
How do I make the line output c("f1", "f22") instead? (Actually,
c(1,22)
would be even better).
strapply in gsu
Dear all,
This problem came up initially while debugging a function, but it
seems to be a more general problem of R. I hope I'm wrong, but I can't
find another explanation. Let me illustrate with the raster package.
For an object "RasterLayer" (which inherits from Raster), there is a
method xyVal
Hi Greg,
thanks for the suggestion:
I have attached some small dataset that can be used to reproduce the
odd behavior of the approxfun-function.
If it gets stripped off my email, it can also be downloaded at:
http://bioinf.gen.tcd.ie/approx.data.Rdata
Strangely, the problem seems specific to the
On Aug 26, 2010, at 10:35 AM, Niels Richard Hansen wrote:
>
>
> On 26/08/10 09.30, Dimitris Rizopoulos wrote:
>> I think you need an I(), i.e.,
>>
>> form <- ~ I(a > 1) - 1
>
> Yes, it solves the concrete problem, but does not really answer
> the question. Let me rephrase. Should the use of t
On Thu, Aug 26, 2010 at 6:57 AM, Immanuel Seeger wrote:
> Dear all,
> I want to equalize a symbolic derivative (of a function with two
> variables) with zero and convert it to a variable, e.g. x.
>
> I'm computing the derivative by: (found it in the archive)
> library(Ryacas)
> f <- function(x,y)
Hi John,
I'm having some trouble sorting out your request, for the following reasons:
1) I don't know what misdist is (can't find it on CRAN or google) or why you
need Rcmdr to run it. Presumably you meant mixdist?
2) relimp is not required by Rcmdr, merely suggested. Rcmdr will work fine
without
Hello Again Gurus and Lurkers:
I¹m trying to build a very user-friendly function which does aov without
having the user type in a formula (which would be tedious in this case).
The idea is to take the response from a PCA score matrix, and the factors
from a list. A simple example is the function
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