.5)
> abline(a=0, b=-0.5)
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/05/2014 02:30 AM, Adrien Bonvin wrote:
Bonjour
Hi everybody,
Firs of all, sorry for my terrible English,
I would like to know if it’s possible to create an “empty plot” in which i
could ad
x27; as shown in the below:
> sprintf("%05d", seq(from=200, to=1200, by=500))
[1] "00200" "00700" "01200"
I like to read the documentation by typing
> help(formatC)
> help(sprintf)
You may find answers what you wish to get. Documentation has been my
sh to make R understand 'NaN' as missing value, it would be a
good choice to use reserved keyword 'NA_character' as shown in the below:
> x <- factor(c("yes", "yes", "no", NA, "yes", "no", NA_character_))
> x
[1] yes yes
Your question is not clear to me.
> x$alpha[1:2]
[1] "a" "b"
> x$alpha[2]
[1] "b"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/5/2014 11:12 AM, Dinesh Chowdhary wrote:
x <- list(seq = 3:7, alpha = c("a", &quo
if x < 0.3. x= 0.4246928
Anyway, move to next!
[1] 0.2535644
Error is printed if x < 0.3. x= 0.5071288
Anyway, move to next!
[1] "error"
Error is printed if x < 0.3. x= 0.1941202
Anyway, move to next!
(continue)
I hope this helps, and thank you so much, Martin Morgan, for your go
It seems that you would like to make a spaghetti plot (in a longitudinal
data analysis). You can use the function 'interaction.plot()'.
> with(my.df, interaction.plot(TIME, ID, X))
I hope this helps.
Chel Hee Lee
On 12/06/2014 02:24 AM, arun wrote:
Not sure whether it is
6.970592 4 6.970592
CID_10.31 CID_10 Acidobacteria1.988448 5 1.988448
CID_10.32 CID_10 Actinobacteria1.644548 6 1.644548
CID_10.33 CID_10 Others 1.582823 7 5.546493
>
I hope this helps.
Chel Hee Lee
On 12/07/2014 08:21 AM, Morway, Eric wrote:
Usin
1jan20051 2005-01-26
10 10 01feb20031 2003-01-27
11 11 29jan20091 2009-01-24
12 12 03mar20081 2008-02-27
>
Leap year is also considered (See the last case 'id=12'). Is this what
you are looking for? I hope this helps.
Chel Hee Lee
On 12/11/2014 9:54 AM, Kuma Ra
"5001,20001" "10001,20001" "15001,20001"
[11] "1,25001" "5001,25001" "10001,25001" "15001,25001" "20001,25001"
> x.idx2 <- upper.tri(x, diag=TRUE)
> x[x.idx2]
[1] "1,1" "1,
NANA NA
RK.Records_CL V.Other
1005317NA NA
1007183NA NA
1008833NA NA
1012281NA NA
1015285 1 NA
1015315 NA NA
1015322NA NA
>
Chel H
t in R. Is this what you are looking for? I hope
this helps. I would also appreciate it if you would provide
reproducible examples next time.
Chel Hee Lee
On 12/18/2014 11:48 AM, Jeff Newmiller wrote:
Read the posting guide. Th
iol.Type'
where 'factor()' is used. I hope this helps.
Chel Hee Lee
On 12/18/2014 08:57 PM, Crombie, Burnette N wrote:
Chel, your solution is fantastic on the dataset I submitted in my question but
it is not working when I import my real dataset into R. Do I need to vectorize
dx], collapse=" ")
+ unlist(strsplit(comb, " "))
+ })
> names(tmp) <- d0$ID
> tmp
$MCZ4325
[1] "C23.2" "T43.2"
$GDR2343
[1] "M20.64" "Y32.1" "M20.64" "T44.2"
$BZD2643
[1] "B83.2" "T43.2" "B83.2&q
0 Out
x8 0 0 0 0 0 B 0
2015-04-07
x1 0
x2 0
x3 0
x4 0
x5 0
x6 A
x7 0
x8 0
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/19/2014 10:45 PM, Kristi Glover wrote:
Hi R User, Would you suggest me
You are welcome, I am glad that I was able to help.
Chel Hee Lee, PhD
Biostatistician and Manager
Clinical Research Support Unit
College of Medicine
University of Saskatchewan
On 12/20/2014 03:53 AM, Kristi Glover wrote:
Thank you Prof. Lee for your code. I am sorry that I noticed that I sent
1, labels=paste("S", 1:6, sep="_"))
> points(x=2, y=4, cex=3)
>
Please see the output. Is this what you are looking for? I hope this
helps.
Chel Hee Lee
On 12/21/2014 01:25 PM, eliza botto wrote:
Dear UseRs,
A point was plotted by the following command
plot(2,4,y
; length(x[unlist(x)])
[1] 1
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 12/22/2014 7:45 AM, najuzz wrote:
#Hi guys,
#I would like to count the number of individuals that receive X=0 troughout
their observational period.
#example dataset:
ID<-c(1,1,1
Dear R users,
1) My problem in short:
Mlogit cannot calculate certain conditional models.
2) My database:
The target was a logistical regression analysis and a probability function
which should include generic coefficients and alternative-specific ones. The
database was a survey, the d
#x27; and 'LC_MESSAGES'
in order to change messages in a given locale? However, it is
questionable. Please see the following example (messages in Japanese):
> Sys.setenv(LANGUAGE="ja_JP.utf8")
> a
エラー: オブジェクト 'a' がありません
Thank you so much. Your explanation is very clear and helpful to
understand. Thank you, again.
Chel Hee Lee
On 1/12/2015 5:56 AM, Prof Brian Ripley wrote:
You don't want to change the locale, rather the message language. That
is not a locale, and e.g.
LANGUAGE=en
is the correct
Dear R users,
1) My problem in short:
Mlogit cannot calculate certain conditional models.
2) My database:
The target was a logistical regression analysis and a probability function
which should include generic coefficients and alternative-specific ones. The
database was a survey, the dependent
This approach may not be fancy as what you are looking for.
> xl <- unlist(x)
> xl[grep("A", names(xl))]
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
>
I hope this helps.
Chel Hee Lee
On 01/16/2015 04:40 AM, Rainer M Krug wrote:
Hi
Consider the fo
222.247934 226.107438 230.00
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/16/2015 02:16 AM, philippe massicotte wrote:
Hi all.
How we evaluate a formula in R?
Ex.:
params <- list(a = 2, b = 3)
x <- seq(1,10, length.out = 100)
func1 <- as.formu
ine(lm(y~x), col="red")
panel.text(1, -1, bquote(R^2 == .(summary(lm(y~x))$r.squared)))
},
grid = TRUE
)
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/17/2015 07:08 AM, Naresh Gurbuxani wrote:
In a conditional xyplot, I would like to add some numeri
3.15
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/19/2015 02:14 PM, Shuhua Zhan wrote:
Dear All,
I'd like to add a list name into the list contents to make a new output. The list is a
list of data.frame derived from summary command in Rqtl. I want to add th
32 33 34
> extr.1(x=x, name="A")
f1.x1.A f1.x2.A f2.x3.A f2.x4.A
11 12 13 14
> extr.1(x=x, name="a")
named numeric(0)
> extr.1(x=x, name="Aha")
f1.x1.Aha f1.x2.Aha f2.x3.Aha f2.x4.Aha
2122232
6.75
6 heart_wt c12.loc49 12 51.2 28.2062.2 3.65
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/20/2015 9:23 AM, Shuhua Zhan wrote:
Hi Chel,
Thank you very much for your help!
I'm sorry I did not post my wanted output correctly. I only want the colnames
o
is helps. I would appreciate if you DO READ the posting guide.
Chel Hee Lee
On 1/20/2015 9:30 AM, Ragia Ibrahim wrote:
Hello,
kindly ,
how to catch this Error
Error in A[[i]] : subscript out of bounds
and check that the list is empty is.null( A[[i]] ) do no twork
thanks in advance
R.I
> x <- c("hola mundo mundo");
> table(unlist(strsplit(x, " ")))
hola mundo
1 2
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/22/2015 8:25 AM, bgnumis bgnum wrote:
Hi all,
I want to cout the different words in a text.
You
7+8)
> new1
[1] 3 3 15 15
>
> grp.id <- c(1,1, 2, 3,3,3, 4,4)
> tapply(X=a, INDEX=grp.id, FUN=sum)
1 2 3 4
3 3 15 15
> aggregate(x=a, by=list(grp.id), FUN=sum)
Group.1 x
1 1 3
2 2 3
3 3 15
4 4 15
> xtabs(formula=a~grp.id)
grp.id
1 2 3 4
> do.call(rbind, lapply(split(data, data$Name), function(x)
x[order(x$CheckInDate),][nrow(x),]))
Name CheckInDate Temp
John John 2014-04-01 99.0
Mary Mary 2014-03-01 98.1
Sam Sam 2014-04-01 97.5
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/23/2015
1
>
I hope this helps.
Chel Hee Lee
On 01/28/2015 07:38 PM, William Dunlap wrote:
with(dat1, ave(integer(length(Date)), Date, FUN=seq_along))
[1] 1 1 2 1 1 2 1 2 1 2 1
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 28, 2015 at 4:54 PM, Morway, Eric wrote:
The two datase
(c,lapply(tapply(ID1, gl(7,2), c), paste, collapse=""))
1234567
"AA" "TG" "CT" "GC" "GT" "CG" "TA"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 01/28/2015 05:5
Hi Bert! yes, you are VERY correct!!! Why am I making this simple thing
so complicated??? ;) Thank you so much for your nice lesson!
Chel Hee Lee
On 01/28/2015 09:59 PM, Bert Gunter wrote:
eek!
Chel Hee,anything that complicated should engender fear and trembling.
Much simpler and more
2 2 b
3 3 c
4 4 d
5 5 e
> eval(parse(text=DFName))[,1]
[1] 1 2 3 4 5
> length(eval(parse(text=DFName))[,1])
[1] 5
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/29/2015 12:34 AM, Jeff Newmiller wrote:
This approach is fraught with dangers.
I reco
Or, you can also do the same job using 'colSums()' as shown in the below:
> colSums(status=="I")
2010 2011 2012
344
>
I hope this helps.
Chel Hee Lee
On 1/28/2015 7:31 PM, JSHuang wrote:
Hi,
I think you need quotation around I like the following:
st
Why don't try this?
> nams<-c("P2O5 ['%']", "D50 [mu*m]")
> pdf("test.pdf")
> for(i in 1:2)
+ plot(1,1,xlab=parse(text=(bquote(.(nams[i])
> dev.off()
Is this what you are looking for? I hope this helps.
On 1/30/2015 7:54 AM, PIKAL Petr wrote:
Thanks
So the key is to put expression(s) dir
> paste(rep(v1, each=3), 1:3, sep="_")
[1] "a_1" "a_2" "a_3" "b_1" "b_2" "b_3" "c_1" "c_2" "c_3"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/30/2015 7:34 AM,
rmat(ex, format="%A")
[1] "Thursday" "Friday"
> weekdays(ex)
[1] "Thursday" "Friday"
>
Is this what you are looking for? I hope this helps.
Chel Hee Lee
On 1/31/2015 3:06 AM, Camilla Timo wrote:
Hi there,
I have a problem, i have a datase
"Rgui" "little"
"win.binary"
path.sep r_arch
";""x64"
>
Chel Hee Lee
On 1/30/2015 9:55 PM, Jim Lemon wrote:
Hi Ahmed,
Hmmm, this seems to work for me (R-3.1.2, Linux)
legend(0,2100, legend=c("2009","
00010
I hope this helps.
Chel Hee Lee
On 1/30/2015 11:52 AM, JS Huang wrote:
Hi,
Here is my implementation. Hope this helps.
b
[1] 1 2 3 4 5
c
[1] 1 2 1 3 5 4
sapply(b,function(x)ifelse(x==c,1,0))
[,1] [,2] [,3] [,4] [,5]
[1,]10000
[2,]
> lapply(1:2, function(x) t(A[rev(1:3),,x]))
[[1]]
[,1] [,2] [,3]
[1,] "g" "d" "a"
[2,] "h" "e" "b"
[3,] "i" "f" "c"
[[2]]
[,1] [,2] [,3]
[1,] "p" "m" "j&qu
lutions). I hope this helps.
Chel Hee Lee
On 2/10/2015 2:29 AM, Rolf Turner wrote:
On 10/02/15 14:04, Ssuhanchen wrote:
Hi!
I want to use R to calculate the variable x which is in a complex
equation
in below:
2
Σ[exp(-x/2)*(x^k)/(2^k*k!)]=0.05
k=0
how to solve this equation to get
A~ha~!! Thank you, Prof. Peter Dalgaard, so much for your wonderful
lesson!!! Learning new things everyday from this R-help mailing list!
Chel Hee Lee
On 2/11/2015 10:37 AM, peter dalgaard wrote:
On 11 Feb 2015, at 17:11 , Chel Hee Lee wrote:
The functional form given in the post
Hi
I am attempting to explore the scale of spatial autocorrelation in a raster
(eventually across a stack of 10 but for now a single layer) and consequently
in a potential sample of points across the landscape (ie. if we wanted to know
what sampling design in terms of distance would minimize aut
Dear R experts --
I never needed to add a dummy column and always query statistical results
by querying summary(model) for GLMER. However, I was recently asked to add
a dummy column for interaction variables when performing GLMER. Could
anyone tell me if it's necessary to add a dummy column for GL
:
https://stackoverflow.com/questions/54892809/r-cairo-pdf-function-does-not-respect-plotting-boundaries
Thank you in advance for any insights into this issue.
Sincerely,
Lee Kelvin
--
Dr Lee Kelvin
Department of Physics
UC Davis
One Shields Avenue
Davis, CA 95616
USA
Ph: +1 (530) 752-1500
Fax: +1 (530)
do to help, please do let me know.
Thank you again,
Best,
Lee
On Monday, 4 March 2019, Paul Murrell
mailto:p...@stat.auckland.ac.nz>> wrote:
Hi
(cc'ed to r-devel where further discussion should probably take place)
Thanks Lee. I see that problem.
There is a "+ 1" in
> LDL <- NA_real_
> is.LDL <- is.na
> v <- c(0.2, 0.28, LDL, 0.9)
> v
[1] 0.20 0.28 NA 0.90
> is.LDL(v)
[1] FALSE FALSE TRUE FALSE
>
Hope this helps.
Chel Hee Lee
On 11/9/2014 12:07 PM, Marc Girondot wrote:
Dear member list,
In many experimental sciences, there
Another (simpler) way that I can think is that
> y * matrix(rep(z,3), ncol=ncol(y), byrow=TRUE)
[,1] [,2]
[1,]00
[2,]6 -3
[3,] 12 -6
I hope this helps.
Chel Hee Lee
On 14-11-19 08:43 AM, Ruima E. wrote:
> Thank you Chel Hee.
>
> Isn't there a sim
> y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2)
> z = matrix(c(12, -6),ncol=2)
> t(apply(y, 1, function(x) x*z))
[,1] [,2]
[1,]00
[2,]6 -3
[3,] 12 -6
I hope this helps.
Chel Hee Lee
On 14-11-19 08:22 AM, Ruima E. wrote:
> Hi,
>
> I have this:
>
> xx <- as.factor(c("AL", "AK", "CA", "FL"))
> xx
[1] AL AK CA FL
Levels: AK AL CA FL
> as.character(xx)
[1] "AL" "AK" "CA" "FL"
I hope this helps.
Chel Hee Lee
On 14-11-22 01:12 AM, Aditya Singh wrote:
c3 6 4
5.c3 5 e c3 7 5
1.c4 1 a c4 4 1
2.c4 2 b c4 5 2
3.c4 3 c c4 6 3
4.c4 4 d c4 7 4
5.c4 5 e c4 8 5
I hope this helps.
Chel Hee Lee
On 11/24/2014 5:12 PM, Dimitri Liakhovitski wrote:
I have the data frame 'df' and my desired solution 'out&#x
uot;
9 "GR.3.8" "GR.3.8"
10 "GR.3.7" "GR.3.7"
11 "GR.3.1" "GR.3.1"
12 "GR.3.8" "GR.3.8"
13 "GR.3.1" "GR.3.8"
15 "GR.3.8" "GR.3.8"
16 "GR.3.1" "GR.3.1"
17 &q
"Merc 450SLC", "Cadillac
Fleetwood",
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic",
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin",
"Camaro Z28&
x)))
grp y
1 2 0
2 3 0
What a surprise! Is this a bug? I would appreciate if you share the
results after testing the codes. Thank you so much for your helps in
advance!
Chel Hee Lee
__
R-help@r-project.org mailing list -- To UNSUBSCRI
I appreciate your kind guidance! I did not read the manual carefully
(it's my fault).
Thank you so much, Prof. John Fox!
Chel Hee Lee
On 01/21/2016 12:52 AM, Fox, John wrote:
Dear Chel Hee Lee,
With the formula method, the default na.action is na.omit; thus,
aggregate(y~grp, dat
quot;, sep = ""))
#xl<-expression(~degree)
xl <- parse(text = paste(Orthophase[i,j], "*degree ~ S",
sep = ""))
printXlab<-paste("Orthophase",Orthophase[i,j],xl,"
")
plot(MyDa
. The variables
are not evaluated:
Hours from Reference Time:RefTimeStringModel span =format(magDt, 3, 2)hrs
oOrthophasei° bBathyphasei°
I don't really understand what state is required for the degrees to
print....
Cathy Lee Gierke
*"We are what we repeatedly do. Excellence, then, is not
I am trying to use the lnam autocorrelation model from the SNA package. I have
it running for smaller adjacency matrices (<1,500) it works just fine but when
my matrices are bigger 4000+. I get the error:
> lnam1_01.adj<- lnam(data01$adopt,x01,ec2001.csr)
Error in optim(rho, n2ll.rho, method =
Hi,
I have one critical question in using R.
I am currently working on some research which involves huge amounts
of data(it is about 15GB).
I am trying to use R in this research rather than using SAS or STATA.
(The company where I am working right now, is trying to switch SAS/STATA to
R)
As far as
S*.
Fourth Edition. Springer-Verlag.
Sincere thanks,
Cathy Lee Gierke
[[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://www.R
to
fit for the optimal splitting criteria at each of the predefined nodes.
Does such a package exist?
Thanks,
Lee
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-proj
to
weight the predictor variables such that rpart will not use an expensive
variable (or will only send limited fractions of the population to the node) if
there is not a comparatively large decrease in misclassifications after
splitting by that variable?
Thanks,
Lee
Hi
I have a dataframe similar to:
>Sample<-c(1,1,1,2,2,2,3,3,3)
>Time<-c(1,2,3,1,2,3,1,2,3)
>Mass<-c(3,3.1,3.4,4,4.3,4.4,3,3.2,3.5)
>mydata<-as.data.frame(cbind(Sample,Time,Mass))
Sample Time Mass
1 1 1 3.0
2 1 2 3.1
3 1 3 3.4
4 2 1 4.0
5 2 2 4
To whom I concern,
I am a recent user of program R
I do not know how to explain this but after I installed the program the
words on the scrpit does R console does not appear clear
but something square(?)/Cube(?) which seems like the words are broken
More suprisingly, the squared words are writte
Hi Everyone,
Sorry to ask what I think is a basic question but I really haven't found my
answer yet in the archives.
I am trying to run a mixed effects model in R using the lme package. My
experiment is such that I am interested in the effects of Temperature (2
levels) and Species (3 levels)
want incorporated into the package for
possible use by users, as well. When I put them in the data folder, I get
errors, and they get converted somehow What is the proper way to put
these data files into the package?
Thanks for your help,
Cathy Lee Gierke
"Our lives begin to end the day we bec
"$(pkg)\", compress=3)" | \
>>R_DEFAULT_PACKAGES=NULL LC_ALL=C $(R_EXE) > /dev/null
>> at the Makefile.win in the src/datasets directory
>> I am using Window XP and tried to compile 2.11.1 version.
>>
>> I can'
am using Window XP and tried to compile 2.11.1 version.
I can't imagine how I can solve this problem. Any hints or suggestions will
be appreciated.
Thank you.
Lee.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
fined reference to `_Unwind_SjLj_Register'
undefined reference to `_Unwind_SjLj_Unregister'
These link errors were happened even after I installed R 2.11.1 and use
Makefile.win
I will really appreciate if you can give me a hint to solve that problems...
Thank you again.
Lee.
On Wed, Jun 2
Thank you.
I finally found what was the problem.
By mistake, I compiled with different version of MinGW compiler. I heard
that new MinGw compiler does not support _Unwind_ kinds of functions
anymore.
Thank you again.
Lee.
On Wed, Jun 23, 2010 at 7:30 PM, Geun Seop Lee wrote:
> Thank
I just installed the R 2.11.1 version on my computer and I encountered a fatal
error: "Unable to restore saved data in .RData" and kick me out of R right
away. I still can run 2.10.2. There is no package called "rattle"
I checked various posts regarding this error. I still can't get it to wo
How do I become a member of R user community?
Albert Lee,
Ph.D. statistician
Confidentiality Notice: This communication, and any file...{{dropped:12}}
__
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PLEASE do read
Dear R-users
I'm trying to use lasso in lars package for subset regression, I have a
large matrix of size 1000x100 and my aim is to select a subset k of the 100
variables.
Is there any way in lars to fix the number k (i.e. to select the best 10
variables)
library(lars)
aa=lars(X,Y,type="lasso"
Hello,
I'm trying to add a column to the following data frame. The new column
will contain "black" when the 5th column(if_TE_related) is
"TE_related", or "orange" when the 4th column is " " (space).
"chromo""MSU_locus" "end5" "end3" "if_TE_related"
"chr04" "LOC_Os04g01006"
Sorry. I made a typo. It is the 5th column (not 4th). Thank you.
Tae-Jin
Begin forwarded message:
> From: Tae-Jin Lee
> Date: February 4, 2011 7:09:38 PM EST
> To: r-help@R-project.org
> Subject: Help!!! from R beginner
>
> Hello,
>
> I'm trying to add a column to
E_related", "black",
"orange")
On Fri, Feb 4, 2011 at 7:09 PM, Tae-Jin Lee wrote:
Hello,
I'm trying to add a column to the following data frame. The new
column
will contain "black" when the 5th column(if_TE_related) is
"TE_related", or &quo
Dear all,
does anybody have experience with building logits in Mlogit?
I want to test the use of a couple of alternative specific variables with a
generic regression coefficient. However, one of them simply does not work. R
says the length of this variable is different.
Problem: If I check the l
Dear all,
does anybody have experience with the calculation of marginal effects
(“effects”) in Mlogit (see Croissant, Package ‘mlogit’, p.8)?
1) Is there a good qualitative explanation available for the listed options
for the argument “type” (“aa”, …)? When do I have to choose aa, ar…? And
w
Ingmar, many thanks. I get that one from R:
Error in model.frame.default(terms(formula, lhs = lhs, rhs = rhs, data =
data), :
variable lengths differ (found for 'X')
X is the variable I have used.
Any comment would be much appreciated.
Best regards!
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ht
Ingmar,
many thanks for your answer.
I give you a smaller version of my program with the isolated "strange"
variable, which I used when trying to elaborate the problem .
[Start of R-Editor quote]
library(foreign)
library(gdata)
library(gtools)
library(gmodels)
library(gplots)
library(xtable)
li
Hi Michael,
many thanks for your comment.
Below the original data as imported from Stata format. I deleted some
columns with explaining variables because they were unnecessary and working;
the model has several explaining variables with generic regression
coefficients.
pid Choice V7.Ch 1
Hi Michael,
I have sent youi the data.
Best,
Lee
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iable with a generic coefficient
(and which is working fine) is 8,436.
Any comment or hint (also on literature, in case my mistakes are too simple
and obvious) would be helpful.
Best regards,
Lee
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Hi Michael,
many thanks for the useful hints which gave me some deeper knowledge of R.
It is definitely much appreciated.
I think I have found the mistake - the problems did not arise from variable
definitions etc. It was an intellectual mistake.
The cause for the problems was that I had to di
I will very appreciate if anyone can provide some materials to draw a
simplex plot of a Dirichlet distribution in R as shown in the page at
http://en.wikipedia.org/wiki/File:Dirichlet_distributions.png .
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https://stat.
Thank you, David. I found some good stuffs for which I have been
looking.
On Sun, 2012-08-26 at 09:01 -0700, David Winsemius wrote:
> On Aug 25, 2012, at 9:00 PM, Chel Hee Lee wrote:
>
> > I will very appreciate if anyone can provide some materials to draw a
> > simplex p
I have some trouble to deal the value of 'NaN'. For example,
> exp(1e3)
[1] Inf
> exp(1e3)*0
[1] NaN
The correct answer should be 0 rather than NaN. I will very appreciate
if anyone can share some technique to get a correct answer.
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#. Live Go...
> >Live: OO#.. Dead: OO#.. Playing
> > Research Engineer (Solar/BatteriesO.O#. #.O#. with
> > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> > --------
Hi all
I am trying to solve a combinatorial optimization problem. Basically, I can
reduce my problem into the next problem:
1.- Given a NxN grid of points, with some values in each cell
2.- Find the combination of K points on the grid such that, the maximum
mean value is obtained
I took the Tr
Hi R people!
I have a directory of .csv files I would like to make into objects
then scatter plots. I have been having varying degrees of progress. I
was able
make an object of all files, loop through it, and make a pdf of the
last file I looped through. I kept renaming the pdf so instead o
Hi,
Using the multicore package and calling sample() in mclapply,
how do I get the results to be reproducible?
I know the random number is seeded by process id and so is
different for each run.
Thanks,
Lik
[[alternative HTML version deleted]]
I have a general question how to understand the message with "non-zero exit
status" when new package is installed. Based on my experience, this message
implies a package dependence. Am I correct to understand this message? Are
there anyone who can provide some reference or explain details about the
Hi.
I run a latent class analysis with polymotous variables.
Because of small sample size, I have to use bootstrap method in order to
select a proper model.
Is there any package or way that I can run a bootstrap method after runing
latent class model with polytomous variables?
Thank
KeunBok Lee
Oh... I see!!! I very appreciate for your clear explanation! Thank you, Uwe
Ligges.
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To Whom it May Concern:
I have been using R version 2.6.2 for awhile now, installed on a
PowerMac G5 running OS 10.4.11 I typically get Quartz graphics
output from R into Adobe Illustrator CS2 simply by copying and
pasting. Upon upgrading to R version 2.7.0, I now receive an error
from Ill
gt;> print(lm3)
>> anova(lm3)
>>
>> When I use lm to fit the data, there are some problems
>> in ??subject*sequence??. I have use GLM in SPSS to
>> fit the same data, and it seems there is no problem.
>>
>> I don??t know where my problem is. How can
Dear Dr. Dalgaard
Sorry for delay reply..
That's exactly what I was looking for - thanks a lot.
Hsin-Ya
Peter Dalgaard wrote:
>
>
>
> Andrew Robinson wrote:
>> In your data, subject is nested within sequence. Was that your
>> intention?
>>
>>
> Presumably yes. This looks like a standar
=x, formulation=y, subject=z, time=t, concentration=c)
A.split<-split(df, list(df$pH ,df$formulation, df$subject) )
A.split []
Best regards,
Hsin-Ya Lee
Henrique Dallazuanna wrote:
>
> Try this:
>
> A.split[[1]]["time"]
> A.split[[1]][["concentration"
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