nlme? Are there any changes planned
to incorporate this feature?
Best wishes, I am looking forward to your feedback
Anna-Lena W�lwer
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arguments "col" or "lty" has not worked:
plot(fit,ylim=c(0,1),xlim=c(0,2000), col=c("red", "blue", "orange"), lty=3)
I would be very grateful for any help! Thank you very much in advance!
Kind regards,
Anna McLean
[[alternative HTML v
e and only used R for the statistics part. I intend
to make a permanent swith to R as I can see the benefits. I am just severely
frustrated by my own inabilities. Therefore, once again many thank´s for your
time!
kind regards
Anna
><º>`•. . • `•. .• `•. . ><º>`•.
Thank you! That was a easy and fast solution!
May I post a follow-up question? (I am not sure if this would rather should be
posted as a new question, but I post it here and then I can re-post it if this
is the wrong place to ask this). I am ever so grateful for your help!
/Anna
erA = factor(rep(LETTERS[1:2], each = 5)),
B1298712 = factor(rep(LETTERS[1:2], each = 5)))
# Many thank's and with kind regards
Anna Zakrisson
><�>`�. . � `�. .� `�. . ><�>`�. . � `�. .� `�. .><�>`�. . � `�. .�
>`�. .><�>
Anna Zakr
column
of the matrix and apply the same function, get the single number and place it
to (2,3) entry of the matrix and so on
how can i do it?
thanks
anna
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in insering this into lattice (I am
not an experienced lattice user). I want to keep the data points in the plot.
Find below dummy data and the script as well as annotations further describing
my question.
thank you in advance!
Anna Zakrisson
mydata<- data.frame(
Year = 1980:2009,
T
A_0 A_1 A_2 B_0 B_1 B_2
1000 10 80 8000 600 220
How do maintain the original order of sr and bond as in my_dat data.frame and
obtain the
above OUTPUT?
Kindly guide.
Anna
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I am using the gRain package but I can't get it to work under R 3.1.0. It is no
longer available in the CRAN.
Does anyone have suggestions for how to get a successful installation?
A
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eciate any help and idea!Thank you!Anna S.
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theme_bw() +
theme(strip.background = element_blank()) +
ggtitle("this plot is ugly but ignore that")+
xlab("my x-axis stuff")+
ylab("my y-axis stuff")+
theme(legend.title=element_blank(), legend.key =
element_blank())
p
Anna Zakrisson Braeunlich
PhD stude
Hi everyone,
I am new to additive modelling and am surprised by the results of a model
I'm working on. I wanted to check with more experienced users to make sure
I'm not misunderstanding something basic.
*Data:* I have 10 replicated runs from an evolutionary simulation model,
measuring the evolv
tle.x = element_text(size=10, colour="black", family="serif",
angle=00))+
theme(axis.title.y = element_text(size=10, colour="black", family="serif",
angle=90))
p
p1 <- p+theme(legend.title=element_blank())
p1
p2 <- p1+ theme(legend.key = element_blank(
table. But I have no idea what the correct
commands are, so that the plot does look nice afterwards.
Thank you very much for your help!
A great new years eve to everybody and a wonderful year 2014.
Greetings,
Anna
for me because outputs
are
vectors and not single numbers.
Thanks in advance for your help!
Simeonidou Anna
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R-hel
t;N")))+
xlab("Station") +
scale_shape(solid = FALSE)+
geom_hline(yintercept=0, linetype=3)+
scale_fill_discrete(guide = guide_legend("my stuff"),
labels=c("cyanobacteria", "seston")) #if I wanted to have
"seston" in italic
x <- levels(unique(AB$time) --> which returns only NA
x <- seq(unique(AB$time) ---> which returns the standard deviation of the
entire column (not the single parts)
What do I do wrong? And how can i fix it?
Thank you so much in advance.
Kind regards,
Anna
addition: Warning message:
In socketConnection(master, port = port, blocking = TRUE, open = "a+b", :
ip-10-158-31-12:11883 cannot be opened
Execution halted
Can you help me?
Best regards
Anna Longari
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columns=1, border=F,
x = 0.2, y = 0.2, corner = c(2, 2),
title="", cex.title=1.3),
ylab = ("Pc"),
xlab="Week",
data= SummPdata,type="o",
tP
doubleYScale(plotN, plotP, add.ylab2 = TRUE) #Why can I not change the axis
color by adding to this argument?
# I want the y1 and y2 axes to be defined not by color, but by shape and
linetype.
# I have managed to draw the shapes (defined by Nc and Pc) by the y1 and y2
axes, but I do not mana
lattice for every second plot. What are your
thoughts on this for the future and are there already some possibilities in
place? I prefer continusing using ggplot2 because it is brilliant.
with kind regards
Anna Zakrisson Braeunlich
PhD student
Department of Ecology, Environment and Plant Sciences
colour="black", family="serif",
angle=00)) +
theme(strip.text.y = element_text(size = 16, colour="black", family="serif",
angle=00)) +
theme(axis.text.x = element_text(size = 16, colour="black", family="serif",
angle=00)) +
theme(axis.text.y
eme(axis.text.x = element_text(size = 20, colour="black", family="serif",
angle=00)) +
theme(axis.text.y = element_text(size = 17, colour="black", family="serif",
angle=00)) +
theme(axis.title.x = element_text(size=20, colour="black", f
t;summary", fun.y = "median", mapping=aes(shape=organism)) +
#must be unfilled
theme(strip.text.x = element_text(size = 12, colour="black", family="serif",
angle=00)) +
theme(axis.text.x = element_text(size = 12, colour="black", family="serif",
angle=90)) +
theme(axis.text.y
gle=90)) +
theme(axis.text.y = element_text(size = 12, colour="black", family="serif",
angle=00)) +
geom_hline(yintercept=0, linetype=3) #draws dotted line at 0
p
# method="lm" is definately wrong,
# I just added it to be a ble to draw some lines at all.
# I
.settings),
key=list(text=list(c("factor2a","factor2b", "factor2c"),
col="black"),
points=list(pch=rep(2:4), cex=1,
col="black"),
x = .35, y = 1.06, corner = c(2, 2)), #legend p
. One sample -
one sediment type.
What is the correct way make the model:
My guess is something like: aov(Count ~ Sediment+Error(Lake/Zone/Sample))
Sincerely,
Anna
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___
:3),
panel=panel.bwplot.intermediate.hh,
key=list(text=list(c("factor2a","factor2b", "factor2c"),
col=c("black")), space="right",
points=list(pch=c(1:3), lty=c(1,3), cex=1,
col=c("
on top
of each other - again, this will look terribel on this data, but will be OK for
mine.
Is this possible in lattice? Maybe using different data frames and make
different plots and superpose them on top of each other? Ideas?
ggplot2?
with kind regards
Anna Zakrisson Braeunlich
PhD student
Dep
par.strip.text=list(custBgCol=bgColors,
custTxtCol=txtColors),
strip=myStripStyle,
scales = list(alternating = TRUE,
x = list(relation = "same", rot=45),
y = list(relation = "same")),
panel.
t;gray","white")), # boxplots
#superpose.fill=list(col=c("black","white")),
superpose.polygon=list(col=c("gray","white")), # legend
par.sub=font.settings)
as adviced by you.
With kind regards
Anna Zakrisson Braeunlich
PhD student
Depa
panel.groups = function(x, y, ..., group.number) {
panel.bwplot(x + (group.number-1.8)/3, y, ...)
})
Anna Zakrisson Braeunlich
PhD student
Department of Ecology, Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden/Sverige
Lives
again, thank you so much! Another day closer to disputation... :-S
Anna
Anna Zakrisson Braeunlich
PhD student
Department of Ecology Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden
Lives in Berlin.
For paper mail:
Katzbachstr. 21
D-10965, B
GREAT! Thank you! Will try this!
Anna
Anna Zakrisson Braeunlich
PhD student
Department of Ecology Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden
Lives in Berlin.
For paper mail:
Katzbachstr. 21
D-10965, Berlin - Kreuzberg
Germany
d of alphabetical. My normal methods do not work.
with kind regards
Anna Zakrisson
# Some dummy data:
mydata<- data.frame(factor1 = factor(rep(LETTERS[1:6], each = 80)),
factor2 = factor(rep(c(1:5), each = 16)),
factor3 = factor(rep(c(1:4), each = 4)),
Hi,
have managed to get rid of the facet labels (so do not spend your time
explaining that to me). Tthere were some old code out there which did not
work. My only remaining issue is how to add the axis labels to the plot
without labels.
Anna
Summ <- ddply(mydata, .(factor3,fact
.
once again, thank you for your time.
Anna
Summ <- ddply(mydata, .(factor3,factor1), summarize,
mean = mean(var1, na.rm = FALSE),
sdv = sd(var1, na.rm = FALSE),
se = 1.96*(sd(var1, na.rm=FALSE)/sqrt(length(var1
Summ$Grouping <-
tweaking grid.arrange. This will come in handy!
With kind regards
Anna Zakrisson Braeunlich
PhD student
Department of Ecology Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden
Lives in Berlin.
For paper mail:
Katzbachstr. 21
D-10965, Berlin
Dear Stephen,
thank you for the tip regarding Incskape. I had never heard of it before. It
looks extremely useful!
With kind regards
Anna Zakrisson Braeunlich
PhD student
Department of Ecology Environment and Plant Sciences
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
Sweden
problem.
I also found:
theme(legend.justification=c(1,0), legend.position=c(1,0))
for the legend justification. Now this starts to look like how I want it to!
Thank you so much for your time!
Best regards
Anna Zakrisson Braeunlich
PhD student
Department of Ecology Environment and Plant Sciences
;white", #Why
is the fill commando not working?
position = "dodge", width = 0.3, size=3) +
geom_line(aes(linetype=factor1), color = "black", size = 0.5) +
geom_errorbar(aes(ymin = mean - sdv , ymax = mean + sdv), width = 0.3,
position = "d
ing this in ggplot2?
I have also tried plotmeans() in the sciplot package, but was unsuccessful.
Sincerely
Anna Zakrisson
library(plotrix)
?brkdn.plot
par(family="serif",font=1)
brkdn.plot("y1", "factor1","factor2", data=mydata,
mct="mean&
d the standard deviations instead of the default
standard errors.
Thank you for taking your time. I have spent a long time trying to solve
this and the frustration slowly approaches anger (in myself) :-)
Yours sincerely
Anna Zakrisson Braeunlich
PhD Student
Department of Systems Ecology
S
I have a code using the gRain package and there cptable is used to get the
conditional probability tables. My problem is that when I want to use this
table I get an error message that the file is too long to keep source. I have
tried to change the option keep.source but without luck. Anyone expe
it possible to not display borderlines between some countries but to
show the borders between others?
Thank you very much for your help!
Anna Fechner
PricewaterhouseCoopers Aktiengesellschaft Wirtschaftsprüfungsgesellschaft
Vorsitzender des Aufsichtsrates
WP StB Dr. Norbert Vogelpoth
Vorstan
Hi!
How Can I create a directory in R to save my files?
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Hello Chris,
I 've found two other issues
with MAE and CRPS, giving warning when running examples.
I've the same issue on my data.
Hope that you could find some time to take a
look here.
Thank you
Anna
> library(ensembleBMA)
Loading required package: chron
> example(M
Hi Marco,
a small example will be helpful to get better to the point.
But I suggest you either to address this question to
R geo mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-geo
and eventually take a look to GeoXp library.
Cheers
Anna
Anna Freni Sterrantino
Department of
pe it helps
Anna
Anna Freni Sterrantino
Department of Statistics
University of Bologna, Italy
via Belle Arti 41, 40124 BO.
Da: SAEC
A: r-help@r-project.org
Inviato: Giovedì 11 Ottobre 2012 0:22
Oggetto: [R] GAM without intercept
Hi everybody,
I am trying to
Hello!
I'd like to know if it is correct to
test with anova two models specified like this:
m1=y~x1+s(x2,by=x3),family="poisson"
m0=y~x1+s(x2),family="poisson"
anova(m1,m0)
Cheers
Anna
Anna Freni Sterrantino
Department of Statistics
University of Bologna, Italy
vi
2,
lwd = 2)
lines(mydata$Ncell[I], pred.gam$fit[I] - 2 * pred.gam$se.fit[I], lty = 2,
lwd = 2)
It all works fine if the model instead is:
ModelSimple <- gam(var ~ s(var1, k=4), data = mydata)
However, I get different results (for obvious reasons) if I predict from
Model or ModelSimple.
Ple
T=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ensembleBMA_5.0.3 chron_2.3-42
loaded via a namespace (and not attached):
[1] tools_2.15.0
Any suggestion?
Anna Freni S
Hello Peter, Duncan, Dan and RList!
How I suspected... I managed to do the
"homework" and forced invbinomal to return zero ( as in Stata)
once that k==n.
Thank you for all you replies!
Cheers
Anna
Da: peter dalgaard
Cc: Duncan Murdoch ;
uniroot(function(x) pbinom(k, n, x) - p, c(0, 1)) :
f() values at end points not of opposite sign
while stata
gen p3=invbinomial(50,50, 0.4)
. display p3
0
. gen p4=invbinomial(50,50, 0.6)
. display p4
0
Thanks
Cheers
Anna
Da: Duncan Murdoch
Cc: Rc
oku.php?id=guides:tutorials:regression:table
I tried to replicate using qbinom
the results obtained in
> invbinomial(10,5, 0.5)
>.54830584
but with no success.
Thank you
Cheers
Anna
Anna Freni Sterrantino
Department of Statistics
University of Bologna, Italy
via Belle A
Hello all!
I would like to create a 3d plot, with the option price explained by
the underlying price and time. Unfortunately, I can't quite get it to
work. I would very much appreciate your help!
Thanks,
Anna
# Black-Scholes Option Graph
library(lattice)
blackscholes <- funct
t of "1G", which disappears when I run
it with the name 'G' instead of '1G'. Am I not allowed to use numerical
values?
Best,
Anna
On Tue, 10 Jan 2012 23:02:04 +0100, Anna Olofsson
wrote:
> Hi,
> I'm pretty new at programming and with the R language.
(I'm using unix) to run
this with source("name_of_file"), but it doesn't work. Shouldn't a plot
come up automatically when I run it? What am I doing wrong? It knows what x
and y is, but I don't get an error of what might be wrong.
> source("name_o
multiply the dependent variable autocorrelation with the
independent variable autocorrelation and then multiply by (N-j)/N where N is
the sample size and j is the lag...calculate z-value...adjust my
p-value...Sincerely
Anna Zakrisson Braeunlich
PhD Student
Department of Systems Ecology
Stockholm
you for your time!
Anna Zakrisson Braeunlich
PhD Student
Department of Systems Ecology
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
E-mail: a...@ecology.su.se
Tel work: +46 (0)8 161103
Mobile: +46-(0)700-525015
Web site: http://www.ecology.su.se/staff/personal.asp?id=163
>&
nt colours/other markings (irrelevant for me)
and identified on the side of the graph.
Thank's!
Anna Zakrisson Braeunlich
PhD Student
Department of Systems Ecology
Stockholm University
Svante Arrheniusv. 21A
SE-106 91 Stockholm
E-mail: a...@ecology.su.se
Tel work: +46 (0)8 161103
Mobile: +4
Hi,
thanks for the suggestion!
I had tried it before, but it did not work - this was probably because I was
using the "image" function to
plot the 2d histogram. When I use directly hist2d and then contour with add=T
it works.
Thanks again
Anna
_
e).
Any suggestions?
Thanks,
Anna
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and provide comm
Sorry guys, I allready found the solution. Excell showed some of the
numbers in the format: 1,90053-E05 and R interpreted it as 1.9... I
changed that in Excel
Cheers, Anna
Am 6. Oktober 2011 17:48 schrieb Uwe Ligges :
>
>
> On 06.10.2011 17:39, Anna Lee wrote:
>>
>> Hello e
s!
Cheers, Anna
--
Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu
benachrichtigen und die E-Mail zu löschen.
This e-mail is confidential. If you have received it in error, please
notify me immediately and
Alright - for the first stock in the month, I pretty much copied Mikkel's
code and got what I wanted. Thanks! Anna
install.packages("surveillance")
library(surveillance)
alldat$year <- isoWeekYear(alldat$mydate)$ISOYear
alldat$months<-months(alldat$mydate)
alldat <-
that
isoWeekYear is great for week and year but not so great for month. If you
have any ideas, please let me know!
Thanks!
Anna
On Tue, Sep 13, 2011 at 6:55 PM, Mikkel Grum wrote:
> The following will get you the first stock in each week. Is that useful?
>
> install.packages("surve
)){
for(i in 1:nrow(divall[[j]])){
workin[[j]]<-which(divall[[j]][,1]<=divall[[j]][i,1]+6 &
divall[[j]][,1]!=divall[[j]][i,1] & divall[[j]][,1]>=divall[[j]][i,1])
}}
If I could get the workin list to work, I would use unique and unlist in
order to fi
project. The thing I most want to understand is how, after specifying a
certain condition, one may save certain data that occurs when that condition
is met. I hope I have been clear enough!
Thank you very much for your help!
Anna
biglist<-list(a=1:4,b=2:6)
lilwin<-list(x=NA,y=2)
lilloss&l
ble occurrences of butterfly species do not behave independently.)
>
>
> On Mon, 29 Aug 2011, Anna Mill wrote:
>
> Hi all,
>>
>> I am trying to do a generalized estimating equation (GEE) with the
>> "geepack"
>> package and I am not 100% sure what ex
1")
or should even almost every count have a special id (e.g. *
id=interaction(month,site)* or *id=interaction(month,type*))
Thanks for your help...
Anna
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nd what I would like to do. I have researched for
a couple of hours regarding this problem but to no avail!
Thanks!
Anna
> otestme values ind
10.001008012 AAPL.UW.Equity
20.015518087 AAPL.UW.Equity
30.013221459 AAPL.UW.Equity
40.012195734
newdate<-NULL
for(i in 1:nrow(thedate)){
mynewdate[i]<-as.Date(thedate[i,1],origin="1899-12-30")
}
print(mynewdate)
}
exdate()
Thank you very much!
Anna
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x27;s the
model:
modPoplar<- nls(Diameter ~ d*(1-exp(-b *Age))^a ,start=list(a=20,b=0.9,d=33))
I attached the graph, too.
Hoping for your answers!
Best, Anna
--
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
- - - - - - - - - -
Telefon: 01577-7805329
E
x27;s the
model:
modPoplar<- nls(Diameter ~ d*(1-exp(-b *Age))^a ,start=list(a=20,b=0.9,d=33))
I attached the graph, too.
Hoping for your answers!
Best, Anna
--
Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zugesandt worden sein, bitte ich Sie, mi
x27;s the
model:
modPoplar<- nls(Diameter ~ d*(1-exp(-b *Age))^a ,start=list(a=20,b=0.9,d=33))
I attached the graph, too.
Hoping for your answers!
Best, Anna
--
Telefon: 01577-7805329
E-Mail: gretschel.a
x27;s the
model:
modPoplar<- nls(Diameter ~ d*(1-exp(-b *Age))^a ,start=list(a=20,b=0.9,d=33))
I attached the graph, too.
Hoping for your answers!
Best, Anna
--
Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail
irrtümlich zugesandt worden sein, bitte ich Sie, mi
Hi Duncan,
I have tried to install a tar.gz package following your instructions
(https://stat.ethz.ch/pipermail/r-help/2008-August/169599.html) but without
success. Here are the steps I followed:
I installed the last version of Rtools and ran Rcmd INSTALL rJava_0.8-8.tar.gz
and got the error m
umb about a suitable number
of samples/predictors or point me to some literature that would help me
understand my problems.
Thanks
Anna
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PLEASE do read the postin
Hello everyone, I know how to add a folder path to my EV path but it only
works for the current R session. Is there a way to add it permanently? Here
is my code:
Sys.setenv(PATH=paste("C:\\Program Files\\Java\\jre1.6.0_13\\bin;",
Sys.getenv(x="PATH"), sep=""))
Thanks a lot!
--
View this message in
mean))
b<-as.vector(tapply(Sporangia,list(Host,Isolate),sd))
c<-sqrt(b/36)
errbar(c(1.5,2.5,3.5,5.5,6.5,7.5),(tapply(Sporangia,list(Host,Isolate),mean)),a+c,a-c,add=T)
Cheers
Anna
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",col=c("grey39","grey64","grey89"),beside=T)
col=c("grey39","grey64","grey89")
legend("topright",inset=.05,title="Host",c("European Larch","Hybrid
Larch","Japan
ne box"), and not
> independent binomial data.
>
sorry for the dumb question: so do you think, that my data is independent
and the model appropriate?
Thanks, Anna
>
>
>
> --
> Peter Dalgaard
> Center for Statistics, Copenhagen Business School
> Solbjerg Plads 3, 2
lagh and Nelder for the definitions, including of 'scaled deviance')),
> but it does change the standard errors.
>
>
> On Mon, 13 Jun 2011, Anna Mill wrote:
>
> Dear all,
>>
>> I am new to R and my question may be trivial to you...
>> I am doing a GLM wit
models attached)?
Thanks a lot for your help!
Anna
first model with binomial error structure:
> success<-c(14,43,44,1,13,28,56,8)
> failure<-c(88,59,58,101,92,77,49,97)
> "fragment"<-c(1,1,1,1,2,2,2,2)
> "type"<-c(1,2,3,4,1,2,3,4)
> y<-cbind(succ
label names and
if not what the best alternative is.
Thanks
Anna
pca<-biodata[,3:10]
model<-prcomp(pca,scale=TRUE)
summary(model)
biplot(model)
--
Dr Anna R. Renwick
Research Ecologist
British Trust for Ornithology,
The Nunnery,
Thetford,
Norfolk,
IP24 2PU,
UK
Tel: +44 (0)1842 750050; Fax:
52 4.30615258
Number of Observations: 1549
Number of Groups: 2
Warning message:
In pt(q, df, lower.tail, log.p) : NaNs produced
Kindly Regards,
Anna-Leena Orsama
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P
llel=TRUE, cpus=Cpu,
type="SOCK"). When running process explorer I see the R-program creating a TCP
connection in the state LISTENER but no ESTABLISHED happens.
Regards
Anna
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Hey List,
does anyone know how I can generate a vector of random numbers from a
given distribution? Something like "rnorm" just for non normal
distributions???
Thanks a lot!
Anna
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data and therefore comparison of means, or non normal
distributed data and therefore comparison of medians?
Cheers, Anna
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Dear List!
I want to compare medians of non normal distributed data. Is it
possible and usefull to calculate 95% confidence intervals for
medians? And if so - how can this be achieved in R?
Thanks a lot!
Anna
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Dennis: thank you so much! I got it now and it works just perfectly.
Thanks a lot to the others too!
Anna
2011/3/21 Dennis Murphy :
> Hi:
>
> To amplify Ista's and David's comments:
>
> (1) You should not be inputting separate vectors into lm(), especially if
> you i
Dennis: Thank you so much! I got it now - it just works perfectly. Thanks a lot!
Anna
2011/3/21 Dennis Murphy :
> Hi:
>
> To amplify Ista's and David's comments:
>
> (1) You should not be inputting separate vectors into lm(), especially if
> you intend to do predict
data
and I thought with the predict function the programm would calculate a
value from the model function for every value of calP... ?
2011/3/20 Ista Zahn :
> Hi Anna,
>
> On Sun, Mar 20, 2011 at 2:54 PM, Anna Lee wrote:
>> Hey List,
>>
>> I did a multiple regressio
xv))
The predicted values are however strange. Now I do not know weather
just the model does not fit the data (actually all coefficiets are
significant and the plot(model) shows a good shape) or wether I did
something wrong with my prediction command. Does anyone have an
idea???
--
Thanks a
Am 16.03.2011 19:29, schrieb Heiman, Thomas J.:
Hi Anna,
AIC and BIC are good criteria for determining degree of model fit..
Sincerely,
tom
Thomas Heiman, PhD
Info Systems Eng, Sr
The MITRE Corporation | Center for Enterprise Modernization
Office: 703-983-2951 | thei...@mitre.org
Dear Bert,
so what can I do to obtain a goodness of fit for a non-linear model if
r² does not work?
And here comes my next question: is it apropriate to comopare a linear
and a non-linear model with anova()?
Thank you so much for answering,
Anna
Am 16.03.2011 18:54, schrieb Bert Gunter
Am 16.03.2011 18:19, schrieb Joshua Wiley:
Dear Anna,
What is your goal in obtaining a value for R^2 ? I believe it is not
provided for a non-linear model, because it does not make much sense.
It certainly will not have the same interpretation as in a linear
model, and all the ways it *could
Am 16.03.2011 18:15, schrieb David Winsemius:
On Mar 16, 2011, at 12:53 PM, Anna Gretschel wrote:
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. P
Dear List,
how can I obtain the value of r suqared for a non-linear model? For
linear models it can be found in the summary() of the model but for
non-linear models I just don't know. Please help!
Anna
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