Thank you.
Does the spatstat or R need bigger memory?
Paul Hiemstra wrote:
>
> CT YU wrote:
>> Hi all,
>> When I use the function envelope(???,pcf,i=???,j=???), there are some
>> errors
>> as followings:
>> Error: cannot allocate vector of size 5.3 Mb
>> In addition: Warning messages:
>> 1: In
Quick question about the usage of glht. I'm working with a data set
from an experiment where the response is bounded at 0 whose variance
increases with the mean, and is continuous. A Gamma error
distribution with a log link seemed like the logical choice, and so
I've modeled it as such.
That's exactly what I needed! Many thanks to Mark, Daniel and Duncan for
your help!
Enrico
On Mon, Apr 14, 2008 at 8:14 PM, Daniel Malter <[EMAIL PROTECTED]> wrote:
> Hi,
>
> type ?plot [Enter]and click the "par" link in the first line. This takes
> you
> to all the parameters you can set for plo
Charles C. Berry wrote:
>
> You want
>
> ?regexpr
>
> Something like
>
> regexpr("Jalapa", as.character( unidad ) ) != -1
>
> HTH,
>
> Chuck
Please, look at this...
> substringJalapa <- regexpr("Jalapa", as.character(conagua$unidad ) )
!= -1
Hubo 50 o más avisos (use warnings() para v
Thanks! Here what I get:
Henrique Dallazuanna wrote:
> Try something like this:
>
> x[grep("Jalapa", x$unidad),, drop = F]
>
> conaguaMexicoSub <- subset(conagua, unidad == x[grep("Jalapa",
x$unidad),, drop = F], select = c(equipo,X101:X309))
Error en x$unidad : $ operator is invalid for atomi
Try something like this:
x[grep("Jalapa", x$unidad),, drop = F]
On Mon, Apr 14, 2008 at 9:10 PM, [Ricardo Rodriguez] Your XEN ICT Team <
[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I have not been able to find an answer to what is a simple question:
>
> conaguaMexicoSub <- subset(conagua, unidad ==
On Tue, 15 Apr 2008, [Ricardo Rodriguez] Your XEN ICT Team wrote:
Hi all,
I have not been able to find an answer to what is a simple question:
conaguaMexicoSub <- subset(conagua, unidad == "Jalapa", select =
c(equipo,X101:X309))
This subset gives me all the rows from conagua where unidad is J
Hi,
type ?plot [Enter]and click the "par" link in the first line. This takes you
to all the parameters you can set for plot commands. The one you are looking
for is:
xaxs
The style of axis interval calculation to be used for the x-axis. Possible
values are "r", "i", "e", "s", "d". The styles are
Hi all,
I have not been able to find an answer to what is a simple question:
conaguaMexicoSub <- subset(conagua, unidad == "Jalapa", select =
c(equipo,X101:X309))
This subset gives me all the rows from conagua where unidad is Jalapa.
But, please, how do I get all the rows where unidad contents
On 14/04/2008 6:56 PM, Enrico Rossi wrote:
> Hello,
>
> If I make a plot, say something simple like
>
> plot( runif(100) )
>
> then the origin (0,0) is not at the bottom-left corner of the box
> surrounding the plot. The axis limits are "padded" slightly. This is
> ordinarily a good feature, bec
Hello,
If I make a plot, say something simple like
plot( runif(100) )
then the origin (0,0) is not at the bottom-left corner of the box
surrounding the plot. The axis limits are "padded" slightly. This is
ordinarily a good feature, because it makes plots look better. But now I
would like to make
Peter:
> Hi Achim: Thanks for the reply! I did notice that robcov() requires
> the X & Y and also a scores vector and that these are not readily
> available under fixed names or at all in output from such functions as
> systemfit and optim. I wonder if it would make sense to have a
> stop-gap f
try
?image
Xiaohui
merca duria wrote:
> I have a matrix with N x N points, that I would like to represent on a map.
>
> The points are an estimation of the bidimensional density of certain
> quantity observed on a geographical region.
> I am interested in a map with several colours displaying s
I have a matrix with N x N points, that I would like to represent on a map.
The points are an estimation of the bidimensional density of certain
quantity observed on a geographical region.
I am interested in a map with several colours displaying several values of
the density
Can you help me about
---
merca duria desea mantener el contacto contigo a través de algunos de
los mejores productos que Google ha lanzado recientemente.
Si ya utilizas Gmail o Google Talk, visita:
http://mail.google.com/mail/b-bb1b3747a4-83a9bc2722-f
Sherri
Is this what you want?
sherri <- data.frame(grp=rep(letters[1:6], 20), value=rnorm(6*20)+6)
temp <- with(sherri, boxplot(value~grp, xaxt='n'))
axis(1, 1:5+0.5, rep('', 5))
Peter Alspach
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Sher
I'm trying to analyze a model with two variables, one is Group with
two levels (male and female), and other is Time with four levels (T1,
T2, T3 and T4). And for the convenience of post-hoc testing I wanted
to consider a model with no intercept for factor Time, so I tried
formula
Group*(Time-1)
H
hi all-
i am creating a boxplot and would like to shift the x-axis tick marks
(named 1:23) so that the ticks are in between the boxes and not centered
in the middle of the boxes. i have searched the help files and google
and have not found what i am looking for.
does anyone know how to do this
Read the help page on '[', then read it again a couple more times (it
can be subtle). I think you want to use fp[[1]] rather than fp[1], the
first will return a single element from the list (a data frame in the
case below), the second returns a list of length 1 (with that element
being a data fram
>> i have a list of many files and want to load them into a list of tables,
>> that
>> can be adressed with a variable 'i'.
> fp <- lapply(files, read.table, header = T)
Thank you that works fine.
But how can i access the data in column fp$foo now?
fp[1]$foo does not work.
ps: thank you all f
>I try to add three different lines (solid, short dash, long dash)
>in to current barplot.
Have a look at the lty parameter. It's documented in ?par. If I
understand your question correctly, you may also be interested in
segments() to actually draw the lines (or maybe abline()).
cu
On Mon, Apr 14, 2008 at 4:47 PM, Gavin Simpson <[EMAIL PROTECTED]> wrote:
> Note that the default is to produce a bray-curtis dissimilarity matrix
> from the input species data. As such, I reproduce this dissimilarity
> matrix as arg 1 to cor and then take the Euclidean distances of the
> coor
The "type" argument to plotting functions determines the type of plot
(plot lines or points or both or ...). To specify the type of line you
need to use the "lty" argument. See the help page for "par" (?par) for
details on the types of line you can use (also "lwd" is the width of the
lines).
B
On Mon, 2008-04-14 at 15:38 -0400, Sarah Goslee wrote:
> Dimensionality doesn't matter. 2, 3, 14 will all work.
>
> What matters is that apparently vegan returns the results in some
> format other than a straight sites x ordination coordinates matrix,
> which is what dist expects. You'll need to c
On Mon, 14-Apr-2008 at 08:37AM +0200, Udo wrote:
|> Zitat von Peter Alspach <[EMAIL PROTECTED]>:
|>
|> > Udo
|> >
|> > Seems you might want merge()
|> >
|> > HTH ...
|> >
|> > Peter Alspach
|>
|> Thank you Peter and Jorge,
|>
|> but as I had written in my last sentence,
|> "Merge doesn´t do
Hi All
I have matrix Like:-
10 20 40 60 100
Glucose UP Down UP No UP
Fructose UP UP Down UP No
Sucrose UP Down UP No Down
Taurine No No No No UP
Cellobiose Up
Dear
I try to make lines smoothly which are added on current barplot. I tried to
use "smo" and it does not work. Has nayone have this experience?
Many Thanks!
Xin
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing lis
Here's an example to make the text 18 point. You can set ps=6 to make small
text.
x=rnorm(100)
opar=par(ps=18) # Make text 18 point
hist(x)
opar
Rob Baer
- Original Message -
From: "Sue Lee" <[EMAIL PROTECTED]>
To:
Sent: Monday, April 14, 2008 2:35 PM
Subject: [R] Histogram Label Fon
See matplot function
On Mon, Apr 14, 2008 at 4:44 PM, Xin <[EMAIL PROTECTED]> wrote:
> Dear
>
> I try to add three different lines (solid, short dash, long dash) in to
> current barplot. I saw there are types of lines ("p","b", ect). However,
> they are not what I request. Has nayone have this
Guillaume Brutel yahoo.fr> writes:
> I am trying to fit a non linear regression model to time series data.
>
> If I do this:
> reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal))
> I get this error message (translated to English from French):
> Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), d
On Mon, 14 Apr 2008, Vorlow Constantinos wrote:
> Dear all,
>
> Wouldn't it be helpful if the
>
> install.views
> install.packages
> update.packages
>
> had a "downloadonly=TRUE" flag which would allow us to download the
> packages and install them later (or put them in a USB stick and take
> them
I think that you can edit the source code of histogram for this:
myhist <- getS3method("plot", "histogram")
body(myhist)[9] <-
parse(text = c(capture.output(body(myhist)[9])[-6],
"} else labels, adj = c(0.5, -0.5), ...)"))
r <- hist(islands, plot = F)
myhist(r, labels = T
Dear
I try to add three different lines (solid, short dash, long dash) in to
current barplot. I saw there are types of lines ("p","b", ect). However, they
are not what I request. Has nayone have this experience?
Many Thanks!
Xin
[[alternative HTML version deleted]]
__
Try this:
fp <- lapply(files, read.table, header = T)
On Mon, Apr 14, 2008 at 4:21 PM, Jonas Stein <[EMAIL PROTECTED]> wrote:
> Hi
>
> i have a list of many files and want to load them into a list of tables,
> that
> can be adressed with a variable 'i'.
>
> something like
>
> files = c("0125um",
Dimensionality doesn't matter. 2, 3, 14 will all work.
What matters is that apparently vegan returns the results in some
format other than a straight sites x ordination coordinates matrix,
which is what dist expects. You'll need to convert it to one, possibly
with as.matrix, or extract the coordin
Hi!
I'm having a trouble changing font size of histogram label.
I have tried help(hist), but I couldn't find anything explain how to fix
label's font size.
Could you help me please?
Thank you.
_
Going green? See the top 12 foods t
Hi
i have a list of many files and want to load them into a list of tables, that
can be adressed with a variable 'i'.
something like
files = c("0125um","2000um","2200um","2500um","2700um")
for (i in 1:5)
{
fp[i] <- read.table( files[i] )
}
but this does not work of course.
Has anyone a g
the below is a three dimensional solution
On Mon, Apr 14, 2008 at 3:09 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> cor(x, dist(mds))^2
> Error in dist(mds) : (list) object cannot be coerced to 'double'
>
>
>
> On Mon, Apr 14, 2008 at 2:40 PM, Sarah Goslee <[EMAIL PROTECTED]> wrote:
> > The
Another option, if you want them in separate figures, is to write a loop
that generates the image file, saves it to the file system, and use
odfInsertPlot to put the file into the document.
This might work better if you have an unknown number of images that you
want to insert.
Max
-Original
Sarah, thanks for your reply.
On Mon, Apr 14, 2008 at 8:32 PM, Sarah Goslee <[EMAIL PROTECTED]> wrote:
> If you ran that code outside ODFWeave, you'd only get one plot,
> so why would you expect to get more within ODFWeave?
No, it depends on the device that is used. If I use PDF or postscript
th
If you ran that code outside ODFWeave, you'd only get one plot,
so why would you expect to get more within ODFWeave?
for (i in 1:3) {
> plot(1,1, main=paste('Plot',i))
> }
You need to add some sort of par() command, or use layout(), to create
a single plot that contains all three of the plot
I am using the function metaMDS with jaccard distances to ordinate a
set of constituent by site matrix. I can post this data if it would
be helpful, but it is large to include in an email. I can also
provide reproducable code if necessary. I would like to get an R^2
value for the axes of the ord
Dear all,
Max, first of all, many thanks for providing the odfWeave package.
My problem: Whenever I have multiple plots in one single chunk of my
ODF file, only the last plot gets shown. The problem can be reproduced
with this toy example (to be used in an ODF file together with
odfWeave -- I'm u
OK, here is another approach.
Here is a function that will adjust the user coordinate system for you.
You give it 2 points in the current coordinates (x1 and y1) and the
values that you want the same 2 points to have in the new coordinate
system (x2 and y2) and it will change the coordinate syste
Try:
by(x, list(x$STNID), function(.x)lm(TEMP ~ DEWP, data = .x))
On Mon, Apr 14, 2008 at 2:03 PM, Ryan Lauritsen <[EMAIL PROTECTED]>
wrote:
> Hi all. I'm brand new to R.
>
> My dataset (stored in MySQL) is a list of weather stations in rows by
> year with various weather variables in columns,
Hi,
I am trying to make a histogram with two x variables,
can I do it in R.
I have two variables which I need to graph as
histograms with frequency on y axis (hopefully the two
shown in different colors). I cannot seem to find how
to do it in R, after reading the help section and some
google stuf
Hi all. I'm brand new to R.
My dataset (stored in MySQL) is a list of weather stations in rows by
year with various weather variables in columns, for example:
STNID YEAR TEMP DEWP
station11990 54 50
station11991 23 10
station11992 34 18
station21990
See textplot function in gplots package
On Mon, Apr 14, 2008 at 1:09 PM, Ng Stanley <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Is there a way to write a table to pdf ? I have checked write.table, it
> only
> writes to text file.
>
>[[alternative HTML version deleted]]
>
> __
What is RG? I suspect it is not a list but a vector and
you operation you wrote does not make sense. Convert it
to a list if you want to store c("a","b") is RG[["AB"]].
G.
On Tue, Apr 15, 2008 at 12:06:14AM +0800, Ng Stanley wrote:
> Hi,
>
> How can I force the assignment ?
>
> > RG[["ABC"]]
Dear all,
I am trying to fit a non linear regression model to time series data.
If I do this:
reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal))
I get this error message (translated to English from French):
Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start =
list(xmid = aux[1],
Hi,
Is there a way to write a table to pdf ? I have checked write.table, it only
writes to text file.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the post
Hi,
How can I force the assignment ?
> RG[["ABC"]] <- c("a", "b")
Error in RG[["ABC"]] <- c("a", "b") :
more elements supplied than there are to replace
On Mon, Apr 14, 2008 at 9:53 PM, Gabor Csardi <[EMAIL PROTECTED]> wrote:
> On Mon, Apr 14, 2008 at 09:47:49PM +0800, Ng Stanley wrote:
> >
Tania Oh wrote:
> Dear Uwe,
> thank you very much for this.
> After reading your solution below, I searched the help pages for
> data.frame, which, factor but I didn't see the option for "drop" in them.
In fact, ?factor links to ?[.factor whcih explains it.
Uwe
> I googled and found "drop"
I think that the problem is with the dev.off commands.
The first time running b there are not a device 2 and device 3 so
dev.off does nothing (or if they did exist then they are turned off).
Then you set the graphics event callback which attaches to the current
graph window (number 3 in general).
Duncan Murdoch stats.uwo.ca> writes:
>
> On 4/14/2008 8:59 AM, Louisa Hay wrote:
> > Hi,
> > I am trying to write R code to produce a
power curve to show how the power of a
> > Wilcoxon-test varies depending on the mean, with data
generated from a
uniform distribution. Any ideas
> how I shoul
CT YU wrote:
> Hi all,
> When I use the function envelope(???,pcf,i=???,j=???), there are some errors
> as followings:
> Error: cannot allocate vector of size 5.3 Mb
> In addition: Warning messages:
> 1: In vector("double", length) :
> Reached total allocation of 502Mb: see help(memory.size)
> 2:
Hi Achim: Thanks for the reply! I did notice that robcov() requires
the X & Y and also a scores vector and that these are not readily
available under fixed names or at all in output from such functions as
systemfit and optim. I wonder if it would make sense to have a
stop-gap function that would
Have you tried plotting it, e.g., like the following:
npts = 51 # or some number
h = seq(0, ???, length=npts)
funh <- rep(NA, npts)
for(i in 1:npts)funh[i] <- fun(h[i])
plot(h, funh)
Hope this helps.
Spencer
Sungsu wrote:
>
> Dear Sp
> > The (imho) unintuitive behaviour is to do with the subsetting function
> > [.factor, not which. There are a couple of workarounds:
> >
> In that case, your intuition needs readjustment
>
> There are other systems which (de facto) drop unused levels by default,
> and it is a real pain t
You forgot to tell us your OS, so it was probably Windows.
If it is Windows, Arial is the default font for the windows() family of
devices, including png(). However, bitmapped plots don't have text or
fonts, just pixels.
As for postscript(), I am unaware of the existence of a PostScript font
I would like to ensure that all the text in my plot is in the arial font, like
in excel and word. I have been saving my graphics to a file using the png()
function, and have tried the postscript() function. In the documentation I
have not been able to find a way of changing the family argumen
On 4/14/2008 8:59 AM, Louisa Hay wrote:
> Hi,
> I am trying to write R code to produce a power curve to show how the power of
> a
> Wilcoxon-test varies depending on the mean, with data generated from a
> uniform distribution. Any ideas how I should go about this?Louisa
I think I gave you enoug
On Mon, Apr 14, 2008 at 09:47:49PM +0800, Ng Stanley wrote:
> Hi,
>
> Two questions:
>
> A) I need to initialize many variables to NULL. So I created variable_names
> <- c("a1", "a2"). What can I do to variable_names so that variable a1 is
> NULL and a2 is NULL ?
for (n in variable_names) assign
Hi,
Two questions:
A) I need to initialize many variables to NULL. So I created variable_names
<- c("a1", "a2"). What can I do to variable_names so that variable a1 is
NULL and a2 is NULL ?
B) How can I check whether an object exist ?
Thanks
Stanley
[[alternative HTML version deleted]]
Hello,
I am trying to test for non-linearity in a set of non-parametric data.
Furthermore, I would like to do so in a multi-variate model. Frankly, I
don't even know if this is possible, and if it is possible, I don't know if
it is implemented in an R package. Any advice would greatly appreciated.
Dear all,
Wouldn't it be helpful if the
install.views
install.packages
update.packages
had a "downloadonly=TRUE" flag which would allow us to download the
packages and install them later (or put them in a USB stick and take
them for installation on anothe PC).
I am behind a firewall (Bank)
Hi all,
When I use the function envelope(???,pcf,i=???,j=???), there are some errors
as followings:
Error: cannot allocate vector of size 5.3 Mb
In addition: Warning messages:
1: In vector("double", length) :
Reached total allocation of 502Mb: see help(memory.size)
2: In vector("double", length)
You might also take a look at the doBy package.
On Apr 14, 2008, at 7:30 AM, zerfetzen wrote:
> Thanks all. I will try to use both tapply and by, and have no idea
> how I
> missed the by function. Thanks again.
_
Professor Michael Kubovy
University of Virginia
Dep
It would be earier to respond to your query if there were a more complete
reference than "Lindkvist 1999".
I've looked at added variable plots, adjusted variable plots, and constructed
variable plots for the Cox model many years ago. The ideas resulted in some
papers, but in practical work
Bill,
We already have erf() and erfc() as examples, and help.search() will find
them. I've added erfinv() and erfcinv() to the examples and concept
index.
I don't see a good reason to add them to e.g. the 'stats' namespace: it is
not as if we wish to encourage people to use them instead of [p
Dear Spencer.
Thank you for your kind reply.
I have n data points observed on the surface of a
torus. I am trying to fit the geodesic line equation
to these points on the surface:
the equation is
âu=h*integrate(((5+cos(v))*sqrt((5+cos(v))^2-h^2))^{-1})
from 0 to vâ.
I wrote the following R
Louisa Hay msn.com> writes:
>
>
> Hi,
> I am trying to write R code to produce a power curve to show
> how the power of a
> Wilcoxon-test varies depending on the mean, with data
> generated from a uniform
> distribution. Any ideas
> how I should go about this?Louisa
Here's a brute-force a
Hi,
I am trying to write R code to produce a power curve to show how the power of a
Wilcoxon-test varies depending on the mean, with data generated from a uniform
distribution. Any ideas how I should go about this?Louisa
_
Amazing p
[EMAIL PROTECTED] wrote:
>> I used "which" to obtain a subset of values from my data.frame.
>> however, I find that there is a "trace" of the values I have removed.
>> Any suggestions would be greatly appreciate.
>>
>> Below is my data:
>>
>> d <- data.frame( val = 1:10,
>> gr
On Mon, Apr 14, 2008 at 08:52:36PM +0800, Ng Stanley wrote:
> Hi,
>
> Didn't make myself clear on A). The twenty OBs are all different, how to
> store them in a vector or list ?
bigOB <- list(OB1, OB2, OB3, ..., OB20)
G.
[...]
--
Csardi Gabor <[EMAIL PROTECTED]>UNIL DGM
_
Hi,
Didn't make myself clear on A). The twenty OBs are all different, how to
store them in a vector or list ?
Your solution to B) is nice and clear. Will read up on the Introduction
also.
On Mon, Apr 14, 2008 at 8:39 PM, Gabor Csardi <[EMAIL PROTECTED]> wrote:
> On Mon, Apr 14, 2008 at 08:32:55
On Mon, Apr 14, 2008 at 08:32:55PM +0800, Ng Stanley wrote:
> Hi,
>
> Two questions:
>
> A) Assuming OB is an object, how do I store 20 of OB in a vector or list ?
replicate(20, OB, simplify=FALSE)
> B) Does R has something similar associative array to Perl ? For example,
> %our_friends = ('bes
Hi,
Two questions:
A) Assuming OB is an object, how do I store 20 of OB in a vector or list ?
B) Does R has something similar associative array to Perl ? For example,
%our_friends = ('best', 'Don', 'good', 'Robert', 'worst', 'Joe');
$our_friends{'cool'} = "Karen";
Thanks
Stanley
[[alter
On 4/14/2008 7:30 AM, zerfetzen wrote:
> Thanks all. I will try to use both tapply and by, and have no idea how I
> missed the by function. Thanks again.
One problem with both of those arises if you are subsetting on several
columns. They will do the calculations for all combinations of all
c
Dear Uwe,
thank you very much for this.
After reading your solution below, I searched the help pages for
data.frame, which, factor but I didn't see the option for "drop" in
them.
I googled and found "drop" associated with the function subset. is
this the help page you were alluding to?
Sor
> I used "which" to obtain a subset of values from my data.frame.
> however, I find that there is a "trace" of the values I have removed.
> Any suggestions would be greatly appreciate.
>
> Below is my data:
>
> d <- data.frame( val = 1:10,
> group = sample(LETTERS[1:5], 10,
On Thu, 10 Apr 2008, Peter Muhlberger wrote:
> Hi Bill: Thanks for the reply! As you've no doubt guessed, I'm not a
> statistician (I'm a social scientist). I hadn't given thought to
> modeling the cluster-based covariance explicitly--interesting
> possibility. My responses are drawn from s
Tania Oh wrote:
> Dear all,
>
> I used "which" to obtain a subset of values from my data.frame.
> however, I find that there is a "trace" of the values I have removed.
> Any suggestions would be greatly appreciate.
>
> Below is my data:
>
> d <- data.frame( val = 1:10,
>
Thanks all. I will try to use both tapply and by, and have no idea how I
missed the by function. Thanks again.
--
View this message in context:
http://www.nabble.com/Equivalent-to-a-BY-command-in-SAS-tp16670452p16676280.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi Oystein,
Maybe this is what you are looking for:
Keyword: merge()
--
table1<-read.table("daily.data")
table2<-read.table("weekly.data")
(maybe you need to create a new common coloumn for daily and weekly data set.
For
Thiemo Fetzer wrote:
As a general rule and practical in many contexts uses the following
steps: (a) Use the R2HTML library to write the results from the summary
to the clipboard (b) Open a spreadsheet and paste the results there.
(c) Edit and shove things around until everything are where you w
Dear all,
I used "which" to obtain a subset of values from my data.frame.
however, I find that there is a "trace" of the values I have removed.
Any suggestions would be greatly appreciate.
Below is my data:
d <- data.frame( val = 1:10,
group = sample(LETTERS[1:5], 10, rep
.45
> 20080410200815 27.45
> 20080411200815 27.45
>
> library(zoo)
> library(xts)
>
> Lines.daily <- "quote_date value
+ 200804071
+ 200804082
+ 200804093
+ 200804104
+ 200804116
+ 20080412
On 4/14/2008 6:35 AM, Chuck Cleland wrote:
> On 4/14/2008 6:21 AM, Louisa Hay wrote:
>>
>> Hi,
>> I am trying to create a power curve to show how the power of a t-test varies
>> depending on the mean. Any ideas how I should go about this?
>> Louisa
>
>Something like this maybe?
>
> mydiffs
On 4/14/2008 6:21 AM, Louisa Hay wrote:
>
> Hi,
> I am trying to create a power curve to show how the power of a t-test varies
> depending on the mean. Any ideas how I should go about this?
> Louisa
Something like this maybe?
mydiffs <- seq(.05,1,.05)
mypower <- vector("numeric", 20)
for(
Louisa Hay wrote:
>
> Hi,
> I am trying to create a power curve to show how the power of a t-test varies
> depending on the mean. Any ideas how I should go about this?
> Louisa
The pt() function supports a non-centrality parameter, so you can do
this calculation directly.
For other situations
Here is one way to get the 'week':
> x <- as.character(seq(20080401, 20080430))
> # get the week
> cbind(x, format(as.Date(x, "%Y%m%d"), "%W"))
x
[1,] "20080401" "13"
[2,] "20080402" "13"
[3,] "20080403" "13"
[4,] "20080404" "13"
[5,] "20080405" "13"
[6,] "20080406" "13"
[7,] "2008040
Hi,
I am trying to create a power curve to show how the power of a t-test varies
depending on the mean. Any ideas how I should go about this?
Louisa
_
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Dear R-help group,
I have a dataset with daily closing prices from a stock exchange (consecutive 5
trading days) from a firm trading a specific commodity. The date variable looks
like:
quote_date
20080411
With the format; mmdd.
Moreover, I have another data set with a (average) weekly pri
> I would like to do looping for this process below to estimate alpha
> beta from gamma distribution:
> Here are my data:
> day_data1 <-
> 123456 789 10 11 12
> 13 14 15 16 17 18 19 20 21 22 23
> 1943 48.3 18.5 0.0 0.0 18.3 0.0
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