Hello:
Is there a way to get a mean from values stored in different rows?
The data looks like this:
YEAR-1, JAN, FEB, ..., DEC
YEAR-2, JAN, FEB, ..., DEC
YEAR-3, JAN, FEB, ..., DEC
What I want is the mean(s) for just the consecutive winter months:
YEAR-1.DEC, YEAR-2.JAN, YEAR-2.FEB
YE
A more portable way (that function only works in some versions of R)
is
as.POSIXct(1317857320, origin="1970-01-01")
possibly with a 'tz' argument if you need to restore the timezone.
On Wed, 5 Oct 2011, jim holtman wrote:
Here is what I use:
unix2POSIXct(1317857320)
[1] "2011-10-05 19:28:40
A bit late, but here is what I always do:
m = merge(bid, ask, tick)
m<- interpNA(m, method="before")
intrepNA can also interpolate NAs in different ways, for example linearly.
Hth,
Ulrich
Am 06.10.2011 03:08, schrieb Robert A'gata:
Hi Roupell,
Yes I am aware of RTAQ function matchTradesQu
I think the problem is your str1 is an unevaluated expression and will
change with the value of i. You should be able to get a fixed title by
this:
par(mfrow = c(2, 1))
for (i in 1:2) {
x <- 1:100
rmse <- sin(x/5) # fake data
plot(x, rmse, main = substitute(list(RMSE(theta), i == z),
On Oct 5, 2011, at 7:45 PM, Eva Powers wrote:
I have 2 dataframes. "mydata" contains numerical data. "mybys"
contains
information on the "group" each row of the data is in. I wish to
aggregate
each column in mydata using the corresponding column in mybys.
corresponding?
Please see th
On Wed, Oct 5, 2011 at 12:27 AM, andrewH wrote:
> Dear folks,
>
> I’m trying to build a function to create and make available some variables I
> frequently use for testing purposes. Suppose I have a function that takes
> some inputs and creates (internally) several named objects. Say,
>
> fun1 <-
Hi:
Here's one way:
m <- matrix(rpois(100, 8), nrow = 5)
f <- function(x) {
q <- quantile(x, c(0.1, 0.9), na.rm = TRUE)
c(sum(x < q[1]), sum(x > q[2]))
}
t(apply(m, 1, f))
HTH,
Dennis
On Wed, Oct 5, 2011 at 8:11 PM, Ben qant wrote:
> Hello,
>
> I'm trying to get the count of value
Hello,
I'm trying to get the count of values in each row that are above and below
quantile thresholds. Thanks!
Example:
> x = matrix(1:30,5,6)
> x
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]16 11 16 21 26
[2,]27 12 17 22 27
[3,]38 13 18 23 28
[4,]4
I ran into a problem with titles on graphs. I wanted a graph with
multiple subplots, with each having a title that involved both
a Greek letter and an identifier for each graph. Below is a
simplified version of code to do this. The graph appears fine,
with the first graph having "i=1" in the ti
Solution: I figured this out on my own (below).
> gnames<-read.csv("/home/victoria/R/gnames.csv",header=FALSE,sep=",")
> gnames
V1V2V3V4V5 V6V7
1 NM_005588 NM_004407 NM_006136 NM_004817 NM_006012 NM_001008693 NM_181435
[snip]
V497
Btw tweaking MatchTradesQuotes should not be an issue and its easy to
accommodate any data format that is passed through in xts object.
At least that's what I did with RTAQ package - used it as a shell to create
unique functions that suit data format for intra-day tick flow from ASX.
Hope th
This game you are playing is quite dangerous, and I recommend that you avoid
this assiduously. Look at the lubricate library or use the chron or Date
classes. Timestamps are rather poor candidates for numeric operations, and the
vagaries of timezones are quite a headache.
---
Hi:
It's a little tricky to read in a data frame 'by hand' without making
NA a default missing value; you've got to trick it a bit. I'm doing
this inefficiently, but if you have the two 'real' data sets stored in
separate files, read.table() is the way to go since it provides an
option for definin
Kevin Wright gmail.com> writes:
>
> Generally, the only way to estimate f1:f2 is if you have all combinations of
> data present for these two factors.
Well, he said it was unbalanced, he didn't say how unbalanced --
i.e. it's not clear (to me) whether there are any completely missing cells
or
Darko - Looking at it carefully. Yes, you're right. It's still native
R code function. I know how to proceed now. Thanks.
On Wed, Oct 5, 2011 at 9:22 PM, Roupell, Darko wrote:
> Btw tweaking MatchTradesQuotes should not be an issue and its easy to
> accommodate any data format that is passed thr
Does this do what you want:
> data1
A B
1 1 a
2 1 b
3 2 c
4 2 d
> data2
D E F
1 x y 1
2 w z 2
> data1.1 <- aggregate(data1$B, list(data1$A), FUN=paste, collapse=',')
> data1.1
Group.1 x
1 1 a,b
2 2 c,d
> merge(data2, data1.1, by.x="F", by.y="Group.1")
F D E x
1 1 x y a,b
2
Here is what I use:
unix2POSIXct(1317857320)
[1] "2011-10-05 19:28:40 EDT"
unix2POSIXct <- function (time) structure(time, class = c("POSIXt",
"POSIXct"))
On Wed, Oct 5, 2011 at 7:38 PM, Mike Williamson wrote:
> Hi,
>
> In short, I would like to know if there is any way to convert a num
Hi Roupell,
Yes I am aware of RTAQ function matchTradesQuotes. But my time series
does not follow the TAQ format like they suggest. So I gave it a try
and find that it doesn't work. In particular, my time series contain
full level 2 order book and trades. I want to do asof join of the book
to the
Hi Chris,
Yes, you're missing something: the colClasses argument to read.csv.
Given a tiny little csv file that looks like this:
1,2,3,"01234"
4,5,6,"00011"
7,8,0,"0"
> testdata <- read.csv("testdata.csv", header=FALSE, colClasses=c(NA, NA, NA,
> "character"))
> testdata
V1 V2 V3V4
1
It is perhaps easier than you think, see the example in gstat:
library(gstat)
?krige
After this is run "meuse" is a SpatialPointsDataFrame:
coordinates(meuse) = ~x+y
Any point shapefile read with readOGR from rgdal (or the alternative
functions in maptools) will also be SpatialPointsDataFrames,
Dear Help-Rs,
I've been dealing with this problem for some time, using a work-around to deal
with it. It's time for me to come clean with my ineptitude and seek a what has
got to be a more streamlined solution from the Help-Rverse.
I regularly import delimited text data that contains numerics
All,
I've read several tutorials re: generating wireframes, but am clearly
missing something. I have data along the lines of:
> tbl [1:10,]
Visits Activity Course.Grade
1 17218.31
2 7 1120.67
3 9 1724.69
4 28 7138.
cuhre is a function in package R2Cuba for multidimensional integration.
There are three kind of variables to integrate. I just use y,w,t to
distinguish them from each other. Is there any problem ?
--
View this message in context:
http://r.789695.n4.nabble.com/cuhre-usage-multidimensional-integra
Dear list,
I have very little experience in dealing with proportions, i am sure this is a
very simple question but i could find no suitable answer beyond doing a chi-sq
test and then using the Marascuilo procedure as a post-hoc analysis.
I am simply wanting to know if the proportions ( i.e the
Hi,
I am trying to read in a rather large list of transactions using the
arules library. It seems in the coerce method into the dgCmatrix, it
somewhere calls unique. Unique.c throws an error when n > 536870912;
however, when 4*n was modified to 2*n in 2004, the overflow protection
should have cha
I have 2 dataframes. "mydata" contains numerical data. "mybys" contains
information on the "group" each row of the data is in. I wish to aggregate
each column in mydata using the corresponding column in mybys.
Please see the example below. What is a more elegant or "better" way to
accomplish
Hi,
In short, I would like to know if there is any way to convert a numeric
into a date, similar to how strptime() can convert a string to a date time
class?
There are some functions, etc. which don't work well with dates, and
tend to force them into numerics. I understand that the numbe
Hi all,
Suppose I have
data1
A B
1 a
1 b
2 c
2 d
and
data2
D E F
x y 1
w z 2
and I want
data2
D E F G
x y 1 a,b
w z 3 c,d
I am trying
data2$G=list(data1$B[data1$A==data2$F,])
How do I correct this approach?
--
View this message in context:
http://r.789695.n4.nabble.com/Subsetting-questi
Another way to build functions "from scratch" :
> func<-'x^2+5'
> funcderiv<- D(parse(text=func), 'x') )
> newtparam <- function(zvar) {}
> body(newtparam)[2] <- parse(text=paste('newz <-
(',func,')/eval(funcderiv)',collapse=''))
> body(newtparam)[3] <- parse(text=paste('return(invisible
Richard M. Heiberger temple.edu> writes:
>
> The next thing to check is this item from doc/manual/R-exts.html
>
> Quoted strings within R-like text are handled specially...
>
> My guess is that the problem is occuring in the .Rd file, not in the .R
> file.
>
> Remove the line, or double t
The next thing to check is this item from doc/manual/R-exts.html
Quoted strings within R-like text are handled specially...
My guess is that the problem is occuring in the .Rd file, not in the .R
file.
Remove the line, or double the "\" characters.
Rich
On Wed, Oct 5, 2011 at 5:37 PM, Jeff
Jeff Breiwick noaa.gov> writes:
>
> Dear R-Group,
>
> I have a function that sorts a data frame and oneo of the lines in the
> function is:
>
> vars <- unlist(strsplit(formc, "[\\+\\-]"))
>
> The function works fine and the above line is always reached. However, when
I
> include the functi
I tried this trick, and clearly things are not going in the right direction.
It seems 'layout' is at the root of my frustration, so I can make two plots
and marge them in adobe illustrator (or something similar).
png("c:/temp/lookat.png",res=120,height=600,width=1200)
layout(matrix(c(1,2),2,2,byr
I found an internal workaround to this to support printing and plot type simple,
tt<-party:::prettytree(cf at ensemble[[1]], names(cf at data at get("input")))
> npt <- new("BinaryTree")
> npt@tree<-tt
> plot(npt)
Error in terminal_panel() :
âctreeobjâ is not a regression tree
> plot(npt, type="
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Martin Batholdy
> Sent: Wednesday, October 05, 2011 2:10 PM
> To: R Help
> Subject: Re: [R] do calculations as defined by a string /
> expandmathematical statements in R
>
> Thank
On 05/10/2011 4:46 PM, Jeff Breiwick wrote:
Dear R-Group,
I have a function that sorts a data frame and oneo of the lines in the
function is:
vars<- unlist(strsplit(formc, "[\\+\\-]"))
The function works fine and the above line is always reached. However, when I
include the function in a packa
# Changing to variable Z since array() is a function
apply(Z.temp <- Z[,,,a:b],1:3,sum)/dim(Z.temp)[4]
# Should work, though it may be more clear to define Z.temp in its own line
M
On Wed, Oct 5, 2011 at 5:10 PM, Martin Batholdy wrote:
> Thanks for all the suggestions!
>
>
>
> Perhaps my post wa
Thanks, Sina! This is very helpful and informative, but still not quite what
I want.
So, here is the thing: When a function returns an object, that object is
available in the calling environment. If it is returned inside a function,
it is available in the function, but not outside of the function
Thanks for all the suggestions!
Perhaps my post was not clear enough.
apply(array,1:2,sum)/dim(array)[3]
and
# reproducible example
x <- 1:1000
dim(x)<-rep(10,3)
# code
apply(x,1:2,sum)
would give me the mean over one whole dimension, right?
The problem with that is, that I just want to c
Hello Greg,
Session info is below. Running Win7 64-bit. I just upgraded my version of
R and tried rerunning the code and got the same odd result. I, too, get an
expected result when I create the plot in the R GUI. The problem crops up
only when I try and create the plot in png() or tiff(). Pe
Dear R-Group,
I have a function that sorts a data frame and oneo of the lines in the
function is:
vars <- unlist(strsplit(formc, "[\\+\\-]"))
The function works fine and the above line is always reached. However, when I
include the function in a package and run "R CMD check pkgname" it gives t
Thanks, Baptiste. I was looking for tableplot() or something like it
and thought textplot() was doing something different. Appreciate the
correction.
Dennis
On Wed, Oct 5, 2011 at 1:29 PM, baptiste auguie
wrote:
> On 6 October 2011 09:23, Dennis Murphy wrote:
>> Hi:
>>
>> One option is the grid
Generally, the only way to estimate f1:f2 is if you have all combinations of
data present for these two factors.
Sometimes it makes sense to include f1:f2 as a random effect in the model
(which does NOT need balanced data) but that is something you have to
decide.
Kevin
On Wed, Oct 5, 2011 at 2
Actually, this may just be a typo in your first post, but if you
actually want to do this calculation:
(array[,,1] + array[,,2] + array[,,3] + array[,,4] + array[,,5] +
array[,,6] + array[,,7] + array[,,8]) / 8
Wouldn't this work?
apply(array,3,sum)/dim(array)[3]
On Wed, Oct 5, 2011 at 4:22 PM
When I copy and paste your code I get what is expected, the 2 subplots line up
on the same y-value. What version of R are you using, which version of
subplot? What platform?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> --
Sorry!! meant: apply(array,1:2,sum)/dim(array)[3]
M
On Wed, Oct 5, 2011 at 4:31 PM, R. Michael Weylandt
wrote:
> Actually, this may just be a typo in your first post, but if you
> actually want to do this calculation:
>
> (array[,,1] + array[,,2] + array[,,3] + array[,,4] + array[,,5] +
> array
On 6 October 2011 09:23, Dennis Murphy wrote:
> Hi:
>
> One option is the gridExtra package - run the example associated with
> the tableGrob() function. Another is the addtable2plot() function in
> the plotrix package. I'm pretty sure there's at least one other
> package that can do this; I thoug
Hi, are you looking for
# reproducible example
x <- 1:1000
dim(x)<-rep(10,3)
# code
apply(x,1:2,sum)
note that ?apply works with many functions...
2011/10/5 Martin Batholdy :
> Dear R-group,
>
>
> is there a way to perform calculations that are defined in a string format?
>
>
> for example I h
For example,
library(agridat)
?gomez.stripsplitplot
Kevin Wright
On Wed, Oct 5, 2011 at 3:13 PM, Dennis Murphy wrote:
> Try Googling 'Three factor ANOVA R'; it didn't take long to find a few
> relevant hits.
>
> Dennis
>
> On Wed, Oct 5, 2011 at 10:56 AM, rafal wrote:
> > I am a student fro
Didn't three of us give you a function (in various flavors) that would
do the mean for variable inputs, reading them from a list? (Though
David's was admittedly much cooler than mine!)
Anyways, look into parse(text=do) with eval() if you want to go the
string route.
Michael
On Wed, Oct 5, 2011 a
Hi:
One option is the gridExtra package - run the example associated with
the tableGrob() function. Another is the addtable2plot() function in
the plotrix package. I'm pretty sure there's at least one other
package that can do this; I thought it was in the gplots package, but
couldn't find one tha
Avoid parsing strings to make expressions. It is easy
to do, but hard to do safely and readably.
In your case you could make a short loop out of it
result <- x[,,,1]
for(i in seq_len(dim(x)[4])[-1]) {
result <- result + x[,,,i]
}
result <- result / dim(x)[4]
Bill Dunlap
Spotfi
Hi,
there are a couple of themes proposed in the wiki, one being white on black,
https://github.com/hadley/ggplot2/wiki/Themes
HTH,
baptiste
On 6 October 2011 04:05, Eugene Kanshin wrote:
> Hello,
> I'm trying to produce some plots in ggplot2 to use them on
> the dark-blue gradient background
Dear R-group,
is there a way to perform calculations that are defined in a string format?
for example I have different variables:
x1 <- 3
x2 <- 1
x4 <- 1
and a string-variable:
do <- 'x1 + x2 + x3'
Is there any way to perform what the variable 'do'-describes
(just like the formula-element
Try Googling 'Three factor ANOVA R'; it didn't take long to find a few
relevant hits.
Dennis
On Wed, Oct 5, 2011 at 10:56 AM, rafal wrote:
> I am a student from Poland. What I am interested in is 3 factor anova with R.
> Could you please help me find an example with using this method with R?
> W
Dennis,
Thank you for your reply. This is a good start for what I want to achieve.
-- Joe
-Original Message-
From: Dennis Murphy [mailto:djmu...@gmail.com]
Sent: Friday, May 20, 2011 10:55 PM
To: Joseph Boyer
Cc: r-help@r-project.org
Subject: Re: [R] Variability plot in R
Here's one at
Hi,
I'm assuming you're using subplot() from Hmisc, but it's a good idea
to specify.
It's not subplot() that's causing the problem, it's layout, or rather the
interaction between the two.
This section run at the command line doesn't work:
layout(matrix(c(1,2),2,2,byrow=TRUE),c(1.5,2.5),respect=T
Perhaps you should start by writing vectorized code; as it stands,
your code suggests you don't understand what simple operations like y
<- x actually do.
More to your question: what are cuhre & crff ? They are not in base R
nor in any packages I have current loaded.
Michael
On Wed, Oct 5, 201
Ista:
But I think the problem here is how to update the model formula. I see no
"simple" way to do that (it can be done straightforwardly enough, but I
wouldn't consider it simple).
... But perhaps what you meant is to update the data argument like this:
cntrl <- data.frame(...) ##response + un
Creating expressions and functions dynamically can be
tricky. Usually I use functions like call(), substitute(),
and formals(); very occasionally I use parse(text=).
Here is one way to make a family of functions that differ
only in the default value their their argument:
> funsA <- lapply(1:3,
Hi:
Is this what you're after?
f <- function(x) !any(x %in% terms_exclude) && any(x %in% terms_include)
db[apply(db[, -1], 1, f), ]
ind test1 test2 test3
2 ind2 227 28.0
4 ind4 3 2 1.2
HTH,
Dennis
On Wed, Oct 5, 2011 at 8:53 AM, natalie.vanzuydam wrote:
> Hi all,
>
> I
Hello,
Below is some example code that should reproduce an error I'm encountering
while trying to create a tiff plot with two subplots. If I run just the
following bit of code through the R GUI the result is what I'd like to have
appear in the saved tiff image:
x<-seq(0:20)
y<-c(1,1,2,2,3,4,5,4
Hi Justin,
On Wed, Oct 5, 2011 at 3:32 PM, justin jarvis
wrote:
> Hi all,
> I am running regressions with many covariates, most of which remain the same
> each time (control variables). Instead of writing 30 demographic variables
> every regression, is there a way I could call them all at once u
lapply(1:10, function(i) Box.test (lfut, lag = i, type="Ljung"))
Add extractors to get statistics as desired.
Michael Weylandt
On Wed, Oct 5, 2011 at 1:09 PM, upananda pani wrote:
> Dear All,
>
> I want to create a loop within a function r. The example follows:
>
> Box.test (lfut, lag = 1, type
Hi all,
I am running regressions with many covariates, most of which remain the same
each time (control variables). Instead of writing 30 demographic variables
every regression, is there a way I could call them all at once using a
variable called, perhaps "demog"?
I have tried:
> demog <- list(ag
Hi,
On Wed, Oct 5, 2011 at 1:44 PM, JeffND wrote:
> Dear folks,
>
> I have a question about the image() function in R. I found the following
> link talking about this
> but the replies didn't help with my situations.
>
> http://r.789695.n4.nabble.com/question-on-image-function-td839275.html#a8392
Consider:
> f <- function(x){ x<- 10;x^2}
> f()
[1] 100
If the argument is not needed, there is no error in omitting it.
R uses "lazy evaluation" -- arguments are not evaluated until needed.
-- Bert
On Wed, Oct 5, 2011 at 8:54 AM, Scott Raynaud wrote:
> It seems I have things set up correctly
Hi,
Searching on http://www.rseek.org for "variance ratio test" turns up the
vrtest package, as does searching for Lo and Mackinlay,
suggesting that's a good place to start.
Sarah
On Wed, Oct 5, 2011 at 2:48 PM, rauf ibrahim wrote:
> Hello,
> I am looking for a code in R for the variance ratio
On 05/10/2011 10:57 AM, honeyoak wrote:
it is possible to dynamically create functions in R using lists? what I want
to do is something like this:
a = list()
for (i in 1:10) a[[i]] = function(seed = i) runif(seed)
so that when I call a[i] I get random draws 1,2,i unfortunately
Dear All,
I want to create a loop within a function r. The example follows:
Box.test (lfut, lag = 1, type="Ljung")
if i want to compute the Box.test for lag 1 to 10, I have to write manually
change each time for different lag. So i wan to write a loop for the lag 1
to 10 and return the statisti
It seems I have things set up correctly. I suspect that the arguments
sshc(100,10) are the isuue. It seems that the 100,10 is not necessary since
the code itself specifies the arguments. It runs and produces a power curve if
I simply type sshc() but it also seems to try to keep running someth
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of behave
> Sent: Wednesday, October 05, 2011 7:07 AM
> To: r-help@r-project.org
> Subject: [R] "stepwise" sum
>
> dear R-Community
>
> is there a function which sums data "stepwis
Hello,
I'd like to output a table to the x11 device, but I can't seem to find an easy
way to do it. Specifically, I'd like to display a 2x2 contingency table
alongside a graphical plot, but can only see how to output to the console. Is
there a library that can do this?
Thanks for any sugges
> cumsum(c(2,1,4,5))
[1] 2 3 7 12
On Wed, Oct 5, 2011 at 10:06 AM, behave wrote:
> dear R-Community
>
> is there a function which sums data "stepwise"
>
> exp:
>
> 2
> 1
> 4
> 5
>
> Desired result
>
> 2 = 2
> 2+1 = 3
> 2+1+4 = 7
> 2+1+4+5 = 12
>
> Is there a built in function for this?
>
> T
I am a student from Poland. What I am interested in is 3 factor anova with R.
Could you please help me find an example with using this method with R?
With all possible countable output for anova as the output presents with 3
factor anova with spss?
I would be glad with any help.
--
View this messa
Hi,
On Wed, Oct 5, 2011 at 10:57 AM, honeyoak wrote:
> it is possible to dynamically create functions in R using lists? what I want
> to do is something like this:
>
> a = list()
> for (i in 1:10) a[[i]] = function(seed = i) runif(seed)
>
> so that when I call a[i] I get random draws 1,
Dear folks,
I have a question about the image() function in R. I found the following
link talking about this
but the replies didn't help with my situations.
http://r.789695.n4.nabble.com/question-on-image-function-td839275.html#a839276
To be simple, I will keep using the example in the above lin
Hi Andrew
I am not sure if I understood your question entirely. You want to store some
objects, but not in the global environment. Correct?! I would do it like
this (although I am sure that there is a more elegant way to do this).
## ---
many thanks. I will try to figure it out.
--
View this message in context:
http://r.789695.n4.nabble.com/about-the-array-transpose-tp3866241p3874870.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
http
Hi All,
This seems to be a bug with RJDBC package and it has been fixed in the
latest RJDBC_0.1-6 version. I would like to try out RJDBC_0.1-6. can you
please guide me to a link where i can find the 64-bit RJDBC_0.1-6.zip . I
could find the 32-bit version at
http://cran.sixsigmaonline.org/bin/win
Thanks Uwe,
The patched 2.13.2 solves this issue.
Best,
Justin M. Balko, Pharm.D., Ph.D.
Research Fellow, Arteaga Lab
Department of Medicine
Division of Hematology/Oncology
Vanderbilt University
777 Preston Research Building
Nashville TN, 37232-6307
Ph: 615-936-1495
-Original Message-
Fr
Hello,
I'm trying to produce some plots in ggplot2 to use them on
the dark-blue gradient background. I am wondering if there is already any
theme/set of options that I can use to change the color scheme and
add transparency.
Thank you very much,
Evgeny.
[[alternative HTML version deleted]]
Hi all,
I'm having some difficulty with lme. I am currently trying to run the
following simple model
anova(lme(x ~ f1 + f2 + f1:f2, data=m, random=~1|r1))
Which is currently producing the error
Error in MEEM(object, conLin, control$niterEM) :
Singularity in backsolve at level 0, block 1
x i
it is possible to dynamically create functions in R using lists? what I want
to do is something like this:
a = list()
for (i in 1:10) a[[i]] = function(seed = i) runif(seed)
so that when I call a[i] I get random draws 1,2,i unfortunately R only
uses the last i . I would also like
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Great, thanks a lot!
Don
On Wed, Oct 5, 2011 at 2:42 AM, Enrico Schumann [via R] <
ml-node+s789695n3873597...@n4.nabble.com> wrote:
> ?expand.grid
>
> Am 05.10.2011 00:21, schrieb darkgaze:
>
> > I don't quite know how to word what I want, but if I have
> >
> > (1, 2, 3); (a, b, c); (x, y)
> >
>
Ok, I chased down all the problems. This is my last output:
> sshc(100,10)
[1] 0.8000 0.7908 0.7844 0.7773 0.7785 0.7989
[1] 5.37 10.29 13.27 13.04 9.66 3.54
[1] " old.abs.dev= 0.0701944484789673"
[1] " abs.dev= 0.034407699378335"
[1] 0.8000 0.8030 0.8057 0.8041 0.8035 0.8180
[1] 5.37 10
Dear all,
I’m analyzing this dataset containing biodiversity indices, measured over
time (Week), and at various contaminant concentrations (Treatment). We have
two replicates (Replicate) per treatment.
I’m looking for the effects of time (Week) and contaminant concentration
(Treatment) on diversi
dear R-Community
is there a function which sums data "stepwise"
exp:
2
1
4
5
Desired result
2 = 2
2+1 = 3
2+1+4 = 7
2+1+4+5 = 12
Is there a built in function for this?
Thx
Dom
--
View this message in context:
http://r.789695.n4.nabble.com/stepwise-sum-tp3874606p3874606.html
Sent from th
Hello,
I am looking for a code in R for the variance ratio test statistic (the
Lo and Mackinlay version or any other versions).
Does anybody have such a code they can share or know a library in which
I can find this function?
Basically I have a number of time series which I need to check for
per
Hi,
I want to ask which way is more effective to further analyse (multiple
comparisons) a mixed model repeated measures anova with 2 fixed factor and 1
random?
anova(lme(expr~treatment*age,random=~1|trial, data)
Is searching for an effect of one factor in each of the subsamples defined
by the seco
Hi all,
I realise that the convention is to provide a working example of my problem
but the data are of a sensitive nature so I'm not able to do that in this
case.
I need to query a database for multiple search terms:
db <- structure(list(ind = c("ind1", "ind2", "ind3", "ind4"), test1 = c(1,
2
On Oct 5, 2011, at 8:14 AM, R. Michael Weylandt > wrote:
(x1+x2+x3)/3
I'm not aware of a "pmean" function but it wouldn't be hard to
homebrew one if you are comfortable with the ... argument
I'll draft one up and send it along
pmean <- function(lis) Reduce("+",lis)/length(lis)
res <- p
Hi everybody,
I used the krige.conv command (geoR package) to create a new data set. The
input was a matrix with three spatial coordinates (x, y, z) in the first
three columns and the value of a variable in the last column. The output
is... a weird sequence of numbers. How can I make this output i
I corrected your code a bit and put it into a function, f0, to
make testing easier. I also made a small dataset to make
testing easier. Then I made a new function f1 which does
what f0 does in a vectorized manner:
x <- array(rnorm(50 * 50 * 50 * 91, 0, 2), dim=c(50, 50, 50, 91))
xsmall <- ar
I took the original code, changed all return()
calls of the form return(n1=v1,n2=v2) to
return(list(n1=v1,n2=v2)) and then sshc(10,100)
chugged away and produced some plots and returned
something with no errors. It took a couple of minutes.
I also changed T->TRUE and F->FALSE, as that makes
the c
On Wed, Oct 5, 2011 at 4:54 PM, Scott Raynaud wrote:
> It seems I have things set up correctly. I suspect that the arguments
> sshc(100,10) are the isuue. It seems that the 100,10 is not necessary since
> the code itself specifies the arguments. It runs and produces a power curve
> if I simply
Hello,
I want to plot one of the trees from a cforest object:
>data(iris)
>cf <- cforest(Species, data=iris)
>From the docs, cf@ensemble contains a list of BinaryTrees:
ensemble: Object of class "list", each element being an object of class "
BinaryTree <../../party/help/BinaryTree%2dclass>". So
Dear list,
I am unsure how to structure my model, i have tried something and it makes
sense but i am unsure if i am interpreting it correctly?
i have a continuous response variable - the observed quantity of evolutionary
history - EH
Then i have a number of species which have a hierarchical st
*Paul,
I use Tinn-r and i didn't see anything hard to configure it. You have to
install the R, then the Tinn-r. After you have to open Tinn-r go to R >>
Configure >> Permanent . This procedure will open the Rprofile.site file, if
you want you can change some parameters and save. After that you hav
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