Hello Jumlong,
For Normal distribution see the help page for pnorm.
For dealing with unknown (empirical) distributions, look at ecdf.
Hope this helps
Michael
On 8 November 2010 16:29, Jumlong Vongprasert wrote:
> Dear all
> I have problem with calculate probability, I have data x1,...
Hi, I would like to create a list recursively and eliminate my for loop :
a<-c()
a[1] <- 1; # initial value
for(i in 2:N) {
a[i]<-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context:
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Whenever you use a recursion (that cannot be expressed otherwise), you
always need a (for) loop.
Apply and the like do not allow to use the intermediary results (i.e. a[i-1]
to calculate a[i]).
So: no, it cannot be avoided in your case, I guess.
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http
Hi Joshua,
> .You can think of a 3d array
> kind of like a journal (I don't know if this is helpful, but I kind of
> like the analogy so...). Each page holds a two dimensional table, so
> I could tell you to look at row 4, column 3 on page 16. Nevertheless,
> at any given point, it is ju
Jason Hwa wrote:
Dear R-help community,
I am working with multidimensional contingency tables and I am having
trouble getting loglm to run on all dimensions without typing out each
dimension.
I have generated random data and provided output for the results I want
below:
d1.c1 <- rnorm(20,
I am saving R graph as a jpeg image and the border is plotted defaultly when
you open the image.How can I avoid the border?
small R code is as follows :-
volume<- seq(100,10,-10)
jpeg(filename = "mygraph.jpg",width=366,height=284,units="px")
{
pie(volume,main=
This is to announce the release of the RGtk2 2.20.x series. RGtk2 is an
interface between R and the GTK+ GUI toolkit and friends. The new release
updates the bindings to support up to GTK+ 2.20 (and remains backwards
compatible to 2.8). Previously, the interface supported only up to GTK+
2.12, whic
Hello,
I have two data.
x<-c(1, 2, 3)
y<-c(4,5,6)
How do i create matrix of 3 by 3 from this two, such that
(1,4) (1,5) (1,6)
(2,4) (2,5) (2,6)
(3,4) (3,5) (3,6)
I tried some thing like this:
xy <- as.data.frame(c(0,0,0), dim=c(3,3))
for(i in 1:3)
for(j in 1:3)
xy[i][j]<-c(x[i
Try
x1<-matrix(1,3,1)%x%x
y1<-y%x%matrix(1,3,1)
Z<-cbind(x1,y1)
And later you need to move towards list and matrix
On Mon, Nov 8, 2010 at 11:15 AM, abotaha wrote:
>
> Hello,
>
> I have two data.
>
> x<-c(1, 2, 3)
> y<-c(4,5,6)
>
> How do i create matrix of 3 by 3 from this two, such that
>
>
Hi:
?Recall
HTH,
Dennis
On Mon, Nov 8, 2010 at 1:25 AM, PLucas wrote:
>
> Hi, I would like to create a list recursively and eliminate my for loop :
>
> a<-c()
> a[1] <- 1; # initial value
> for(i in 2:N) {
>a[i]<-a[i-1]*b - c[i-1] # b is a value, c is another vector
> }
>
>
> Is it poss
Hi,
First, create a MATRIX of the correct size:
xy <- matrix(nrow=3, ncol=3)
Then, in your for loop, you need to index each cell correctly, like this:
xy[i,j]
Finally, if you want to assign "1,4" to each cell, you need to paste
x[i] and y[j] together, like this:
xy[i,j]<-paste(x[i],y[j], sep=
Hi:
If you want the literal character strings, this works:
x<-c(1, 2, 3)
> y<-c(4,5,6)
> outer(x, y, function(x, y) paste('(', x, ',', y, ')', sep = '') )
[,1][,2][,3]
[1,] "(1,4)" "(1,5)" "(1,6)"
[2,] "(2,4)" "(2,5)" "(2,6)"
[3,] "(3,4)" "(3,5)" "(3,6)"
However, I have the sense y
Hello everyone,
I apologize if this question is out of the scope of this forum
I have a small dataset with a dependent continuous variable which I want to
compare amongst 3 levels of a factor, which I analyzed using a
Kruskall-Wallis test. I am wondering which posthoc multiple comparison I
should
Hi all,
I would like to compute ergodic mean using MCMC output from WinBUGS. I
tried using CODA package, but it seems that it is not implemented yet.
Could anyone help me to compute this?
Kind regards,
Raquel
--
Raquel Rangel de Meireles Guimarães
Doutoranda em Demografia
raq...@cedeplar.ufmg
The Rserve documentation at
http://rosuda.org/Rserve/doc.shtml#start
states that even when making multiple connections to the Rserve,
Windows won't separate workspaces physically and share environments,
which will obviously cause problems and should therefore not be used.
Are there any alternativ
Thanks a lot guys...all of your solution are useful, and good.
once again thanks.
--
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Sent from the R help mailing list archive at Nabble.com.
Dear R Help,
I am trying to get fields showing the last day of each month for a monthly
closing project. In order to find the last day of the previous month, I
subtract the number of days from the current month. For all months my code
works; however, for October, my code doesn't work...it returns
Benjamin Williams bwmcct.com> writes:
> I am trying to get fields showing the last day of each month for a monthly
> closing project. In order to find the last day of the previous month, I
> subtract the number of days from the current month. For all months my code
> works; however, for October,
Josh,
thanks so much for your detailed reply. This is exactly what I was looking
for.
--
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Sent from the R help mailing list archive at Nabble.com.
_
Dear r-user,
I've wrote a small function to visualize the creation of a histogram.
Therefore I create a vector with random values and let them plot in
histograms. I better give a small example in R-Code:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),m
On Mon, Nov 8, 2010 at 7:06 AM, Benjamin Williams wrote:
> Dear R Help,
>
> I am trying to get fields showing the last day of each month for a monthly
> closing project. In order to find the last day of the previous month, I
> subtract the number of days from the current month. For all months my
Nice thing about R is there is more than one way of doing something:
> x
name ip Bsent Breceived
1 a 1 0.00 0.00
2 a 2 1.43 19.83
3 a 1 0.00 0.00
4 a 2 1.00 1.00
5 b 1 0.00 2.00
6 b 3 0.00 2.00
7 b 2 2.00 0.00
8 b
Could be both. Do you have perl installed and is it on a path that R can find.
On Mon, Nov 8, 2010 at 1:32 AM, Ralf B wrote:
> Hi all,
>
> I tried to run Rserve: I installed it from CRAN using
>
> install.packages("Rserve")
>
> and tried to run it from the command line using:
>
> R CMD Rserve
>
You might want to look at statconnWS (available from rcom.univie.ac.at).
Warning: This project is not open source.
On 11/8/2010 12:40 PM, Ralf B wrote:
> The Rserve documentation at
>
> http://rosuda.org/Rserve/doc.shtml#start
>
> states that even when making multiple connections to the Rserve,
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 10-11-08 08:05 AM, wolfgang.pola...@gmail.com wrote:
> Thanks, what if a I want to add to the plot a second column of observations,
> like a type h line from the x,y plane?
> Wolfgang
see segments3d.
rgl is somewhat more do-it-yourself/constru
Alternatives to Windows platform are Linux. Try Ubuntu download from
http://www.ubuntu.com/getubuntu/downloadmirrors if you are a Linux newbie to
use Rserve
Also see Rapache at http://rapache.net/ for using Apache Web Server and R
together.
The third R interface on web is Rweb see
http://www.jsta
Dear r-user,
I've wrote a small function to visualize the creation of a histogram.
Therefore I create a vector with random values and let them plot in
histograms. I better give a small example in R-Code:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),m
On 08-Nov-10 12:56:26, Etienne Stockhausen wrote:
> Dear r-user,
> I've wrote a small function to visualize the creation of a histogram.
> Therefore I create a vector with random values and let them plot in
> histograms. I better give a small example in R-Code:
>
> i=0
> funVec=rnorm(1000)
> temp=
On Nov 8, 2010, at 5:13 AM, vikrant wrote:
I am saving R graph as a jpeg image and the border is plotted
defaultly when
you open the image.How can I avoid the border?
small R code is as follows :-
volume<- seq(100,10,-10)
jpeg(filename =
"mygraph.jpg",width=366,height=284,unit
Dear R Help List,
I have the following data set:
eg.
1, date11, f1
1, date12, f2
1, date13, f3
[...]
2, date21, fi
2, date22, fj
[…]
f1 – fn are various levels of a factor variable.
Each ID may contain 1 to many entries. These represent basically semantic data
(triplets).
I want to construc
Hola Raquel,
As Ben already told you, ergodic means are pretty easy to compute from
scratch. If you are lazy, or you want something more, like estimated
standard errors, you can check out the functions ergMean and mcmcMean in
package dlm.
Best,
Giovanni Petris
On Sun, 2010-11-07 at 14:34 -0200,
If you want a conditional distribution of x11 given x1,..x10, you need
to make some assumptions about the joint distribution. It seems from the
original post that items 1. and 2. refer to the marginal distribution of
each observation.
In general, it would help to specify the question a little mor
On Nov 8, 2010, at 4:30 AM, Nick Sabbe wrote:
Whenever you use a recursion (that cannot be expressed otherwise), you
always need a (for) loop.
Not necessarily true ... assuming "a" is of length "n":
a[2:n] <- a[1:(n-1))]*b + cc[1:(n-1)]
# might work if b and n were numeric vectors of length
Umesh-
You should contact the package maintainer or a SIG with a question about
a package. RMark isn't posted on CRAN so it is unlikely most folks will
even know about it. There is a support forum for MARK and RMark on the
phidot site where you got them. I'll look at your data and problem an
Hi,
I am running simulations that does multiple comparisons to control.
For each simulation, I need to model 7 nls functions. I loop over 7 to do
the nls using try
if try fails, I break out of that loop, and go to next simulation.
I get warnings on nls failures, but the simulation continues to
http://r.789695.n4.nabble.com/file/n3032045/rsv1.txt rsv1.txt
I am very grateful to David's suggestion, here , I upload my dataset
"rsv1.txt", also the question,
ks<-ken.sto(rsv1,per="TRUE",per.n=0.3,va="FALSE",sav="FALSE")
it does not work, all results are NULL, i do not known why it is ?
Estaré ausente de la oficina desde el 08/11/2010 y no volveré hasta el
16/11/2010.
Responderé a su mensaje cuando regrese, si es urgente enviar mail a
INNOVACION
Visítenos en: www.lineadirecta.com
"Este mensaje y los documentos que, en su caso, lleve anexos, pueden contener
información con
Hi All,
I need some help in putting text in a stacked barplot. The barplot is filled
with 5 levels and now I would like to put text to each level in the stacked
barplot. However, it seems that the code that I am using is not placing the
text at the correct hight (centered at each fill) in the barp
Please post some data (fake data is fine) for stuffa. The example doesn't
execute as is.
Rich
On Mon, Nov 8, 2010 at 10:27 AM, Ashraf Yassen wrote:
> Hi All,
>
> I need some help in putting text in a stacked barplot. The barplot is
> filled
> with 5 levels and now I would like to put text to
Hi,
I'm try to compute the minimum sample size needed to have at least an 80% of
power, with alpha=0.05. The problem is that empirical proportions are really
small: 0.00154 in one case and 0.00234. These are the estimated failure
proportion of two medical treatments.
Thomas and Conlon (1992)
Hello,
I have a simple xyplot with rlm lines.
I would like to add the a and b coefficients (y=ax+b) of the rlm calculation
in each panel.
I know I can do it 'outside' the xyplot command but I would like to do all
at the same time.
I found some posts with the same question, but no answer.
Is it
Hi,
I would like to use the function "extract" from package raster and i get the
following error:
m01e <- extract(marsh01, p)
Error: could not find function "extract"
marsh01 is a raster object and p is an intersectExtent object that is not null.
The package is installed and loaded, i use a
Hi Monica,
Evidently, there is not an "extract" function in your search path
(base R, loaded packages, etc.). Given that you are talking about a
function from a package (rather than base R), it would probably help
us more if you mentioned what version of raster you have installed.
You can get thi
Not with R, but look for G*Power3, a free tool for power calc,
includes FIsher's test.
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni
wrote:
>
>
> Hi,
> I'm try to compute the minimum sample size needed to have at least an 80% of
I want to create a graph to express the idea of the area under a pdf curve,
like
http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg
Thank you for any help.
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-sh
My impression is that varclus from the Hmisc library is very
convenient to use when you want to get an idea of the correlation
among predictor variables in a regression setting, but if you want
to perform cluster analysis in general, you may be better off using
a different function in R, such as hc
On Nov 8, 2010, at 11:16 AM, Mitchell Maltenfort wrote:
Not with R,
Really?
require(sos)
findFn("power exact test")
found 54 matches; retrieving 3 pages
2 3
These look on point:
http://finzi.psych.upenn.edu/R/library/statmod/html/power.html
http://finzi.psych.upenn.edu/R/library/binom/htm
On Nov 8, 2010, at 11:18 AM, Wu Gong wrote:
I want to create a graph to express the idea of the area under a pdf
curve,
like
http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg
I think you could have titled this request:
Yet Another Request to Do My Searching For Me:
http://search.
Hi all,
I have a matrix of a mountain of form 21x21 and values in them are height
(Z). Using the persp function I can view this mountain in 3D.
Now, I am trying to find a measure to find the centre of gravity (maybe
centroid?) of this mountain. Any idea what would be the best way to go?
--
View
Hi,
I don't have access to the article, but must presume that they are doing
something "radically different" if you are "only" getting a total sample size
of 20,000. Or is that 20,000 per arm?
Using the G*Power app that Mitchell references below (which I have used
previously, since they have a
Weighted mean of x and y coordinates (sorry for the pun :)), that is
something like
n = 21
y = matrix( c(1:n), n, n)
x = matrix( c(1:n), n, n, byrow = TRUE)
# These are the Center of mass coordinates:
xCenter = sum(x * Z)/sum(Z);
yCenter = sum(y * Z)/sum(Z);
If you also need the z coordinate, it
Hi Joshua,
OK, my session info is this:
R version 2.11.0 (2010-04-22)
x86_64-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_Unite
1. The for loop is probably as or more efficient than recursion,
especially for large n (comments/corrections on this claim from
cogniscenti are welcome);
2. One can **Always** express a for loop recursively -- that is the
nature of computer languages (comments/corrections again welcome).
3. Her
Yep, it is 20.000 per arm, sorry. The reference it's about an application of
the method, and I cannot download the paper with the main algorithm, so I don't
know exactly how they did.
Thanks everybody for the rich and interesting suggestions. Through free web
software (PS, others) I fou
Dear all,
I am trying (!!!) to generate pdfs that have 8 plots on one page:
df = data.frame(
day = c(1,2,3,4),
var1 = c(1,2,3,4),
var2 = c(100,200,300,4000),
var3 = c(10,20,300,4),
var4 = c(10,2,3,4000),
var5 = c(10,20,30,40),
Users of R 2.11.1 (let alone 2.11.0) x64 Windows were asked to update
in August, and we switched off the last remnants of support when
2.12.0 was released.
Please do as the posting guide asked you to do before posting: update
to R 2.12.0 (or 2.12.0 patched), and update all your packages.
lat
Look at power.examp in the TeachingDemos package. If that plot is not good
enough, you can steal the code and modify to your specifications.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r
Hi all,
I have a couple of questions that are general statistics questions rather
than being R-specific.
I'm interested in figuring out how to compute something like standard error
for difference scores (in particular, differences scores of reaction times).
Does anyone know if there is a standar
Hi,
How to rbind these vectors from a list?:
> l <- list(a = c(1, 2), b = c(1, 2, 3))
> l
$a
[1] 1 2
$b
[1] 1 2 3
> do.call(rbind, l)
[,1] [,2] [,3]
a121
b123
Warning message:
In function (..., deparse.level = 1) :
number of columns of result is not a multiple of ve
Try this:
t(sapply(l, '[', 1:max(sapply(l, length
On Mon, Nov 8, 2010 at 5:05 PM, johannes rara wrote:
> Hi,
>
> How to rbind these vectors from a list?:
>
> > l <- list(a = c(1, 2), b = c(1, 2, 3))
> > l
> $a
> [1] 1 2
>
> $b
> [1] 1 2 3
>
> > do.call(rbind, l)
> [,1] [,2] [,3]
> a1
What class of object / structure do you exactly want
in the end? A matrix, a data.frame, a vector?
johannes rara wrote:
Hi,
How to rbind these vectors from a list?:
l <- list(a = c(1, 2), b = c(1, 2, 3))
l
$a
[1] 1 2
$b
[1] 1 2 3
do.call(rbind, l)
[,1] [,2] [,3]
a121
b
Thanks, data.frame or matrix.
-J
2010/11/8 Erik Iverson :
> What class of object / structure do you exactly want
> in the end? A matrix, a data.frame, a vector?
>
> johannes rara wrote:
>>
>> Hi,
>>
>> How to rbind these vectors from a list?:
>>
>>> l <- list(a = c(1, 2), b = c(1, 2, 3))
>>> l
>
So what do you want the matrix to
look like, since the number of columns
will be different between the two rows?
johannes rara wrote:
Thanks, data.frame or matrix.
-J
2010/11/8 Erik Iverson :
What class of object / structure do you exactly want
in the end? A matrix, a data.frame, a vector?
Hi:
Is this what you had in mind?
library(reshape2)
df <- df[, -10] # last variable is superfluous
dm <- melt(df, id = 'day')
head(dm)
day variable value
1 1 var1 1
2 2 var1 2
3 3 var1 3
4 4 var1 4
5 1 var2 100
6 2 var2 200
xyplot(va
This is the ideal result (data.frame):
> result
names X1 X2 X3
1 a 1 2 NA
2 b 1 2 3
>
2010/11/8 Erik Iverson :
> So what do you want the matrix to
> look like, since the number of columns
> will be different between the two rows?
>
>
>
> johannes rara wrote:
>>
>> Thanks, data.fram
On Nov 8, 2010, at 1:00 PM, Giulio Di Giovanni wrote:
> Yep, it is 20.000 per arm, sorry. The reference it's about an
> application of the method, and I cannot download the paper with the
> main algorithm, so I don't know exactly how they did.
> Thanks everybody for the rich and interesting s
Then one solution is to use
rbind.fill from the plyr package.
johannes rara wrote:
This is the ideal result (data.frame):
result
names X1 X2 X3
1 a 1 2 NA
2 b 1 2 3
2010/11/8 Erik Iverson :
So what do you want the matrix to
look like, since the number of columns
will be diffe
I have tried it, but it does not seem to work with vectors, only data.frames
> do.call(rbind.fill, l)
NULL
>
-J
2010/11/8 Erik Iverson :
> Then one solution is to use
> rbind.fill from the plyr package.
>
> johannes rara wrote:
>>
>> This is the ideal result (data.frame):
>>
>>> result
>>
>> na
Hello!
Hope there is a nifty way to speed up my code by avoiding loops.
My task is simple - analogous to the vlookup formula in Excel. Here is
how I programmed it:
# My example data frame:
set.seed(1245)
my.df<-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0))
my.df<-my.df
Try this:
merge(my.df, my.lookup)
On Mon, Nov 8, 2010 at 5:43 PM, Dimitri Liakhovitski <
dimitri.liakhovit...@gmail.com> wrote:
> Hello!
> Hope there is a nifty way to speed up my code by avoiding loops.
> My task is simple - analogous to the vlookup formula in Excel. Here is
> how I programmed
Dimitri -
While merge is most likely the fastest way to solve
your problem, I just want to point out that you can use
a named vector as a lookup table. For your example:
categories = my.lookup$category
names(categories) = my.lookup$names
creates the lookup table, and
my.df$category = catego
Does anyone recommend a more efficient way to "roll" values in a time series
dataset?
I merged a bunch of different time series datasets (10's of thousands of them)
whose observation dates and sampling interval differ. Some time series
observations are reported at the beginning of the month,
If you are willing to shift the c vector by 1 and have 1 (the initial value) as
the start of c, then you can just do:
cumsum( cc * b^( (n-1):0 ) ) / b^( (n-1):0 )
to compare:
cc <- c(1, rnorm(999) )
b <- 0.5
n <- length(cc)
a1 <- numeric(100)
a1[1] <- 1
system.time(for(i in 2:n ) {
a1
Hi:
Look into the zoo package and its rollapply() function. The package is
designed to handle irregular and multiple series.
HTH,
Dennis
On Mon, Nov 8, 2010 at 12:16 PM, Richard Vlasimsky <
richard.vlasim...@imidex.com> wrote:
> Does anyone recommend a more efficient way to "roll" values in a t
On Mon, Nov 8, 2010 at 3:16 PM, Richard Vlasimsky
wrote:
> Does anyone recommend a more efficient way to "roll" values in a time series
> dataset?
>
> I merged a bunch of different time series datasets (10's of thousands of
> them) whose observation dates and sampling interval differ. Some time
Here's an example of a PDF with the tails shaded:
http://stackoverflow.com/q/3494593/37751
-JD
On Mon, Nov 8, 2010 at 10:18 AM, Wu Gong wrote:
>
> I want to create a graph to express the idea of the area under a pdf curve,
> like
>
> http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg
>
>
I wrote this a bit ago, its far from perfect, but might give you
ideas/serve your purpose.
HTH,
Josh
plot.dist <- function(alpha, from = -5, to = 5, n = 1000, filename = NULL,
alternative = c("two.tailed", "greater", "lesser"),
distribution = c("normal", "t", "F", "chisq", "binomial"),
colour =
Hello R users,
Here is my question about generating random sample. How to set the random seed
to recreate the same random numbers? For example, 10 random numbers is
generated from N(0,1), then "runif(10)" is used.What if I want to get the same
10 random numbers when I run runif(10) again? Is i
Thanks a lot - extremely heplful!
While I'll definitely try to use merge in the future, in my situation
I run into problems with memory (files are too large).
However, Phil's suggestion is perfect for me - sped me up considerably!
Thank you, again!
Dimitri
On Mon, Nov 8, 2010 at 2:51 PM, Phil Spec
Use "set.seed()"
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the word... imbue it
?set.seed is what you're looking for
Xiaoxi Gao wrote:
Hello R users,
Here is my question about generating random sample. How to set the random seed to
recreate the same random numbers? For example, 10 random numbers is generated from
N(0,1), then "runif(10)" is used.What if I want to get the
Hi Xiaoxi,
Take a look at the following:
> set.seed(123)
> rnorm(10)
[1] -0.56047565 -0.23017749 1.55870831 0.07050839 0.12928774 1.71506499
[7] 0.46091621 -1.26506123 -0.68685285 -0.44566197
> rnorm(10)
[1] 1.2240818 0.3598138 0.4007715 0.1106827 -0.5558411 1.7869131
0.4978505
[8]
Hello all,
I think I am missing something about the sorting parameter in the "match"
command/
Here is an example:
a1 <- data.frame(name = c("D", "B", "C", "A", "A", "C"))
a2 <- data.frame(name = c("A", "B", "C", "D"), num = 1:4)
a1
a2
merge(a1, a2, sort = F, by.x = T)
The result is:
name n
You can use the
normal.and.t.dist()
function in the HH package, available from CRAN with
install.packages("HH")
This works on all platforms. There is menu access to the function through
Rcmdr if you install
install.packages("RcmdrPlugin.HH")
If you are on Windows, you can additionally control tha
Dear Jari Oksanen,
I am Trinadh Kumar, a student of Biotechnology from Texas Tech University.
I have a question for you regarding constrained ordination graphs.
I had previously plotted constrained ordination graphs using distance based
redundancy analysis for species data on patients.
The species
Missing: a closer reading of the help page --
Value
A data frame. The rows are by default lexicographically sorted on the
common columns, but for sort = FALSE are in an unspecified order.
So sort = FALSE says unspecified. If you want the original order,
then add a column to the dataframe with th
y=sapply(1:nrow(x), function(i) sapply(1:ncol(x), function(j)
getrc(i, j)))
Arvin Zolfaghari, Ph.D.
Aon Benfield | Aon Benfield Analytics
55 Bishopsgate, London, EC2N 3BD, UK
t: +44 20 7522 8241 | f: +44 20 7522 3846| M: +44 78 8940 6637
e: arvin.zolfagh...@aonbenfield.com | w:
Ted -
If you want to retain the exact order of the rows of one
of the merged matrices, I think you have to merge them
"by hand":
cbind(a1,num=a2[match(a1$name,a2$name),'num'])
name num
1D 4
2B 2
3C 3
4A 1
5A 1
6C 3
> Date: Mon, 8 Nov 2010 06:18:49 -0800
> From: monte.shaf...@gmail.com
> To: r-help@r-project.org
> Subject: [R] try (nls stops unexpectedly because of chol2inv error
>
> Hi,
>
> I am running simulations that does multiple comparisons to control.
>
> Fo
Regarding unusual combinations of factors in categorical data.
Are there any R packages that can be used to identify the outliers i.e.
unusual combinations in categorical datasets ?
Thanks.
Notice of Confidentia
Hi,
On Mon, Nov 8, 2010 at 2:25 PM, Alan Chalk wrote:
> Regarding unusual combinations of factors in categorical data.
where all variables are categorical?
> Are there any R packages that can be used to identify the outliers i.e.
> unusual combinations in categorical datasets ?
"outlier" or "u
Hi all,
I have the following data in abc.dat
===
50 0 1 0 0
55 114 0 1
60 786 0 3
6522 324 2 3
7058 1035 1 7
7530 2568 034
80 9 293615 162
8527 216946 365
90
Here is an example of what I think the original poster wanted to
achieve.
> rnorm(10)
[1] -1.2165869 -0.4698460 -0.4209811 -1.4856052 0.3765774 -1.3822470
[7] 0.2818458 0.5500957 -1.1474455 -1.2221257
> x <- .Random.seed
> runif(10)
[1] 0.5610780 0.5911841 0.5868183 0.3833801 0.7397059 0.497
Tena koe Tal
sort: logical. Should the results be sorted on the 'by' columns?
Thus it is clear what sort=TRUE does, sort=FALSE does not do this. That
doesn't mean it leaves the result in the same order as x (your a1), although
many people (including me at first) assume it does. It is easy en
try this:
> x <- read.table('clipboard')
> x
V1 V2 V3 V4 V5
1 50 01 0 0
2 55 1 14 0 1
3 60 7 86 0 3
4 65 22 324 2 3
5 70 58 1035 1 7
6 75 30 2568 0 34
7 80 9 2936 15 162
8 85 27 2169 46 365
9 90 80 1439 212 432
10 95 236 16
Typing too fast; last line should be:
barplot(t(x.m[, 2:5]), names.arg = x.m[,1], las=2)
On Mon, Nov 8, 2010 at 5:42 PM, casperyc wrote:
>
> Hi all,
>
> I have the following data in abc.dat
>
> ===
> 50 0 1 0 0
> 55 1 14 0 1
> 60 7 86
Reduce(function(x1,x2)b*x1-x2,c,init=1,accum=TRUE)
might be what you are looking for.
This is not fully tested, so you should test it before
you want to use it.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi all,
What's the equivalent to length(unique(x)) == 1 if want to ignore
small floating point differences? Should I look at diff(range(x)) or
sd(x) or something else? What cut off should I use?
If it helps to be explicit, I'm interested in detecting when a vector
is constant for the purpose of
how about
all.equal(x,rep(mean(x),length(x)))
or
all.equal(x,rep(mean(x),length(x), tolerance=...)
albyn
On Mon, Nov 08, 2010 at 06:45:00PM -0600, Hadley Wickham wrote:
> Hi all,
>
> What's the equivalent to length(unique(x)) == 1 if want to ignore
> small floating point
hi people
I need to make an application with Java and R.
I installed the library rJava using the command R - install.packages
(rJava), and i configured my $ R_HOME = / Library / Frameworks /
R.framework / Resources.
My SO is OSX.
When I run an exemple in eclipse, it gives me the following err
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