I am not sure if you really want what you explain that you want
To multiply each row with a vector you can use e.g.
t(apply(ret,1, function(y) pos*y))
or
t(t(ret)*pos)
but then you will get the same matrix just multiplied by a vector.
If you want to perform matrix multiplication then just
re
>
> On Mon, 28 Jan 2008 13:38:51 -0600, Roger Koenker wrote:
>
> Howard Wainer (Graphical Discovery, PUP, 2005, p 20) gives
> this dubious honor to Playfair (1759- 1823). Nightingale (1820-
> 1910) was far too enlightened for this sort of thing, see for example
> her letter to Galton about
port <- as.vector(ret %*% port)
should do it.
Bill Venables
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Megh Dal
Sent: Wednesday, 30 January 2008 5:20 PM
To: [EMAIL PROTECTED]
Subject: [R] Multiplying each row of a big matrix with a vector
I have
I have a big matrix 'ret'. I want to multiply each row of it with a 2nd vector
'pos', resulting result, I want to save in a vector named 'port'. I wrote
following code:
> pos
[1] 2593419 2130220 6198197 1673888 198 1784732 2052120 -7490228
-5275000
> dim(ret)
[1] 500 9
x[c(2,4)] work as well
Henrik Bengtsson a écrit :
> x[1:n]
>
> /H
>
> On Jan 29, 2008 5:07 AM, <[EMAIL PROTECTED]> wrote:
>
>> Seems strange to me to define an operator relatively to a very special case.
>> I have to admit that I do not use 1:1e7 every day :-)
>>
>> Wouldn't it be more appropr
Dear Sophie,
> I wonder if anyone can please offer any advice on a model
> including 2 fixed effects and 1 random effect, as well as a
> covariate?
>
> The experimental design is as follows:
> I have a two by two factor design, where the two factors, Age
> (A) and Group size (G), both have 2
Dear R-community,
I created 5 different xyplots and graphed all of them with the print command on
one page (e.g.
print(graph1, split=c(1,1,1,5), more = T)
... print(graph5, split=c(1,5,1,5), more =T)
Using the above commands separates each graph by a white space. However, since
the gr
Dear Ronny,
> I have sucsessfully used a linear mixed effect model, lme,
> (REML). The results are satisfactory, but I have problems
> with sorting out how to report the result in a scientific paper.
>
> Is there a genearal guidance for reporting lme results
> available (web page, book or ar
pellegrini <[EMAIL PROTECTED]> wrote in news:[EMAIL PROTECTED]:
> There is something that I do not get with scale function.
>
> Say that I have a vector v <- c(1,2,3,4,5,6),
>
> scale(v,center=TRUE,scale=TRUE) gives a new vector with 0 as mean
> and 1 as standard deviation.
>
> but,
>
> scale
By doing "cbind(ID,TREAT,T1,T2)" you are coercing everything to
character and therefore factors in the dataframe. Why didn't you just
say:
data<-data.frame(ID,TREAT,T1,T2)
On Jan 29, 2008 9:02 PM, kamokoi <[EMAIL PROTECTED]> wrote:
>
> R-helpers,
>
> Thanks in advance for your help. I am an R
R-helpers,
Thanks in advance for your help. I am an R newbie and I am having trouble
figuring out the easiest/most efficient way to assign classes to columns in
a newly created data frame. R seems to want to convert everything to a
factor when I use the cbind function to compile vectors into a
Dear Michael
If I understand you correctly, you already have an estimate of the
measurement error? It would seem that if you can estimate the error, then
this estimate comes with a standard error.
For example, suppose that you have a nonlinear model where one of the
predictors is a sum of two var
On 1/29/2008 6:23 PM, Kasper Peeters wrote:
> Hi,
>
> I am looking for a package which allows me to draw arrows (cylinders
> with cones on them, basically) in 3d images produced with rgl. Is such
> a thing available?
The shapes3d demo in rgl draws cones. You could use the same ideas to
draw cy
One suggestion you have probably already considered, but: why not R CMD
BATCH the script? You can set up an executable to do that. No need to
compile the actual R code.
If you want to hide the code and/or it has interaction, you can also set it
up on a web or intranet server for them, calling the
On Wed, 30 Jan 2008, Peter Jepsen wrote:
>
> I am trying to find an R function to compute 'direct adjusted survival'
> with standard errors. A SAS-macro to do this is presented in Zhang X,
> Loberiza FR, Klein JP, Zhang MJ. A SAS macro for estimation of direct
> adjusted survival curves based on a
Hi,
I am looking for a package which allows me to draw arrows (cylinders
with cones on them, basically) in 3d images produced with rgl. Is such
a thing available?
I did notice the 'compositions' package, which contains
'arrow3D'. However, it relies on 'tensorA', which fails to install on
my mach
Hello,
I am trying to find an R function to compute 'direct adjusted survival'
with standard errors. A SAS-macro to do this is presented in Zhang X,
Loberiza FR, Klein JP, Zhang MJ. A SAS macro for estimation of direct
adjusted survival curves based on a stratified Cox regression model.
Comput
Is this what you want?
> numSim <- 15
>
> genx<-double(numSim)
> N <- rep(20, numSim)
> A <- F <- M <- numeric(numSim)
>
> result <- lapply(1:5, function(.x){
+ for (i in 1:numSim) {
+ PN<-(runif(N[i], 0, 1))
+ A[i]<-sum(ifelse(PN>0.2, 1, 0))
+ PF<- runif((A[i]*0.5), 0,
I would support this advice. Stacked bar charts are generally not recommended
because comparisons are difficult. Having 12 groups increases the difficulty.
Best have a look at Cleveland's work before you go down your currently intended
path. Try delving into "Elements of Graphing Data" and "V
I am trying to make a decision tree using rpart. The function runs very
quickly considering the size of the data (1742, 163). When I call the
summary command I get this:
> summary(bookings.cart)
Call:
rpart(formula = totalRev ~ ., data = bookings, method = "class")
n=1741 (1 observation deleted
I am trying to make a decision tree using rpart. The function runs very
quickly considering the size of the data (1742, 163). When I call the
summary command I get this:
> summary(bookings.cart)
Call:
rpart(formula = totalRev ~ ., data = bookings, method = "class")
n=1741 (1 observation deleted
I would question if a stacked area chart is really the best way to display
information on 12 groups. You can put a lot of information into the plot, but
the viewer will probably experience information overload and not be able to get
much useful information out of the plot. Stacked area plots r
On 29-Jan-08 18:52:30, John Kane wrote:
> Any idea who did? I have it in my list of quotes and
> attributed to you.
>
> Anon is so boring as a source.
It was Jim Lemon, earlier in this thread:
Date: Tue, 29 Jan 2008 22:16:55 +1100
Ted.
> --- Gabor Grothendieck <[EMAIL PROTECTED]>
> wrote:
>
>
The 2 tests are testing different hypotheses. The KS test is testing
the null that the data comes from a normal with mean 0 and sd
0.11908969, this can be false if the data is not normal or if the mean
is not 0 or if the sd is not 0.0119 or any combination of the 3.
The Shapiro test was not given
Richard Longland wrote:
> Thanks for the help. This substitute command might be what I'm after,
> but, the whole string is being read in from a file. So I can't
> explicitly write the thing out as you did in your example, as with your
> example in R-news.
>
> My rules are:
> Convert any ",g" or
well if you want to find the spectral density aka what frequencies
explain most of the variance then I would suggest the spectral
density. This can be implemented with spec.pgram(). This is
conducted with the fast fourier transform algorithm.
a<-ts(data, frequency = 1) #make the time serie
Paul Sweeting wrote:
> Hi
>
> Sorry to bother the list again, but no-one has so far been able to suggest
> any help for the query below. As an added incentive, I have been asked "why
> don't you do this in Stata? It's just a case of adding a flag in the
> regression..."
>
> I'm loathe to start l
Thanks for the help. This substitute command might be what I'm after,
but, the whole string is being read in from a file. So I can't
explicitly write the thing out as you did in your example, as with your
example in R-news.
My rules are:
Convert any ",g" or "g," to ",gamma" or "gamma,"
(This com
I can add that I've recently uploaded the gRain package for "gRAphical
Independence Networks", aka probabilistic networks/Bayesian networks.
Regards
Søren
Fra: [EMAIL PROTECTED] på vegne af Domenico Vistocco
Sendt: ti 29-01-2008 19:07
Til: Ralf Finne
Cc: r-help
On Jan 29, 2008 2:30 PM, Patrick Connolly <[EMAIL PROTECTED]> wrote:
> On Sun, 27-Jan-2008 at 07:38PM -0500, Gabor Grothendieck wrote:
>
> |> On the PC there is a builtin GUI but not on UNIX and there are
> |> some packages that are OS specific in which case you might
> |> get more or less selectio
see ?set.seed
On 29/01/2008, Majnu John <[EMAIL PROTECTED]> wrote:
> Hello All,
>
> I used sample(..., replace = TRUE) repeatedly to generate 100
> bootstrap samples of a data that I'm working with. If I re-run my codes,
> it'll generate a different set of 100 bootstrap samples.
>
> Is there a
Hello All,
I used sample(..., replace = TRUE) repeatedly to generate 100
bootstrap samples of a data that I'm working with. If I re-run my codes,
it'll generate a different set of 100 bootstrap samples.
Is there any way that I can get the same set of 100 bootstrap
re-samples at each re-run
On Sun, 27-Jan-2008 at 07:38PM -0500, Gabor Grothendieck wrote:
|> On the PC there is a builtin GUI but not on UNIX and there are
|> some packages that are OS specific in which case you might
|> get more or less selection but probably more. Also depending
|> on the specific system you may have gr
Hi,
I have written the following code which works fine
step<-5
numSim<-15
N<-double(numSim)
A<-double(numSim)
F<-double(numSim)
M<-double(numSim)
genx<-double(numSim)
for (i in 1:numSim) {
N[i]<-20
PN<-(runif(N[i], 0, 1))
A[i]<-sum(ifelse(PN>0.2, 1, 0))
PF<- runif((A[i]*0.5
spec.pgram does accept multivariate time series, and I have used this
with success. I just don't know if there is an implementation of
cross-spectrum in R. Or what exactley Phase vs frequency is telling
me.
On Jan 29, 2008 2:06 PM, Donald McKenzie <[EMAIL PROTECTED]> wrote:
> See the help for sp
Search the archives and find who posted it.
On Jan 29, 2008 1:52 PM, John Kane <[EMAIL PROTECTED]> wrote:
> Any idea who did? I have it in my list of quotes and
> attributed to you.
>
> Anon is so boring as a source.
>
> --- Gabor Grothendieck <[EMAIL PROTECTED]>
> wrote:
>
>
> > I did not write t
"Ng Stanley" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> This should be elementary, but I can't find any resource on creating a
> legend for heatmap. Please help.
Take a look at:
* heatmap.2 in gplots package
* heatplot in made4 Bioconductor package
* heatmap_2 in Heatplus
Any idea who did? I have it in my list of quotes and
attributed to you.
Anon is so boring as a source.
--- Gabor Grothendieck <[EMAIL PROTECTED]>
wrote:
> I did not write that.
>
> On Jan 29, 2008 9:05 AM, S Ellison
> <[EMAIL PROTECTED]> wrote:
> > >>> "Gabor Grothendieck" <[EMAIL PROTECTED]>
On 1/29/08, Dr. Ottorino-Luca Pantani <[EMAIL PROTECTED]> wrote:
> Dear R users,
> I would like to exactly reproduce a figure like the 1.5 or 1.9 or 4.13
> from the book
> Mixed effects models in S and S-Plus.
> Not for the sake of it, but because I have my own data I would like to
> plot in that f
I am having a hard time interpreting the phase and coherency plots.
x is two timeseries that occur at the same time i.e.
a b
1 11.2 12.3
16 11.3 12.4
31 11.4 12.5
46 11.5 12.6
...etc
even though my example is does not show this they are oscillating at
m
I know that there are packages implementings the probabilistic expert
systems (the so-called probabilistic networks or bayesian networks).
You find (at least) the following packages:
bnlearn (bayesian network structure learning)
deal (learning bayesian networks with mixed variables)
G1DBN (for dy
Dear R community,
I would like to introduce the package "dtw" for computing Dynamic Time
Warping (DTW) alignments and related algorithms on timeseries.
The DTW algorithm computes the time axis stretch which optimally maps one
timeseries (query) onto another (template); it outputs the resulting
cu
Richard Longland wrote:
> Hi,
>
> Although I understand how to use expression and paste to add
> superscripts and symbols to my graphs, I have a problem with adding
> superscripts to an existing string.
>
> For example, I read in the following from a separate file:
> 25Mg(p,g)
>
> I want to co
The 'sandwich' package has sandwich standard errors of various sorts for a
generalized linear models. I don't know if it handles the proportional
odds model.
-thomas
On Tue, 29 Jan 2008, Paul Sweeting wrote:
Hi
Sorry to bother the list again, but no-one has so far been able to sug
Hi
Sorry to bother the list again, but no-one has so far been able to suggest
any help for the query below. As an added incentive, I have been asked "why
don't you do this in Stata? It's just a case of adding a flag in the
regression..."
I'm loathe to start learning another stats package, so if
How do one dertermine the optimal number of rescaling cycles to use when
performing DCA using the decorana function (library vegan)?
Thanks for your time,
Marc
===
Marc Bélisle
Professeur adjoint
Chaire de recherche du Canada en écologie spatiale et en écologie du paysage
Départe
Also, see ?aggregate:
> with(A, aggregate(no_of_accidents, by=list(SEX=sex), FUN='sum'))
SEX x
1 F 34
2 M 83
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Attiglah, Mama
> Sent: Tuesday, January 29, 2008 7:21 AM
> To: [EMAIL PROTECTED]; r
Rainer Hurling gwdg.de> writes:
>
>
> Am 29.01.2008 00:04 (UTC+1) schrieb Roger Bivand:
> > On Mon, 28 Jan 2008, Rainer Hurling wrote:
> >
> >> Roger,
> >>
> >> I tried your suggestions and ran into same problems as Pieter did before.
> >>
> >
> > Rainer:
> >
> > I don't think that we know wh
Try this
X = rbind(y1, y2)
colnames(X) = as.character(x)
barplot( X, density=c(35,40), col=c("red","blue"), beside=FALSE)
barplot( X, density=c(35,40), col=c("red","blue"), beside=TRUE)
Alternatively,
barplot( y1, ylim = c(0, max(y1,y2)), col="blue", beside=FALSE)
par(new=TRUE)
barplot( y2
On Jan 29, 2008 10:15 AM, Gustaf Granath <[EMAIL PROTECTED]> wrote:
> Hi,
> I have measured two response variables (y1, y2) at each treatment level
> (x = 0, 1.5 or 3). Now I would like to show the y1 and y2 against x in a
> bar plot. However, y1 and y2 differ in scale so I need two y-axises, one
>
Dear R users,
I would like to exactly reproduce a figure like the 1.5 or 1.9 or 4.13
from the book
Mixed effects models in S and S-Plus.
Not for the sake of it, but because I have my own data I would like to
plot in that fashion
(no colors)
If I write
plot(ergoStool)
I can get a good informat
Dear R Users,
I am posting a smaller portion of a question I posted last week. Code I have
written
takes weeks to run, and I am trying to make all the lines as efficient as
possible. Here is
one part of the code.
I have some data that looks like
[,1] [,2] [,3] [,4]
[1,]1313
[
Hi,
I have measured two response variables (y1, y2) at each treatment level
(x = 0, 1.5 or 3). Now I would like to show the y1 and y2 against x in a
bar plot. However, y1 and y2 differ in scale so I need two y-axises, one
on the left side and one on the right side (and I dont want to
standardiz
Dear All,
I wonder if anyone can please offer any advice on a model including 2 fixed
effects and 1 random effect, as well as a covariate?
The experimental design is as follows:
I have a two by two factor design, where the two factors, Age (A) and Group
size (G), both have 2 levels (old or you
Bernd Weiss wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> skestin schrieb:
> | I suppose it's very simple but I can't find the way to generate a
> sequence of
> | characters, e.g. from "a" to "z".
> | Could you please help me with this?
>
> ?letters
> letters
>
"letters" is oka
Have a look at the reshape package. With your data as
data.frame xx :
library(reshape)
dd <- melt(xx, id=c("sex", "age",
"region"),measured=c(no_of_accidents)); dd
cast(dd, sex~variable, sum)
--- [EMAIL PROTECTED] wrote:
> Hello,
>
> I'm struggling with an elementary problem with R. I
> have
Thanks Hadley for your comment. I've got a minimum of 12 species and
thus just using different greys won't do it. I've tried it with two
background fills, i.e. white and a medium grey, and then used various
angles for the shading lines, even plotted it twice to obtain some sort
of cross shadi
On Jan 29, 2008 9:12 AM, Markus Didion <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I would like to create a stacked area chart to show the development of
> species biomass over time. Since it is intended for publication I need
> to prepare in black and white. I have tried to modify the "stackedPlot
> I'm struggling with an elementary problem with R. I have a simple data
> frame such as this one giving the number of accidents subdivided by sex,
> age and region.
If you're trying to do pivot tables in R, I'd recommend looking at the
reshape package, which was designed specifically to tackle th
On 1/29/2008 7:39 AM, skestin wrote:
I suppose it's very simple but I can't find the way to generate a sequence of
characters, e.g. from "a" to "z".
Could you please help me with this?
If you want the standard collation sequence, use the letters variable,
as others have said.
If you want the
Hi,
Although I understand how to use expression and paste to add
superscripts and symbols to my graphs, I have a problem with adding
superscripts to an existing string.
For example, I read in the following from a separate file:
25Mg(p,g)
I want to convert that to superscripts etc. (in LaTex form
Dear all,
I would like to create a stacked area chart to show the development of
species biomass over time. Since it is intended for publication I need
to prepare in black and white. I have tried to modify the "stackedPlot"
function submitted to this list by Christian Lasarczyk on Tue, 16 Aug
Hello everybody,
There is something that I do not get with scale function.
Say that I have a vector v <- c(1,2,3,4,5,6),
scale(v,center=TRUE,scale=TRUE) gives a new vector with 0 as mean and 1
as standard deviation.
but,
scale(v, center=FALSE,scale=TRUE) gives a new vector with a standard
d
Hello,
I have sucsessfully used a linear mixed effect model, lme, (REML). The results
are satisfactory, but I have problems with sorting out how to report the result
in a scientific paper.
Is there a genearal guidance for reporting lme results available (web page,
book or article)?
Best r
Hi all T gurus,
I would like to test if my dataset is indeed from N(0, 0.011908969).
K.S. test gives following result:
> ks.test(data, "pnorm", 0, 0.011908969)
One-sample Kolmogorov-Smirnov test
data: data
D = 0.1092, p-value = 1.318e-05
alternative hypothesis: two-s
[EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]
k:
Three or four solutions have already been offered. Here is (yet)
another:
> Atxt <- "
+ sex age region no_of_accidents
+ 1 F young north 10
+ 2 F young south 12
+ 3 F old north 5
+ 4 F
Thanks a lot! Its working fine
Pratap
Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
Try this:
lapply(apply(combn(ncol(x),2), 2, function(y)x[,y]),
function(z)t.test(z[,1], z[,2]))
On 29/01/2008, nalluri pratap
wrote:
> Hi Users,
>
> This is regarding the paired t-test. I have 5 variab
MY NAME IS DR DONALD WHITE OF WHITEGO AGENCY, LONDON .
I WANT TO MAKE A RESERVATION FOR A GROUP OF
CHRISTIAN DELEGATES COMING FOR A SEMINAR/CONFERENCE IN YOUR
CITY FROM 1ST OF APRIL TILL THE 15TH OF APRIL 2008.
GET BACK TO ME VIA EMAIL IF THE DESIRED DATES ARE AVAILABLE.
REGARDS,
+447045713899
I cleaned up the environment and restarted everything
and it does work. But I have no idea what it is what
is different now. Anyway, it works!
Thanks a lot Gabor!
Best,
Werner
> I can't reproduce your error:
>
> > library(sqldf)
> > sqldf("select * from warpbreaks limit 6")
> breaks woo
Try this:
lapply(apply(combn(ncol(x),2), 2, function(y)x[,y]),
function(z)t.test(z[,1], z[,2]))
On 29/01/2008, nalluri pratap <[EMAIL PROTECTED]> wrote:
> Hi Users,
>
> This is regarding the paired t-test. I have 5 variables (say)
> Data$v1,Data$v2,Data$v3,Data$v4,Data$v5 in my data frame. Now
I did not write that.
On Jan 29, 2008 9:05 AM, S Ellison <[EMAIL PROTECTED]> wrote:
> >>> "Gabor Grothendieck" <[EMAIL PROTECTED]> 29/01/2008 12:35:27
> >>>
> > As is common in human affairs, even
> > the illusion of understanding is preferred to a lofty digression
> upon
> > why the audience does
>>> "Gabor Grothendieck" <[EMAIL PROTECTED]> 29/01/2008 12:35:27
>>>
> As is common in human affairs, even
> the illusion of understanding is preferred to a lofty digression
upon
> why the audience does not understand.
Gabor,
This should be in fortunes; it is depressingly accurate.
But it doesn't
If I understand your question, you can try something like this:
cor(data.frame(lapply(split(x, x$state), "[", 3)))
On 29/01/2008, Serguei Kaniovski <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I cannot figure out how to use "tapply" to compute the correlation matrix
> in the variable "x" between the s
Türkiye'de Yayýnlanan Ýlk Ses Markasý !
Siemens Aktiengesellschaft firmasýnýn notalar þeklinde olan ses markasý
12.11.2007 tarihli 147 numaralý Resmi Marka Bülteni'nde ilan edildi. Bu marka
baþvurusu Türkiye'de ilk ses markasý baþvurusudur. Ses markasýný TPE'nin bu
link
x[1:n]
/H
On Jan 29, 2008 5:07 AM, <[EMAIL PROTECTED]> wrote:
> Seems strange to me to define an operator relatively to a very special case.
> I have to admit that I do not use 1:1e7 every day :-)
>
> Wouldn't it be more appropriate to define a a:b operator numeric (that
> is preserving the init
Seems strange to me to define an operator relatively to a very special case.
I have to admit that I do not use 1:1e7 every day :-)
Wouldn't it be more appropriate to define a a:b operator numeric (that
is preserving the initial class of a and b) and in specific case that
need optimization, chang
Hi Users,
This is regarding the paired t-test. I have 5 variables (say)
Data$v1,Data$v2,Data$v3,Data$v4,Data$v5 in my data frame. Now, I need to
perform a paired t-test on all the possible 10 pairs.How do I set up the pairs
table directly and pass those variables in to t-test.
Thanks
> letters
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q"
"r" "s" "t" "u" "v" "w" "x" "y" "z"
> LETTERS
[1] "A" "B" "C" "D" "E" "F" "G" "H" "I" "J" "K" "L" "M" "N" "O" "P" "Q"
"R" "S" "T" "U" "V" "W" "X" "Y" "Z"
>
Hope that will help.
Mama
-
Mama Attiglah, PhD
Adva
Maybe this is what you'd like.
library(ggplot2)
freqs <- data.frame(value= c(
0.000,1.204,1.301,1.362,1.447,1.505,1.602,1.653,1.756,1.806,1.903,1.959,
2.053,2.107,2.204,2.258,2.354,2.408,2.505,2.559,2.656,2.709,2.806)
,
tp1= c( 8,1,0,13,0,6,0,25,0,5,0,15,0,4,0,7,0,0,0,1,0,0,0)
,
--
X-USF-Spam-Status: hits=-2.6 tests=BAYES_00
X-USF-Spam-Flag: NO
Hello,
Sequences of letters can be generated by typing
LETTERS #for capital letters or
letters #for small letters
Best wishes
Christoph
skestin schrieb:
> I suppose it's very simple but I can't find the way to generate a sequ
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
skestin schrieb:
| I suppose it's very simple but I can't find the way to generate a
sequence of
| characters, e.g. from "a" to "z".
| Could you please help me with this?
?letters
letters
HTH,
B.
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Even easier than you think... G.
> letters
[1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "r" "s"
[20] "t" "u" "v" "w" "x" "y" "z"
On Tue, Jan 29, 2008 at 04:39:11AM -0800, skestin wrote:
>
> I suppose it's very simple but I can't find the way to generate a sequence of
I suppose it's very simple but I can't find the way to generate a sequence of
characters, e.g. from "a" to "z".
Could you please help me with this?
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Is
?pyramid.plot
in the plotrix package of any help?
Plotting three of these on the same device is possible after
par(mfrow=c(1,3))
>>> Tom Willems <[EMAIL PROTECTED]> 29/01/2008 11:43:46 >>>
dear R-ussers,
I would like to creeate a graph, i wich my data is presented as
verticaly
oriented hi
http://en.wikipedia.org/wiki/Pie_chart
cites papers using the polar area diagram prior to Nightengale although
it does say that many sources credit it to her.
On Jan 29, 2008 6:16 AM, Jim Lemon <[EMAIL PROTECTED]> wrote:
> Rolf Turner wrote:
> > ...
> > I have been for many years under the impres
I can't reproduce your error:
> library(sqldf)
> sqldf("select * from warpbreaks limit 6")
breaks wool tension
1 26A L
2 30A L
3 54A L
4 25A L
5 70A L
6 52A L
> R.version.string # Vista
[1] "R version 2.6.2 a
A = read.table("clipboard", header=TRUE)
> A
sex age region no_of_accidents
1 F young north 10
2 F young south 12
3 F old north 5
4 F old south 7
5 M young north 24
6 M young south 30
7 M o
Hi Pietro,
Depending on the actual structure you want in the output, you can use
some of the functions in the apply family, particularly tapply,
aggregate, or by. Something like :
R> tapply( d[["no_of_accidents" ]], d[[ "sex" ]], sum )
F M
34 83
R> aggregate( d["no_of_accidents" ], d[ "sex" ],
My phd-student works in stata using cluster option in a regression model like
this example:
(from link: http://www.ats.ucla.edu/stat/stata/faq/clusterreg.htm)
"""
/* model 2 -- same as svy: regress with psu */
regress math homework, cluster(schid)
"""
I need to put som extra "blinded" data into
Hello,
I'm struggling with an elementary problem with R. I have a simple data
frame such as this one giving the number of accidents subdivided by sex,
age and region.
sex age region no_of_accidents
F young north 10
F young south 12
F o
Hi all,
I am struggling to compile a massive R code that I have written through
some years, aiming to provide automated investment strategies to my
Portfolio Managers. You may ask me why do I not rewrite the code in C or
C++; the answer is it is really massive and that will take me lots of
time.
dear R-ussers,
I would like to creeate a graph, i wich my data is presented as verticaly
oriented histograms, wich give the frequency of the measured values,
grouped per used measurement methode.
So the X axis should hold the grouping variable and the Y axis a continuos
variable. well not real
Hello,
I cannot figure out how to use "tapply" to compute the correlation matrix
in the variable "x" between the states? The data is in long format, e.g.:
state,year,x
Alabama,2001,0.45
Alabama,2002,0.47
Alabama,2003,0.48
Alabama,2004,0.44
Arizona,2001,0.34
Arizona,2002,0.32
Arizona,2003,0.38
Ar
Rolf Turner wrote:
> ...
> I have been for many years under the impression that the pie chart
> was invented by Florence Nightingale. Am I misinformed?
>
Hi Rolf,
Yes, you have been misled. Dear old Florence invented a related
illustration now usually referred to as a "polar area diagram". It w
Hi,
sqldf sounds like a very useful package but I don't
even get the example to run:
> a1s <- sqldf("select * from warpbreaks limit
6")
Error in combine(FUN(...)) : argument "value" is
missing, with no default
>
I am using R 2.6.1 on Windows Vista Business and have
updated all packages.
S
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Am 29.01.2008 00:04 (UTC+1) schrieb Roger Bivand:
> On Mon, 28 Jan 2008, Rainer Hurling wrote:
>
>> Roger,
>>
>> I tried your suggestions and ran into same problems as Pieter did before.
>>
>
> Rainer:
>
> I don
Romain Francois wrote:
> Hi,
>
> Not sure I understand the question, but maybe you need ?unique or
> ?duplicated
>
Or maybe ?union, ?setdiff, ?intersection. Or ?merge.
> Cheers,
>
> Romain
>
> mohamed nur anisah wrote:
>
>> hello!!
>>
>> say that i have the values of x1 and x2. my
Hi,
Not sure I understand the question, but maybe you need ?unique or
?duplicated
Cheers,
Romain
mohamed nur anisah wrote:
> hello!!
>
> say that i have the values of x1 and x2. my x1 has longer length than x2.
> how am i going to remove the similar values of x1 and x2. Any suggestion?
Hans W. Borchers wrote:
> Ralf Finne syh.fi> writes:
>
>> Hi R-users
>> Is there any functions in R that can implement "expert systems"?
>> The aim of an expert system is to produce a probable diagnosis
>> for a patient with certain symptoms.
>> In the classical expert system a mumber of "exper
hello!!
say that i have the values of x1 and x2. my x1 has longer length than x2. how
am i going to remove the similar values of x1 and x2. Any suggestion?? Thanks
in advance!!
Cheers,
Anisah
e f
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