G'day all,
On Wed, 23 Jul 2008 20:48:59 -0400
Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 23/07/2008 8:17 PM, cindy Guo wrote:
> > Yes, I know. I mean if I want to generate 100 numbers from
> > N(0,1)I((0,1),(5,10)). There are two intervals (0,1) and (5,10).
> > Then the function will give 50
How to convert the strings into matrix ?
(strings <- strsplit(c("1 2 3", "1 2", "1 3 4 5"), " "))[[1]][1] "1" "2" "3"
[[2]][1] "1" "2"
[[3]][1] "1" "3" "4" "5"
_
Easily edit your photos like a pro with Photo Gallery.
[[al
Hi all:
I wanna know how to expoort ".xls" file. And in the exported ".xls" file, how
to set different cells with different color?
Thanks a lot!
My best.
--
***
Xin Meng
Capitalbio Corporation
National Engineering Research
Le mer. 23 juil. à 14:29, Angelo Scozzarella a écrit :
Hi,
how can I treat data organised in classes and frequencies?
Ex.
class frequency
20-23 9
23-25 7
26-28 5
29-31 5
32-34 3
It depends w
This can be done readily with zoo. See zoo's 3 vignettes
and its help files for commands used below.
# setwd("...directory where we have write permission...")
# can alternately manually download these two files
download.file("https://stat.ethz.ch/pipermail/r-help/attachmen
At 10:23 PM +0200 7/23/08, Tobias Verbeke wrote:
Angelo Scozzarella wrote:
what the command for the point-biserial correlation?
Have a look at the polycor package by John Fox.
http://cran.r-project.org/web/packages/polycor/index.html
HTH,
Tobias
A point-biserial correlation is just a Pear
Hello -
Jia Ying Mei wrote:
Hi,
I know this can be done in Stata (which is quite messy) but I wanted to
know if it can be done in R. So lets say I have a merged data set (I
used the merge function by date for the attached two files), where all
the missing values are filled with NAs (which is
Hi,
I know this can be done in Stata (which is quite messy) but I wanted to
know if it can be done in R. So lets say I have a merged data set (I
used the merge function by date for the attached two files), where all
the missing values are filled with NAs (which is what the all.x=TRUE does).
On 23/07/2008 8:17 PM, cindy Guo wrote:
Yes, I know. I mean if I want to generate 100 numbers from
N(0,1)I((0,1),(5,10)). There are two intervals (0,1) and (5,10). Then the
function will give 50 numbers in the first interval and 50 in the other.
No, it doesn't handle that case at all. I didn't
Michal Figurski wrote:
Thank you all for your words of wisdom.
I start getting into what you mean by bootstrap. Not surprisingly, it
seems to be something else than I do. The bootstrap is a tool, and I
would rather compare it to a hammer than to a gun. People say that
hammer is for driving na
Ryacas is one package that checks this. Look at this file:
http://ryacas.googlecode.com/svn/trunk/R/yacasInstall.R
On Wed, Jul 23, 2008 at 5:53 PM, Philip James Smith
<[EMAIL PROTECTED]> wrote:
> Hi R-People:
>
> I use 2 machines: a machine with a Windows XP operating system, and another
> with
This problem can be easily solved analytically:
we want to minimize sum(res(i) -a*st(i) -b*mod(i))^2 subject to a+b=1,a,b>=0,
so we want to minimize
f(a) = sum((res(i)-mod(i)) - a*(st(i)-mod(i)))^2 for 0<=a<=1
Define Xi = res(i) - mod(i), Yi = st(i) - mod(i), then
f(a) = sum(Xi - a*Yi)^2
f(0
Using the built in BOD data frame:
library(prettyR)
str(describe(BOD)) # note Numeric component
BOD.describe.df <- as.data.frame(describe(BOD)$Numeric)
library(xtable)
xtable(BOD.describe.df)
On Wed, Jul 23, 2008 at 3:46 PM, jcarmichael <[EMAIL PROTECTED]> wrote:
>
> Hello everyone. I am new
A cleaner approach than attach()---because you don't need to remember
to detach()---is:
my.df<-data.frame(cbind(x=runif(100), y=rnorm(100)))
head(my.df)
with(my.df, plot(y~x))
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Bo
On Wed, 23 Jul 2008, Abiel Reinhart wrote:
I would like to run an aggregation on a zoo object that has multiple series
in it, with one of more series having NA values. The problem is that by
default the aggregate function will produce an NA value in each aggregated
period that contains an NA. Fo
I would like to run an aggregation on a zoo object that has multiple series
in it, with one of more series having NA values. The problem is that by
default the aggregate function will produce an NA value in each aggregated
period that contains an NA. For instance, if I run aggregate(x,
as.yearmon(i
Here is one way to do it:
> y <- textConnection("UNIQID UniGene Gene 1_SL 2_SL 17_SL 18_SL 38_SL
+ 1175390 Hs.10095 MLLT1 -0.00595 0.62315 0.85315 1.11215 -0.195
+ 1175392 Hs.10101 C1orf166 -0.4945 -0.04025 0.1299 -0.00575 -0.1824
+ 1187428 Hs.101014 CEP57 0.60085 0.2564 -0.42885 -0.57635 -0.1473
On Wed, 23 Jul 2008, John Theal wrote:
EViews and Gretl give comparable (and I am inclined to presume, correct)
results. R on the other hand, has the exogenous regressor with a negative
coefficient. If I use other data I encounter the same problem - agreement
between EViews and Gretl, disagree
Gustaf Rydevik wrote:
On Wed, Jul 23, 2008 at 4:08 PM, Michal Figurski
<[EMAIL PROTECTED]> wrote:
Gustaf,
I am sorry, but I don't get the point. Let's just focus on predictive
performance from the cited passage, that is the number of values predicted
within 15% of the original value.
So, the pr
From: Kaposi-Novak, Pal
Sent: Wednesday, July 23, 2008 5:07 PM
To: jim holtman
Subject: RE: [R] average replicate probe values
Dear Dr Holtman,
Thank you very much for your response.
What I want is avarege data points in a data.frame from probes which re
Thanks a lot, that resolve my probleme!
--
View this message in context:
http://www.nabble.com/Flip-matrix-from-file--tp18614800p18619853.html
Sent from the R help mailing list archive at Nabble.com.
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Hi the list,
When I developpe my code, I put the file that define all the functions
in ./MyPack/R/prog.R and the tests in ./MyPack/tests/testsProg.R
The test files start by
load("../data/myData.rda")
source("../R/prog.R")
#test function number one...
But "R CMD check MyPack" does not find the
on 07/23/2008 04:53 PM Philip James Smith wrote:
Hi R-People:
I use 2 machines: a machine with a Windows XP operating system, and
another with a Linux Ubuntu OS. I transport my code between these 2
machines. However, pathnames to data files always need to be "adjusted"
to account for the OS t
On 24/07/2008, at 9:53 AM, Philip James Smith wrote:
Hi R-People:
I use 2 machines: a machine with a Windows XP operating system, and
another with a Linux Ubuntu OS. I transport my code between these 2
machines. However, pathnames to data files always need to be
"adjusted" to account for
Dear forum,
I was wondering if one could create multilevel responses when at the same
time controlling for everything (i.e. for the multilevel structure, the
fixed and/or the random effect parameters). What I mean is the following: in
an ordinary glm setting you can simulate a binary respons
Hi R-People:
I use 2 machines: a machine with a Windows XP operating system, and
another with a Linux Ubuntu OS. I transport my code between these 2
machines. However, pathnames to data files always need to be "adjusted"
to account for the OS that I'm working on.
Here is my question:
How do
Fernando,
I don't have time to do all that you asked, but here is some code that makes
violin plots with mean, median, and 95% CI. I like this plot very much, even
if boxplot purists think it is horrible :-)
I think the boxplot was developed before we had computing power. Now we can
show the det
Dear friends - I'm on winXP, R 2.71 - I have with some help dveloped
this multivariate normal model, which gives very plausible results in
WinBUGS even without any
initial values specified. However, when I then try to run the same model
via the bugs function in R2WinBUGS with inits specified as
Take a look at the systemfit package. A constraint on a parameter, like the one
you described, could be implemented using R*beta0 = q, where R is R.restr and q
is q.restr in the function call. I haven't tried it but it's easy to test with
some data.
Horace
-Original Message-
From: [E
There is a R function to calculate weighted mean : weighted.mean() under
stats package. Is there any direct R function for calculating weighted
variance as well?
[[alternative HTML version deleted]]
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On 7/23/2008 4:22 PM, Duncan Murdoch wrote:
On 7/23/2008 3:41 PM, cindy Guo wrote:
Hi, I want to generate random samples from truncated normal say
Normal(0,1)Indicator((0,1),(2,4)). It has more than one intervals. In the
library msm, it seems to me that the 'lower' and 'upper' arguments can only
Angelo Scozzarella wrote:
what the command for the point-biserial correlation?
Have a look at the polycor package by John Fox.
http://cran.r-project.org/web/packages/polycor/index.html
HTH,
Tobias
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You might look at the summary method in Dr. Harrell's Hmisc package, I believe
this would give you almost exactly what you want.
latex(summary(gender ~ age + bmi, data=yourdata, method='reverse'))
This will get you quartiles, but if you read the documentation
(?summary.formula) you can see how
Here is one way of doing it:
# generate 200 x 2500 samples and then check them
c <- qnorm(0.05)
i <- 2500
iter <- 200
x <- matrix(rnorm(i * iter, 0, 1) < c,nrow=i) # TRUE/FALSE according
to condition
# now create the counts
x.c <- colSums(x)
# write out to a file 1 value on each line
cat(x.c, fil
Try see doBy package.
Below you have a code on brute force.
kind regards,
miltinho
brazil
===
number.of.records<-20
my.df<-data.frame(v1=sample(21:34,number.of.records, replace=T),
v2=runif(number.of.records), v3=runif(number.of.records))
my.df$v1<-as.numeric(as.character(my.df$v1))
my.df$v
Hi Angelo,
Look carefully at package vcd; and at log-linear models (+ glm(...,
family=poisson)). For overdispersion there are more advanced methods.
HTH, Mark.
Angelo Scozzarella wrote:
>
> Hi,
>
> how can I treat data organised in classes and frequencies?
>
> Ex.
>
> class frequen
I have figured out the problem. Thanks.
Sincerely,
Yanwei Zhang
Department of Actuarial Research and Modeling
Munich Re America
Tel: 609-275-2176
Email: [EMAIL PROTECTED]
-Original Message-
From: Zhang Yanwei - Princeton-MRAm
Sent: Wednesday, July 23, 2008 3:32 PM
To: Zhang Yanwei - Prin
On 7/23/2008 3:41 PM, cindy Guo wrote:
Hi, I want to generate random samples from truncated normal say
Normal(0,1)Indicator((0,1),(2,4)). It has more than one intervals. In the
library msm, it seems to me that the 'lower' and 'upper' arguments can only
be a number. I tried rtnorm(1,mean=0,sd=1, l
Hello everyone. I am new to R, so please bear with me. I am trying to find
an easy way to export descriptive statistics and other information about my
data frame to a LaTeX format.
I have found the describe function in PrettyR to be very helpful in
producing results in the exact format I'm look
Jim, thank you very much This looks like it works.
(I just need to study the code and see if I can understand precisely why
and how it works.)
Eric
Have you reviewed the old
xlisp machinery that used to
be in R? Check out the RXLisp
library at
http://www.omegahat.org/RXLisp/
and see if this will work. The
PDF
http://www.omegahat.org/RXLisp/examples.pdf
reviews calling xlisp from R.
HTH
Steve McKinney
-Original Message-
From: [E
Hi, I want to generate random samples from truncated normal say
Normal(0,1)Indicator((0,1),(2,4)). It has more than one intervals. In the
library msm, it seems to me that the 'lower' and 'upper' arguments can only
be a number. I tried rtnorm(1,mean=0,sd=1, lower=c(0,2),upper=c(1,4)) and it
didn't w
Dear list,
In my previous email, the model I'd like to estimate is
rea=a*st+b*mod+error, where a+b=1 and a,b>0. My apologies for the
misunderstanding.
Thanks for all your help,
Jorge
On Wed, Jul 23, 2008 at 3:35 PM, Jorge Ivan Velez <[EMAIL PROTECTED]>
wrote:
>
> Dear list,
>
> I have a data
Dear list,
I have a data set which looks like myDF (see below) and I'd like to estimate
the linear model rea=a*rea+b*mod+error, where a+b=1 and a,b>0. I tried
mymodel=lm(rea~ st + mod-1, data=myDF)
summary(mymodel)
but I couldn't get what I'm looking for. Also, I tried
RSiteSearch("constrained
Sorry if I did not state clearly.
Put it another way. If the variance of the observation is proportional to the
predictor, that is, var(y_i)=x_i*sigma^2, what should be specified in the
"weights" argument in the lm function?
fit=lm(y~x,weights=???)
Sincerely,
Yanwei Zhang
Department of Actuari
Hello there,
Here's a question regarding p-values on clusters produced by hierarchical
cluster analysis. A web search led me to the program pvclust to tackle this
problem. But when I run the problem I get strange results. The 'AU'
(approximately unbiased) p-values are very different from the '
yes, that is
thanks
milton ruser pisze:
I hope this helps
my.df<-read.table(stdin(), head=T, sep=" ")
V5 V5.1 V5.2 V5.3 V5.4 V5.5
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.846
Try this:
with(x, barplot(frequency, names.arg = class))
On Wed, Jul 23, 2008 at 3:29 PM, Angelo Scozzarella
<[EMAIL PROTECTED]> wrote:
> Hi,
>
> how can I treat data organised in classes and frequencies?
>
> Ex.
>
> class frequency
>
> 20-23 9
> 23-25
Sorry in the previos replay I forgot monday' scale
so weight should be:
weight <- c(0.91,1,1,1,1,1.21,1.22)
names(weight) <- c("mon","tue","wed","thu","fri","sat","sun")
- Original Message -
From: "Gabriela Cendoya" <[EMAIL PROTECTED]>
To: "Robin Williams" <>
Cc: <>
Sent: Wednesday, Ju
Hi,
what the command for the point-biserial correlation?
Thank
Angelo Scozzarella
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and pro
Try this:
addmargins(as.matrix(my.df), 2)
On Wed, Jul 23, 2008 at 2:43 PM, Rheannon <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I have the following data frame (DF):
>
> V5V5.1V5.2 V5.3 V5.4 V5.5
> 2 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.86
All that is needed is
my.df[sample(nrow(my.df), size=), ]
On Wed, 23 Jul 2008, milton ruser wrote:
I hope this helps
my.df<-read.table(stdin(), head=T, sep=" ")
V5 V5.1 V5.2 V5.3 V5.4 V5.5
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06
Hi,
Thanks, but I'm not sure I know what you mean by "insert a ENTER after
the last value". Do you mean press enter after I write in the last value?
Jia Ying Mei
milton ruser wrote:
setwd("c:\\temp")
#you need to insert a ENTER after your last value
> df1<-read.table("testinput.txt", hea
Hi Robin:
I think you can avoid the loops doing this:
my.df<-data.frame(d.o.w=sample(c("mon","sat","sun"), 20, replace=T),
admissions=rnorm(20))
weight <- c(1,1,1,1,1,1.21,1.22)
names(weight) <- c("mon","tue","wed","thu","fri","sat","sun")
my.df$NewAdm <- my.df$admissions * w
I hope this helps
my.df<-read.table(stdin(), head=T, sep=" ")
V5 V5.1 V5.2 V5.3 V5.4 V5.5
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.718
Like This?
my.df<-read.table(stdin(), head=T, sep=" ")
V5 V5.1 V5.2 V5.3 V5.4 V5.5
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
-5890
Hi
I'm looking for solution or function which I can use to sample data
frame, to obtain new (smaller) data frame similar to sample() function
Jarek Jasiewicz
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PLEASE do r
I don´t know what you means by "treat", but may be you want something like
my.df<-read.table(stdin(), head=T, sep=",")
class,frequency
20-23,9
23-25,7
26-28,5
29-31,5
32-34,3
rownames(my.df)<-my.df$class
barplot (my.df$frequency)
axis(1, 1:dim(my.df)[1], rownames(my.df))
Cheers,
miltinho astronau
Hi all,
I met with a problem about the weighted least square regression.
1. I simulated a Normal vector (sim1) with mean 425906 and standard deviation
4.
2. I simulated a second Normal vector with conditional mean b1*sim1, where b1
is just a number I specified, and variance proportional to
Greetings. This is very basic but we can't figure it out. The following
simple code counts how many values land in the "tail" (below a quantile) of
a standardized normal distribution of random numbers. First two commands
are inputs:
> c <-0.05
> i <-2500
Second two commands are formulas I want
Hello,
I have the following data frame (DF):
V5V5.1V5.2 V5.3 V5.4 V5.5
2 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
3 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
4 -5890.18905 -6019.84665 -6211.06
Ok, thanks!
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
> See ?title where its mentioned that calls are ok and a formula
> is a call. Actually the first ~ is not needed since its already
> a formula.
>
> On Tue, Jul 22, 2008 at 1:12 PM, Mike Prager <[EMAIL PROTECTED]> wrote:
> > "Gabor Grot
Hi again,
It seems one line in my function came chopped for whatever reason
so the if else statement below should be:
if (pval==0) p_value <- "<2.2e-16" else p_value <- pval
Thanks,
Monica
> Date: Wed, 23 Jul 2008 13:24:45 -0500> From: [EMAIL PROTECTED]> Subject: Re:
> [R] Warnin
Hi,
how can I treat data organised in classes and frequencies?
Ex.
class frequency
20-23 9
23-25 7
26-28 5
29-31 5
32-34 3
Thanks
Angelo Scozzarella
A few things that will help you, if not now then in the future:
1) Preallocate the result object. This allow you to avoid using
cbind()/rbind(), which constantly creates a new large copy in each
iteration. That will eventually bite you if you have a lot of data.
In your case you know the number
I second that - quantile regression seems to be what you want.
on 07/23/2008 10:10 AM Ben Bolker said the following:
Firas Swidan gmail.com> writes:
Hi,
I am having difficulties in finding ways to analyse scatter plots and
quantitatively differentiate between them.
Try quantile regress
Hi.
On Wed, Jul 23, 2008 at 9:23 AM, Chris82 <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I have a problem to flip a 200x200 matrix, which is imported by a .asc file.
> I want to flip the matrix like in a created example below:
>
>> b <- matrix(1:9,3,3,byrow=T)
>> b
>
> [,1] [,2] [,3]
> [1,]1
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Michal Figurski
> Sent: Wednesday, July 23, 2008 10:22 AM
> To: r-help@r-project.org
> Subject: Re: [R] Coefficients of Logistic Regression from
> bootstrap - how to get them?
>
> Thank you all for your
Hi Erik,
Thanks for your answer. I did print the p-value just before the statement and
it is only one value – or at least I see only one value ….. that is strange.
I am doing a test and I want to see what is it's p-value ….for example my
function "myf" is like that:
myf <- function(m, se, con
On 7/23/2008 12:20 PM, Vadim Organovich wrote:
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal
a.string. Is this right?
No, cat(a.string) should print the original a.string. shQuote(a.string)
adds quotes so that a.string can be used in the shell.
How do I check that, and what quantification does it provide?
Thanks,
Firas.
On Wed, 2008-07-23 at 21:20 +0530, Bogaso Cristofer wrote:
> Have u checked skewness of data?
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
> Behalf Of Firas Swidan
> Sent: 23 July
Just to clarify,
if you have two data.frame, one with your data, other with data-admissions,
just use
data.merge<-merge(my.df, data.weigth, by.x="data", by.y="data", all=T)
miltinho
On 7/23/08, milton ruser <[EMAIL PROTECTED]> wrote:
>
> Hi Rob Williams
>
> I think it is one way of you do the j
Hi Rob Williams
I think it is one way of you do the job.
Cheers,
miltinho astronauta
brazil
my.df<-data.frame(cbind(data=sample(c("mon","sat","sun"), 20, replace=T),
values=rnorm(20)))
my.df$values<-as.numeric(my.df$values)
data.weigth<-read.table(stdin(), head=T, sep=",")
data,weigt
Monica -
Monica Pisica wrote:
Hi,
I am using an if else statement inside a function …. If I use that
function I have no problems …. If I use the function with the if else
statement inside a second function I get the following waring:
Warning message: In if (pval == 0) p_value <- "< 2.2e-16" el
jjdow <- sample(c('Sun', 'Mon', 'Tues'), 10, replace=TRUE)
jjdata <- data.frame(admission=1:10, d.o.w.=jjdow)
#> jjdata
# admission d.o.w.
#1 1 Tues
#2 2 Tues
#3 3 Tues
#4 4Sun
#5 5Mon
#6 6Mon
#7 7Sun
#8
One more try and I quit: This is what happens if someone does not
sent a sample of data you have to create things on the fly without
testing.
days <- list(list("Sat", 1.21), list("Sun", 1.22), list("Mon", 0.91))
for (i in days){
.subset <- data$d.o.w == i[[1]] # subset of data that matches
My fingers slipped on the keyboard. Here what they intended to write.
This sets up a list of the data and the matches on a subset for
processing
days <- list(list("Sat", 1.21), list("Sun", 1.22), list("Mon", 0.91))
for (i in days){
.subset <- data$d.o.w == i[[1]] # subset of data that matc
You would have something that looks like this:
if (data$d.o.w == "Sat") data$admission <- round(data$admission * 1.21)
if (data$d.o.w == "Sun") ...
On Wed, Jul 23, 2008 at 11:44 AM, Robin Williams <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> Again I have searched the net and so on, without find
Hello,
I have a problem to flip a 200x200 matrix, which is imported by a .asc file.
I want to flip the matrix like in a created example below:
> b <- matrix(1:9,3,3,byrow=T)
> b
[,1] [,2] [,3]
[1,]123
[2,]456
[3,]789
> b1 <- apply(t(b),1,rev)
> b1
Dear Patrizio and all,
Do you know a similar approach to this, so that *any* function is sequentially
applied on a matrix?
Particularly, instead of wanting to sum columns every 8 rows, how could we
apply a linear regression to the columns of a matrix every 8 rows?
ie, if we have a matrix say of
Thank you all for your words of wisdom.
I start getting into what you mean by bootstrap. Not surprisingly, it
seems to be something else than I do. The bootstrap is a tool, and I
would rather compare it to a hammer than to a gun. People say that
hammer is for driving nails. This situation is a
Hi,
I am using an if else statement inside a function …. If I use that function I
have no problems …. If I use the function with the if else statement inside a
second function I get the following waring:
Warning message:
In if (pval == 0) p_value <- "< 2.2e-16" else p_value <- pval :
the con
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal
a.string. Is this right?
I am not sure cat() correctly prints strings which are generated by
triple-shQuote():
> shQuote(shQuote("a"))
[1] "\"\\\"a\\\"\""
> cat(shQuote(shQuote(shQuote("a"))), '\n')
"\"\
I have written some programs in Common Lisp and I have been using SAS to pipe
those programs to my lisp compiler in batch mode by using the %xlog and
%xlst SAS commands. I wonder if there is in R a similar way to pipe commands
to LISP so that all my work would be concentrated in R even when I have
Oh, you are rigth Marc, thanks.
Another option is:
pats.df <- do.call(rbind.data.frame, pats)
On Wed, Jul 23, 2008 at 1:10 PM, Marc Schwartz
<[EMAIL PROTECTED]> wrote:
> on 07/23/2008 09:42 AM Henrique Dallazuanna wrote:
>>
>> Try this:
>>
>> pats.df <- do.call(rbind, pats)
>
> Henrique,
>
> Tak
Hi, I have a huge data set to deal with. Sometimes I got warning message
about limitation of memory, sometimes R reads in the data but it is very
slow.
My question is, is there anything I can do to lift the memory limitation (I
did memory.limit(4095), anything else I can do?). I know I can also
on 07/23/2008 09:42 AM Henrique Dallazuanna wrote:
Try this:
pats.df <- do.call(rbind, pats)
Henrique,
Take note of the object that results from this:
> do.call(rbind, pats)
shape shape.col shape.lty cell.fill back.fill label label.size
[1,] 0 "black" 1 "white" "white"
Hello all,
I have been using R's time series capabilities to perform analysis for quite
some time now and I am having some questions regarding its reliability. In
several cases I have had substantial disagreement between R and other packages
(such as gretl and the commercial EViews package).
I
On Wed, Jul 23, 2008 at 4:03 PM, Shubha Vishwanath Karanth
<[EMAIL PROTECTED]> wrote:
> Hi R,
>
>
>
> Let,
>
>
>
> x=1:80
>
>
>
> I want to sum up first 8 elements of x, then again next 8 elements of x,
> then again another 8 elements. So, my new vector should look like:
>
> c(36,100,164,228,29
Dear Michael,
Perhaps,
data.frame(do.call(rbind,lapply(pats,function(x) t(as.matrix(x,ncol=10)
HTH,
Jorge
On Wed, Jul 23, 2008 at 9:23 AM, Michael Friendly <[EMAIL PROTECTED]> wrote:
> For a function that takes an argument as a list of lists of parameters, I'd
> like to be able to conv
Have u checked skewness of data?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Firas Swidan
Sent: 23 July 2008 18:38
To: r-help@r-project.org
Subject: [R] Quantitative analysis of non-standard scatter plots.
Hi,
I am having difficulties in finding ways
Maybe:
tapply(x, gl(10, 8), sum)
or
unlist(lapply(split(x, gl(10, 8)), sum))
On Wed, Jul 23, 2008 at 11:03 AM, Shubha Vishwanath Karanth
<[EMAIL PROTECTED]> wrote:
> Hi R,
>
>
>
> Let,
>
>
>
> x=1:80
>
>
>
> I want to sum up first 8 elements of x, then again next 8 elements of x,
> then again a
At 7:33 PM +0530 7/23/08, Shubha Vishwanath Karanth wrote:
Hi R,
Let,
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8 elements. So, my new vector should look like:
c(36,100,164,228,292,356,420,484,548,612)
I used:
aggregate(x,list(rep(1:10
Hi all,
Again I have searched the net and so on, without finding an answer to this
surely simple problem. A short bit of code would be appreciated.
I have a object named `data' with the following column headings.
Date, maxitemp, minitemp, admissions, d.o.w.
Where d.o.w. is day of the week,
Dear Shubha,
Try this:
x=1:80
tapply(x,rep(1:10,each=8),sum)
1 2 3 4 5 6 7 8 9 10
36 100 164 228 292 356 420 484 548 612
HTH,
Jorge
On Wed, Jul 23, 2008 at 10:03 AM, Shubha Vishwanath Karanth <
[EMAIL PROTECTED]> wrote:
> Hi R,
>
>
>
> Let,
>
>
>
> x=1:80
>
>
>
> I want to
> x=1:80
>
> I want to sum up first 8 elements of x, then again next 8 elements of x,
> then again another 8 elements. So, my new vector should look like:
>
> c(36,100,164,228,292,356,420,484,548,612)
>
> I used:
>
> aggregate(x,list(rep(1:10,each=8)),sum)[-1]
>
> or
>
> rowsum(x,group=r
on 07/23/2008 09:03 AM Shubha Vishwanath Karanth wrote:
Hi R,
Let,
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8 elements. So, my new vector should look like:
c(36,100,164,228,292,356,420,484,548,612)
I used:
a
Michael Friendly yorku.ca> writes:
>
> For a function that takes an argument as a list of lists of parameters,
> I'd like to be able to convert that
> to a data.frame and vice versa, but can't quite figure out how.
>
[snip data]
pats.df <- do.call("rbind",pats)
pats2 <- apply(pats.df,1,
Many Thanks Jorge... That was one more way...Is it possible if I can do
this without using rep(1:10,each=8) or the groupingbecause I feel
the number 8 here is fixed... If there is some technique of tracking the
position of first 8, then next 8... don't know whether I am clear in
conveying...
try
?unlist
it may help
regards
+-
| Patrizio Frederic
| Research associate in Statistics,
| Department of Economics,
| University of Modena and Reggio Emilia,
| Via Berengario 51,
| 41100 Modena, Italy
|
| tel: +39 059 205 6727
| fax: +39 059 2
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