Hi all,
This message is in response to:
http://www.mail-archive.com/r-h...@stat.math.ethz.ch/msg93690.html
Basically, in SAS you can retrieve the partial correlation coefficients using:
proc reg data = ch11q02;
model y = x1 x2/pcorr1 pcorr2;
run;
quit;
Is there a comparable way to retrieve
Hi,
how can I do a data step in R which can be best described in SQL terminology as
an outer join of rows (eg cases) and inner join of columns (variables)? In case
of conflicting non-missing values the first dataset has higher always priority.
Missing values should always be replaced by valid
einsundeins floriansense at gmail.com writes:
Hello everyone,
this is my first post here and I hope I signed up correctly and someone will
take me by the hand and help me out. I am new to R and cannot figure out
what to do here...
... I want to have an User Interface that requests
Thanks for your help.
Unfortunately when I try it this generates an error as follows:
randomSamples-lapply(1:2000,function(){
+ meta_comp[sample(nrow(meta_comp),nTimes),]})
Error in FUN(1:2000[[1L]], ...) : unused argument(s) (1:2000[[1]])
jholtman wrote:
nTime - 15 # how many samples
Hi,
if you like to use sql you can use the sqldf package.
Another possibility is the basic function merge.
HTH Christian
Hi,
how can I do a data step in R which can be best described in SQL terminology
as an outer join of rows (eg cases) and inner join of columns (variables)? In
Hi R users,
I try to build a function to compute odds ratio and relative risk however
something wrong. I stuck for many hours but I really don't know how to solve
it. Would someone please give me a hint?
OR.RR-function(x){
+ x - as.matrix(any(dim(x)==2))
+
On Tue, Sep 29, 2009 at 4:29 AM, Chunhao Tu tu_chun...@yahoo.com wrote:
Hi R users,
I try to build a function to compute odds ratio and relative risk however
something wrong. I stuck for many hours but I really don't know how to solve
it. Would someone please give me a hint?
Chunhao Tu wrote:
Hi R users,
I try to build a function to compute odds ratio and relative risk however
something wrong. I stuck for many hours but I really don't know how to solve
it. Would someone please give me a hint?
OR.RR-function(x){
+ x - as.matrix(any(dim(x)==2))
+
Did you run debug over your function?
Load the library debug, and then run mtrace over your function.
library(debug)
? mtrace
hth
On Tuesday 29 September 2009 04:29:37 Chunhao Tu wrote:
Hi R users,
I try to build a function to compute odds ratio and relative risk however
something wrong. I
I have two types of survival data for Drosophila
cohort. For example:
uncensored
age - c (0, 2, 4, 6, 8, 10)
alive1 - c(10, 9, 6, 3, 1, 0)
alive 2 - c(10, 9, 4, 1, 1, 0)
and censored
age - c(0, 2, 4, 6, 8, 10)
alive1 - c(10, 8, 3, 1, 1, 0)
escaped1 - c(0, 1, 1, 0, 0, 0)
alive2 - c(10, 7,
Ping-Hsun Hsieh wrote:
Hi,
I am trying to plot my dataset, consisting of one column with numeric values
and one column with group IDs.
The set is similar to the following df.
df - NULL
for ( i in 1:20)
{
tmp1 - runif(1000,0,5)
tmp2 - cbind(tmp1,i)
df - rbind(df,tmp2)
}
Now I would
On 09/29/2009 01:29 PM, Chunhao Tu wrote:
Hi R users,
I try to build a function to compute odds ratio and relative risk however
something wrong. I stuck for many hours but I really don't know how to solve
it. Would someone please give me a hint?
OR.RR-function(x){
+ x-
Or with plyr there's a more flexible approach:
res.aov - dlply(warpbreaks, .(tension), function(x) aov(breaks ~ wool, data=x))
# aov results are stored in a list, you can directly extract what you want with
l*ply
l_ply(res.aov, function(x) print(summary(x)))
ldply(res.aov, function(x)
Nair, Murlidharan T wrote:
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Sunday, September 27, 2009 1:17 PM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] error while plotting
Nair, Murlidharan T wrote:
I am getting the
Dear List,
I have the following data:
test - data.frame(date =
as.Date(c('2007-01-01','2008-03-24','2003-03-02','2008-05-03','2002-05-23','2001-06-30','2005-12-04')),
nr = c(2000,2000,2000,2001,2002,2003,2003))
test
date nr
1 2007-01-01 2000
2 2008-03-24 2000
3 2003-03-02
Hey guys,
im sort of a beginner with R, but here's what i need to do.
i need to perform a time series analysis on a set of financial data that i've
been given. im trying to look at the ACF and PACF and fit it to a particular
model (i think its the ARIMA model because i've read that financial
Hi List,
I'm trying to perform some spectral analysis on a time series, but I've
got a few data points missing, so they have NA values. Unfortunately,
the fft function doesn't seem to like this, and gives a completely empty
result.
Other than making up values to fill in the gaps, is there any
Hi all,
I'm running an analysis with the random forest tool. It's being applied to a
data matrix of ~60,000 rows and between about 40 and 200 columns. I get the
same error with all of the data files (Cannot allocate vector of size
428.5MB).
I found dozens of threads regarding this problem, but
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
\preformatted{
\# vertex1name
vertex2name [optionalWeight]
vertex3name [optionalWeight]
}
Here, the first vertex of
Is this what you want?
test[order(test$nr, -as.integer(test$date)),]
Xavier
- Mail Original -
De: Stefan Uhmann stefan.uhm...@googlemail.com
À: r-help@r-project.org
Envoyé: Mardi 29 Septembre 2009 11h27:20 GMT +01:00 Amsterdam / Berlin / Berne
/ Rome / Stockholm / Vienne
Objet: [R]
Dear all,
Given mypi
mypi - c(0.1,0.2,0.2,0.1,0.3,0.4,0.4,0.4,0.4,0.2)
I want to create myfreq as follows
mypi myfreq
0.1 2
0.2 3
0.2 3
0.1 2
0.3 1
0.4 4
0.4 4
0.4 4
0.4 4
0.2 3
where myfreq is frequency of its corresponding observation. How to do that?
Thank you,
Regards,
A. Kudus
Hello,
Â
Could someone help me please and to tell how to get the probability from
empirical distribution (not parametric) for each data value (R function).
For example, for normal distribution there is such a function like:
Â
âpnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)â
Â
Hi,
How would you compare several (simple) regression lines in R? Have you
heard of any implementation of the Johnson-Neyman procedure or
anything of this sort? Any suggestions?
Best,
Marcin
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On 9/29/2009 6:06 AM, abdul kudus wrote:
Dear all,
Given mypi
mypi - c(0.1,0.2,0.2,0.1,0.3,0.4,0.4,0.4,0.4,0.2)
I want to create myfreq as follows
mypi myfreq
0.1 2
0.2 3
0.2 3
0.1 2
0.3 1
0.4 4
0.4 4
0.4 4
0.4 4
0.2 3
where myfreq is frequency of its corresponding
2) See packages Epi and epitools
-Johannes
2009/9/29 Dmitry Gospodaryov gospodar...@rambler.ru:
I have two types of survival data for Drosophila
cohort. For example:
uncensored
age - c (0, 2, 4, 6, 8, 10)
alive1 - c(10, 9, 6, 3, 1, 0)
alive 2 - c(10, 9, 4, 1, 1, 0)
and censored
age -
davew wrote:
Hi all,
I'm running an analysis with the random forest tool. It's being applied to a
data matrix of ~60,000 rows and between about 40 and 200 columns. I get the
same error with all of the data files (Cannot allocate vector of size
428.5MB).
I found dozens of threads
From Writing R Extensions:
‘#’, ‘_’ and ‘’ must not be escaped.
Uwe Ligges
Gábor Csárdi wrote:
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
\preformatted{
\# vertex1name
Hi Xavier,
thank you for your suggestion, it's not exactly what I need. However I
made some progress using tapply, but it does not give me the 'correct'
result (see index - that's what I want to obtain!) when NAs are involved:
test - data.frame(date =
Uwe, thanks, but this does not help, I still get:
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
! You can't use `macro parameter character #' in vertical mode.
argument ...ike the following: \begin {alltt} ##
Apologies for the misunderstanding. I can come up with a solution that might
suit your needs:
library(plyr)
out - ddply(test, .(nr), function(x) data.frame(date=x$date,
index=rank(-as.integer(x$date
out[is.na(out$nr) | is.na(out$date), index] - NA
Xavier
- Mail Original -
De:
Add a dummy argument to the function:
randomSamples-lapply(1:2000,function(dummy){
+ meta_comp[sample(nrow(meta_comp),nTimes),]})
On Mon, Sep 28, 2009 at 9:39 PM, ewaters ewat...@nchecr.unsw.edu.au wrote:
Thanks for your help.
Unfortunately when I try it this generates an error as follows:
Hello,
I have written a C++ code and I would like to use it in R! Is this possible?
Many thanks in advance,
Stepho
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The function ecdf(x) computes the empirical cdf from data in the vector x.
You can plot it with plot(ecdf(x)), or compute the emp. cdf at new values,
e.g.
my.cdf - ecdf(x)
my.cdf( 0:3 ) # computes the emp. cdf at 0,1,2,3
John
On Tue, Sep 29, 2009 at 05:27:54AM -0700, Stefo Ratino wrote:
Hello,
I have written a C++ code and I would like to use it in R! Is this possible?
Yes, see e.g. http://cran.r-project.org/package=Rcpp
Dirk
--
Three out of two people have difficulties with fractions.
Thanks to all of you who replied. You've provided some great leads. I
really appreciate it.
Gary
On Mon, Sep 28, 2009 at 10:01 AM, Gary Lewis gary.m.le...@gmail.com wrote:
Hi - Many organizations now make their data available as XML via a
REST web service architecture. Is there any R package
Hi, R-users,
I met a problem:
Items:[Anna 'moi =) akku loppu joskus 4ltä. Kestää kauan nää..'\tAmer, Tuusula
(0:20)\t20\t12\t16\t00\t00\t11]/Anne 'Ei jakoa,uus päivä muistio et 4n niin
peruin. Hups'\t (0:16)\t0\t12\t18\t00\t00\t11/Elina 'Konsertissa. En tod.
vastaa teille'\tEtu-Töölö,
An interesting, and topical, example of multivariate data for
classroom illustrations are the American college football rankings.
Starting at the end of October (or week 8, the 8th week of the
football season) a set of rankings called the BCS (Bowl Championship
Series) will be published. This is
Thanks for the responses
@Patrik Burns
I'm going to try running on a 64 bit machine. Unfortunately R isn't
installed properly on it yet and our admin guy is away, so it'll have to
wait.
@ Uwe Ligges
Unless the program suddenly starts generating masses and masses of data, I
don't think this is
Try this:
Str - :[Anna 'moi =) akku loppu joskus 4ltä. Kestää kauan
nää..'\tAmer, Tuusula (0:20)\t20\t12\t16\t00\t00\t11]/Anne 'Ei
jakoa,uus päivä muistio et 4n niin peruin. Hups'\t
(0:16)\t0\t12\t18\t00\t00\t11/Elina 'Konsertissa. En tod. vastaa
teille'\tEtu-Töölö, Helsinki
On Sep 29, 2009, at 8:39 AM, Dirk Eddelbuettel wrote:
On Tue, Sep 29, 2009 at 05:27:54AM -0700, Stefo Ratino wrote:
Hello,
I have written a C++ code and I would like to use it in R! Is this
possible?
Yes, see e.g. http://cran.r-project.org/package=Rcpp
And section 5 of the Writing R
Dear List,
I just googled to find out if notched box plots are possible with
ggplot2, but couldn't find a answer to it.
boxplot() has the option: notch = TRUE, e.g.: boxplot(mpg$hwy, notch=TRUE)
My example code (taken from the net) is:
require(ggplot2)
qplot(class, hwy, fill=factor(year),
On Tue, Sep 29, 2009 at 6:58 AM, xavier.char...@free.fr wrote:
Apologies for the misunderstanding. I can come up with a solution that might
suit your needs:
library(plyr)
out - ddply(test, .(nr), function(x) data.frame(date=x$date,
index=rank(-as.integer(x$date
out[is.na(out$nr) |
Hi guys,
I still did not solve my problem properly! I have to compare the values of two
lists of 250 numbers as a result of using the ?by function!
List1 of 250
$ 0 : num [1:28] 22 11 31...
$ 1 : num [1:15] 12 14 9 ...
..
..
..
- attr(*, dim)= int 250
- attr(*, dimnames)=List of 1
i am having a problem in saving plots in pdf. i have this code below and it
only shows me the last plot. i tried keeping my devices open by removing
dev.off() from the code but the pdf file won't open
for (i in 1:n) {
.
.
.
pdf(D:/research/plot.pdf)
plot(mon, mu, type ='o')
dev.off()
}
my for
Hi all,
I ran in to a problem while trying to install RMAGEML package.
/usr/bin/ld: cannot find -ljava . (please look at the bottom for
details of the error)
I have set the following in ~/.bashrc file.
export JAVA_HOME=/usr/lib/jvm/java-1.6.0-openjdk-1.6.0.0.x86_64
export
Hello,
Â
Could someone help me please and to tell how to get the probability from
empirical DENSITY (not parametric) for each data value (R function).
For example, for normal distribution there is such a function like:
Â
âdnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)â
Hi,
Try opening and closing the device outside the loop,
pdf(D:/research/plot.pdf)
for (i in 1:n) {
plot(mon, mu, type ='o')
}
dev.off()
HTH,
baptiste
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Dear Steve,
I don't think that ggplot2 has that option. Hadley, please correct me if
I'm wrong.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Lina, check whether something like
data.frame(density(rnorm(10))[1:2])
contains the information you want. Otherwise, try to be (much) more specific in
what you want so that we do not need to guess (and of course provide minimal,
self-contained, reproducible code). That has a higher chance to
Gábor Csárdi wrote:
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
\preformatted{
\# vertex1name
vertex2name [optionalWeight]
vertex3name [optionalWeight]
}
Hello,
I have a question about function arima.sim
I tried to somulate a AR(1) process, with no innovation, no error term.
I used this code:
library(forecast)
e=rnorm(100,mean=0,sd=0)
series=arima.sim(model=list(ar=0.75),n=100,innov=e)+20
Then I tried to applicate ti this series auto.arima
My data is called xc and has more than 15 variables.
When I used summary(xc) it gave me the detail description of each
variable.
Summary(xc)
Y1x1 x2
x3 ..
Min. :0. Min. : 1.000 Min. : 1.000 Min. : 1.000
1st
I have the following code which I wanted to convert using for loop
previous code:
R_ev1 - R_event[R_event (rx[1] + 300*0) R_event = (rx[1] + 300*1)]
R_ev2 - R_event[R_event (rx[1] + 300*1) R_event = (rx[1] + 300*2)]
R_ev3 - R_event[R_event (rx[1] + 300*2) R_event = (rx[1] + 300*3)]
R_ev4 -
something like this should work (not exactly sure what you mean by 'compare')
result - lapply(List1, function(.lst1){
lapply(List2, function(.lst2, .lst1){
all(.lst2 %in% .lst1)
}, .lst1 = .lst1)
})
On Tue, Sep 29, 2009 at 10:34 AM, Christina Rodemeyer
On Tue, Sep 29, 2009 at 5:36 PM, Duncan Murdoch murd...@stats.uwo.ca wrote:
Gábor Csárdi wrote:
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
\preformatted{
\# vertex1name
Through converting a miRNAs file from FASTA to character format I get a vector
which looks like the following:
nml
[1] hsa-let-7a MIMAT062 Homo sapiens let-7a
[2] hsa-let-7b MIMAT063 Homo sapiens let-7b
[3] hsa-let-7c MIMAT064 Homo sapiens let-7c
jim holtman wrote:
Add a dummy argument to the function:
randomSamples-lapply(1:2000,function(dummy){
+ meta_comp[sample(nrow(meta_comp),nTimes),]})
Or
df = data.frame(x=1:10, y=10:1)
replicate(10, df[sample(nrow(df), 5),], simplify=FALSE)
Martin
On Mon, Sep 28, 2009 at 9:39 PM, ewaters
I have read all the help on axis scales, which seems to be much harder to
deal with than it should be, given how common the need to alter axes.
Anywayyeah, suppress the axis using yaxt then call and define the new
scale using axis().
The problem is, this doesn't change the actual axis,
Hi,
On Sep 29, 2009, at 12:03 PM, mau...@alice.it wrote:
Through converting a miRNAs file from FASTA to character format I
get a vector which looks like the following:
nml
[1] hsa-let-7a MIMAT062 Homo sapiens let-7a
[2] hsa-let-7b MIMAT063 Homo sapiens let-7b
[3] hsa-let-7c
On Tue, 29 Sep 2009, mau...@alice.it wrote:
Through converting a miRNAs file from FASTA to character format I get a vector
which looks like the following:
nml
[1] hsa-let-7a MIMAT062 Homo sapiens let-7a
[2] hsa-let-7b MIMAT063 Homo sapiens let-7b
[3] hsa-let-7c MIMAT064 Homo
Not sure exactly what you want, but does using 'ylim' do what you want:
plot(1:5, sample(100:200,5), type='b', ylim=c(0,450))
On Tue, Sep 29, 2009 at 12:08 PM, chipmaney chipma...@hotmail.com wrote:
I have read all the help on axis scales, which seems to be much harder to
deal with than it
On 29-Sep-09 16:03:31, mau...@alice.it wrote:
Through converting a miRNAs file from FASTA to character format I get
a vector which looks like the following:
nml
[1] hsa-let-7a MIMAT062 Homo sapiens let-7a
[2] hsa-let-7b MIMAT063 Homo sapiens let-7b
How about this:
x - c(sadfdsaf 24353245, 'wqerwqer 6577', 'xzcv 6587')
sub(^([^[:space:]]+)[[:space:]].*, \\1, x)
[1] sadfdsaf wqerwqer xzcv
On Tue, Sep 29, 2009 at 12:03 PM, mau...@alice.it wrote:
Through converting a miRNAs file from FASTA to character format I get a
vector which
Dear Thierry,
Thank you very much for the fast reply.
Is there a way not to plot outliers in the ggplot2 boxplots? In a way
that the scale of the y-axis is rendered as if there was no outlier?
(not just the the ouliers removed with the same scale of the
y-axsis...)
[ example for a boxplot:
That code is way more than a _minimal_ example, and its not
_reproducible_ either, so just a comment:
Have you considered creating a vector rather than separate event
variables?
R_ev[1:16] - R_event (rx[1] + 300*(0:15) R_event = (rx[1] +
300*(1:16)
?cut # would also appear to be a
Thanks, Uwe.
You code does what I would like to have!
Best,
Mike
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Tuesday, September 29, 2009 1:36 AM
To: Ping-Hsun Hsieh
Cc: r-help@r-project.org
Subject: Re: [R] plot error -- figure margins too large
Thanks David,
Yes, I am talking about the MASS package.Thank you for pointing out that
these scale the same. My question is, how do I get from the V1 data:
V1
1 164.4283
2 166.2492
3 170.5232
4 156.5622
5 127.7540
6 136.7704
7 136.3436
to the other set of data:
+ 1 -2.3769280
+ 2
On 9/29/2009 11:57 AM, Gábor Csárdi wrote:
On Tue, Sep 29, 2009 at 5:36 PM, Duncan Murdoch murd...@stats.uwo.ca wrote:
Gábor Csárdi wrote:
Dear All,
I have the following in a .Rd file:
...
human readable (not binary) format. The format itself is like
the following:
Try this:
sapply(xc, summary)
On Tue, Sep 29, 2009 at 12:42 PM, Ashta sewa...@gmail.com wrote:
My data is called xc and has more than 15 variables.
When I used summary(xc) it gave me the detail description of each
variable.
Summary(xc)
Y1 x1
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Henrique
Dallazuanna
Sent: Tuesday, September 29, 2009 9:57 AM
To: Ashta
Cc: R help
Subject: Re: [R] Summary
Try this:
sapply(xc, summary)
This fails if there are NA's
A alternative in cases where is there NA's shoul be:
sapply(sapply(df, summary), '[', 1:7)
On Tue, Sep 29, 2009 at 2:28 PM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Henrique
If you want to see how Venables and Ripley get their lda distances,
then this is a quick path to the (uncommented) source:
1) methods(lda)
[1] lda.data.frame* lda.default*lda.formula*lda.matrix*
2) since you call involved a formula I looked first at:
getAnywhere(lda.formula)
A single
I have a dataset. Initially, it has 25 levels for a certain factor,
Description.
However, I then subset it, because I am only interested in 2 of the 25
factors. When I subset it, I get the following. The vector lists only the
two factors, yet there remain 25 levels:
Quadrats.df$Description
hi,
on the mac you have the opportunity to start a script directly instead
of opening it in the script editor.
So when I doubleclick on an .R - file the code is executed immediately.
I haven't found such an option in the preferences of the R program for
windows.
Is there also a way to do
Try this;
do.call(rbind, sapply(gg2, '[', 1))
On Tue, Sep 29, 2009 at 2:43 PM, Carlos Hernandez carlos.u...@gmail.com wrote:
Dear All,
I´m using the following code:
all1-gg2[[1]][[1]]; for(i in 1:48){ all1 - rbind(all1,gg2[[i]][[1]]) }
to create a new matrix that contains all the matrices
Dear All,
I´m using the following code:
all1-gg2[[1]][[1]]; for(i in 1:48){ all1 - rbind(all1,gg2[[i]][[1]]) }
to create a new matrix that contains all the matrices in a list called gg2.
gg2 is a list that looks like
gg2
[[1]]
[[1]][[1]]
matrix one
[[2]]
[[2]][[1]]
matrix two
.
.
.
[[48]]
On Tue, Sep 29, 2009 at 6:47 PM, chipmaney chipma...@hotmail.com wrote:
I have a dataset. Initially, it has 25 levels for a certain factor,
Description.
However, I then subset it, because I am only interested in 2 of the 25
factors. When I subset it, I get the following. The vector lists
On Tue, Sep 29, 2009 at 7:56 AM, Douglas Bates ba...@stat.wisc.edu wrote:
An interesting, and topical, example of multivariate data for
classroom illustrations are the American college football rankings.
Starting at the end of October (or week 8, the 8th week of the
football season) a set of
Thank you for your quick reply! It works perfectly!
On Tue, Sep 29, 2009 at 7:51 PM, Henrique Dallazuanna www...@gmail.comwrote:
Try this;
do.call(rbind, sapply(gg2, '[', 1))
On Tue, Sep 29, 2009 at 2:43 PM, Carlos Hernandez carlos.u...@gmail.com
wrote:
Dear All,
I´m using the
plot(y~x, type='p', col=3, xlim=c(50,1000), ylim=c(0,80), xlab='', ylab='',
data=example)
abline(a=11, b=0.04, col = 2)
--
Thanks,
A.B.
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Felix,
Thanks, that did the trick! Lattice is a lot less intuitive than basic
plotting!
Also, another person suggested using gap.plot from the plotrix package to put a
break in the graph. I am surprised Lattice doesn't have something similar
since it seems like a common problem when you
On Sep 29, 2009, at 1:43 PM, Carlos Hernandez wrote:
Dear All,
I´m using the following code:
all1-gg2[[1]][[1]]; for(i in 1:48){ all1 - rbind(all1,gg2[[i]]
[[1]]) }
Looks to me that you would be getting a duplicate copy of the first
matrix, but aside from that what problems are you
Dear All!
I'm looking for equivalent of Matematica function Which which works as
follows:
z = Which[x10,0.3, 10=x20,0.5, 20=x100,1]
where x is a vector
I can replace it with custom function with set of ifelse but I'm looking
for simpler and faster (much faster) solution
best wishes
Jarek
Hello All,
I'm a new R user and have a question about what in SAS would be called
macro variable substitution. Below is some R code that doesn't work,
but I think it will illustrate what I'd like to do.
readfunc-function(x) {
x - read.table(paste(x,.csv,sep=), header=TRUE,sep=,)
}
Hi,
I guess you want ?assign
See also this page for a working example,
http://wiki.r-project.org/rwiki/doku.php?id=guides:assigning-variable-names
HTH,
baptiste
2009/9/29 David Young dyo...@telefonica.net:
Hello All,
I'm a new R user and have a question about what in SAS would be called
well thanks, when I post the mail I thought I got too simple example
which may be really replaced by cut, but I thought about little more:
let say:
z = Which[x10,x/3, 10=x20,0.5, 20=x100,x^2/(x-1)]
where there are both values and formula
sorry for mismatch and thanks for quick answer
Jarek
One option, see ?assign and write a wrapper around it, using the pos
argument.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of David Young
Sent: Tuesday, September 29, 2009 10:58 AM
To: r-help@r-project.org
Subject: [R]
On Sep 29, 2009, at 1:49 PM, Martin Batholdy wrote:
hi,
on the mac you have the opportunity to start a script directly
instead of opening it in the script editor.
So when I doubleclick on an .R - file the code is executed
immediately.
I haven't found such an option in the preferences of
On Tue, Sep 29, 2009 at 8:05 PM, David Winsemius dwinsem...@comcast.netwrote:
On Sep 29, 2009, at 1:43 PM, Carlos Hernandez wrote:
Dear All,
I´m using the following code:
all1-gg2[[1]][[1]]; for(i in 1:48){ all1 - rbind(all1,gg2[[i]][[1]]) }
Looks to me that you would be getting a
Hi,
I want to compile R with command completion. But I don't find such an
option in configure. Can somebody let me know how to enable command
completion in an R session?
Regards,
Peng
__
R-help@r-project.org mailing list
well function arguments are in square brackets. z is result (new
vector). I show Matematica syntax, but cannot explain what I expect. Sorry
The example is wrong because it can be replaced by R cut function. The
arguments are: condition,action and can be replaced by ste of ifelse
formulas:
Dear all
I have a data set for which PCA based between group analysis (BGA) gives
significant results but CA-BGA does not.
I am having difficulty finding a reliable method for deciding which ordination
technique is most appropriate.
I have been told to do a 1 table CA and if the 1st axis
Thanks for the response. I'm sorry I didn't provide the code or data example
earlier. I was using the polynomial fitting technique of this form;
test - lm(x[,34] ~ I(x[,1]) + I(x[,1]^2) + I(x[,1]^3))
for the original fitting operation. I also tried to use;
lm(y ~ poly(x,3,raw=TRUE))
with the
On 30/09/2009, at 3:45 AM, Lina Rusyte wrote:
Hello,
Could someone help me please and to tell how to get the probability
from empirical DENSITY (not parametric) for each data value (R
function).
For example, for normal distribution there is such a function like:
“dnorm(q, mean = 0, sd =
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jarek Jasiewicz
Sent: Tuesday, September 29, 2009 11:36 AM
To: Erik Iverson
Cc: R-help@r-project.org
Subject: Re: [R] Equivalent for Matematica function Which...
well
On Sep 29, 2009, at 2:36 PM, Jarek Jasiewicz wrote:
well function arguments are in square brackets. z is result (new
vector). I show Matematica syntax, but cannot explain what I expect.
Sorry
The example is wrong because it can be replaced by R cut function.
The arguments are:
No. The forest object is too large as is. I didn't think it's worth
the extra memory to store them. They were never kept even in the
Fortran/C code.
Andy
From: Chrysanthi A.
Sent: Monday, September 28, 2009 5:20 PM
To: r-help@r-project.org
Subject: [R] how to visualize gini coefficient in
Hello,
I have
srv - Surv(sample(1:10), sample(0:1, 10, replace = TRUE))
srv
[1] 1 10 2+ 8 6+ 7+ 3 5+ 4+ 9+
srv.char - as.character(srv)
srv.char
[1] 1 10 2+ 8 6+ 7+ 3 5+ 4+ 9+
Is there an inverse to as.character(srv). That is, I would like
identical(srv,
William Dunlap pisze:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jarek Jasiewicz
Sent: Tuesday, September 29, 2009 11:36 AM
To: Erik Iverson
Cc: R-help@r-project.org
Subject: Re: [R] Equivalent for Matematica function
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Tuesday, September 29, 2009 4:52 AM
To: Nair, Murlidharan T
Cc: r-help@r-project.org
Subject: Re: [R] error while plotting
Nair, Murlidharan T wrote:
-Original Message-
From: Uwe Ligges
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