Are there any particular licences under which R packages must be released or is
it the discretion of the author? The same question if the package is to be
destined for CRAN?
Kind regards,
Nathan
--
Dr. Nathan S. Watson-Haigh
OCE Post
William Dunlap wrote:
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Wacek Kusnierczyk
Sent: Thursday, May 28, 2009 12:31 PM
To: Caroline Bazzoli
Stavros Macrakis-2 wrote:
What is the recommended way of checking whether an RODBC connection is
open?
Since odbcValidChannel is not exported from namespace RODBC, I suppose I
shouldn't be using it.
This is the best I could come up with, but it seems a bit 'dirty' to be
using a
Steve Jaffe wrote:
hmm, that is what I was afraid of. I considered that but thought to myself,
surely there must be an easier way. I wonder why this feature isn't
available. It's there in scripting languages, like perl, but also in
hardcore languages like C++ where std::sort and sorted
Naoki Irie-3 wrote:
I am using rgl.sphere to visualize scatter plot data in three dimensional
space. However, as I can not see the labels of each data point directly in
RGL window, I usually look for the values of x, y, z axis to find out the
label (or line number of the data point).
Duncan Murdoch wrote:
On 28/05/2009 6:06 PM, Steve Jaffe wrote:
hmm, that is what I was afraid of. I considered that but thought to
myself,
surely there must be an easier way. I wonder why this feature isn't
available. It's there in scripting languages, like perl, but also in
hardcore
Stavros Macrakis wrote:
I agree that it is surprising that R doesn't provide a sort function with a
comparison function as argument. Perhaps that is partly because calling out
to a function for each comparison is relatively expensive; R prefers vector
operations.
That said, many useful
Thanks all for the on- and off-list responses. For a relevant
discussion see the normality tests thread [1], and specifically
another excellent overview by Thomas on the robustness of the t-test
[2].
Best,
Liviu
[1] http://thread.gmane.org/gmane.comp.lang.r.general/86180
[2]
Licences are considered to apply to 'distribution' (I'm not entirely
sure if that is what you mean by 'released'), but some also apply to
usage.
The primary requirements for distribution, especially over CRAN, are
that the licence be clear and usable. Examples of problems
- licences that
Hi there,
I´ve a question concerning the nnet package in the area of survival analysis:
what is the final value, which is computed to fit the model with the following
nnet-c
all:
net - nnet(cat~x,
data=d,
size=2,
decay=0.1,
censored=TRUE,
maxit=20,
Hi all ,
I am facing some problems with 'ls' command. Whenever I use it I
get the following error :
ls(KEGG.db)
Error in as.environment(pos) :
no item called KEGG.db on the search list
ls(pkgname)
Error in as.environment(pos) :
no item called KEGG.db on the search list
and this
I am not sure what you want. The error msg is clear to me. You can use
search() to search what are in the search path, for example.
search()
[1] .GlobalEnvpackage:stats package:graphics
[4] package:grDevices package:utils package:datasets
[7] package:methods Autoloads
Hi all
Does anyone here know of an efficient algorithm to find the longest
identical subsequence across multiple time series? Say I have two time
series A and B (not necessarily of identical length), and I wish to find the
beginning and ending index of the longest common subsequence across both
IP addresses are very (very!) difficult to parse and sort correctly
because there are all sorts of supported formats. Try to use something
like PostgreSQL instead: it is already implemented there. But if you
are sure all your data is of the n.n.n.n form, then something along the
lines of the
The value includes the weight decay term: it is a penalized fit. See
the documentation pages 245 and 247. (nnet is support software for a
book, so the book is the primary documentation.)
On Fri, 29 May 2009, Andrea Weidacher wrote:
Hi there,
I?ve a question concerning the nnet package in
Hello everybody,
I am trying to a read a binary file with different formats. I use the readBin
function so I can read bytes, short and double numbers depending on the bytes
per element in the byte stream. But now I need to read bit by bit, and join
them in groups of ten because every ten bits
Allan Engelhardt wrote:
IP addresses are very (very!) difficult to parse and sort correctly
because there are all sorts of supported formats. Try to use something
like PostgreSQL instead: it is already implemented there. But if you
are sure all your data is of the n.n.n.n form, then
normalizedip - function(ipstring){
ipsepstring - strsplit(ipstring,\\.)[[1]]
cat(sapply(ipsepstring,function(x)
sprintf(%03i,as.numeric(x))),sep=.)
}
normalizedip(1.2.3.55)
yields
001.002.003.055
and therefore should allow you to sort in correct order.
edwin Sendjaja wrote:
Hi,
In my humble opinion, forget R and use more appropriate tools, like a
very small C program that reads 10 bytes at a time and outputs 8 integers.
Then use R to read the resulting file.
Ciao!
mario
Inigo Pagola Barrio wrote:
Hello everybody,
I am trying to a read a binary file with
Peter Dalgaard wrote:
Allan Engelhardt wrote:
IP addresses are very (very!) difficult to parse and sort correctly
because there are all sorts of supported formats. Try to use something
like PostgreSQL instead: it is already implemented there. But if you
are sure all your data is of the
I want to create a neural network, and then everytime it receives new data,
instead of creating a new nnet, i want to use a backpropagation algorithm
to adjust the weights in the already created nn.
I'm using nnet package, I know that nn$wts gives the weights, but I cant
find out which weights
I have a number of signals whose ACF may or may not cross the zero line.
Whether roots exist or not is a piece of useful information for me.
Likewise, if any root exists, I'd like to know its lag value.
Unluckily R function ACF does not seem to provide such information.
I wonder whether someone
Hi. Have you looked at graphviz?
I have previously used R to generate graphviz diagrams. I create a
template graphviz diagram and then use R to parse this template,
substitute appropriate values and then outputto a temporary file. I
then use a system() call to run graphviz on thisoutputted
I've not ever tried something like this. You didn't quite answer the
question though. Do you need interactive sessions, or are users
choosing from a number of batch jobs?
If the latter, then perhaps you would be better forsaking Rcmdr (which
I have not used) and instead consider a shell-based, or
Hello Madan,
I am rather novice as well, so I went the ODBC way.
If you define an ODBC connection to Oracle, with System DSN dsn-name,
the code to get the results of a query into, say, mydata is like :
library(RODBC) # First of all
channel - odbcConnect(dsn-name, uid=someuser) # double quotes
Inigo Pagola Barrio wrote:
Hello everybody,
I am trying to a read a binary file with different formats. I use the readBin
function so I can read bytes, short and double numbers depending on the bytes
per element in the byte stream. But now I need to read bit by bit, and join
them in groups
Thomas Levine wrote:
Ah, that makes sense. But now another two issues have arisen.
Firstly, the error bars look like confidence intervals, and I'm pretty
sure that they are but does some document verify this? I suppose I
could check the code too.
Secondly, I just read about how dynamite plots
Peter Dalgaard wrote:
Allan Engelhardt wrote:
[...]
Getting rid of the conversions including the matrix(unlist) combo is
left as an exercise (it's too hot here)
Here's one way:
con - textConnection(as.character(a$ip))
o - do.call(order,read.table(con,sep=.))
close(con)
I need your help for the following purpose :
I would like to create regularly spaced points on a multisegment line (a
'psp class' object).
A function of the spatstat library( = pointsOnLines() ) is dedicated to
that objective but the problem is that the probability of falling on a
On behalf of the editorial team, I am happy to inform you that the first
issue of The R Journal is now available:
http://journal.r-project.org
I would like to thank team members Heather Turner, Peter Dalgaard, and John Fox
for extraordinary efforts in producing this first number of the Journal.
On 29-May-09 11:23:15, Pascal LAFFARGUE wrote:
I need your help for the following purpose :
I would like to create regularly spaced points on a multisegment
line (a 'psp class' object).
A function of the spatstat library( = pointsOnLines() ) is dedicated
to that objective but the problem
losemind wrote:
Of course, in MLE, if we collect more and more observations data, MLE
will perform better and better.
But is there a way to find a bound on parameter estimate errors in
order to decide when to stop collect data/observations, say 1000
observations are great, but 500
Dear Group members
I was wondering whether there is any interface to use processing
(www.processing.org) to visualize R analyses?
If so could somebody point me to relevant ressources?
If not, are there any attempts to build such an interface / package?
Best Regards,
Simon
--
Nur bis
The updated package has been submitted to CRAN and will propagate to
mirrors over the next day or so.
It is maintained on R-Forge at http://r-forge.r-project.org/projects/writexls
, where downloads are available as well.
Package: WriteXLS
Version: 1.7.1
Description: Cross-platform perl
Hi R users,
Someone knows how to replace Infinite value by zero. I have a vector with
some Inf value and I want to substitute these values by zero to get the mean
of the components of the vector.
Any idea?
Many thanks,
Marlene.
[[alternative HTML version deleted]]
Hi everyone,
Could anyone tell me what means the follow error message
Error in xj[i] : invalid subscript type 'closure'
It happens when I run the function plm, like this:
ff-totaccz~lactivoz+varvolnegz
ss-plm(ff,data=regaccdis,na.omit)
Error in xj[i] : invalid subscript type 'closure'
Hello !
Â
Iâm trying to estimate a system of equation (demand and supply) using the
systemfit package. My program is:
Â
library(systemfit)
demand - tsyud ~ tsyud1 + tsucp + tspo + tssn
supply - tscn ~ tsyn + tsqn + tsksn + tsucp
system - list(demand=eqdemand, learning = eqsupply)
labels -
I am running R on Windows XP
I first tried to install rJava and got this message
utils:::menuInstallLocal()
package 'rJava' successfully unpacked and MD5 sums checked
updating HTML package descriptions
then i tried this
require(rJava)
Loading required package: rJava
Error in if
Dear Marlene,
Have a look at is.infinite(). Replacing them by zero is not a very
bright idea if you want the mean of the resulting vector. Have a look at
the example below.
#create a vector
x - rnorm(100, mean = 1000)
#replace 20 values with Inf
x[sample(seq_along(x), 20, replace = FALSE)] - Inf
On Thu, May 28, 2009 at 7:35 PM, Ronggui Huang ronggui.hu...@gmail.comwrote:
Dear all,
I want to set tool-tips for a gtkButton. I use the following code
which works under Windows. However, it doesn't work under Linux. Any
hints? Thanks.
Unfortunately, on platforms besides Windows, the
Thanks a lot Andrea, it works!
Marlene.
2009/5/29 Andrea Weidacher da0...@yahoo.de
Hi Marlene,
try this:
x[which(x==Inf)] - 0
Andrea.
--
*Von:* marlene marchena marchenamarl...@gmail.com
*An:* R-help@r-project.org
*Gesendet:* Freitag, den 29. Mai 2009,
You can figure out which weights go with which connections with the
function summary(nnet.object) and nnet.object$wts. Sample code from
Venables and Ripley is below:
# Neural Network model in Modern Applied Statistics with S, Venables and
Ripley, pages 246 and 247
library(nnet)
hi all,
I often have a data frame like this example
data.frame(sq=c(1,1,1,2,2,3,3,3,3),area=c(1,2,3,1,2,3,1,2,3),habitat=c(garden,garden,pond,field,garden,river,garden,field,field))
for each sq I have multiple habitats each with an associated area.
I want to aggregate the data frame so that
Although, in this case, the which() is unnecessary.
x[x==Inf] - 0
See the online help, that is,
help('[')
and also the documentation, An Introduction to R, available online at CRAN.
-Don
At 2:58 PM +0100 5/29/09, marlene marchena wrote:
Thanks a lot Andrea, it works!
Marlene.
Hello
Im am working with a biological data including variables called Habitat and
Site, example:
Habitat Site
Forest Low
Forest Low
Forest High
Forest High
I want to tell R that the Site variable is nested within the Forest variable
(that it is not a new variable).
Does
Hi all,
I have a binary matrix with NAs included. Each row and column
includes at least one NA, so I don't want to omit them. Is there a
way to sum across rows and columns, ignoring the NAs but not deleting
the row or column? If not, I suppose I can write a loop function, but
I have learned
Try this:
as.data.frame.table(xtabs(area ~ habitat + sq, DF), responseName =
area.sum)[c(2:3, 1)]
sq area.sum habitat
1 10 field
2 13 garden
3 13pond
4 10 river
5 21 field
6 22 garden
7 20pond
8 2
Sometimes I get confused with R's documentation. It seems the documents is not
maintained and updated well. Anyone has similar feeling? I don't mean to offend
anyone. I hope R would get better and better. But documentation is really one
very important factor which could attract people coming or
marlene marchena marchenamarlene at gmail.com writes:
Hi R users,
Someone knows how to replace Infinite value by zero. I have a vector with
some Inf value and I want to substitute these values by zero to get the mean
of the components of the vector.
Any idea?
Many thanks,
use:
sum(x,na.rm=T)
like this:
my.row.sums - apply(my.matrix,1,sum,na.rm=T)
If you want to do across columns then use 2 instead of 1. Look at ?apply
and ?sum.
Adrian
On Fri, May 29, 2009 at 7:46 AM, Wade Wall wade.w...@gmail.com wrote:
Hi all,
I have a binary matrix with NAs included.
Josef.Kardos at phila.gov writes:
I finally realized I didn't have Java installed on my computer, so I
downloaded Java and tested that it works.
I then attempted to reinstall rJava and load the package, but got this;
utils:::menuInstallLocal()
package 'rJava' successfully unpacked
Wade Wall wrote:
Hi all,
I have a binary matrix with NAs included. Each row and column
includes at least one NA, so I don't want to omit them. Is there a
way to sum across rows and columns, ignoring the NAs but not deleting
the row or column? If not, I suppose I can write a loop function,
On Fri, May 29, 2009 at 11:01 AM, Zheng, Xin (NIH) [C]
zheng...@mail.nih.gov wrote:
Sometimes I get confused with R's documentation. It seems the documents is
not maintained and updated well. Anyone has similar feeling? I don't mean to
offend anyone. I hope R would get better and better. But
Zheng, Xin (NIH) [C] wrote:
Sometimes I get confused with R's documentation. It seems the documents is not
maintained and updated well. Anyone has similar feeling? I don't mean to offend
anyone. I hope R would get better and better. But documentation is really one
very important factor which
Ok, pls take a look at '?median'. Can you see However, the default method
makes use of 'sort' and 'mean'? Then let's look at 'median.default'. I do see
'sort'. But where is 'mean'? At first glance I didn't catch the exact point of
its algorithm. Why not say more clearly that make use of partial
Hi, Does anyone know what the mean value of a lognormal distribution in base-2
is? I am simulating stochastic population growth and if I were working in
base-e, I would do:lambda - 1.1 #multiplicative growth rates - 0.6
#stochasticity (std. dev)lognormal - rlnorm(10, log(lambda) - (s^2)/2,
I apologize for the poor formatting of my previous post as the line breaks got
stripped. I hope this is easier to read.
Hi,
Does anyone know what the mean value of a lognormal distribution in base-2 is?
I am simulating stochastic population growth and if I were working in base-e, I
would
Zheng, Xin (NIH) [C] wrote:
Sometimes I get confused with R's documentation. It seems the
documents is not maintained and updated well. Anyone has similar
feeling? I don't mean to offend anyone. I hope R would get better and
better. But documentation is really one very important factor which
Which version of R is this? From the BUGS section in NEWS for R 2.9.0
patched:
o median.default() was altered in 2.8.1 to use sum() rather
than mean(), although it was still documented to use mean().
This caused problems for POSIXt objects, for which mean() but
A closure is a standard R function (that is not a primitive function).
This is presumably from package plm. You called plm() with three
arguments, that is subset=na.omit by the matching rules. I think
you intended na.action=na.omit. na.omit is a 'closure'.
On Fri, 29 May 2009, Cecilia
I'm not sure I understand the max.col spec or its rationale. In particular:
* What is the significance and effect of assuming that the entries are
probabilities, as they do not seem to be limited to the interval [0,1]?
* In what contexts is it useful for max.col to consider numbers within a
Thanks for your information. I am curious why in 2.8.1 the document was not
changed. That's easier than writing something in NEWS.
But I DO see 'sum' rather than 'mean' in 2.9.0 now. It's not reverted yet as
the NEWS saying. And it's still documented to use 'mean'.
-Original Message-
On Fri, May 29, 2009 at 11:01 AM, Zheng, Xin (NIH) [C]
zheng...@mail.nih.gov wrote:
Sometimes I get confused with R's documentation. It seems the documents is
not maintained and updated well. Anyone has similar feeling? I don't mean to
offend anyone. I hope R would get better and better. But
On Fri, May 29, 2009 at 3:47 AM, simon_mail...@quantentunnel.de wrote:
Dear Group members
I was wondering whether there is any interface to use processing (
www.processing.org) to visualize R analyses?
Nothing direct, to my knowledge.
As far as I know, Processing is a high-level language
Hi list
I try to remove all object less one, this object is called index. I have
many object with different names and pattern option in ls function may not
remove this object why any word in index object repeats with other object.
Anybody can give me one advise for this question
Thank
I am trying split a string and use one part of it to label graphs. I am using
strsplit for that. While I am able to split it, how do I access the separated
parts.
filName-chrI_2223
part- strsplit(filName,\\_)
part
[[1]]
[1] chrI 2223
part[1]
[[1]]
[1] chrI 2223
I looked up the help archive
Thanks a lot for your answers. I can try to implement backpropagation myself
with that information.
But there isnt a function or method of backpropagation of error for new
examples of training only to change the already created neural net?
I want to implemennt reinforcement learning...
Thanks in
Try this:
unlist(part)[1]
or
part[[1]][1]
On Fri, May 29, 2009 at 2:09 PM, Nair, Murlidharan T mn...@iusb.edu wrote:
I am trying split a string and use one part of it to label graphs. I am
using strsplit for that. While I am able to split it, how do I access the
separated parts.
Not that I know of.
If you do come across any, let me know, or better still, email r-help.
Good luck with what you are trying to do.
Jude Ryan
From: Filipe Rocha [mailto:filipemaro...@gmail.com]
Sent: Friday, May 29, 2009 1:17 PM
To: Ryan, Jude
Cc:
Dear R People:
I have the following vector and am using the paste command:
ya
[1] 57 2 8
paste(stuff,ya,sep= )
[1] stuff 57 stuff 2 stuff 8
What I want to have is
stuff 57 2 8
I also tried:
yb - paste(cat(stuff,ya))
stuff 57 2 8 yb
character(0)
I have the feeling that it's really
Nothing like posting your snag to make you solve it!
The answer is simply:
c(stuff,ya)
[1] stuff 572 8
Sorry,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
On Fri, 2009-05-29 at 12:39 -0500, Erin Hodgess wrote:
Dear R People:
I have the following vector and am using the paste command:
ya
[1] 57 2 8
paste(stuff,ya,sep= )
[1] stuff 57 stuff 2 stuff 8
What I want to have is
stuff 57 2 8
I also tried:
yb - paste(cat(stuff,ya))
fernando espindola wrote:
Hi list
I try to remove all object less one, this object is called index. I have
many object with different names and pattern option in ls function may not
remove this object why any word in index object repeats with other object.
Anybody can give me one advise
Noted with thanks.
Regards
Ronggui
2009/5/29 Michael Lawrence mflaw...@fhcrc.org:
On Thu, May 28, 2009 at 7:35 PM, Ronggui Huang ronggui.hu...@gmail.com
wrote:
Dear all,
I want to set tool-tips for a gtkButton. I use the following code
which works under Windows. However, it doesn't
Let's see if I understand this. Do I iterate through
x - factor(x, levels(c(levels(x), NA), exclude=NULL)
for each of the few hundred variables (x) in my data frame?
I tried to do this all at once and failed:
ToyData
Data1 Data2 Data3 Weight
101 Sam Red Banana1.1
102 Sam
fernando espindola wrote:
Hi list
I try to remove all object less one, this object is called index. I have
many object with different names and pattern option in ls function may not
remove this object why any word in index object repeats with other object.
Anybody can give me one advise for
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter way to do
this with
Dear R users,
Suppose I have a vector that consists of characters like ABC, A02, RCA,
etc., and there are about 700 of possible characters. For example,
x - c(ABC, ABC, ABC, A02, ABC, RCA, ABC, ABC)
I'd like to get a frequency matrix that looks something like this:
ABC 6
A02 1
RCA 1
I
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter way to do
this with
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter way to do
this with
You can try
rm(list = ls()[!(ls() %in% index)]).
-Christos
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Luc Villandre
Sent: Friday, May 29, 2009 2:06 PM
To: fernando espindola
Cc: r-help@r-project.org
Subject: Re: [R]
On Fri, May 29, 2009 at 7:50 AM, DanielWC daniel.carsten...@gmail.com wrote:
Hello
I am working with a biological data including variables called Habitat and
Site, example:
Habitat Site
Forest Low
Forest Low
Forest High
Forest High
I want to tell R that the
I think I am using the improved version of setdiff(...) that handles
data.frames, so I think some odd behavior was expected but this one is escaping
me.
It appears that the the addition of duplicate entries is not caught by the
setdiff(...). Is this expected behavior?
If so, is there
Not sure what you mean by permutations here. I think what you mean is
that given a matrix m, you want a matrix whose rows are c(i,j,m[i,j]) for
all i and j. You can use the `melt` function in the `reshape` package for
this. See below.
Hope this helps,
-s
library(reshape)
About this;
m - matrix( 1:9, nr = 3 )
dimnames( m ) - list( c(d, e, f), c(a, b, c) )
data.frame( row = rep( rownames(m), ncol(m)), col = rep(
colnames(m), each = nrow(m)), data = as.vector(m) )
row col data
1 d a1
2 e a2
3 f a3
4 d b4
5 e b5
6 f
Oh, I should have mentioned that the result of melt is a data.frame, not a
matrix. You can convert with as.matrix if you like.
I should also have shown that dimnames are carried along:
m - matrix(1:4,2,2,dimnames=list(x=c('a','b'),y=c('x','y')))
m
y
x x y
a 1 3
b 2 4
melt(m)
x y
On Fri, 2009-05-29 at 11:08 -0700, Ian Coe wrote:
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix.
Ian Coe wrote:
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter
Dear All,
I am observing a strange behavior and searching the archives and help
pages didn't help much.
I have a csv with a variable number of fields in each line.
I use
dataPoints - read.csv(inputFile, head=FALSE, sep=;,fill =TRUE);
to read it in, and it works. But - some lines are long and
Hi R-helpers,
I want to determine the coefficients of the following
regression for several subsets, and I want to save it in a
dataframe:
The data is in «regaccdis», «regaccdis$caedois» is the
column that defines the subsets and the function I have
runned is
Sorry about the redundant emails. I was having some email issues and
did not realize they would all go through.
Thanks to everybody who answered my question.
Regards,
Ian
CONFIDENTIALITY NOTICE: This e-mail communication (inclu...{{dropped:23}}
Ian Coe wrote:
Hi,
Is there a way to convert a matrix into a vector representing all
permutations of values and column/row headings with native R functions?
I did this with 2 nested for loops and it took about 25 minutes to run
on a ~700x700 matrix. I'm assuming there must be a smarter
Dear R People:
Could someone recommend a baby book on path analysis and SEM, please?
Or if someone has an example that they use in the classroom setting,
that would be very cool too.
Thanks in advance,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and
x - c(ABC, ABC, ABC, A02, ABC, RCA, ABC, ABC)
table(x)
x
A02 ABC RCA
1 6 1
table(x)[c(ABC, A02, RCA)]
x
ABC A02 RCA
6 1 1
On May 29, 2009, at 2:15 PM, liujb wrote:
Dear R users,
Suppose I have a vector that consists of characters like ABC, A02,
RCA,
etc., and there are
Not sure it is really matrix to vector but here are a few of attempts:
abM-matrix(1:9, nrow=3)
rownames(abM) - letters[1:3]
colnames(abM) - letters[4:6]
data.frame( cols=colnames(abM)[col(abM)[1:9]], rows= rownames(abM)
[row(abM)[1:9]], vals=abM[1:9])
cols rows vals
1da1
2
Try this:
m - matrix(1:9, 3, dimnames = list(R = letters[1:3], C = LETTERS[1:3]))
as.data.frame.table(m, responseName = Value)
R C Value
1 a A 1
2 b A 2
3 c A 3
4 a B 4
5 b B 5
6 c B 6
7 a C 7
8 b C 8
9 c C 9
On Fri, May 29, 2009 at 2:08 PM, Ian Coe
Hi,
I've been trying to install the statnet package on my Solaris 10/Intel system.
I've been having problems with one of the dependencies, sna. During the
compile, it always fails and complains about __builtin_isnan (please see below
for output). I have tried installing this using Sun Studio
Dear Erin,
Although it's very old now, I like Duncan's Introduction to Structural
Equation Modeling (Academic Press, 1975); for a more complete treatment,
although it too is pretty old, Bollen, Structural Equations with Latent
Variables (Wiley, 1989). I have notes and other materials from a short
Dear Jason,
On Fri, May 29, 2009 at 2:48 PM, Jason Rupert jasonkrup...@yahoo.com wrote:
I think I am using the improved version of setdiff(...) that handles
data.frames, so I think some odd behavior was expected but this one is
escaping me.
It appears that the the addition of duplicate
Dear Erin,
I agree with John that one should be careful about taking SEMs too seriously.
For another good book on sem and path models, look at John Loehlin's book.
Address = {Mahwah, N.J.},
Author = {Loehlin, John C},
Edition = {4th},
Publisher = {L. Erlbaum
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