maybe not the optimal but it works:
Hi,
I have three original curves as follows,
n-seq(20,200,by=10)
t-c(0.1138, 0.1639, 0.2051, 0.2473, 0.2890, 0.3304, 0.3827, 0.4075,
0.4618, 0.4944,
0.5209, 0.5562, 0.5935, 0.6197, 0.6523, 0.6771, 0.6984, 0.7209, 0.7453)
es-c(0.3682, 0.4268, 0.5585,
Albert Crosby acrosby at sdale.org writes:
I'm new to R - and trying to create a plot similiar to the spider plot at
http://www.statgraphics.com/eda.htm#radar .
I can't figure out several things... most of which I would think would be
straightforward
How can I change the lines
My R version is R 2.7.0, but the png still can not be used when I embed R in
MyApp
Dirk Eddelbuettel wrote:
Wrong list, so re-directing to r-help. Consider r-sig-debian for Debian
questions too, but subscribe or else your posts bounce.
On 4 June 2008 at 07:18, Rongrong wrote:
|
|
On Wed, Jun 04, 2008 at 07:36:04PM -0600, Manli Yan wrote:
no,the id is variable of a table,such as:
treatment id age response
low 1 50 20
low 1 60 30
high5 50 30
high5 60 40
...
I want to rearranage the table
On Thu, Jun 05, 2008 at 09:05:25AM +0200, Philipp Pagel wrote:
a bad idea. Alternatively, this should work:
foo$id - cumsum(as.logical(diff(foo$id))) + 1
I should have explicitly said that this only works after sorting, of
course.
cu
Philipp
--
Dr. Philipp Pagel
Lehrstuhl für
Or (more simply?) install it from the bash prompt using:
R CMD INSTALL Rmpi_0.5-5.tar.gz
--configure-args=--with-mpi=/opt/openmpi/include/
(or whatever your path to openmpi might be).
I missed this as well and confused myself for a bit, but it is mentioned in the
R news article on Rmpi H.Yu
I have followed the instructions of numerous references in trying to package
my own functions. It appears as though i have all my paths and programs
aligned but i get the following error message when i R CMD CHECK my package:
*checking if this is a source bundle.ERROR
Only *source* packages
Dear all:
Subjects were measured two times (t1 and t2) on different variables (v1 ...
vn).
Between t1 and t2 there was an experimental manipulation. I computed two PCAs
for time-points t1 and t2.
Is it possible to combine both PCAs in order to get only one set of
eigenvectors? Due to the
Dear All
I have one a question to ask you SARIMA model ( 0,0,1)(1,0,1).
Could you help me to write the SARIMA equation as a data below ?
Constant 43.557 t = 10.09
Ma -0.37 t = -4.806
SAR 0.991 t = 48.098
SMA 0.915 t = 9.487
Thanks to Dirk and Duncan
Dirk's suggestion of the suppressMessages function unfortunately doesn't
quite do the job. The warning message isn't printed, but it is picked up
by the check function and leads to a whinge.
Duncan is right - I had an older version of maps and installing the
new
Dylan,
You can write r sin(x + alfa) as a sin(x) + b cos(x). Then you just have
to fit a linear model. r = sqrt(a^2 + b^2) and tan(alfa) = b / a
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en
I think you should specify your grouping factor:
g a vector or factor object giving the group for the corresponding
elements of x. Ignored if x is a list.
batlett.test(xx, groupingfactor)
Hope this helps.
Birgit
hanen wrote:
i'm trying to test the homogeneity of variance of 92 samples
i'm trying to test the homogeneity of variance of 92 samples each one
contains 3 observations.
to use bartlett.test function i have created a (3,92) matrix (named xx):
bartlett.test(xx)
this message appears:
Erreur dans bartlett.test.default(xx) :
l'argument g est manquant, avec aucune
Albert Crosby wrote:
I'm new to R - and trying to create a plot similiar to the spider plot at
http://www.statgraphics.com/eda.htm#radar .
I can't figure out several things... most of which I would think would be
straightforward
How can I change the lines for each series plotted
Thanks for your reply.
I use following coding to invoke R 2.7.0 process on Debian4:
// argv[0] is my current work path
char* av[] = {argv[0],--silent,--no-save, --gui=X11};
Rf_initialize_R(4, av);
R_Interactive = FALSE;
setup_Rmainloop();
I have installed
Birgitle [EMAIL PROTECTED] 05/06/2008 10:33:20
I think you should specify your grouping factor:
Yes, the message says the grouping factor is missing.
But xx should not be a matrix.
Your x data should be in a (1-dimensional) vector of length 276, and so should
your grouping factor.
Toy
Of course this is a trivial task:
text.in - readLines(file.in)
matrix.in - strsplit(text.in, [ \t]+)
# matrix.out - transpose matrix.in
# loop over the lines of matrix.out creating text.out
writeLines(text.out, file.out)
What is the _most elegant_ way to write the transposition of
matrix.in
On Wed, 4 Jun 2008 13:40:58 -0400,
Gary Lewis (GL) wrote:
Hi - Writing to see if someone can suggest whether a problem warrants a bug
report. It concerns the use of stepFlexclust in the flexclust package. The
problem concerns the size of clusters returned.
Versions: R-2.7.0 on
Hi Stefan,
Is it possible to combine both PCAs in order to get only one set of
eigenvectors?
Yes there is: statis() in the ade4 package is probably what you want. In
short, it does a k-table analysis that will give you a common
ordination/position. It also shows how each time-set deviates
Hi everybody,
my question is simple (yet I found no direct answer on this mailing list):
What are the consequences (legalwise) when I use R to do some statistical
analysis and outputs on my website (e.g. a user uploads his data, clicks
button Summary and as an ouput graphics/text generated from
I know this is fairly basic, but I must have somehow missed it in the
manuals.
I have two vectors, often of unequal length. I would like to compare
them for identity. Order of elements do not matter, but they should
contain the same.
I.e: I want this kind of comparison:
if (1==1) show(yes)
Alberto Monteiro wrote:
Of course this is a trivial task:
text.in - readLines(file.in)
matrix.in - strsplit(text.in, [ \t]+)
You probably want
write.table(t(read.table(file.in)), file = file.out, row.names = FALSE,
col.names = FALSE)
Uwe Ligges
# matrix.out - transpose matrix.in
#
Soil inf. wrote:
I have followed the instructions of numerous references in trying to package
my own functions. It appears as though i have all my paths and programs
aligned but i get the following error message when i R CMD CHECK my package:
*checking if this is a source bundle.ERROR
On 6/5/2008 7:33 AM, stephanos wrote:
Hi everybody,
my question is simple (yet I found no direct answer on this mailing list):
What are the consequences (legalwise) when I use R to do some statistical
analysis and outputs on my website (e.g. a user uploads his data, clicks
button Summary and as
This may help:
a - c('a')
b - c('a','b','c')
c - c('a','b','d')
all(a %in% b)
[1] TRUE
all(b %in% a)
[1] FALSE
all(b %in% c)
[1] FALSE
d - c('b', 'c')
all(d %in% b)
[1] TRUE
What you probably want to insure that the vectors contain the same elements
is:
if (all(v1 %in$ v2) all(v2 %in%
probably you want to look at:
?any
?all
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
Of course you could use readLines() and post process by using
appropriate functions such as grep(), strsplit() and friends later on.
Uwe Ligges
vittorio wrote:
Dear friend,
In an R program running permanently on a server I would like to read hour by
hour the temperature in *C and the
On 6/5/2008 6:38 AM, Karin Lagesen wrote:
I know this is fairly basic, but I must have somehow missed it in the
manuals.
I have two vectors, often of unequal length. I would like to compare
them for identity. Order of elements do not matter, but they should
contain the same.
I.e: I want this
Uwe Ligges wrote:
Of course this is a trivial task:
text.in - readLines(file.in)
matrix.in - strsplit(text.in, [ \t]+)
You probably want
write.table(t(read.table(file.in)), file = file.out, row.names =
FALSE, col.names = FALSE)
Unfortunately, it doesn't work. In fact,
Hello Karin
This is your code :
a = c(a)
b = c(b,c)
c = c(c,b)
if (a==a) show(yes) else show(blah)
if (a==b) show(yes) else show(blah)
if (b==c) show(yes) else show(blah)
Have a look at the conditions (a==b) and (b==c)
a==b
[1] FALSE FALSE
b==c
[1] FALSE FALSE
They are size 2.
I think R
Uwe Ligges wrote:
You probably want
write.table(t(read.table(file.in)), file = file.out, row.names =
FALSE, col.names = FALSE)
Ok, almost there. I forgot to tell (because I didn't know)
that the strings were separated by tabs (I just thought of
splitting by _any_ space), and that the
On Thu, Jun 05, 2008 at 10:26:08AM -0200, Alberto Monteiro wrote:
Uwe Ligges wrote:
You probably want
write.table(t(read.table(file.in)), file = file.out, row.names =
FALSE, col.names = FALSE)
Ok, almost there. I forgot to tell (because I didn't know)
that the strings were
Jim Price price_ja at hotmail.com writes:
My apriori apologies if this is felt to be the wrong list for this issue -
although it starts with R, it's a combination of programs that creates the
problem.
Currently we are using windows metafile format for in-text tables for
reports created
Dear R-Users,
I am trying to use paste to paste the character ' around a character
vector.
Paste appears to place spaces ( ) around this. Anyway this can be done
without the unintended introduction of space characters ?
Please see example below.
Many thanks,
Tolga
charlist-c(a,b,c)
dear all,
in the package pwr , there is the fonction power.anova.test which permit to
obtain the power for a one-way ANOVA...but I'm looking for a way to compute
the power of a multiway ANOVA.( find the 1-beta). Is it possible?
do you have some ideas ?
regards
[[alternative HTML
Hello,
I have been using mgcv to run GAM hurdle models, analyzing
presence/absence data with GAM logistic regressions, and then analyzing
the data conditional on presence (e.g. without samples with no zeros)
with GAMs with a negative binomial distribution.
It occurs to me that using the
Jim Price [EMAIL PROTECTED] wrote:
[...]
Currently we are using windows metafile format for in-text tables for
reports created in Word. However, we've discovered some artifactual lines
being created in our final output once the Word document is changed to PDF.
The process is as follows:
I have a dist object containing 1 row that is only NA (not very intelligent
to have bas dataset with one NA speciesanyway).
I would like to delete this row from this object.
It may be not a difficult problem but I can not find a solution presently.
So I would be very happy if somebody could
Let V = (v11, v12, ..., v1n, v21, v22, ..., v2n) be the matrix of data with
2*n columns. Now, you simply do PCA on this data matrix.
See the following paper, which has a related example:
Gabriel, K. R. and Odoroff, C. L. (1990). Biplots in biomedical research.
Statistics in Medicine, 9,
[EMAIL PROTECTED]:
charlist-c(a,b,c)
lapply(charlist, function (y) paste(',y,'))
Use 'sep='. And using 'sapply' instead of 'lapply' gives a nicer input.
The following works.
$ sapply(charlist, function(y) paste(',y,', sep=))
a b c
'a' 'b' 'c'
--
Karl Ove Hufthammer
That's currently not supported (because no one up to now had wanted to
do it).
Andy
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Monday, June 02, 2008 4:21 PM
To: r-help@r-project.org
Subject: [R] Random Forests
Hi All,
I'm by no means an expert on anything related to M$ products... However
I was recently forced (reluctantly at that) to put together a PPT
presentation which included multiple R graphics. So I thought I'd share
what I found to produce decent looking graphics.
I created all the graphics
Karl Ove Hufthammer:
Use 'sep='. And using 'sapply' instead of 'lapply' gives a nicer input.
The following works.
$ sapply(charlist, function(y) paste(',y,', sep=))
a b c
'a' 'b' 'c'
But a simpler solution is to use sQuote(). Note that by default this may use
directional (‘curly’)
Many thanks.
Is there a way to give me the number of the row, if I have the row name?
B.
mel-10 wrote:
Birgitle a écrit :
I have a dist object containing 1 row that is only NA (not very
intelligent
to have bas dataset with one NA speciesanyway).
I would like to delete this row
On Thu, 5 Jun 2008, John Poulsen wrote:
Hello,
I have been using mgcv to run GAM hurdle models, analyzing presence/absence
data with GAM logistic regressions, and then analyzing the data conditional
on presence (e.g. without samples with no zeros) with GAMs with a negative
binomial
Birgitle a écrit :
I have a dist object containing 1 row that is only NA (not very intelligent
to have bas dataset with one NA speciesanyway).
I would like to delete this row from this object.
newDistObject = distObject[-unwantedRow, ]
hih
__
Additonally I got this error message
TestDist = Dist.HalbDisGow88[-147,]
Fehler in Dist.HalbDisGow88[-147, ] : falsche Anzahl von Dimensionen
(Error in Dist.HalbDisGow88[-147, ] : wrong number of dimensions
Birgit
mel-10 wrote:
Birgitle a écrit :
I have a dist object containing 1 row
I have (below) an attempt at an R script to find the limit distribution
of
a continuous-time Markov process, using the formulae outlined at
http://www.uwm.edu/~ziyu/ctc.pdf, page 5.
First, is there a better exposition of a practical algorithm for doing
this? I have not found an R package that
.. or identical(sort(unique(a)),sort(unique(b)))
-- Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of jim holtman
Sent: Thursday, June 05, 2008 5:01 AM
To: Karin Lagesen
Cc: r-help@r-project.org
Subject: Re:
Dear List,
I'm wondering whether it is possible to use R to make spherical plots.
I have 3d data: azimuth, elevation and a certain variable Y.
I want to plot Y in terms of azimuth and elevation such that it seems to be a
contour plot overlaid on a sphere (but projected on a plane, of
Hello All,
I wanted to perform a fourier transform on high frequency financial data. I
have searched and have not found much on this topic for R. I was wondering
if anyone has used any libraries for it or have come across any papers I may
read.
Many Thanks,
Neil Gupta
[[alternative
Have still some problems
match(name, names(object))
[1] NA
which(names(object)==Iname)
integer(0)
TestDist = object[-name,]
Fehler in -name : ungültiges Argument für unären Operator
Error in -name: invalid argument for unären (don`t know what this means)
operator
Thanks
Birgit
mel-10
Birgitle a écrit :
Many thanks.
Is there a way to give me the number of the row, if I have the row name?
B.
a= object
'w' = name
match('w', names(a))
# or
which(names(a)=='w')
# or
but deletion should work with the number and/or with the name.
I've been trying to create a list of function where function[[i]] should
actually return the value of i.
trying:
func - vector(list,2)
i - 1
while (i = 2)
{
func[[i]] - function()
{
i
}
i - i+1
}
however only returned the last value of i for each function. Any help how to
achieve the solution
If you want power spectral density then spec.pgram() and a wrapper to this
spectrum should do the trick (MASS which is a part of R base I think)
add the argument log=no if you don't want to see the power on a log scale
also remember that the smoothed periodogram is a consitent estimator of the
Is this what you want:
func - vector(list,2)
i - 1
while (i = 2)
+ {
+ func[[i]] - local({
+ z - i
+ function(){ z }
+ })
+ i - i+1
+ }
func[[1]]()
[1] 1
func[[2]]()
[1] 2
Your solution was using 'i' which picked up the last definition of 'i'.
'local' create a
Hoping someone can help me with some graphics. Here is my data set.
A B
1
5
1
5
1
6
2
7
3
7
4
7
5
8
6
9
6
10
7 11
Using this code, I split the data into two groups, each of size 5, and made a
scatterplot.
Bill-read.table('something
Hi,
Does anybody know how to install the
Arfima 1.01. from OX? They have mentioned it in fARMA package in R but im not
succeeded to install it yet.
please help me if some one knows about it.
nbsp;
Thanks
[[alternative HTML version deleted]]
hi, hi all,
There are several methods to analyse two (e.g. coinertia, procuste)
or K tables (e.g. statis, mfa, gpa, etc.) in the packages ade4, FactoMiner, etc.
Otherwise analyses on instrumental variables can be an interesting way to
explore your dataset (e.g. within/between analyses, etc.)
I am developing an application that uses Eclipse framework on Windows.
I have to call R from Java. I downloaded rJava package and installed
it. I set the classpath in the Run Dialog in Eclipse to the directory
where the JRI.jar is. I set the Path environmental variable to the
directory where
On 6/5/2008 10:50 AM, Andreas Posch wrote:
I've been trying to create a list of function where function[[i]] should
actually return the value of i.
trying:
func - vector(list,2)
i - 1
while (i = 2)
{
func[[i]] - function()
{
i
}
i - i+1
}
however only returned the last value of i for each
I also use R PDF graphic files; however, note that they can be
imported directly into Word, at least on my version of Word which is
Word 2007 on Windows Vista SP1, using Insert | Object
(_not_ Insert | Picture) eliminating the intermediate step.
On Thu, Jun 5, 2008 at 9:56 AM, Soukup, Mat [EMAIL
-Original Message-
From: [EMAIL PROTECTED] on behalf of Neil Gupta
Sent: Thu 6/5/2008 15:21
To: R-help@r-project.org
Subject: [R] Fourier Transform
Hello All,
I wanted to perform a fourier transform on high frequency financial data. I
have searched and have not found much on this topic
Good morning,
I am a new R user and I am trying to learn how to use it.
I am trying to solve this problem.
I have a dataframe df of daily securities (for a year) earnings as
follows:
SEC_ID DAY EARNING
IT001 200701015.467
IT001 20070102
Check out the three vignettes (i.e. pdf documents in the zoo package). e.g.
Lines - SEC_ID DAY EARNING
IT001 200701015.467
IT001 200701025.456
IT001 200701034.954
IT001 200701043.456
IT002
Replace cov(z) with cov(z, use = pair) is there are missing values
as there are here.
On Thu, Jun 5, 2008 at 11:54 AM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Check out the three vignettes (i.e. pdf documents in the zoo package). e.g.
Lines - SEC_ID DAY EARNING
I would start by creating a matrix that held the
returns with rows being the dates and columns
being the securities. You can do this by something
along the lines of:
days - as.character(df[, 'DAY'])
sec - as.character(df[, 'SEC_ID']
earningmat - array(NA, c(length(unique(days)),
I'm in need of a version of R that will run on fedora 9 and haven't been able
to find one. I need this in order to run Bioconductor. Any advice?
--
View this message in context:
http://www.nabble.com/Need-to-run-R-on-Fedora-9.-tp17674187p17674187.html
Sent from the R help mailing list archive
The native *.dll can't be found for any reason.
For a first start simply put the native lib in your Eclipse Java project.
If you are developing a plugin or a RCP application
a must read is this article:
http://www.eclipsezone.com/articles/eclipse-vms/
explaining how to load and place native
On 06/05/08 09:28, RobertsLRRI wrote:
I'm in need of a version of R that will run on fedora 9 and haven't been able
to find one. I need this in order to run Bioconductor. Any advice?
Why not say (as root)
yum install R
?
Worked for me. The point is that there is now a Fedora rpm for R.
2008/5/27 Mihalicza Péter [EMAIL PROTECTED]:
Dear List and Hadley,
I would like to have a boxplot with ggplot2 and have the outlier values
labelled with their name attribute. So I did
library(ggplot2)
dat=data.frame(num=rep(1,20), val=c(runif(18),3,3.5),
name=letters[1:20])
On Wed, Jun 4, 2008 at 2:03 PM, Thompson, David (MNR)
[EMAIL PROTECTED] wrote:
Hello,
A few questions about the following examples:
1. Why do the two plotting versions not produce the same result?
Because version one has two layers, and version two has one?
2. Is the 'scale_x_continuous'
I have looked at math plot and still can not figure out how to get a degree
symbol into the y label. I would like degrees C
thanks
Stephen
--
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for us is puff us up and make us
Sorry, my last post came out unreadable. I'll try again. Here is the data.
Think of them as transposed columns. A: 1,1,1,2,3,4,5,6,6, 7B:
5,5,6,7,7,7,8,9,10,11 Split the data into two groups, each of size 5, and
make a scatterplot. Bill-read.table('something here')
Try:
RSiteSearch(plotmath degree character)
On Thu, Jun 5, 2008 at 1:18 PM, stephen sefick [EMAIL PROTECTED] wrote:
I have looked at math plot and still can not figure out how to get a degree
symbol into the y label. I would like degrees C
thanks
Stephen
--
Let's not spend our time and
Try this:
d - dist(USArrests)
ix - which(Iowa == labels(d))
d1 - as.dist(as.matrix(d)[-ix, -ix])
On Thu, Jun 5, 2008 at 9:39 AM, Birgitle [EMAIL PROTECTED] wrote:
I have a dist object containing 1 row that is only NA (not very intelligent
to have bas dataset with one NA speciesanyway).
Does this do what you want?
Bill - data.frame(A= c(1,1,1,2,3,4,5,6,6,7),
B=c(5,5,6,7,7,7,8,9,10,11),
group=rep(1:2, each=5) )
with(Bill, plot(A,B, pch=group) )
mns - with(Bill, tapply(B, group, mean))
tmp - par('usr') # or range(Bill$A)
med - with(Bill, mean(
See,
http://tolstoy.newcastle.edu.au/R/help/06/05/27989.html
HTH,
Chuck
On Thu, 5 Jun 2008, biologeeks wrote:
dear all,
in the package pwr , there is the fonction power.anova.test which permit to
obtain the power for a one-way ANOVA...but I'm looking for a way to compute
the
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I have a data set on quarterly institutional
ownership of equities; some of them have had recent IPOs, some have not
(I have a binary flag set). The total dataset size is 700k+ rows.
My goal is this:
On Thu, 5 Jun 2008, [EMAIL PROTECTED] wrote:
I have (below) an attempt at an R script to find the limit distribution
of
a continuous-time Markov process, using the formulae outlined at
http://www.uwm.edu/~ziyu/ctc.pdf, page 5.
First, is there a better exposition of a practical algorithm for
Negative indexing is often handy, but I'm in need of an appropriate
idiom
for handling cases in which the index set can be null:
x - rnorm(5)
a - 1:5
s - rep(FALSE,5)
y - x[-a[s]]
# I'd like y == x but instead one has x[-a[s]] == x[a[s]] ==
numeric(0), which is rather
# unfortunate --
Thanx Hadley,
More questions inline.
-Original Message-
From: hadley wickham [mailto:[EMAIL PROTECTED]
Sent: June 5, 2008 01:09 PM
To: Thompson, David (MNR)
Cc: r-help@r-project.org
Subject: Re: [R] ggplot questions
On Wed, Jun 4, 2008 at 2:03 PM, Thompson, David (MNR)
[EMAIL
Maybe you should provide a minimal, working code with data, so that we all
can give it a try.
In the mean time: take a look at the Rprof function to see where your code
can be improved.
Good luck
Bart
Daniel Folkinshteyn-2 wrote:
Hi everyone!
I have a question about data processing
On 6/06/2008, at 1:08 AM, biologeeks wrote:
dear all,
in the package pwr , there is the fonction power.anova.test which
permit to
obtain the power for a one-way ANOVA...but I'm looking for a way to
compute
the power of a multiway ANOVA.( find the 1-beta). Is it possible?
do you have
Here are a few possibilities. The first three
use seq_along and the last two use if and length:
x[ !seq_along(x) %in% a[s] ]
x[ !match(seq_along(x), a[s], 0) ]
x[ setdiff(seq_along(x), a[s]) ]
if (length(a[s])) x[-a[s]] else x
x[ if (length(a[s])) -a[s] else TRUE ]
On Thu, Jun 5, 2008 at
I have a question.
I have a data set (about 100,000 observations). How would I get the
distribution curve graph? This is like, if I use hist(x, freq=TRUE,
breaks=1000) to get the histogram, now the question is, I don't need the
histogram itself, I just need the curve that connects the top of
I managed to use an example (see attachment) of clever regression
routines. I customized it to suit my needs.
The initial model I try to fit consists of the first 10 powers of time
(time the observation was recorded) and the first 10 powers of the
phase. In fact my files record patients'
I have a multivariate data set, and I would like to get the S-estimates of
the scatter (robust estimates of variance-covariance matrix). Which
library/function should I use?
Thank you very much!
--
View this message in context:
Dear R-Users,
Is there a way to create a matrix of data frames in R ?
I am pulling in data frames from a SQL database into R using RODBC. I
would like to pull in a sequence of data frames which are indexed across
two dimensions, and store each data frame as an element in a matrix.
I did try
See ?density
On Thu, Jun 5, 2008 at 4:56 PM, rlearner309 [EMAIL PROTECTED] wrote:
I have a question.
I have a data set (about 100,000 observations). How would I get the
distribution curve graph? This is like, if I use hist(x, freq=TRUE,
breaks=1000) to get the histogram, now the question
would a density plot do? try
plot(density(x))
if you are specifically after the histogram tops rather than a density
estimate, then get the hist object with plot=F, then look at the counts
attribute:
histobj = hist(x, freq=TRUE, breaks=1000, plot=F)
plot(histobj$counts)
hope this helps.
on
I've tried to tackle a similar question at the request of a coworker.
Unfortunately, it is difficult to read in HTML code because it lacks
character that can consistently be used as a delimiter. The only
guideline I can offer is that any text you're interested in is going to
be between a and a .
On Thu, Jun 5, 2008 at 4:22 PM, [EMAIL PROTECTED] wrote:
Dear R-Users,
Is there a way to create a matrix of data frames in R ?
Using builtin BOD and women:
m - matrix(list(BOD, women, women, BOD), 2)
m[[1,2]] # women
height weight
1 58115
2 59117
3 60120
4
Thanks, I'll take a look at Rprof... but I think what i'm missing is
facility with R idiom to get around the looping, and no amount of
profiling will help me with that :)
also, full working code is provided in my original post (see toward the
bottom).
on 06/05/2008 03:43 PM bartjoosen said
i know this is an R mailing list :) but... i'll recommend you try python
with the beautifulsoup module - makes html processing a cinch.
another thing to note is that wunderground provides very handy RSS feeds
for every location, so rather than parsing the html page (with it's
associated
Staying in R, the XML package in conjunction with the XPATH query
language is likely to be your friend.
library(XML)
html=htmlTreeParse(http://www.wunderground.com/global/stations/16239.html;,
useInternal=TRUE)
xpathApply(html, //[EMAIL PROTECTED]'tempf' and
+@pwsid='LIRA']/@value,
Hallo,
which function can i use to do (baseline) logistic regression +
goodness
of fit tests?
so far i found:
# logistic on binary data
lrm combined with resid(model,'gof')
# logistic on binary
allFits - lmList(y ~ t|id, data=table1, pool=FALSE)
allCoefs - sapply(allFits, coef) ## preferred by me
or
allCoefs - list(length(allFits))
for(i in 1:length(allFits)) allCoef[[i]] - coef(allFits[[i]])
--Matt
From: Manli Yan [mailto:[EMAIL PROTECTED]
Sent:
Apologies, the second method should have been
allCoefs - vector(list, length(allFits))
for(i in 1:length(allFits)) allCoefs[[i]] - coef(allFits[[i]])
--Matt
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Austin, Matt
Sent: Thursday, June 05, 2008
I just discovered what seems to me to be a slight funny in respect
of formal argument names. If I define a function
foo - function(a,b){ ... whatever ...}
then ``inside'' foo() the exists() function will return TRUE
from ``exists(a) whether an object named ``a'' exists or not.
But
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