Dear R-helpers,
In the following example I compute ret and returns the SAME way. In ret I
use compute returns for EACH column and in returns I do it for the whole
data frame. Could someone please tell me why I see a lagged result,by which
I mean ret and returns are different by one lag.
This is all in the help page for pdf ... including a request not to
falsely blame R and two workarounds for the broken viewers.
On Thu, 28 Jul 2011, Duncan Murdoch wrote:
On 11-07-28 1:22 PM, selwyn quan wrote:
Hi,
Am using R 2.13.1 on Linux (Fedora). Is anybody else having problems with
On Thu, Jul 28, 2011 at 10:54 PM, Jonathan Callahan
jonat...@mazamascience.com wrote:
One of their requirements (hard or soft?) is that each module in the
framework be compiled with statically linked libraries into a standalone
binary that can be run from the command line.
But if they insist
hello,
i'm searching the r-coding to plot wilcoxon signed test. i already searched
everywhere for a month, but i didn't found it. Could anyone help me, please.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
sq == selwyn quan selw...@stanford.edu
on Thu, 28 Jul 2011 15:59:14 -0700 writes:
sq On Fri, 29 Jul 2011, Rolf Turner wrote:
On 29/07/11 09:51, selwyn quan wrote:
On Thu, 28 Jul 2011, Duncan Murdoch wrote:
On 11-07-28 1:22 PM, selwyn quan wrote:
Hi,
thank you for the reply! I'm sorry I added so little information.
I thought DateHour was a character class but it was a factor. So I converted
to character and order worked as you described.
For this time I only wanted to convert DateHour to a number because I
thought I had to do that in
The package xgrid 0.1-11 is now available on CRAN. The package provides
functions to distribute and collate results from
simulation studies and other computationally expensive tasks to
Apple Xgrid clusters from within R.
More specifically, the routines within the package facilitate access to
Thanks for a patient reply...
I feel really stupid about this other problem dput(). What I didn't
understand is how I can make the same thing with my data. You use rnorm and
rpois which generates random poisson ditribution and normal distribution
purhaps just to create a data.frame? Should I
Good morning to all,
I am encountering a blocking issue when using the function 'breackpoints'
from package 'strucchange'.
*Context:*
I use a data frame, 248 observations of 5 variables, no NA.
I compute a linear model, as y~x1+...+x4
x4 is a dummy variable (0 or 1).
I want to check this model
(oops, forgot to cc. the list)
On Jul 29, 2011, at 08:09 , Ashim Kapoor wrote:
Dear R-helpers,
In the following example I compute ret and returns the SAME way. In ret I
use compute returns for EACH column and in returns I do it for the whole
data frame. Could someone please tell me why I
Hi:
To add to Peter Dalgaard's comments, a look at head(ret) and
head(returns) would also have clued you in:
head(ret, 1)
GOOG.OpenGOOG.High GOOG.Low GOOG.Close GOOG.Volume
2007-01-03 1.377067e-05 3.184271e-05 3.378641e-05 7.049545e-05 3.01405e-09
Dear all,
Quite often I have the situation that I've multiple response variables
and I create Linear Models for them in a function. The following code
illustrates my usual approach:
---8---
set.seed(123)
dat - data.frame(x = rep(rep(1:3, each = 3), 4), y = rep(1:3, 12))
Am Donnerstag, den 28.07.2011, 09:12 +0200 schrieb Martin Maechler:
PM == Paul Menzel paulepan...@users.sourceforge.net
on Wed, 27 Jul 2011 23:53:51 +0200 writes:
[…]
PM So my newest suggestion is to add a comment to
PM plot(sin, -pi, 2*pi)
PM in `?plot`. Like
Dear Bruce,
It's doable with ggplot2, but image() is probably a better solution.
To use ggplot2, you will need to convert your array into a data.frame where
each row has the information for one cell (x, y and colour)
library(ggplot2)
#create some dummy data
dataset - expand.grid(x =
Michel:
I am encountering a blocking issue when using the function 'breackpoints'
from package 'strucchange'.
*Context:*
I use a data frame, 248 observations of 5 variables, no NA.
I compute a linear model, as y~x1+...+x4
x4 is a dummy variable (0 or 1).
I want to check this model for
Hi,
I got two data point vectors. Now I want to make a ks.test(). I you print
both vectors you will see, that they fit pretty fine. Here is a picture:
http://www.jochen-bauer.net/downloads/kstest-r-help-list-plot.png
As you can see there is one histogram and moreover there is the gumbel
On Thu, 28 Jul 2011, seanstcl...@verizon.net wrote:
I am running the ctree function in R.
My data has about 10 variables, many of which are categorical. 2 of the
categorical variables have many levels (one has 900 levels, another has
1,000 levels). As an example, 1 of these
David Winsemius dwinsemius at comcast.net writes:
On Jul 28, 2011, at 1:07 PM, Hans W Borchers wrote:
maaariiianne marianne.zeyringer at ec.europa.eu writes:
Dear R community!
I am new to R and would be very grateful for any kind of help. I am
a PhD
student and need to fit a
Achim,
Thank you so much for this prompt answer. Really appreciated !
Anyway, I am still a bit lost... don't you mind if I ask you somme
additional questions?
* *one standard approach is to employ a HACcovariance matrix*
I did many researches but I never found this recommendation. The only
Im starting to use the GMM estimator with panel data in R. I´ve read the
document «Panel Data Econometrics in R: The plm Package» (Croissant and
Millo).
In Stata before using the functions lag() or diff() we must sort the data by
individual and by time. I would like to know if I have to do
On Jul 29, 2011, at 3:18 AM, Ungku Akashah wrote:
hello,
i'm searching the r-coding to plot wilcoxon signed test. i already
searched everywhere for a month, but i didn't found it.
You need to work on your searching skills. It should have taken a few
seconds:
#
??wilcoxon (Only four
Dear All
I would like to produce interaction boxplots and this seems to work:
par(mfrow=c(2,2))
A=sample(rnorm(50,50,10))
B=sample(rnorm(50,100,10))
Test=merge(A,B,by=0)#by=0 where 0 is the row.names
TreatA=(gl(2,50,100,labels=c(High,Low)))
TreatB=rep(gl(2,25,50,labels=c(High,Low)),2)
Hi
Dear All
I would like to produce interaction boxplots and this seems to work:
par(mfrow=c(2,2))
A=sample(rnorm(50,50,10))
B=sample(rnorm(50,100,10))
Test=merge(A,B,by=0)#by=0 where 0 is the row.names
TreatA=(gl(2,50,100,labels=c(High,Low)))
Thank you David for your answer.
The lattice package was not a good example. The thing is the same errors occurs
with any package I try to install.
Thank you
Miguel
On Jul 28, 2011, at 4:31 PM, David Winsemius wrote:
On Jul 28, 2011, at 3:08 PM, Miguel Leal wrote:
I'm having problems
On Thu, Jul 28, 2011 at 6:54 PM, Eduardo M. A. M.Mendes
emammen...@gmail.com wrote:
Hello
Many thanks.
Sorry for not replying earlier but I have followed advice and read as much
documentation as I could.
I managed to get a zoo object the way I want
dadoszoo : 'zoo' series from 2010-06-27
Hello,
I would be very grateful if somebody more knowledgeable then me could assist
me in the following.
I have two (three actually but for simplicity I will say two) models which I
would like to fit jointly as a state space object. Here are the equations:
(1)
w = a1 + b1*(p) + e1
a1 = a1[t-1]
On Jul 28, 2011, at 6:56 PM, Alexander James Rickett wrote:
I'm trying to get ready to submit a package to CRAN, but in order for the
package to install on OS X, I need to temporarily set the environment
variable NOAWT=1. I put 'export NOAWT=1' in my package's 'configure' script,
and 'R
Hello dear subscribed Users,
this is my first post, so please forgive me for any inconveniences.
The following problem: I have a dataframe containing a factor column.
For each column i would like to compare means as parted by the factor.
Using the normal t.test function I have already achieved
Hi!
My name is Martin and I have a problem concerning the boxplot function
in R. I want my boxes to be limited by the 1st and 3rd quartile and NOT
the 'hinges' values that are the default setting in R.
Do anyone knows if there is any command that I could do to change this
default setting?
Hi,
Interpretation problem ! so what i did is by using the:
fit1 - fitdist(vectNorm,beta)
Warning messages:
1: In dbeta(x, shape1, shape2, log) : NaNs produced
2: In dbeta(x, shape1, shape2, log) : NaNs produced
3: In dbeta(x, shape1, shape2, log) : NaNs produced
4: In dbeta(x, shape1, shape2,
Hi,
I´m trying to move to ubuntu at all, but I don´t like R in Ubuntu because it
runs in the terminal, so i can´t access directly to some options R gots in
Windows.
Is there any R can i install in ubuntu which runs in hos own prompt or
terminal??
Thanks
-
Mario Garrido Escudero
PhD student
Thanks very much again,
I´m reading some papers and articles about this issue and I think i´m
starting to understand the problem.
And thanks for the link to Professor Fox about the non-sequential Anova.
I'll be back with more doubts. I'm sure of that.
-
Mario Garrido Escudero
PhD student
Dear all,
I want to work with R in an optimization problem and I was wondering if
there are any available solvers. I found RCplex package but downloading it,
I get problems. Why? Is it not free? are there any other solvers available?
many thanks!
--
View this message in context:
Hi,
I have a dataframe that I imported from a .txt file by:
skogTemp - read.delim2(Skogaryd_shoot_data.txt, header=TRUE, fill=TRUE)
and the data are factors, how can avoid factors from the beginning? Although
the file contains both characters and numbers.
I tried to convert some of the columns
Hello,
Is there a function to calculate TLS - Total least squares (orthogonal
regression) with R?
Thank you
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-calculate-Total-least-squares-in-R-Orthogonal-regression-tp3703863p3703863.html
Sent from the R help mailing list
Just a hunch I can't test from my phone, but in your final lapply you are
passing a function of x that has no x in it, so I wouldn't be surprised if R
was unhappy about that.
Change the latter M's to x and see if that helps.
Cheers
Michael Weylandt
On Jul 29, 2011, at 6:30 AM, Gunnar
struc.test - breakpoints(y~x1+x2+x3+x3+x4, data=D)
*I get an error message:*
Erreur dans chol2inv(qr.R(fm$qr)) :
l'?l?ment (5, 5) est nul, donc l'inverse ne peut ?tre calcul?
(sorry for the french version, I don't know how to get the message
english translation in R).
My first
On Jul 29, 2011, at 8:13 AM, Miguel Leal wrote:
Thank you David for your answer.
The lattice package was not a good example. The thing is the same
errors occurs with any package I try to install.
Read all of the message, this time for meaning. You appear to have
ignored my suggestion
Hi Hadley,
Thanks for the idea, but this was a typo...
My code is the following:
D - data.frame(CPU=pred.cor2$CPU, PREP=PREP, BRG=BIZ$JOBPREPLOTRULE_BRG,
CLOG=res.WIP, WE=DUMMY)
model.mes - CPU~PREP+BRG+CLOG+WE
stab.model - Fstats(model.mes, data = D, from = 0.1,
vcov = function(x,
Hi:
Here are a couple of attempts, one with base graphics and one with lattice.
d - data.frame(TreatA = gl(2, 50, 100, labels = c('High', 'Low')),
TreatB = rep(gl(2, 25, 50, labels = c('High', 'Low')), 2),
Test = c(rnorm(50, 50, 10), rnorm(50, 100, 10)))
lbl -
On Jul 29, 2011, at 8:20 AM, Dail wrote:
Hello,
Is there a function to calculate TLS - Total least squares (orthogonal
regression) with R?
Searching the archives should bring up plenty of on target advice.
--
David Winsemius, MD
West Hartford, CT
Dear R folks,
wanting to compare different implementations of a solution I want to
script it to iterate over the different implementations. Is there a way
to do this in the R shell/command line?
$ more /tmp/iterf.r
f1 - function(n = 10,
l = 10)
On Fri, Jul 29, 2011 at 10:03 AM, gaiarrido gaiarr...@usal.es wrote:
Hi,
I´m trying to move to ubuntu at all, but I don´t like R in Ubuntu because
it
runs in the terminal, so i can´t access directly to some options R gots in
Windows.
Is there any R can i install in ubuntu which runs in hos
David Winsemius dwinsemius at comcast.net writes:
On Jul 29, 2011, at 8:20 AM, Dail wrote:
Is there a function to calculate TLS - Total least squares (orthogonal
regression) with R?
Searching the archives should bring up plenty of on target advice.
Also, please don't cross-post to
On Jul 29, 2011, at 9:11 AM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
Just a hunch I can't test from my phone, but in your final lapply
you are passing a function of x that has no x in it, so I wouldn't
be surprised if R was unhappy about that.
Change the latter M's to x
On Jul 29, 2011, at 8:05 AM, Petr PIKAL wrote:
Hi
Dear All
I would like to produce interaction boxplots and this seems to work:
par(mfrow=c(2,2))
A=sample(rnorm(50,50,10))
B=sample(rnorm(50,100,10))
Test=merge(A,B,by=0)#by=0 where 0 is the row.names
Start here:
http://cran.r-project.org/web/views/Optimization.html
HTH,
Dennis
On Fri, Jul 29, 2011 at 1:29 AM, sasanthi chrys...@gmail.com wrote:
Dear all,
I want to work with R in an optimization problem and I was wondering if
there are any available solvers. I found RCplex package but
I wish place the following axis label in such a manner that some of the
text is plain and the scientific name is in italics (i.e. a mixture of
two font types)
Using plot:
mtext(Total Landings of Pecten maximus (tonnes),font,=3, side=2,
line=3)
makes everything italic, but how do I
Dear R-list,
I have a plot with y-axis corresponding to wind measurments
and x-axis with date-time information.
When I want to identify some extrem wind events in the
wind-curve, I use locator() to get the exact
date-information, by clicking in the points in graph I´m
interested in.
I get in the
Hi:
The argument stringsAsFactors = FALSE will suppress character data
from being converted to factors. For more control, see the arguments
colClasses and as.is in ?read.table - they should carry over to
read.delim[2] but don't quote me on that :)
HTH,
Dennis
On Fri, Jul 29, 2011 at 1:45 AM,
Hi Paul,
how about this
for (i in 1:2) { print( do.call(paste(f,i,sep=),list(2, 3) )) }
or using get
for (i in 1:2) { print( get(paste(f,i,sep=))(2, 3) ) }
cheers
Am 29.07.2011 15:28, schrieb Paul Menzel:
Dear R folks,
wanting to compare different implementations of a solution I want to
Martin,
I don't know of an easy way to make that change, but do note:
The two ‘hinges’ are versions of the first and third quartile,
i.e., close to ‘quantile(x, c(1,3)/4)’. The hinges equal the
quartiles for odd n (where ‘n - length(x)’) and differ for even
n. Whereas the
On Jul 29, 2011, at 9:28 AM, Paul Menzel wrote:
Dear R folks,
wanting to compare different implementations of a solution I want to
script it to iterate over the different implementations. Is there a
way
to do this in the R shell/command line?
You might consider examining the short and
Dear Eik,
Am Freitag, den 29.07.2011, 15:46 +0200 schrieb Eik Vettorazzi:
how about this
for (i in 1:2) { print( do.call(paste(f,i,sep=),list(2, 3) )) }
or using get
for (i in 1:2) { print( get(paste(f,i,sep=))(2, 3) ) }
works great for me. Thank you very much. Regarding search machines
Hi Philip,
have a look at ?plotmath and try
mtext(expression(paste(Total Landings of~~italic(Pecten
maximus),(tonnes))), side=2,line=2)
~~ is used for extra space between plain and italic font, but that
might be a matter of taste.
hth.
Am 29.07.2011 15:34, schrieb Philip Boulcott:
I wish
I'm using rioja 0.5-6 with R 2.12.1 / x86_64-pc-linux-gnu on
Kubuntu 11.04.
When I run this example from the rioja reference manual:
library(rioja)
# compare diatom data from core from Round Loch of Glenhead
# with SWAP surface sample dataset
data(RLGH)
data(SWAP)
result -
Thanks very much,
and...Which do you recommend?
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola
Universidad de Salamanca
--
View this message in context:
http://r.789695.n4.nabble.com/R-in-Linux-Ubuntu-tp3703329p3704107.html
On Jul 29, 2011, at 10:10 AM, Eik Vettorazzi wrote:
Hi Philip,
have a look at ?plotmath and try
mtext(expression(paste(Total Landings of~~italic(Pecten
maximus),(tonnes))), side=2,line=2)
~~ is used for extra space between plain and italic font, but that
might be a matter of taste.
And the
Hi Martin,
As Sarah said, I do not know of any way to change the default, but
you can certainly do it manually each time. Here is an example:
## generate some data
set.seed(10)
x - rnorm(101)
## store boxplot results
s - boxplot(x, plot = FALSE)
## look at them
s
## replace the stats with lower
Dear Mario,
Have a look at Eclipse with the StatET plugin. One of the benefits is that is
available for both Linux and Windows which simplifies things when you need to
switch often between Windows and Linux (e.g. Windows at work and Linux at home).
Best regards,
Thierry
On Fri, Jul 29, 2011 at 3:49 PM, gaiarrido gaiarr...@usal.es wrote:
Thanks very much,
and...Which do you recommend?
I am using emacs + ESS - but as you have mentioned options, I assume you
would like to e.g. manage package installation from the GUI - which you can
do with both mentioned GUIs.
Can someone help me out with a small problem?
I've started using netcdf files recently, and I want to extract the grid id
and also the coordinates from a HIRHAM netcdf file.
I know how to extract a slice of dataset both in space and in time and I
also know the area that this file should cover,
Well,if you would have had a look at ?POSIXct the documentation would
have given
you quite a comprehensive explanation of the meaning of the values of
your time vector as they are the values these POSIXct values are
actually stored in (they are seconds since 1970-01-01). Additionally you
Mark,
The R-help list is not telepathic. I asked entirely reasonable follow-up
questions intended to elicit from you a more extensive explanation of
your problem.
On Fri, Jul 29, 2011 at 10:26 AM, m...@statcourse.com wrote:
1. I did not receive your reply, which, not incidentally, does not
Dear list,
I want to plot a sample depth curve over a barplot. It would be perfect if
the argument inside in the barplot function would be functional, cause I
could just add this curve to the actual barplot, but it seems like it is not
(not yet implemented). Argument inside would allow not to
On Jul 29, 2011, at 10:44 AM, Ana wrote:
Can someone help me out with a small problem?
I've started using netcdf files recently, and I want to extract the
grid id
and also the coordinates from a HIRHAM netcdf file.
I know how to extract a slice of dataset both in space and in time
and I
I am not familiar with the HIRHAM netcdf files. Do you have the netcdf library
installed and can you print the output of the ncdump -h command on one of the
files?
-Roy
On Jul 29, 2011, at 7:44 AM, Ana wrote:
Can someone help me out with a small problem?
I've started using netcdf files
This problem may be more related to netcdf files than to RAnd the
answer to your question very much depends on which netcdf convention
the AWI folks who created your data followed. Additionally it depends on
the Software you use to read these ncdf files.
Regarding the grid locations
Thanks to several who responded--I should have looked (doh!) in the
documentation. play3d() and movie3d() were just the ticket!
On 7/28/2011 4:48 PM, Jean V Adams wrote:
Dale,
I do not have the same color problem when I run your code on my PC.
The colors in both devices look the same.
On Fri, Jul 29, 2011 at 7:44 AM, Ana rrast...@gmail.com wrote:
Can someone help me out with a small problem?
I've started using netcdf files recently, and I want to extract the grid id
and also the coordinates from a HIRHAM netcdf file.
I know how to extract a slice of dataset both in space
Dear Collin,
as always, a reproducible example code would help us to understand what
you want to do. This way I can only guess
And my guess would be that it is much easier to use:
par(new=TRUE)
and to superimpose the barplot with whatever curve you want to include.
You may need to set
Hi Roy and David,
My main problem is how to interpret the coordinate system.
If I try to open it directly in a GIS software that supports necdf
format, the location is far away from where it should be (plots
rlong,rlat).
All I want to do is to get a data.frame with a column for long,
another for
Hi Ana:
On Jul 29, 2011, at 8:03 AM, Ana wrote:
netcdf precip.DMI.HS1 {
dimensions:
rlat = 84 ;
rlon = 90 ;
time = 10800 ;
time_bnds = 2 ;
variables:
float lat(rlat, rlon) ;
lat:standard_name = latitude ;
lat:long_name = latitude
à data=read.table(http://statcourse.com/research/boston.csv;, , sep=,,
header = TRUE)
à library(rpart)
à fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT)
Please: Show me the tree.
Mark
Original Message
Subject: Re: [R] help
Well, the CF ncdf convention (for example) suggests to use variables
with identical names as the dimensions as coordinate variables. In your
case the dimensions are called rlat/rlong so the software uses these
rotated axes. In your file you however also have variables lat and long
and they
Thanks Jannis,
I am including an example code of what I am trying to do:
par(mar=c(4,5,3,1))
barplot(Value[,1],space=0, col=grey30,axis.lty=2)
barplot(sampledepth[,1]/2, space=0, col=grey30, inside=FALSE,
add=TRUE)
sampledepth is divided by two to have the same axis scale as Value
float lat(rlat, rlon) ;
lat:standard_name = latitude ;
lat:long_name = latitude ;
lat:units = degrees_north ;
float lon(rlat, rlon) ;
lon:standard_name = longitude ;
lon:long_name = longitude ;
Ah, now we're getting somewhere. Think how easy this would be if
you'd provided this information initially.
On Fri, Jul 29, 2011 at 11:11 AM, m...@statcourse.com wrote:
Ø data=read.table(http://statcourse.com/research/boston.csv;, , sep=,,
header = TRUE)
Ø library(rpart)
Ø fit=rpart (MV~
Mark,
The below is not directly reproducible. There is no MV or PT
variable in the dataset you reference. I am assuming you meant:
dat - read.table(http://statcourse.com/research/boston.csv;, ,
sep=,, header = TRUE)
fit - rpart(MEDV ~ CRIM + ZN + INDUS + CHAS + NOX + RM + AGE + DIS +
RAD +
data=read.table(http://statcourse.com/research/boston.csv;, ,
sep=,, header = TRUE)
library(rpart)
fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+
PT+B+LSTAT)
predict(fit,data[4,])
plot only reveals part of the tree in contrast to the results on obtains
Hi,
I have a question about a special recursive filter problem.
What I have:
- given variables:
x: time series with rather randomly occuring '0' and '1'
wait: non negative integer
- a working but ineffectiv implementation (see below)
How the implementation works (what I want):
The filter
Hi all,
table() did the trick, and very efficiently, too! Thanks for the
advice,
Dave
On Thu, Jul 28, 2011 at 5:39 PM, David Winsemius dwinsem...@comcast.netwrote:
On Jul 28, 2011, at 4:24 PM, David Warren wrote:
Hi all,
I'm working with a sizable dataset that I'd like to
I understand, that eta^2 presents a useful measure of effect sizes.
However, the power-calculation-tool G*Power I intend to use, requires an
Effect size f which is calculated like this: Variance explained by
special effect / Variance explained by special effect + Variance within
groups
I run a
Your plot only shows part of the tree provided by print(fit).
I would like to plot the entire tree.
Original Message
Subject: Re: [R] help with plot.rpart
From: Joshua Wiley [1]jwiley.ps...@gmail.com
Date: Fri, July 29, 2011 8:25 am
To:
Dear colleagues,
I'm using R64 (2.13) on Mac OS 10.6.8 and I've encountered a problem with the
recode function in Rcommander. The application cannot deal with apostrpohes (
' ) do not. I've got a factor from the 2008 Canada Election study (highest
level of schooling) and some of the values
The rlat and rlon variables are the grid coordinate system, which
appears to be some non-standard conical projection. Climate
scientists!
Actually this file, as it contents state, follow the CF convention, which is
widely used in climate science, oceanography and atmospheric science
Hi! Thank you all. I finally understood the way to solve the problem
from Barry's mail.
From the file created with coordinates(Lat,Long) using fan, everything
looks ok (wgs84). I will now try Bryan's approach in R.
Ana
On Fri, Jul 29, 2011 at 4:22 PM, Barry Rowlingson
On Fri, Jul 29, 2011 at 11:26 AM, m...@statcourse.com wrote:
data=read.table(http://statcourse.com/research/boston.csv;, ,
sep=,, header = TRUE)
library(rpart)
fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+
PT+B+LSTAT)
predict(fit,data[4,])
plot
On Fri, Jul 29, 2011 at 4:32 PM, Roy Mendelssohn
roy.mendelss...@noaa.gov wrote:
Actually this file, as it contents state, follow the CF convention, which is
widely used in climate science, oceanography and atmospheric science (see
http://cf-pcmdi.llnl.gov/documents/cf-conventions). The
I'm not sure I understand what your filter intends to do, but could this not
be done more efficiently with logicals and the which? You also might need
the cumsum() function and diff() with the optional lag argument if I've
misunderstood your filter.
Specifically try this:
res =
Thank you for all your help. The unexpected solution to my predict()
problem was to use the Misc menu and remove all objects before
proceeding further. The final digits of the trees continue to be
clipped but I can live with that.
Mark
__
Hello!
Hope you could help me split the strings.
I have a set of strings:
x-c(name_a1_2.5.o,name_a2_2.53.o,name_a3_bla_1.o)
I need to extract from each string:
1. Its unique part that comes before the last _, i.e.: a1,a2,a3_bla.
2. The part that comes after the last _ and before .o at the end,
Hi, everyone.
I need to run lm with the same response vector but with varying predictor
vectors. (i.e. 1 response vector on each individual 6,000 predictor vectors)
After looking through the R archive, I found roughly 3 methods that has been
suggested.
Unfortunately, I need to run this task
On Jul 29, 2011, at 11:32 AM, Simon Kiss wrote:
Dear colleagues,
I'm using R64 (2.13) on Mac OS 10.6.8 and I've encountered a problem
with the recode function in Rcommander.
It's probably in a package named 'car'.
library(car)
?recode
The application cannot deal with apostrpohes ( ' )
I wasn't at my normal computer yesterday, so I didn't run the example. I
thought rgl was one of those color palette generator packages. So, my
suggestion of using animation was completely off base (oops).
But I did notice two other things:
1) your color column changes from a factor to a
Oh darn, I missed the recursive-ness entirely. You condition on the filtered
series, not the signal itself.
In that case, I have a solution which is pretty fast, but not particularly
R-esque.
In effect your filter just says, take x but if you see a 1, sit out for the
next wait periods. This
Hi Janis,
thank you for you answer.
Actually already I did exactly what you said. However, when I convert
the number of seconds from the first measurement (some day in July
2010) then a I get a date in the year 2050 which is impossible. The
latest date possible is 29March2011.
my value in
On Jul 29, 2011, at 12:21 PM, cristabel.duran wrote:
Hi Janis,
thank you for you answer.
Actually already I did exactly what you said. However, when I
convert the number of seconds from the first measurement (some day
in July 2010) then a I get a date in the year 2050 which is
try this:
x-c(name_a1_2.5.o,name_a2_2.53.o,name_a3_bla_1.o)
strsplit(sub(.*?([^_]+)_([^-]+).o$, \\1 \\2, x), ' ')
[[1]]
[1] a1 2.5
[[2]]
[1] a2 2.53
[[3]]
[1] bla 1
On Fri, Jul 29, 2011 at 12:08 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
Hope you could help
Hi,
If you are doing repeated fits in this specialized case, you can
benefit by skipping some of the fancy overhead and going straight to
lm.fit.
## what you had
data.residuals - t(apply(predictors, 1, function(x)( lm(regress.y ~
-1 + as.vector(x))$residuals)))
## passing the matrices directly
1 - 100 of 146 matches
Mail list logo
