Janet Choate-2 wrote:
Thanx for clarification on stating my problem, Charlie.
I am attempting to merge to files, i.e.:
hi39 = merge(comb[,c(hillID,geo)], hi.h39, by=c(hillID))
if this is relevant or helps to explain:
the file 'comb' is 3 columns and 1127 rows
the file 'hi.h39' is 5
Dear List Subscribers,
I am working on the following problem and was wondering if
there is some command or set of commands to solve it:
Thank you in advance,
Eric
1. The dataset Cancer0 may have multiple dates of treatment (DATE) for
each patient (ID) with a given disease (SITE).
This should do the trick:
data$date - as.Date(data$date, '%m/%d/%Y')
data$month - format(data$date, '%Y-%m')
by(data$rammday, data$month, sum)
Hope that helps,
Jason
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Stephan
Hi,
The code below creates a value, x$a, which depending on how you access it
evaluates to its initial value, or to what it's been changed to. The last
two lines should, I would have thought, evaluate to the same value, but they
don't.
f - function () {
x - NULL;
x$a - 0;
Have you tried
plot(pic)
-Original Message-
From: Berwin A Turlach ber...@maths.uwa.edu.au
Sent: Wednesday, March 31, 2010 10:15 PM
To: James Rome jamesr...@gmail.com
Cc: r-help@r-project.org r-help@r-project.org
Subject: Re: [R] pdf files in loops
On Wed, 31 Mar 2010 22:06:48 -0400
Jeff Brown wrote:
Hi,
The code below creates a value, x$a, which depending on how you access it
evaluates to its initial value, or to what it's been changed to. The last
two lines should, I would have thought, evaluate to the same value, but they
don't.
f - function () {
x -
Hi,
It might be too late but I'll answer anyway.
First, don't forget to follow the posting guide and provide some sample
data, your code, and/or exactly what you're looking for. And as Jannis
said, search also the archive.
Here are some thoughts about what you can do (if I understand you
Thank you both for your advice. I'll follow up on it, but it is good
to know that this is a known effect.
Claus
On Wed, Mar 31, 2010 at 3:02 PM, Stephan Kolassa stephan.kola...@gmx.de wrote:
Hi Claus,
welcome to the wonderful world of collinearity (or multicollinearity, as
some call it)! You
Jeff Brown wrote:
Sorry for spamming. I swear I had worked on that problem a long time before
posting.
But I just figured it out: I have to change the values, which are
represented as integers, not strings. So the following code will do it:
df - data.frame (
a = factor( c( bob,
Hi,
I want to produce some rules (for example with JRip, M5Rules, OneR, PART,
ect.) with R and then work with produced rules.
Does anyone know how to work with the rules?
Regards,
fascob
--
View this message in context:
http://n4.nabble.com/Rule-Induction-tp1747834p1747834.html
Sent from
James Rome wrote:
On 3/31/2010 10:01 PM, Berwin A Turlach wrote:
G'day James,
On Wed, 31 Mar 2010 21:44:31 -0400
James Rome jamesr...@gmail.com wrote:
I need to make a bunch of PDF files of histograms.
[...]
What am I doing wrong?
Many thanks! The missing empty line was the problem, and M-q works great.
Hans-Peter
Esc-q or M-q will wrap a whole paragraph. The only problem is that you need
to make sure the paragraph you want to wrap is separated from other
__
Hi R users:
I found that I cannot stack() a data.frame with factors.
db1-data.frame(replicate(6,factor(sample(c(A,B),6,replace=TRUE
str(db1)
db2-stack(db1)
db2
db2 does not have any row.
How can I stack them by the variables X1,X2,...,X6?
Thank you for your help.
Kenneth
Kenneth Roy Cabrera Torres krcabrer at une.net.co writes:
Hi R users:
I found that I cannot stack() a data.frame with factors.
db1-data.frame(replicate(6,factor(sample(c(A,B),6,replace=TRUE
str(db1)
db2-stack(db1)
db2
db2 does not have any row.
How can I stack them by the variables
Hi Roslina
In order to get the X coordinates of your barplot, assign the barplot to a
variable:
x.coor - barplot(something)
Now use the x.coor for your line plot.
Cheers,
Tal
Contact
Details:---
Contact me:
Dear all,
I am new R user and I am sure that this question has been asked quite often
and I have also googled it and read about it! I understood that in order to
read excel sheet into R you need to open it and saved it as csv or text, is
this true? or you can use read.delim2 and read.csv2 to do
On 2010-04-01 3:53, Ken Knoblauch wrote:
Kenneth Roy Cabrera Torreskrcabrerat une.net.co writes:
Hi R users:
I found that I cannot stack() a data.frame with factors.
db1-data.frame(replicate(6,factor(sample(c(A,B),6,replace=TRUE
str(db1)
db2-stack(db1)
db2
db2 does not have any row.
How
Dear Cheba
Please, install the package xlsReadWrite
I suppose that read.csv is to read csv files, not xls ones.
cheers
milton
On Thu, Apr 1, 2010 at 6:19 AM, cheba meier cheba.me...@googlemail.comwrote:
Dear all,
I am new R user and I am sure that this question has been asked quite often
Hello,
I am trying to give different colors to boxes in a violin plot obtained
via bwplot from lattice package using a color palette from RColorBrewer:
require(RColorBrewer)
MyPalette - brewer.pal(6, Set3)
A call to:
boxplot(count ~ spray, data = InsectSprays, col = MyPalette)
yields
On Thu, Apr 1, 2010 at 3:05 AM, Peter Dalgaard pda...@gmail.com wrote:
Jeff Brown wrote:
Sorry for spamming. I swear I had worked on that problem a long time before
posting.
But I just figured it out: I have to change the values, which are
represented as integers, not strings. So the
See this link for a number of alternatives:
http://rwiki.sciviews.org/doku.php?id=tips:data-io:ms_windows
On Thu, Apr 1, 2010 at 6:19 AM, cheba meier cheba.me...@googlemail.com wrote:
Dear all,
I am new R user and I am sure that this question has been asked quite often
and I have also
Hello,
I'm using rpart function for creating regression trees.
now how to measure the fitness of regression tree???
thanks n Regards,
Vibha
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Thank you. Unfortunately this recommendation does not solve my problem and I
don't know why. Here is my test-code:
pfad-paste(C:/...,xls,sep=.)
xl - COMCreate(Excel.Application)
xl[[Visible]] - FALSE
wkbk - xl$Workbooks()$Open(pfad)
sh - xl$ActiveSheet()
A3R -
Thank you guys, both solutions work great! Seems I have two new
packages to investigate :)
Regards,
Tony Breyal
On 31 Mar, 14:20, Tony B tony.bre...@googlemail.com wrote:
Dear all,
Lets say I have the following:
x - c(Eve: Going to try something new today..., Adam: Hey @Eve, how are
you
vibha patel wrote:
Hello,
I'm using rpart function for creating regression trees.
now how to measure the fitness of regression tree???
thanks n Regards,
Vibha
If the sample size is less than 20,000, assume that the tree is a
somewhat arbitrary representation of the relationships in the data
Try this;
On Thu, Apr 1, 2010 at 7:00 AM, koj jens.k...@gmx.li wrote:
Thank you. Unfortunately this recommendation does not solve my problem and I
don't know why. Here is my test-code:
pfad-paste(C:/...,xls,sep=.)
xl - COMCreate(Excel.Application)
xl[[Visible]] - FALSE
Hi,
I have a dichotomous variable (Q1) whose answers are Yes or
No.
Also I have 2 categorical explanatory variables (V1 and V2)
with two levels each.
I used logistic regression to determine whether there is an
effect of V1, V2 or an interaction between them.
I used the R and SAS, just for
Hi Silvano,
this is FAQ 7.17
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-does-the-output-from-anova_0028_0029-depend-on-the-order-of-factors-in-the-model_003f
hth.
Silvano schrieb:
Hi,
I have a dichotomous variable (Q1) whose answers are Yes or No.
Also I have 2 categorical explanatory
On Apr 1, 2010, at 8:19 AM, Silvano wrote:
Hi,
I have a dichotomous variable (Q1) whose answers are Yes or No.
Also I have 2 categorical explanatory variables (V1 and V2) with two
levels each.
I used logistic regression to determine whether there is an effect
of V1, V2 or an interaction
Please be more precise.
You seem not to use the rcom (note the lowercase name) package,
but the RDCOMClient package which is not available from CRAN
but as part of the Omegahat project.
rcom has a command comCreateObject and RDCOMClient has a command COMCreate.
The code you supplied will
Try this also:
df[] - as.numeric(unlist(df))
df
On Thu, Apr 1, 2010 at 2:53 AM, Jeff Brown dopethatwantsc...@yahoo.com wrote:
Sorry for spamming. I swear I had worked on that problem a long time before
posting.
But I just figured it out: I have to change the values, which are
represented
Thank you. I figured that out myself last night. I always forget that
read.table does not actually read data into a matrix.
GTM MatLab toolbox comes with a nice guide to use the package which may as
well become an R vignette.
Anyway, I got the singular matrix warnings myself and do not know
Hi,
I need to generate the time series of the production, but as I'm new to this
topic I am not able to do that. This is what the time series should be:
PROD(t)=PROD(t,T)
PROD(t-1)=PROD(t-1,T)
PROD(t-2)=PROD(t-1)*PROD(t-2,T-1)/PROD(t-1,T-1)
PROD(t-3)=PROD(t-2)*PROD(t-3,T-2)/PROD(t-2,T-2)
...
...
I'd like to access data in my R session from elsewhere via HTTP. My
particular use case would be R on Linux, accessing data from Windows,
all on the same intranet.
Naively, when I say this, I imagine something like:
theData - big.calculation.returning.data.frame()
startHttpServer(port=8675)
Hello all,
I am trying to do factorial regression using lm() like this (example):
model-lm(y ~ x1 + x2 + x3 + x4 + x1*x2*x3*x4)
The final term 'x1*x2*x3*x4' adds all possible interactions between
explanatory variables to the model. i.e. x1:x2, x1:x2:x3, etc, etc. Now, the
issue is that some of
I am having a total brain fart... complete and total. This is part R, Part
Statstitics, Part My Brain is on vacation apparently.
Ok I have a time series I need to LOG and DIFF for ARIMA with Regressors.
Say 100 data points.
Obviously when I diff the series once I get 98 data points now. So what
Hi,
I would like to know if there is some debugger in R where I can check that I
am not using or not doing calculation on my previously stored objects. I
can't use rm (list = ls()) to remove all the objects since I am using a for
loop for reading 500 files and making making common calculation for
Hi Parthiban,
I urge you to rethink your approach, or at least proceed with extreme
caution. Lower-order terms involved in higher-order interactions may not be
what you think they are. And there are serious problems with stepwise model
selection.
I encourage you to read a good regression
Look at the 1st example on the help page for the updateusr function in the
TeachingDemos package for a way to align a barplot and information added to it.
See the axis function for creating your own custom axis.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
I have a line graph that is flat and has sharp peaks (mass spectrometry
data). I would like to plot my data and label the peaks with their x-axis
value. I know it is possible to have a function to find local maxima and
then to use these values in a label command but I am hoping there is a
Hi Roslina,
here is a simple example:
barplot.x.location - barplot(c(12:17))
points(x = barplot.x.location, y = rep(10, length(aa)), col = 2, pch = 7)
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Hi Ayush,
Could you supply with a simple code to try to give an answer on ?
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
Rob,
Please look again at Romain's reply to you on 19th March. He informed you
then that Rcpp has its own dedicated mailing list and he gave you the link.
Matthew
R_help Help rhelp...@gmail.com wrote in message
news:ad1ead5f1003291753p68d6ed52q572940f13e1c0...@mail.gmail.com...
Hi,
I'm a
I am drawing a density histogram, and want to label the plots with the
mean using ltext(). But I need the x,y coordinates to feed into ltext,
and I can't calculate them easily from my data. Is there a way to get
the x and y ranges being used for the plot, so I can put the text at the
correct
Dear R users,
I have problem with bagging survial tree after finding the final tree.
f-rpart(Surv(time ,dead )~ x1 +x2+ x3+x4+x5+x6, data=crp)
f.prun-prune(f,cp=0.036701)
# final tree i have 3 endnote including
#the variable x3 and x4
how can i use the bagging code i use
like this but it
I have a simple problem: I need to load a ERSI shapefile of US states
and check whether or not a set of points are within the boundary of
these states. I have the shapefile, I have the coordinates but I'm
having a great deal of difficulty bringing the two together. The
problem is the various GIS
Snippet of my code:
library(foreign)
function1 - function(y,t){
###do some matrix operations ##
}
function2 - function(y){
y1 = permute(y)
F1 = function1(y1)
}
setwd(C:\\Results\\) ## Read Multiple Files
files.total = list.files()
for (j in files.total){
table1 = read.table(j)
### do
Thanks Philippe,
I will look at sciviews - I have never used it before.
I use Rcmdr, and I find I am often selecting and running the same lines of
code. I just thought if we could name them then easily call that region it
would be useful.
Cheers
--
View this message in context:
The thread has been handled in Rcpp-devel. Rob posted there 7 minutes
after posting on r-help.
FWIW, I think the problem is fixed on the Rcpp 0.7.11 version (on cran
incoming)
Romain
Le 01/04/10 17:47, Matthew Dowle a écrit :
Rob,
Please look again at Romain's reply to you on 19th March.
One way I can think of is putting your loop inside a function (let's say
func3) and then use:
library(debug)
mtrace(func3)
func3()
# and this at the end:
mtrace.off()
And see the steps...
Tal
Contact
Details:---
Contact me:
I have code that creates the same matrices , a and b. The code creating
a is not in a function and the code creating b is in a function. I would
like to do operations on b like I can on a, however I can't figure out
how I can return a matrix (or data frame) from a function. Thanks for your
help!
On Apr 1, 2010, at 9:56 AM, n.via...@libero.it wrote:
Hi,
I need to generate the time series of the production, but as I'm new
to this
topic I am not able to do that. This is what the time series should
be:
PROD(t)=PROD(t,T)
PROD(t-1)=PROD(t-1,T)
On 2010-04-01 9:53, James Rome wrote:
I am drawing a density histogram, and want to label the plots with the
mean using ltext(). But I need the x,y coordinates to feed into ltext,
and I can't calculate them easily from my data. Is there a way to get
the x and y ranges being used for the plot, so
On Apr 1, 2010, at 11:53 AM, James Rome wrote:
I am drawing a density histogram, and want to label the plots with the
mean using ltext(). But I need the x,y coordinates to feed into ltext,
and I can't calculate them easily from my data. Is there a way to get
the x and y ranges being used for
He could have posted into this thread then at the time to say that.
Otherwise it appears like its open.
Romain Francois romain.franc...@dbmail.com wrote in message
news:4bb4c4b8.2030...@dbmail.com...
The thread has been handled in Rcpp-devel. Rob posted there 7 minutes
after posting on r-help.
On Apr 1, 2010, at 12:16 PM, Greg Gilbert wrote:
I have code that creates the same matrices , a and b. The code
creating
a is not in a function and the code creating b is in a function.
I would
like to do operations on b like I can on a, however I can't
figure out
how I can return a
On Wed, 31 Mar 2010, Parminder Mankoo wrote:
Hello All,
Does anyone know why length(fit1$time) length(fit2$n) in survfit.coxph
output? Why is the predicted time length is not the same as the number of
samples (n)?
In fact it is not true that length(fit1$time) length(fit2$n), since
Ashley,
This appears to be your first post to this list. Welcome to R. Over 2 days
is quite a long time to wait though, so you are unlikely to get a reply now.
Feedback: since nlrq is in package quantreg, its a question about a package
and should
be sent to the package maintainer. Some
Since Frank has made this somewhat cryptic remark (sample size 20K)
several times now, perhaps I can add a few words of (what I hope is) further
clarification.
Despite any claims to the contrary, **all** statistical (i.e. empirical)
modeling procedures are just data interpolators: that is, all
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Jeff Brown
Sent: Wednesday, March 31, 2010 6:45 PM
To: r-help@r-project.org
Subject: [R] Scope and assignment: baffling
Hi,
The code below creates a value, x$a, which
Incidentally, there is nothing new or radical in this; indeed, John Tukey,
Leo Breiman, George Box, and others wrote eloquently about this decades ago.
And Breiman's random forest modeling procedure explicitly abandoned efforts
to build simply interpretable models (from which one might infer
The code you presented is very close to object oriented in the style
of the proto package. For example,
library(proto)
# generate a single object p
p - proto(a = 0,
geta = function(.) .$a,
incra = function(.) .$a - .$a + 5)
p$geta()
p$a # same
p$incra()
p$geta()
# Or if you
Thanks Dennis for the thorough explanation and correction on the design.
John
--- On Thu, 4/1/10, Dennis Murphy djmu...@gmail.com wrote:
From: Dennis Murphy djmu...@gmail.com
Subject: Re: [R] trying to understand lme() results
To: array chip arrayprof...@yahoo.com
Cc: R-help@r-project.org
Date:
Hi,
I have a time series problem that I would like some help with if you have
the time. I have many data from many sites that look like this:
Site.1
datetimelevel temp
2009/10/01 00:01:52.0 2.8797 18.401
2009/10/01 00:16:52.0 2.8769
Dear fellows,
I need a R script that compare known curves (e.g. logistic, exponential)
with my curve. That curve was generated fitting data of forest cover
(hectares) measured in 10 road distances (buffers).
I´d like that comparison should be done using AICc to select the best model,
that is,
You can adapt the following example
da - data.frame(y=rnorm(50), x=1:50)
plot(y~x, data=da)
abline(h=c(-2,2), lty=3)
with(da, text(x[abs(y)2], y[abs(y)2], label=x[abs(y)2], pos=2))
Walmes.
-
..ooo0
Dear all,
I have a data.frame like below and I need
to plot horizonal with error bar only for upper limit.
On the code below I am able to plot the bars within
groups, but I need (1) change from vertical to horizonal
plot and (2) add the error bar.
any hint are welcome.
milton
Wow, those are much more elegant. Thanks!
Peter suggests:
df[] - lapply(df, factor, levels=allLevels, labels=seq_along(allLevels))
Henrique suggests:
df[] - as.numeric(unlist(df))
In both of those cases, why is the [] needed? When I evaluate df vs. df[],
they both look the same, but
Simon
I need a R script that compare known curves (e.g. logistic, exponential)
with my curve. That curve was generated fitting data of forest cover
(hectares) measured in 10 road distances (buffers).
I´d like that comparison should be done using AICc to select the best model,
that is, the
On 31.03.2010 10:21, (Ted Harding) wrote:
On 30-Mar-10 23:23:09, Jim Lemon wrote:
Hi all,
The gurus may pour scorn on me for not knowing this, but I happened
to mistype if as is in the heat of debugging a function. As I
scanned the debugged function with some satisfaction, I noticed the
On 30.03.2010 23:00, R P Herrold wrote:
On Tue, 30 Mar 2010, Prof Brian Ripley wrote:
A hint: I have seen this exact error message with a build configured
to use the system's zlib 1.2.4 (which has been out for about 2 weeks),
and this incompatibility is noted in the current R 2.11.0 alpha's
On 30.03.2010 20:57, David Winsemius wrote:
On Mar 30, 2010, at 12:42 PM, Driss Agramelal wrote:
## Hello everyone,
##
## I am trying to execute 150 times a lm regression using the 'for' loop,
with 150 vectors for y,
##
## and always the same vector for x.
##
## I have an object with 150
I need a two sample t.test between M and F. The data are arranged in one
column, x. Cant seem to figure how to run a two sample t.test. Not really
sure what this output is giving me, but there should be no difference
between M and F in the example, but summary p-value indicates this.
How can I
I'm currently doing a lot of simple GIS work in R, including points
in polygon queries. My .Rprofile file has
require(maptools)
require(rgdal)
With that as a starting point, I find that the data structures play
well together.
Define a coordinate reference system object with
crs.ll -
I am trying to learn R, and am having problems with doing a simple linear
regression. I loaded data from a fixed width file, using wd=c(...), and
read.fwf(...) and I can read in the file ok and it comes in as vectors in
columns, which is what I expected. The problem is when I try to do a linear
On Thu, 1 Apr 2010, Uwe Ligges wrote:
herrold, before:
The true fault causing the noted message seems to be that there is a
faulty compression/decompression occuring in a carried library --
would't it be better to not bundle a frozen library, and to rather
simply 'flag' packages needed a
Hello,
Marlin Keith Cox wrote:
I need a two sample t.test between M and F. The data are arranged in one
column, x. Cant seem to figure how to run a two sample t.test. Not really
sure what this output is giving me, but there should be no difference
between M and F in the example, but summary
Marlin -
Consider the following:
df = data.frame(x=rep(1:4,2),y=rep(c('M','F'),c(2,2)))
t.test(x~y,data=df)
Welch Two Sample t-test
data: x by y
t = 4.899, df = 6, p-value = 0.002714
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence
On 01/04/2010 2:59 PM, Bruce Kindseth wrote:
I am trying to learn R, and am having problems with doing a simple linear
regression. I loaded data from a fixed width file, using wd=c(...), and
read.fwf(...) and I can read in the file ok and it comes in as vectors in
columns, which is what I
Bruce,
You don't tell us what class of data your y is. Assuming y is defined
in your enviroment, what does
class(y)
and
str(y)
tell you? You'll most likely have to fine-tune your data reading
process, or do some post-processing to make sure the y object (and x for
that matter) are the
Thanks Walmes,
that does indeed work well for labeling all points greater than some
specified value. But the problem is that while my peaks are very sharp there
is more than one point along the line as it slopes up. This method will
label those points as well , making the data look cluttered.
Hi Everyone,
I am using R 2.10.1. lmer function works properly, however pvals.fnc
() does not despite the fact that I uploaded:
- library(lme4)
- library(coda)
- library(languageR)
This is the error message I get
pvals.fnc(lexdec3.lmerE2, nsim=1)$fixed
Error in
Good comments Bert. Just 2 points to add: People rely a lot on the tree
structure found by recursive partitioning, so the structure needs to be
stable. This requires a huge samples size. Second, recursive
partitioning is not competitive with other methods in terms of
predictive
Hi,
I have text (see below) that is aligned nicely when printed in the
command window in R but when plotted using text function, it does not
show alignment unless I use the family=mono in the call to the text
function. Is there a way to specify a different font while maintaining
the alignment?
On 01/04/2010 3:39 PM, Tighiouart, Hocine wrote:
Hi,
I have text (see below) that is aligned nicely when printed in the
command window in R but when plotted using text function, it does not
show alignment unless I use the family=mono in the call to the text
function. Is there a way to specify a
Suppose I have a binary outcome (disease/no disease and all subjects had the
same period of exposure) and 2 or 3 (categorical) predictors.
I can obviously build a logistic regression model which describes the data,
possibly including interaction terms, on a relative scale:
You can set the number of extreme points to be labeled instead of define a
cutoff. Look:
da - data.frame(y=rnorm(50), x=1:50)
plot(y~x, data=da)
abline(h=c(-2,2), lty=3)
with(da, text(x[abs(y)2], y[abs(y)2], label=x[abs(y)2], pos=2))
da - da[order(da$y),]
plot(y~x, data=da)
# five small and big
ozge gurcanli gurcanli at cogsci.jhu.edu writes:
Hi Everyone,
I am using R 2.10.1. lmer function works properly, however pvals.fnc
() does not despite the fact that I uploaded:
- library(lme4)
- library(coda)
- library(languageR)
This is the error message I get
thanks again walmes. but the new problem would be that not all of the peaks
are the same intensity. therefore the top five datapoints from my highest
peak have greater intensity values than the highest point in the
second-highest peak.
but this is once again helpful. i found that there is a
Hi,
One option with Grid graphics,
m -
matrix(c( 1667,3,459,
2001, 45, 34,
1996, 2,5235),
dimnames=list(c(Eric Alan, Alan,John David)),
ncol=3, byrow=T)
## install.packages(gridExtra, repos=http://R-Forge.R-project.org;)
library(gridExtra)
XLSolutions is proud to announce our April 2010 R/S-PLUS Advanced
Programming course in USA
*** R/S Systems: Advanced Programming
*** S/R-PLUS Programming 3: Advanced Techniques and Efficiencies.
More on website
http://www.xlsolutions-corp.com/rplus.asp
Ask for group discount and reserve
The discussion of Leo Breiman's paper in Statistical Science: Statistical
Modeling - The Two cultures, is a must read for all statisticians doing
prediction modeling. Especially see the exchange between Cox and Breiman (I
call this the Cox-Breiman duel).
Ravi.
I would like to revisit a problem that was discussed previously (see
quoted discussion below). I am trying to do the same thing, using a
string to indicate a column with the same name. I am making foo a
string taken from a list of names. It matches the row where item =
5, and picks the
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Euan Reavie wrote:
I would like to revisit a problem that was discussed previously (see
quoted discussion below). I am trying to do the same thing,
I meant to add that I'm guessing from this:
What it seems to be doing is converting the text
Aulacoseira_islandica to a number (25, for some reason
is that foo is a factor, but we have no way of knowing without a
reproducible example.
Euan Reavie wrote:
I would like to revisit a problem
Dear R users,
i'm using a custom function to fit ancova models to a dataset. The data are
divided into 12 groups, with one dependent variable and one covariate. When
plotting the data, i'd like to add separate regression lines for each group
(so, 12 lines, each with their respective individual
Thank you very much. Since I have never heard of blotter before, now I am
really excited. It seems exactly what I have been searching. Would be really
grateful if you could share some codes/examples regarding to this. I did not
happen to find the help file for the package.
Thanks again.
--
A link to a collection of tutorials and videos on R.
Tutorials: http://www.dataminingtools.net/browsetutorials.php?tag=rdmt
Videos: http://www.dataminingtools.net/videos.php?id=8
__
R-help@r-project.org mailing list
Thank you for your reply Matthew,
There are many things I could say about the myriad difficulties I have had
in progressing with R, none worth stating here. For the record I have read
the posting guidelines, at least to the extent possible in the time
available.
For what its worth, I think I
Perhaps something like this:
library(zoo)
library(chron)
# read in data
Lines1 - datetimelevel temp
2009/10/01 00:01:52.0 2.8797 18.401
2009/10/01 00:16:52.0 2.8769 18.382
2009/10/01 00:31:52.0 2.8708 18.309
2009/10/01 00:46:52.0
1 - 100 of 110 matches
Mail list logo