Hello,
You can find some hints there:
http://geography.uoregon.edu/geogr/topics/index.html
Regards
Le 12/06/28 14:26, iverson a écrit :
Hi guys
i need some help to build choropleth.
Basically i am trying to colour regions on the map by population.
I possess the shape file of the country, a
I believe chambers book has an example using S4 classes. which isnt much
help for a beginner.
you can do it the old school way. build it by hand with one vector of
pointers and another data structure of leaf nodes. there is nothing magical
about a tree. you can build one in basic or fortran. painf
Hi Kat,
On Thu, Jun 28, 2012 at 8:22 AM, kat_the_great wrote:
> Dear R Users:
>
> I'm a STATA user converting to R, and I'd like to be to do the following.
> #Assign var_1 and var_2 a value
> 10->var1
> 20->var2
>
> #Now I'd like to print the values of var_1 and var_2 by looping through
> var_1 a
On Thu, 28-Jun-2012 at 06:27PM +1200, Patrick Connolly wrote:
|> On Tue, 26-Jun-2012 at 11:57PM -0500, Erin Hodgess wrote:
|>
|> |> Hello again:
|> |>
|> |> Here is a solution to the dates without leading zeros:
|> |>
[...]
|>
|> |> This is not particularly elegant, but it does the trick.
|>
Hi guys
i need some help to build choropleth.
Basically i am trying to colour regions on the map by population.
I possess the shape file of the country, and also the population data,
however, i am having trouble to create the plot, below is my code:
population=read.csv("nz2.csv")
population
n
Hi All,
I have a whole lot of *.raw files in my working folder and I am doing the same
analysis on each of those and want to save all the results in a single file. I
am making some mistake here and can't figure out how to solve it.
Say, the *.raw files are ABCD.raw, EFGH.raw, IJKL.raw ...
Th
On Tue, 26-Jun-2012 at 11:57PM -0500, Erin Hodgess wrote:
|> Hello again:
|>
|> Here is a solution to the dates without leading zeros:
|>
|> pou1 <- function(x) {
|> #Note: x is a data frame
|> #Assume that Column 1 has the date
|> #Column 2 has station
|> #Column 3 has min
Hi David,
I believe you left out saying what color scheme you would prefer, so
it's hard to help, but you might be a start by looking at the
appropriate parts of Hadley's website here: http://had.co.nz/ggplot2/,
particularly the "scales" section.
Best,
Michael
On Wed, Jun 27, 2012 at 11:07 AM, D
Dear all!
Thanks for clarification.
OV
To: Rolf Turner
Sent: Wednesday, June 27, 2012 1:33 PM
Subject: Re: [R] chisq.test
Hi Rolf,
Thanks for spotting the mistake.
A.K.
- Original Message -
From: Rolf Turner
.org>
Sent: Wednesday, June 2
You can use main = unique(d$Subject) to solve this problem.
HTH,
b.
On 27 June 2012 08:49, Marcel Curlin wrote:
> Well at this point I have what I need (rough plot for data exploration) but
> the simplicity of the first approach is quite elegant and it has become a
> learning project. I have su
Hello all,
Please consider the following
library(lattice)
Colors. <-rep(brewer.pal(7, "Dark2"),2)
color <- 1
Data.X.. <- data.frame(UnitArea = c(rnorm(1000), rnorm(1000)), Type =
c(rep("Base",1000),rep("Log",1000)))
histogram( ~ UnitArea | Type, data =
Hello,
I have R script my_script.R . It loads a large input file "data.txt" and
then outputs a large matrix "out". (the command of running it on cluster
using my.cmd is put at the end)
Now I have 30 different input files "data1.txt", "data2.txt" ... and
"data30.txt" and want to generate and save
Dear R-help,
I am writing some simulation code to create multiple sets of time-to-event
clinical trial data (for use in meta-analysis). Within each trial, I want to
apply censoring via simulation of uniform variables (with minimum zero and
maximum the median outcome time for that particular trial)
take a look at
library(zoo)
example(plot.zoo)
which shows one way to do this,
Best,
Michael
On Wed, Jun 27, 2012 at 6:50 PM, denissearchundia
wrote:
> hi
>
> i try to do a graph who shows 2 time series at the same time
>
> thanks!
>
> --
> View this message in context:
> http://r.789695.n4.n
Hello,
I am running cluster analysis, and am attempting to create a graph of my
clusters. I keep on getting an error that says that my figure margins are too
large.
d <- file.choose()
d <- read.csv(d,header=TRUE)
mydataS <- scale(d, center = TRUE, scale=TRUE)
#Converts mydataS from a
How do you want your output? If you want a character vector of
lowercase letters, just use gsub("[^::A-Z::]","", t) to substitute ""
for all non-uppercase characters. strsplit() can lead you to a vector
of the single-letter upper case letters.
Best,
Michael
On Wed, Jun 27, 2012 at 2:15 PM, mdvaan
Dear R Users:
I'm a STATA user converting to R, and I'd like to be to do the following.
#Assign var_1 and var_2 a value
10->var1
20->var2
#Now I'd like to print the values of var_1 and var_2 by looping through
var_1 and var_2 in such a manner:
while(y<3){
print(var_y)
y+1->y
}
In STATA, the
I need to look through a dataset with two factor variables, and depending
on certain criteria, create a new variable containing the data from one of
those other variables.
The problem is, R keeps making my new variable an integer and saving the
data as a 1 or 2 (I believe the levels of the factor)
Please help:
I am using qplot as below and want to specify a different color scheme for race
but dont know how, can someone show me.
Thanks in advance
Code and input file below:
library(ggplot2)
library(gridExtra)
d<-read.table("results", header=TRUE, fill=TRUE)
plot2<-qplot(X,Y,data=d
hi
i try to do a graph who shows 2 time series at the same time
thanks!
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___
A very important thing is home game or guest game. Why did I mention Almeria
- Sevilla earlier? Almeria are heroes at home. At least, they were in
2009/2010. They scored almost in every home game, no matter who is there at
the other end of it - Barcelona or Real Madrid.
http://www.oneclicksports.
t <- "TheWeatherIsVeryNice"
How do I extract the upper case letters? - TWIVN
Thanks!
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__
HI,
Try this:
dat1 <- read.table(text="
subject time.ms V3
1 1 stringA
1 12 stringB
1 22 stringC
2 1 stringB
2 14 stringC
2 25 stringA
", sep="",header=TRUE)
dat2<-aggregate(dat1$time.ms,list(dat1$subject),max)
colnames(dat2)<-c("subject","time.ms")
m
that's an old thread but checkout the colorout package - it is awesome and it
does exactly what you're asking for !
http://cran.r-project.org/web/packages/colorout/
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Hi,
I'm wondering if I can do this in R. I have a data set with questions
(x-value) and respondents (y-value). If the respondent answered the
question they get a 1 and if the respondent didn't answer the question they
have a 0. Is there anyway where I can graph something similar to a grid
where
How can I make a reference i this case? I want to make a reference to
'Artikel XXX'
For example
In The Artikel '' there is two tables.
..
Litterature
Artikel XXX
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Hi Gerrit
I did the following: access the function via
graphics:::sunflowerplot.formula, and then commented some lines to see,
where the problem lies. I guess that it is the chunk with "mf <- eval(m,
parent.frame())", but to be honest, I do not know why. My workaround is as
follows:
sunflowerplot
Hi,
I can see the mistake in the code as "$" before Y.
dat1<-as.factor(c("1/2/2011","1/4/2011","1/4/2011"))
dat1
[1] 1/2/2011 1/4/2011 1/4/2011
Levels: 1/2/2011 1/4/2011
as.Date(as.character(dat1), "%m/%d/%Y")
[1] "2011-01-02" "2011-01-04" "2011-01-04"
> as.Date(as.character(dat1),"%m/%d/$Y
I can make kurtosis but now skewness curve. How can I do that? I have used
the following code.
x<-(1:601-301)/50
x
dt3<-dt(x,3)
pt3<-pt(x,3)
# t distribution has mean 0; variance f/(f-2); skewness 0; excess kurtosis
6/(f-4)
dn01<-dnorm(x,0,1)
pn01<-pnorm(x,0,1)
dlogis<-dlogis(x,0,1)
plogis<-pl
Hello,
I seem to be having a similar problem, but with glmer models. Here,
model.avg() doesn't return anything but coefficient values:
fl19<-glmer( corrFLEDGE ~ INFECTION * rsLD * bin.age + (1 | year) + (1 |
RINGNO),data=corrmalaria,family=poisson)
fledglings<-dredge(fl19)
top.fledglings <- ge
Thanks Pablo for your answer, it was very insightful, but I guess I got
something wrong.
I formed a survey design as:
> library(survey)
> mydesign <- svydesign(ids=~vill_neigh_code+clust, strata=~strat,
> weights=~sweight, data=mydata)
where
strat: stratum (urban or (sub-county) rural).
clust:
Hi I am new with R
I Have to build a binary tree with R. I'm very confused was wondering if
anyone had any R sample code they would share.
Any bady can help me?
Bye
Giuseppe
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Hi List,
I am hoping some of you have experience with the spdep package.
I have had success with lm.morantest and lm.LMtests.
I am trying to run a spatial error model and I keep getting the following
error
Error in res[i, listw$neighbours[[i]]] <- listw$weights[[i]] :
NAs ar
Hi I am new with R
I Have to build a binary tree with R. I'm very confused was wondering if
anyone had any R sample code they would share. I've come across a lot of C++
code(nothing in R) and this is not helping.
Any bady can help me?
Bye
Giuseppe
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Hello
In (survfit(Surv(Time,Status)~1))
I want to have status=0 as Failure and status=1 as Censore.
Changing above formula to (survfit(Surv(Time,Status)~0)) doesnot help!!
What should i do?
Thank You.
_
Best Regards
Niloofar.Javanrouh
MSc Of BioStatistics
[[alternat
Hi Petr,
Appreciate your feedback and sorry for the delay in responding. The
following is the description of problem from start:-
We have a set of sensors in XY plane arranged in more or less a rectangular
grid and we know their (x,y) co-ordinate. Now these sensors send data and
from that data
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not show the R² of the regression line.
When I run
I would like to compare the incidence rates of three groups. They are
supposed to have different risks so I would like to test whether there is a
increasing trend in the incidence rates. Does R or any packages provide a
trend test for incidence rates? I checked epiR and epitools. It seems they
do n
Not being the expert I have looked at papers like 'simpleR { Using R for
Introductory Statistics' by John Verzani. Some sample Code is in these
papers. Looks like I have to search for ideas to code a programming
language like Java by taking input from R(r-java). Can R be used
directly to hit web u
why cann't you use 'try'?
you can get the size of the file with 'file.info' and then make a decision to
read or not.
Sent from my iPad
On Jun 27, 2012, at 23:02, Jie Tang wrote:
> hi ,
> I am reading a series of files by the command shown as below.
> cop_x_data<-read.table(flnm(i)
> ,skip=2,a
hi ,
I am reading a series of files by the command shown as below.
cop_x_data<-read.table(flnm(i)
,skip=2,allowEscapes=TRUE,blank.lines.skip=TRUE)
the first two line of the files are headfile and I skip them by skip=2.
and sometimes the data file is null and there is no any data
in the file except
Hello,
Probably you should try this (thanks Google):
> install.packages("Rbbg", repos = "http://r.findata.org";)
Regards
Le 27/06/2012 22:54, yoda55 a écrit :
I have the following error message when I try to install RBloomberg.
Les packages binaires téléchargés sont dans
C:\Users\b
On Wed, 27 Jun 2012, Cand2d wrote:
Hi to all,
I am trying to run breakpoints() on a fairly large sample (>10.000
observations). The process is very slow, any idea on how to speed this
up? I have tried the hpc="foreach" parameter, but this didn't work at
all when I tried to run it on a smalle
Well at this point I have what I need (rough plot for data exploration) but
the simplicity of the first approach is quite elegant and it has become a
learning project. I have succeeded in formatting the overall plot OK but
have not been able to solve the problem of titles or any kind of
label/legen
I have the following error message when I try to install RBloomberg.
Les packages binaires téléchargés sont dans
C:\Users\bloom\AppData\Local\Temp\RtmpktX4UK\downloaded_packages
Message d'avis :
packages ‘quantstrat’, ‘RBloomberg’, ‘rsproxy’, ‘VaR’ are not available (for
R version 2.15.1)
Hi to all,
I am trying to run breakpoints() on a fairly large sample (>10.000
observations). The process is very slow, any idea on how to speed this up? I
have tried the hpc="foreach" parameter, but this didn't work at all when I
tried to run it on a smaller sample.
breakpoints(x ~ x.l1 + x.l2
Hi,
I am trying to estimate a multivariate P-GARCH model for two factors x&y. I
have selected p-garch to study the leverage effects. Is there any toolkit in
R that can help me do this?
Thanks,
Andy
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Hello R users,
I have used the package adehabitat to create a brownian bridge for my
radio-tracked data from one animal (using the kernelbb function). I have worked
through page 17 of the manual using the code:
data <- kernelbb(x, sig1 = 6.23, sig2 = 58, grid = 100)
image(data)
This produces a
Here are 2 approaches:
Use logspline density estimates (logspline package) rather than kernel
density estimates, this can give you a function to pass to integrate
or other tools, the estimates may be a little different from the
kernel density estimates.
If you need to use kernel density estimates
On Wed, Jun 27, 2012 at 8:11 PM, jim holtman wrote:
> If you look, half of the time is spent in the 'findSubsets" function
> and the other half in determining where the differences are in the
> sets. Is there a faster way of doing what findSubsets does since it
> is the biggest time consumer. Th
On Wed, Jun 27, 2012 at 05:36:08PM +0300, Adrian Duşa wrote:
> Dear R-help list,
>
> Part of a program I wrote seem to take a significant amount of time,
> therefore I am looking for an alternative approach.
> In order to explain what is does:
>
> - the input is a sorted vector of integer numbers
Hello Xi,
Have you tried replacing the for loop by an apply construct, e.g., lapply or
sapply? In my experience these functions are more efficient than for. At any
rate, if you succeed with, say, lapply, there are some R packages that support
parallel processing versions. I believe the packa
I have the following matrix operation
A %*% B %*% A
Where these matrices have the following dimensions and class attributes.
> dim(A)
[1] 5764 5764
> class(A)
[1] "dgCMatrix"
attr(,"package")
[1] "Matrix"
> dim(B)
[1] 5764 5764
> class(B)
[1] "dgCMatrix"
attr(,"package")
[1] "Matrix"
Now, wh
If you look, half of the time is spent in the 'findSubsets" function
and the other half in determining where the differences are in the
sets. Is there a faster way of doing what findSubsets does since it
is the biggest time consumer. The setdiff might be speeded up by
using 'match'.
On Wed, Jun
Hi Jim,
On Wed, Jun 27, 2012 at 7:27 PM, jim holtman wrote:
> One place to start is to use Rprof to see where time is being spent.
> I used the sample you sent and this is what I got:
>
>
> 0 16.7 root
> 1. 16.2 system.time
> 2. . 16.1 testfoo
> 3. . . 16.1 setdiff
> 4. . . . 8.2 a
Hello,
No, the Ljung-Box test wouldn't be inappropriate in that case. First you
detrend the series and then test for serial independence. It's even
usual to do so. I would use the default values for lag and fitdf. But
use type="Ljung", the Box-Pierce test is nowadays seldom used in
pratice, i
One place to start is to use Rprof to see where time is being spent.
I used the sample you sent and this is what I got:
0 16.7 root
1. 16.2 system.time
2. . 16.1 testfoo
3. . . 16.1 setdiff
4. . . .8.2 as.vector
5. . . . .8.2 findSubsets
6. . . . . .6.4 increment
Or just avoid the formula version:
> with(iris, sunflowerplot(Sepal.Width, Sepal.Length, xlab="A"))
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
> From: r-help-
It seems to be a bug when you specify an x-label.
# No xlab= works fine, but uses variable names to label x and
# y axes
> sunflowerplot(Sepal.Length~Sepal.Width, data=iris)
# These all throw the error message
> sunflowerplot(Sepal.Length~Sepal.Width, data=iris, xlab="A")
> sunflowerplot(Sepal.L
On 2012-06-26 23:02, John wrote:
On Wed, 27 Jun 2012 16:58:29 +1200
Rolf Turner wrote:
On 27/06/12 08:54, arun wrote:
Hi,
The error is due to less than 5 observations in some cells.
NO, NO, NO It's not the observations that matter, it is
the ***EXPECTED COUNTS***. These
am trying to produce two dot plot figures in ggplot2. So far the first one
(p) in the program below is working fine.
However when I want to move to a faceted plot (p1) I seem to lose my ordering
or, more likely, I'm just getting an ordering I am not expecting and I always
have trouble under
How about
> dates <- as.factor(c("1/2/2011", "1/4/2011", "1/4/2011", "1/4/2011",
"1/6/2011", "1/7/2011", "1/8/2011", "1/9/2011", "1/10/2011"))
> dates
[1] 1/2/2011 1/4/2011 1/4/2011 1/4/2011 1/6/2011 1/7/2011 1/8/2011
[8] 1/9/2011 1/10/2011
Levels: 1/10/2011 1/2/2011 1/4/2011 1/6/2011 1/7/
On Jun 27, 2012, at 6:58 AM, Mohan Radhakrishnan wrote:
Homework ? You mean the question is at a high level.
I use R. It is just the simulation part that I am investigating.
Several
papers on Capacity Planning that I am working on are useful for
analyzing log traffic using R for GOF tests e
Dear R-help list,
Part of a program I wrote seem to take a significant amount of time,
therefore I am looking for an alternative approach.
In order to explain what is does:
- the input is a sorted vector of integer numbers
- some higher numbers may be derived (using a mathematical formula)
from l
On 27/06/2012 15:17, Mayank Bansal wrote:
It would be difficult for me to give the package.
Then, following the posting guide, you need to find an example you can give.
Can you explain me what does this error
"/usr/lib/R/bin/INSTALL: line 34: 9964 Done echo 'tools:::.install_packages()' 9965
It would be difficult for me to give the package.
Can you explain me what does this error
"/usr/lib/R/bin/INSTALL: line 34: 9964 Done echo 'tools:::.install_packages()'
9965 Segmentation fault | R_DEFAULT_PACKAGES= LC_COLLATE=C "${R_HOME}/bin/R"
$myArgs --slave --args ${args}"
generally mean.
The
On Wed, Jun 27, 2012 at 4:36 AM, Christof Kluß wrote:
>
> for example the same model with intercept R² = 0.6, without intercept R²
> = 0.9 and higher. In my definition of R², R² has to be equal or less
> without intercept
>
> I do not know what R shows, but in the summary of the model without
> in
-Original Message-
From: mb...@sun.ac.za
Sent: Wed, 27 Jun 2012 08:32:24 +0200
To: jrkrid...@inbox.com, ke...@math.montana.edu
Subject: RE: [R] plotting two histograms on one plot with hist function
Thank you for all the advice! After mailing the question, I actually
Hi Xi,
Maybe you should try to "parallelize" your calculations.
See package "parallel".
http://stat.ethz.ch/R-manual/R-devel/library/parallel/doc/parallel.pdf
Arnaud
On Mon, Jun 25, 2012 at 8:07 PM, Xi wrote:
> Dear All,
>
> I have been searching online for help increasing my R code more
effici
On Jun 27, 2012, at 13:15 , Uwe Ligges wrote:
>
> 1 - crossprod(residuals(model)) / crossprod(y - mean(y))
And the reason why that is not used in R:
> y<- rnorm(100,10,1)
> x <- 1:100
> model <- lm(y~x-1)
>
> 1 - crossprod(residuals(model)) / crossprod(y - mean(y))
[,1]
[1,] -27.600
try this:
sapply(estIID50, "[[", 'para')
Best,
Dimitris
On 6/27/2012 2:17 PM, Al Ehan wrote:
Sorry, one more simple question. how do I pick, from the generated
lapply, the $para for each X. say here I have 2 generated list from
previous function. how do I take only the $para for each X?
Tha
Sorry, one more simple question. how do I pick, from the generated lapply,
the $para for each X. say here I have 2 generated list from previous
function. how do I take only the $para for each X?
Thank you so much for your kindness.
$X1
$X1$type
[1] "gev"
$X1$para
xi alpha ka
p-value > 0.05 means that the 95% confidence intervals span zero.
Use "confint" to get the CI. It is described in "?model.avg".
cheers,
kamil
Dnia 2012-06-27 12:00, Robertson, Andrew pisze:
Dear R users,
Recent changes to the MuMIn package now means that the model averaging command
(model.av
It's working I am excited than ever!!
Thank you so much
On Wed, Jun 27, 2012 at 12:45 PM, Dimitris Rizopoulos <
d.rizopou...@erasmusmc.nl> wrote:
> you will have to check what function pargev() returns as a result. I would
> guess that is probably a list. In any case, you could use someth
Hello,
Try the following.
estIID50 <- apply( IID50, 2, function(x) pargev(lmom.ub(x)) )
And see ?apply
Hope this helps,
Rui Barradas
Em 27-06-2012 12:31, Al Ehan escreveu:
Hi R-users,
I'm trying to repeat the same procedure to 1000 data set. I know this is
very easy, but I got stuck findin
you will have to check what function pargev() returns as a result. I
would guess that is probably a list. In any case, you could use
something like the following:
estIID50 <- lapply(IID50, function (m) pargev(lmom.ub(m)))
I hope it helps.
Best,
Dimitris
On 6/27/2012 1:31 PM, Al Ehan wrote:
Hi R-users,
I'm trying to repeat the same procedure to 1000 data set. I know this is
very easy, but I got stuck finding the right and fastest way in running it.
IID50=Riidf[1:50,1:1000] #where IID50 is a dataframe consist of 1000 time
series(as column) and 50 time scales (row).
#what I tried to
On 27.06.2012 10:36, Christof Kluß wrote:
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not s
Hello, R-help,
does anybody have already a work-around for the problem that the formula
version of sunflowerplot() throws an error when provided with a value for
xlab (or ylab) different from NULL:
sunflowerplot( Sepal.Length ~ Sepal.Width, data = iris, xlab = "A")
Error in model.frame.defa
Homework ? You mean the question is at a high level.
I use R. It is just the simulation part that I am investigating. Several
papers on Capacity Planning that I am working on are useful for
analyzing log traffic using R for GOF tests etc. But I am not able to
find many papers on using R or r-java
Hello,
Here's a solution using aggregate and merge. I've kept it in two steps
for clarity.
d <- read.table(text="
subjecttime.ms V3
1 1 stringA
1 12 stringB
1 22 stringC
2 1stringB
2 14 stringC
2 25 stringA
", header=TRUE)
ag <- aggregate(t
Hello,
Maybe the link below is of some use.
http://stats.stackexchange.com/questions/14061/area-under-the-pdf-in-kernel-density-estimation-in-r
Hope this helps,
Rui Barradas
Em 27-06-2012 01:13, pilaw escreveu:
Hello,
I need density function so that I can find expected value (using
integra
Hello,
Homework?
There's a no homework rule but see inline.
Em 27-06-2012 10:20, Mohan Radhakrishnan escreveu:
Hi,
I am looking for some assistance with these requirements. We
are trying to simulate web requests to hit a web applications. Let's
assume I have a url to hit that requir
Hello,
My guess would be that solve() does not take advantage of the
special structure of the matrix and that you may want a sparse matrix
representation.
Take care
Oliver
On Tue, Jun 26, 2012 at 1:56 PM, Paul Rathouz wrote:
> Hi -- I am wondering why the time to solve (invert)
Hi,
I am looking for some assistance with these requirements. We
are trying to simulate web requests to hit a web applications. Let's
assume I have a url to hit that requires a username and password.
1. These web requests should have exponential inter arrival times.
2. Gene
Hello,
I'm afraid you're wrong, this has nothing to do with leading zeros. Just
see:
x <- c("1/2/2011", "1/4/2011", "1/4/2011", "1/4/2011", "1/6/2011",
"1/7/2011",
"1/8/2011", "1/9/2011", "1/10/2011")
as.Date(x, "%m/%d/%Y")
y <- factor(x)
str(y)
as.Date(as.character(y), "%m/%d
for example the same model with intercept R² = 0.6, without intercept R²
= 0.9 and higher. In my definition of R², R² has to be equal or less
without intercept
I do not know what R shows, but in the summary of the model without
intercept it does not show the R² of the regression line.
When I run
On Jun 27, 2012, at 09:33 , Christof Kluß wrote:
> Hi
>
> is there a command that calculates the correct adjusted R-squared, when
> I work without intercept? (The R-squared from lm without intercept is
> false.)
When people say that, they are usually implying that a "correct" R-squared can
be
On 06/27/2012 10:33 AM, Christof Kluß wrote:
> Hi
>
> is there a command that calculates the correct adjusted R-squared, when
> I work without intercept? (The R-squared from lm without intercept is
> false.)
Hi! This answer in R-FAQ might help you:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Wh
Hi
>
> How could I select the rows of a dataset that have the maximum value in
> one variable and to do this nested in another variable. It is a
dataframe
> in long format with repeated measures per subject.
> I was not successful using aggregate, because one of the columns has
You could do
On 27.06.2012 09:33, Christof Kluß wrote:
Hi
is there a command that calculates the correct adjusted R-squared, when
I work without intercept? (The R-squared from lm without intercept is
false.)
Then we need your definition of your version of "correct" - we know the
definition of your versi
Hi
is there a command that calculates the correct adjusted R-squared, when
I work without intercept? (The R-squared from lm without intercept is
false.)
Greetings
Chrsitof
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