Hi
If you want sensible response you need to ask sensible question and more
importantly provide working example or at least an example that produces
the error you want get rid of.
with
test.functional.t(res.em1,res.em2,mint,maxt,se.m=0,points=300)
I get
Error in seq(mint, maxt, length.out
Dear all,
I would like to place into a same plot two plots. Their y values though are
different a lot (first series has around -100 y values and the second one
slightly over 1). The x value are the same.
I would like to have in the same plot (under the same x,y axis the two plots).
Is it
Bernardo Powaga bspowaga at gmail.com writes:
I would like to fit a generalized linear model for the binomial family
with some non standard link functions. For instance, this is the
Aranda-Ordaz link:
η = ln( ( (1 - π)^-α - 1 )/α)
snip
Is there any way to tell glm() to add this
Try this:
set.seed(123)
x - 1:5
y1 - rnorm(5)
y2 - rnorm(5)
par(mar = c(5,4,5,4))
plot(x, y1, xlab = x-axis, ylab = y-axis, type = l)
par(new = TRUE)
plot(x, y2, axes = FALSE, yaxs = i, type = l,
ann = FALSE, col = 2)
axis(4, las = 2)
mtext(z-axis, 4, 2)
I hope it helps.
Best,
Dimitris
Dear Petr
I am sorry about the code being incomplete. I also take note of not using R
keywords and functions as variable names. I was able to overcome my problem
using unlist(). Thanks.
Regards
Aparna
On Thu, Mar 1, 2012 at 3:57 PM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
If you want
Hi,
We are exploring to integrate a solution that can analyse data from MySql, and
generate graphical reports showing trends/future predictions.
I have just started to look into R-Project. Please let me know if R-Project
provides this functionality. If yes please direct me to any
Ackbar:
have a look at ur.ers directly. The coefficients can be recovered from the slot
'testreg', i.e.,
example(ur.ers)
slotNames(ers.gnp)
coef(ers.gnp@testreg)
RTFM: help(ur.ers) and help(ur.ers-class)
Best,
Bernhard
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
Hello Keith,
see ?Acoef for retrieving the coefficients. Incidentally, in the package dse
simulation methods are made available.
Best,
Bernhard
Dr. Bernhard Pfaff
Director
Global Asset Allocation
Invesco Asset Management Deutschland GmbH
An der Welle 5
D-60322 Frankfurt am Main
Tel: +49
Hi Neeraj
Check out the task view of some of the many options for trending, time series
analysis and visualisation at
http://cran.r-project.org/web/views/TimeSeries.html
Best wishes
Chris
Chris Campbell
MANGO SOLUTIONS
Data Analysis that Delivers
+44 1249 705450
-Original
On 03/01/2012 07:13 PM, Alaios wrote:
Dear all,
I would like to place into a same plot two plots. Their y values though are
different a lot (first series has around -100 y values and the second one
slightly over 1). The x value are the same.
I would like to have in the same plot (under the
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to implement the R code.
Consider the following data frame:
df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
c(34,12,23,25,34,42,48,29,30,27))
What I want to
Dear R users,
Is it possible to write the following list to a text-file?
List:
[[1]]
[1] 500
[[2]]
[1] 1
[[3]]
[,1] [,2] [,3] [,4] [,5]
FID12345
Var20211
I would like to have the textfile look like this:
500
1
FID 1 2 3 4 5
Var 2 0 2 1 1
On Mar 1, 2012, at 12:52 AM, Bernardo Powaga wrote:
Hello R users,
I would like to fit a generalized linear model for the binomial
family with some non standard link functions. For instance, this is
the Aranda-Ordaz link:
η = ln( ( (1 - π)^-α - 1 )/α)
I know how to define a new link
But there are some important reasons to use Excel. In my work there
are a lot of people that I have to send the equivalent of a data.frame
to who want to look at the data and possibly slice/dice the data
differently and then send back to me updates. These folks do not know
how to use R, but do
Perhaps something like
sink(outtext.txt)
lapply(LIST, print)
sink()
You could replace print with cat and friends if you wanted more
detailed control over the look of the output.
Michael
On Thu, Mar 1, 2012 at 5:28 AM, t.galesl...@ebh.umcn.nl wrote:
Dear R users,
Is it possible to write the
yes, e.g.
require(rms)
f - ols( ) # ols is a wrapper for lm
g - Function(f)
g(age=30) # get predicted mean at age=30 and defaults for other variables
(medians modes)
Frank
Pascal Oettli-2 wrote
Hi Keith,
Do you mean as predict.lm can do?
Regards,
Pascal
Thank everyone for your help.
Problem solved. I'm getting more used with vectorization with your help,
Regards,
Phil
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Sent from the R help mailing list archive at Nabble.com.
Dear R helpers,
I have some difficulties in using 'break' function with loop, and the
followings are my script. What I try to do is (1) permute 'or' first; (2) doing
t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I get 10
permutations; (4) redo everything again but this time
Hi everyone. I have recently installed R 2.14.1 on my 64-bit Windows 7
laptop. I am attempting to include some favourite functions in the
Rprofile.site file to run at R start-up as I did with my previous 32-bit
XP machine. I have edited the Rprofile.site file in C:\Program
I have never seen 'break' used in an 'ifelse'; you probably meant to
use an 'if' statement there.
On Thu, Mar 1, 2012 at 8:24 AM, Tsai, Pei-Chien
pei-chien.t...@kcl.ac.uk wrote:
Dear R helpers,
I have some difficulties in using 'break' function with loop, and the
followings are my script.
On 12-02-28 2:11 PM, Chris Hane wrote:
Hello,
I am trying to paste together a formula to use in the mob function of
party. This means the formula will be of the form y ~ x1+ ...+xM | z1+..zN.
I am doing some preliminary fits of y ~ x1+ ...+xM, then want to add the
conditional part of the
On 12-03-01 8:24 AM, Tsai, Pei-Chien wrote:
Dear R helpers,
I have some difficulties in using 'break' function with loop, and the
followings are my script. What I try to do is (1) permute 'or' first; (2) doing
t-test if this 'or' pass criteria 1 (k=1); (3) end the loop when I get 10
Vinicius,
Vinicius Magalhães wrote
+ n - length(*x.ts*)
+ for (p in 0:maxord[1]) for (d in 0:maxord[2]) for (q in 0:maxord[3])
+ for (P in 0:maxord[4]) for (D in 0:maxord[5]) for (Q in 0:maxord[6])
+ {
+ fit - arima(*x.ts*, order=c(p,d,q),
+ seas = list(order=c(P,D,Q),
+
If I have two factors, v1 and v2 and I want to have a stacked bar graph of
the two variables side by side I could do
barplot(cbind(table(v1),table(v2)))
if v1 and v2 have the same number of categories.
If they don't have the same number of categories this won't work.
I'm sure there's a simple
Hello,
jholtman wrote
I have never seen 'break' used in an 'ifelse'; you probably meant to
use an 'if' statement there.
On Thu, Mar 1, 2012 at 8:24 AM, Tsai, Pei-Chien
lt;pei-chien.tsai@.acgt; wrote:
Dear R helpers,
I have some difficulties in using 'break' function with loop, and the
I am running the following line to download data from the US Energy
Information Administration. This function has worked successfully for me in
the past but yesterday gave the error/warning messages below.
If I simply type http://ir.eia.gov/wpsr/psw09.xls; (no quotes) into a
browser, the file is
It might be a local network or OS issue: this works fine for me on my
personal Mac
download.file(url = http://ir.eia.gov/wpsr/psw09.xls;, destfile =
~/herewego.xls, mode = wb)
Someone with more Windows knowledge may have to help you out, but
often using IE settings (activated by the setInternet2
Many thanks Jim and Chris for your helpful answers!
J
On Wed, Feb 29, 2012 at 11:48 AM, Chris Campbell
ccampb...@mango-solutions.com wrote:
Hi Jason
If you close an R session and save without choosing a filename, a file
called .RData will be created. Open a new session and type getwd().
Thank you so much Jim, Duncan, and special thanks to Rui!!
The modified script from Rui works perfectly and I really learned a lot from
all your suggestions.
Regards,
Amber
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Sent from
Jon,
You could create new variables with the combined levels just for the
purpose of plotting.
Assume I have data.frame bpt
str(bpt)
'data.frame': 12 obs. of 2 variables:
$ V1: Factor w/ 3 levels low,med,high: 1 1 1 1 2 2 2 2 3 3 ...
$ V2: Factor w/ 6 levels 1,2,3,4,..: 1 1 2 2 3 3 4 4 5 5
On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to implement the R code.
Consider the following data frame:
df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10),
On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to implement the R code.
Consider the
Hello Everyone
Both the MCMCpack and the bayesm libraries allow us to make draws from the
Inverse Wishart distribution.
But I wanted to find out how exactly is the Inverse Wishart distribution
parameterized in these libraries.
The reason I ask is the following:
Now its generally standard to
Hi Richard,
clearly there is a problem with latin ligature because the word resulting
from my ask with findFreqTerms give me some wordsU+FB01n
U+FB01nancement
U+FB01nancier U+FB01nancièreU+FB01nancières
U+FB01nanciersU+FB01xe
where U+FB01 is a code for
Hello,
syrvn wrote
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to implement the R code.
Consider the following data frame:
df - data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value =
Hello,
I have a spreadsheet of pairs of coordinates and I would like to plot a line
along which curves/arcs connect each pair of coordinates. The aim is to
visualise the pattern of point connections.
Thanks! Ian
--
View this message in context:
Hi
I would like to know how I can change the name of a model for each
trainning cycle of a model.
I work with the RSNNS package and to build a neural network, I used :
for (i in 5:30)
model_ANN - mlp(X, Y, size=n,) # where size is the number of neurons
in the hidden layer
but I need to
Sorry, correction:
The second index matrix is the matrix of elements not in the first,
not another combination, this time 3 out of 10.
Change this in my first post
inxmat - with(DF, combn(ID, 3))
meansDist2 - apply(inxmat, 2, function(jnx) f(jnx, DF$value, 45))
(i2 - which(meansDist2 ==
Hi everyone.
I'm using segmented package to find break point in a bi-linear relationship.
In a particular case, I find 1 pointcut (so 2 slopes).
I would like to know if it is possible to retrieve information in the
segmented object that could let me to plot 1 particular segment with a
different
Hello,
consider the following data.frame:
test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
c(1,1,NA,1,NA))
test
n v pattern
1 1 6 1
2 2 5 1
3 3 7 NA
4 4 5 1
5 5 3 NA
I tried to use apply and the adply function to set v to
I have a data frame with a number of observed and predicted values by
classification as shown below:
Count Volume FCLASS
1 55000 6 Grade Separated
2 43000 39000 Grade Separated
3 26000 26500 Major Arterial
4 19500 2 Major Arterial
...
On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to implement the R code.
Consider the
Dear gurus,
Im a newbie, and I want to ask a very general question.
Assume that I have a set of numbers as follows,
1, 1, 2, 10, 100, 10,1
From these, I need to identify which number is the most different as
compared to others. (in this case, it will be 100, since its way larger
than the other
Hi,
On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
Hello,
consider the following data.frame:
test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
c(1,1,NA,1,NA))
test
n v pattern
1 1 6 1
2 2 5 1
3 3 7 NA
4 4 5
Your criteria did not make sense since in both cases pattern == 1, so
I chose to set to NA if pattern == 1
test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
+ c(1,1,NA,1,NA))
test
n v pattern
1 1 6 1
2 2 5 1
3 3 7 NA
4 4 5 1
5 5 3 NA
# set v to NA
Here is another way of doing it:
x - read.table(text =Count Volume FCLASS
+ 1 55000 6 'Grade Separated'
+ 2 43000 39000 'Grade Separated'
+ 3 26000 26500 'Major Arterial'
+ 4 19500 2 'Major Arterial', as.is = TRUE)
result - sapply(split(x,
On Mar 1, 2012, at 12:30 PM, Suranga Kasthurirathne wrote:
Dear gurus,
Im a newbie, and I want to ask a very general question.
Assume that I have a set of numbers as follows,
1, 1, 2, 10, 100, 10,1
From these, I need to identify which number is the most different as
compared to others. (in
Google is your friend! -- as usual.
If you had searched on glm with regularization you would have bumped
into the glmnet R package, which I think is what you're looking for.
-- Bert
On Wed, Feb 29, 2012 at 6:22 PM, Dmitriy Lyubimov dlie...@gmail.com wrote:
Hello,
Thank you for probably not
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of mails
Sent: Thursday, March 01, 2012 8:11 AM
To: r-help@r-project.org
Subject: [R] Delete rows from data.frame matching a certain criteria
Hello,
consider the following
Hi,
On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:
Hi,
On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
Hello,
consider the following data.frame:
test - data.frame(n = c(1,2,3,4,5), v = c(6,5,7,5,3), pattern =
c(1,1,NA,1,NA))
snip
So basically the result
Hello, BODY { font-family:Arial, Helvetica,
sans-serif;font-size:12px; }
Does any one know if there are any functions/packages available in R
for robust fitting of ARMA time series models (e.g., similar to the
function arima.rob() in S-PLUS)?
Many thanks and
You're all correct: I copied in the wrong thing. My apologies!
On Thu, Mar 1, 2012 at 1:00 PM, Ista Zahn istaz...@gmail.com wrote:
Hi,
On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:
Hi,
On Thu, Mar 1, 2012 at 11:11 AM, mails mails00...@gmail.com wrote:
Hello,
consider the following
Thank you.
On Thu, Mar 1, 2012 at 9:58 AM, Bert Gunter gunter.ber...@gene.com wrote:
Google is your friend! -- as usual.
If you had searched on glm with regularization you would have bumped
into the glmnet R package, which I think is what you're looking for.
-- Bert
On Wed, Feb 29, 2012
On Mar 1, 2012, at 1:02 PM, Sarah Goslee wrote:
You're all correct: I copied in the wrong thing. My apologies!
On Thu, Mar 1, 2012 at 1:00 PM, Ista Zahn istaz...@gmail.com wrote:
Hi,
On Mar 1, 2012, at 12:38 PM, Sarah Goslee wrote:
Hi,
On Thu, Mar 1, 2012 at 11:11 AM, mails
Hello to everyone.
I need your help. I´m trying to fit the same *glm.nb* to a different data
set and i am getting these errors in some of the data. Sometimes, one data
set has two of these errors when fitting the model.
1.- Error en while ((it - it + 1) limit abs(del) eps) { :
valor ausente
On 01-03-2012, at 19:03, isabe...@ghement.ca wrote:
Hello, BODY { font-family:Arial, Helvetica,
sans-serif;font-size:12px; }
Does any one know if there are any functions/packages available in R
for robust fitting of ARMA time series models (e.g., similar to the
BODY { font-family:Arial, Helvetica, sans-serif;font-size:12px; }
Hi Berend,
Many thanks for your prompt reply. I followed your instructions but
couldn't find what I was looking for. I was hoping that someone who's
already worked with such a function could be able to
On 01.03.2012 14:39, Ross Bowden wrote:
Hi everyone. I have recently installed R 2.14.1 on my 64-bit Windows 7
laptop. I am attempting to include some favourite functions in the
Rprofile.site file to run at R start-up as I did with my previous 32-bit
XP machine. I have edited the Rprofile.site
On 01-03-2012, at 19:33, isabe...@ghement.ca wrote:
Hi Berend,
Many thanks for your prompt reply. I followed your instructions but couldn't
find what I was looking for. I was hoping that someone who's already worked
with such a function could be able to point it out to me. My Google
?plot
On 01.03.2012 16:15, hendersi wrote:
Hello,
I have a spreadsheet of pairs of coordinates and I would like to plot a line
along which curves/arcs connect each pair of coordinates. The aim is to
visualise the pattern of point connections.
Thanks! Ian
--
View this message in context:
Le jeudi 01 mars 2012 à 07:07 -0800, Mickael R problem a écrit :
Hi Richard,
clearly there is a problem with latin ligature because the word resulting
from my ask with findFreqTerms give me some wordsU+FB01n
U+FB01nancement
U+FB01nancier U+FB01nancière
R 2.14.0
OS X
Marc's proposed solution (appearing at the end of this email) is perfect --
thanks so much
However, some questions remain:
1. The following works:
plot(1, type=n)
text(1, 1, expression(symbol(\342)))
but this does not work (TEXT appears in a symbol font)
On Thu, Mar 1, 2012 at 12:07 PM, Doran, Harold hdo...@air.org wrote:
Typically this list doesn't support general statistical questions and
unfortunately I don't have a better recommendation. It may be more helpful
for you to work with a statistician than seek help here.
My point is simply
Hello, again.
Petr Savicky wrote
On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
Hello,
I am stuck with selecting the right rows from a data frame. I think the
problem is rather how to select them
then how to
Hi,
Thanks for your help,
This worked very well:
na.action=na.roughfix
Kevin
On Sun, Feb 26, 2012 at 3:10 PM, Weidong Gu anopheles...@gmail.com wrote:
Hi,
You can set na.action=na.roughfix which fills NAs with the mean or
mode of the missing variable.
Other option is to impute missing
Hi there,
I am trying to find an example how to use Rscript
Let's suppose I want to pass 3 arguments (I don't want [options] and -e
[expressions] as described in help)
*on the command line
myRscript.R -arg1=value1 -arg2=value2 -arg3=value3
*In the script
#! /path/to/Rscript
args =
Hello,
How do I get the dates of all Fridays between two dates?
thanks,
Ben
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi Dennis,
There are some subtleties in the way that 'symbol' is handled in plotmath.
There is no symbol() function, per se, though there is an as.symbol() function,
which is used differently.
'symbol' as used in plotmath, is telling R to plot the character using a symbol
font, which is the
Thank you Rui, that helped a lot. The correct values show up when I'm using
the following code. Now fun(Temp,v) returns a matrix, and Temp and v
stay the same. But I'd like to use the reduced vectors in some
calculations..can they be extracted in some way so that I have them
separately
Of course, just use
x - fun(Temp, v)
x$Temp # To get back temp
x[[Temp]]
x$v # To get back v
x[[v]]
Michael
On Thu, Mar 1, 2012 at 3:15 PM, babyluck madr...@gmx.ch wrote:
Thank you Rui, that helped a lot. The correct values show up when I'm using
the following code. Now fun(Temp,v)
Inelegant, but here's one way:
d1 - Sys.Date()
d2 - Sys.Date() + 100
library(lubridate)
d - seq(d1, d2, by = day)
d[wday(d)==6]
Michael
On Thu, Mar 1, 2012 at 3:02 PM, Ben quant ccqu...@gmail.com wrote:
Hello,
How do I get the dates of all Fridays between two dates?
thanks,
Ben
Not paying close attention to detail, I entered the equivalent of
pstr-c(b1=200, b2=50, b3=0.3)
when what I wanted was
pnum-c(b1=200, b2=50, b3=0.3)
There was a list thread in 2010 that shows how to deal with un-named vectors,
but the same
lapply solution doesn't seem to work here i.e.,
On Mar 1, 2012, at 2:02 PM, Ben quant wrote:
Hello,
How do I get the dates of all Fridays between two dates?
thanks,
Ben
Days - seq(from = as.Date(2012-03-01),
to = as.Date(2012-07-31),
by = day)
str(Days)
Date[1:153], format: 2012-03-01 2012-03-02
Thank you very much!! Exactly how I wanted it :)
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Sent from the R help mailing list archive at Nabble.com.
__
Great thanks!
ben
On Thu, Mar 1, 2012 at 1:30 PM, Marc Schwartz marc_schwa...@me.com wrote:
On Mar 1, 2012, at 2:02 PM, Ben quant wrote:
Hello,
How do I get the dates of all Fridays between two dates?
thanks,
Ben
Days - seq(from = as.Date(2012-03-01),
to =
On Mar 1, 2012, at 2:40 PM, Dennis Fisher wrote:
R 2.14.0
OS X
Marc's proposed solution (appearing at the end of this email) is
perfect -- thanks so much
However, some questions remain:
1. The following works:
plot(1, type=n)
text(1, 1, expression(symbol(\342)))
but this
On Thu, Mar 01, 2012 at 11:53:00AM -0800, statquant2 wrote:
Hi there,
I am trying to find an example how to use Rscript
Let's suppose I want to pass 3 arguments (I don't want [options] and -e
[expressions] as described in help)
*on the command line
myRscript.R -arg1=value1 -arg2=value2
On Thu, Mar 1, 2012 at 12:28 PM, John C Nash nas...@uottawa.ca wrote:
Not paying close attention to detail, I entered the equivalent of
pstr-c(b1=200, b2=50, b3=0.3)
when what I wanted was
pnum-c(b1=200, b2=50, b3=0.3)
There was a list thread in 2010 that shows how to deal with un-named
Hello,
My direct desire is a good (fast) way to fill values forward until there is
another value then fill that value foward in the data xx (at the bottom of
this email). For example, from row 1 to row 45 should be NA (no change),
but from row 46 row 136 the value should be 12649, and from row
On Thu, Mar 01, 2012 at 03:28:31PM -0500, John C Nash wrote:
Not paying close attention to detail, I entered the equivalent of
pstr-c(b1=200, b2=50, b3=0.3)
when what I wanted was
pnum-c(b1=200, b2=50, b3=0.3)
There was a list thread in 2010 that shows how to deal with un-named
Gotta love R.
Thanks to Bill Dunlap, Peter Langfelder and Jim Holtman for no less than 3 different
solutions.
JN
On 12-03-01 04:25 PM, Peter Langfelder wrote:
pstr-c(b1=200, b2=50, b3=0.3)
split = sapply(strsplit(pstr, split = =), I);
pnum = as.numeric(split[2, ]);
names(pnum) =
That's nice.
Please read the posting guidelines and get back to us with some information on
what the data looks like an what you are doing.
For example do you just want lines or do you want a smoother, etc?
John Kane
Kingston ON Canada
-Original Message-
From: ir...@cam.ac.uk
dear Phil,
plot.segmented() accepts vectorized 'col', 'lty' and 'lwd' arguments. Then,
par(mfrow=c(1,2))
plot(o.seg,col=2:3,lty=2:3,lwd=c(1,2))
plot(z,y)
plot(o.seg,col=2:3,lty=1,linkinv=T,add=T,lwd=2)
hope this helps you,
vito
On Thu, 1 Mar 2012 08:57:39 -0800 (PST), Filoche wrote
Hi
Hi!
I'm running R version 2.13.0 (2011-04-13)
Platform: i386-pc-mingw32/i386 (32-bit)
When i type in the command:
sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))
R returns the value:
-5.551115e-17
Why doesn't R return zero in this case? There shouldn't be any rounding
error in a simple sum.
Thanks,
No it's an outlier problem, I think.
If you have a fairly small number of sets of these numbers simple visual
inspection of a boxplot for each set would probably acomplish what you want.
Try this in R for an example. Just paste the next two lines into R
xx - c(1, 1, 2, 10, 100, 10,1)
Of course there's rounding error: your computer can't
store those decimal numbers precisely. See R FAQ 7.31 for
details.
See also:
sum(10*c(-0.2, 0.8, 0.8, -3.2, 1.8)) / 10
Sarah
On Thu, Mar 1, 2012 at 4:49 PM, Mark A. Albins kamoko...@gmail.com wrote:
Hi!
I'm running R version 2.13.0
On Thu, Mar 01, 2012 at 02:31:01PM -0700, Ben quant wrote:
Hello,
My direct desire is a good (fast) way to fill values forward until there is
another value then fill that value foward in the data xx (at the bottom of
this email). For example, from row 1 to row 45 should be NA (no change),
In base ten, using any fixed number of digits, compute
1/3 + 1/3 + 1/3
(doing the divisions before the additions).
Why isn't it 1?
1/5 has the same sort of problem in base two.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
On Thu, Mar 01, 2012 at 01:49:44PM -0800, Mark A. Albins wrote:
Hi!
I'm running R version 2.13.0 (2011-04-13)
Platform: i386-pc-mingw32/i386 (32-bit)
When i type in the command:
sum(c(-0.2, 0.8, 0.8, -3.2, 1.8))
R returns the value:
-5.551115e-17
Why doesn't R return zero in
Hi all, I am running an -MNP- multinomial probit model package using R. It
gives me the following objection instead of giving me the results:
Erreur : impossible d'allouer un vecteur de taille 137.9 Mo (in english: cannot
allocate a 137.9 Mb vector memory).
I have already increased the memory
That is great! Thank you very much.
Ben
On Thu, Mar 1, 2012 at 2:57 PM, Petr Savicky savi...@cs.cas.cz wrote:
On Thu, Mar 01, 2012 at 02:31:01PM -0700, Ben quant wrote:
Hello,
My direct desire is a good (fast) way to fill values forward until there
is
another value then fill that
A) you will generally get a better response when your question includes
reproducible code/sample data, and a clear identification of the desired final
result.
B) in most cases like this, a proliferation of names is not as useful as the OP
(you) thinks it is. Much better is to build a list of
On Thu, Mar 01, 2012 at 04:55:55PM -0500, Sarah Goslee wrote:
Of course there's rounding error: your computer can't
store those decimal numbers precisely. See R FAQ 7.31 for
details.
See also:
sum(10*c(-0.2, 0.8, 0.8, -3.2, 1.8)) / 10
Hi.
This is 0. This works without rounding for one
Hi,
I'm trying to figure out the formula used by ggplot2 to calculate the
width of a bar in geom_bar so that I can use that elsewhere in the
plot. My code looks like this:
ggplot(xAll, aes(Date)) +
geom_bar(subset = .(Direction == Up), aes(y = Change, fill =
Time), stat = identity) +
On 2012-03-01 13:52, John Kane wrote:
No it's an outlier problem, I think.
If you have a fairly small number of sets of these numbers simple visual
inspection of a boxplot for each set would probably acomplish what you want.
Try this in R for an example. Just paste the next two lines into R
Dear expeRts,
I would like to colorize the backgrounds of a pairs plot according to the
respective panel number. Here is what I tried (without success):
count - 0
mypanel - function(x, y, ...){
count - count+1
bg. - if(count %in% c(1,4,9,12)) #FDFF65 else NA
points(x, y, cex=0.5,
Rich's pointers deals with lattice/grid graphics. Does anyone have a
solution for base graphics?
Thanks
Frank
Richard M. Heiberger wrote
Frank,
This can be done directly with a variant of the panel.axis function.
See function panel.axis.right in the HH package. This was provided for me
Dear R People:
I have been using xlsReadWrite to read Excel files and am very pleased
with it. Thank you xlsReadWrite People!
My question is: is there a function, similar to get.hist.quote, to
download Excel files from the web, please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
On Thu, Mar 1, 2012 at 6:43 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Dear R People:
I have been using xlsReadWrite to read Excel files and am very pleased
with it. Thank you xlsReadWrite People!
My question is: is there a function, similar to get.hist.quote, to
download Excel files
Thanks J!
I was thinking about the url, download.file functions too
Just wanted to make sure that there wasn't something special for Excel.
Thanks,
Erin
On Thu, Mar 1, 2012 at 6:53 PM, J Toll jct...@gmail.com wrote:
On Thu, Mar 1, 2012 at 6:43 PM, Erin Hodgess erinm.hodg...@gmail.com wrote:
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