On Thu, 4 Mar 2004, BXC (Bendix Carstensen) wrote:
> > On Wed, 3 Mar 2004, Peter Dalgaard wrote:
> >
> > > Torsten Hothorn <[EMAIL PROTECTED]> writes:
> > >
> > > > Could anyone give me a a hint how I can convert 13264300800 to
> > > > 2003/02/11 again, please?
> > >
> > > > ISOdate(1582,10,14)
Hi
On 3 Mar 2004 at 11:25, Grace Conlon wrote:
> In R, How to read Excel file and access the data item?
> Thank you.
Quite strightforward way is to copy your data to notepad, save to
a txt file and read this txt file to r by appropriate read.table()
command.
Or you can use coppying through cl
Mohammad Rahimi wrote:
Hi folks,
I guess it is a very premitive question.
If i have already created an image.
image(input)
and i want to add a line from (x0,y0) to (x1,y1) on that
image. how i can sketch a line superimposed on that.
See ?lines
Uwe Ligges
Best regards
-m
_
Cline Julia wrote:
I have an SPSS data file that I would like to analyze using RAqua. Is
there any way for me to open the file using RAqua. I can probably get a
hard copy of the file contents from my collaborator, but I'd rather not
have to deal with all that input.
Julie Cline
Hi,
Could any one point me to the projection, and parameters if necessary, that
would show each country/continent with it's area accurately refelcted on the
plot? E.g. aitoff vs. albers vs. bonne vs. cylequalearea vs. guyou - they
don't all look the same to mee but some of the documentations sugges
Hi,
This may already be in the R-help email archive but I can't seem to view
msgs older than 2002/2001
In trying to work out which map projection I wanted to use I wrote a loop to
print them out, along with the info that is in the documentation of
'mapproj'. If there is interest I'd like to c
Dear all,
I want to derive from a data set that I have a set of 9
interpolation functions using approxfun() and store
them in an R object. The data has some structure that I would
like to reflect in the storage, so ideally I would store them in
a data.frame. So far I failed.
Here is what I tr
Hi,
I am having a new problem that how I can set the probability scale
from 0 to 1 when I using on probability scale. Sometimes I get the maximum
probabilty more than 100%.
Thanks!
Best Regards,
WeiQiang
In response to a posting from WeiQiang Li:
> Hi,
> I am facing the problem th
You can avoid the loop in your calculation. Since Friday is
day of the week 5, if the day of the week d of chron date x
is greater than 5 then Friday was d-5 days ago; otherwise, if d-5 is
zero or negative then we have to add 7 to it to ensure that its
in the past. Thus:
prevFriday <- functi
Hi,
The searchable mail archive at http://maths.newcastle.edu.au/~rking/R/ is
very useful. Unfortunately it seems that many of the emails are not
available, many show up in the search results but then return a "file not
found" when following the link.
Is anyone else experiencing this, and is ther
You can ensure the name gets set appropriately like this:
> aggregate(list(Contract=df$Contract), list(ID=df$ID), max)
ID Contract
1 011
2 023
3 032
---
Date: Wed, 03 Mar 2004 21:46:27 -0600
From: Marc Schwartz <[EMAIL PROTECTED]>
To: Greg Blevins <[EMAIL PROTEC
month.day.year takes a chron object and produces a list
with three components named month, day and year. with
executes its second argument in an environment where the
components of the list of the first argument are variables
so arg 2 can refer to the names month, day and year.
---
Date: Thu,
Hello All,
I am new to R and want to use R to scale my data
before analysis. I need to do two fold scaling: a).
scale on a per object (chip) basis, scale to a mean of
0 and stddev to 1; b). scale on a per feature basis:
scale data linearly to the interval [0,1].
Could somebody help me with this?
Say your data frame is `dat'. Try:
tapply(dat$Contract, dat$ID, max)
(The output is not a data frame, but that shouldn't be a problem...)
HTH,
Andy
> From: Greg Blevins
>
> Hello R experts,
>
> The following problem outstrips my current programming knowledge.
>
> I have a dataframe wi
On Wed, 2004-03-03 at 21:19, Greg Blevins wrote:
> Hello R experts,
>
> The following problem outstrips my current programming knowledge.
>
> I have a dataframe with two fields that looks like the following:
>
> ID Contract
> 01 1
> 01 1
> 02 2
> 02 3
> 02 1
> 03 2
>
On Wed, Mar 03, 2004 at 06:58:02AM -0500, Gabor Grothendieck wrote:
>
> Here are three different ways (using x as defined in your
> post):
>
> with( month.day.year(x), day.of.week(month,day,year) )
>
> do.call( "day.of.week", month.day.year(x) )
>
> as.numeric(x-3)%%7 # uses fact that chr
How about something like ... (if your data frame is called the.data)
summarized <- as.data.frame(levels(the.data$ID))
names(summarized) <- "ID"
summarized$Contract <- as.numeric(tapply(the.data$Contract, the.data$ID, max))
Andrew
On Wednesday 03 March 2004 19:19, Greg Blevins wrote:
> Hello
Subject of the message: Re: Your music
Recipient of the message: Jenny Ong Pei Pei
Prohibited attachment: mp3music.pif
Privileged/Confidential Information may be contained in this message.
If you are not the addressee indicated
How does this compare with R of the qr decomposition? spencer graves
John Fox wrote:
Dear Amin,
I have a function (created just for demonstration, and reproduced below) for
finding the row-echelon form of a matrix. I'm sure that many list members
could produce something that's better numericall
Hello R experts,
The following problem outstrips my current programming knowledge.
I have a dataframe with two fields that looks like the following:
ID Contract
01 1
01 1
02 2
02 3
02 1
03 2
03 2
03 2
03 1
03 1
03 1
etc...
I would lik
I have an SPSS data file that I would like to analyze using RAqua. Is
there any way for me to open the file using RAqua. I can probably get
a hard copy of the file contents from my collaborator, but I'd rather
not have to deal with all that input.
Julie Cline
On Wed, 3 Mar 2004 18:14:25 -0700
Christof Bigler <[EMAIL PROTECTED]> wrote:
> I have a spatial data set with ordinal response variable containing
> four levels. I would like to know if and how spatial autocorrelation
> can be taken into account when ordinal logistic regression is used
> (e.g.
If you really want interpolation, should you be using spline() rather than
smooth.spline()? The later is for smoothing data observed with noise, not
for interpolation.
Andy
> From: W. C. Thacker
>
> Dear R listers,
>
> When using smooth.spline to interpolate data, results are generally
> good.
Dear Amin,
I have a function (created just for demonstration, and reproduced below) for
finding the row-echelon form of a matrix. I'm sure that many list members
could produce something that's better numerically, but this should be OK at
least for toy problems.
John
- snip -
I have a spatial data set with ordinal response variable containing
four levels. I would like to know if and how spatial autocorrelation
can be taken into account when ordinal logistic regression is used
(e.g. the function lrm from the Design package).
Thanks for your help!
Christof
___
There are a number of ways to do this:
1. select range including headers in Excel and
copy to clipboard (via ctrl-C) and in R:
mydata <- read.table("clipboard", header=T)
2. In Excel File | Save As and select csv as type. In R:
mydata <- read.csv("/myfile.csv")
3. RODBC package
This package im
In tseries package. You might have to download & install the package If
library(tseries) doesn't work.
HTH
Manoj
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erin Hodgess
Sent: Thursday, March 04, 2004 2:53 AM
To: [EMAIL PROTECTED]
Subject: [R] get.h
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> [EMAIL PROTECTED]
> Sent: Wednesday, March 03, 2004 6:27 PM
> To: Peter Dalgaard
> Cc: [EMAIL PROTECTED]
> Subject: Re: [R] read.spss and time/date information
>
>
>
>
> On Wed, 3 Mar 2004, Peter
"Sivakumar Mohandass" <[EMAIL PROTECTED]> writes:
> Dear all,
>
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its quality both on the screen and
> on paper. Coul
Peter,
These things can be a little mysterious at the beginning. Lots of things
in R are arranged in groups called packages (or libraries). Some basic
ones come preinstalled with R, some you have to install yourself. If you
type
library()
at your R prompt you should get a list of all p
Hi Sivakumar,
Also, just of potential note: in case you are working with Windows and
generate postscript files, you will not see the EPS image on-screen when it
is inserted for instance in a Word document. Rest assured that if you print
the document, it will show up. All journal editors can see s
On 3 Mar 2004 at 15:42, Peter Flom wrote:
> Hello
>
> I am running R 1.8.1 on a Windows platform
>
> I am attempting to fit an ordinal logistic regression model, using the
> polr function, as described in Venables and Ripley. But when I try
It would help if you started by saying
library(MAS
> help.search("polr")
Help files with alias or concept or title matching 'polr' using
regular expression matching:
polr(MASS) Proportional Odds Logistic Regression
Type 'help(FOO, package = PKG)' to inspect entry 'FOO(PKG) TITLE'.
>
Did you ask "library(MASS)" first?
ho
"Peter Flom" <[EMAIL PROTECTED]> writes:
> Hello
>
> I am running R 1.8.1 on a Windows platform
>
> I am attempting to fit an ordinal logistic regression model, using the
> polr function, as described in Venables and Ripley. But when I try
>
> model4 <- polr(ypsxcat~committed + as.factor(sex)
The R postscript() driver will produce eps files. See ?postscript.
If the journal requires that eps files include a preview image
(something that will display the image on the computer screen) then
that can be added after the fact with, for example, Adobe
Illustrator. R's driver does not produc
I get a message that the polr function was not found.
Did you remember to load the MASS library first?
library(MASS)
Hadley
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PLEASE do read the posting guide! http:/
help.search() is your friend:
polr(MASS) Proportional Odds Logistic Regression
HTH,
Andy
> From: Peter Flom
>
> Hello
>
> I am running R 1.8.1 on a Windows platform
>
> I am attempting to fit an ordinal logistic regression model, using the
> polr function, as described in Venab
Peter Flom wrote:
I am running R 1.8.1 on a Windows platform
I am attempting to fit an ordinal logistic regression model, using the
polr function, as described in Venables and Ripley. But when I try
model4 <- polr(ypsxcat~committed + as.factor(sex)
+ as.factor(drugusey) + anycsw + as.factor(sex
On Wed, 3 Mar 2004, Sivakumar Mohandass wrote:
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its quality both on the screen and
> on paper. Could some one please
Andrew Robinson and Andy Jaworski both pointed out that polr is in
library MASS. MASS seems to be missing from my installation, and the
former Andrew suggested that I reinstall it.
When I tried help.search('polr'), nothing was found, and when I tried
library(MASS) or require(MASS) the package was
See my reply at the end of the email.
> > I'd like to generate thousands of normal numbers from my C function using
> > the C API functions provided R. I have two options:
> >
> > 1. double norm_rand(); (page 61 of R extension 1.8.1)
> > 2. double rnorm(double mu, double sigma); (page 58 of R ex
See ?postscript.
Sean
On 3/3/04 3:35 PM, "Sivakumar Mohandass" <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its qual
Patrick Hausmann wrote:
>
> Dear R-list,
>
> I try to calculate the 10- and 90-Percentile from grouped data.
> Using 'lappy' it works fine but I have no success with 'gapply':
>
> df <- data.frame(test = runif(100))
> df <- df[order(df[,1]),]
> z <- rep(1:10, each = 10)
> lapply(split(df, z),
On Wed, 2004-03-03 at 14:35, Sivakumar Mohandass wrote:
> Dear all,
>
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its quality both on the screen and
> on paper.
Grace Conlon wrote:
>
> In R, How to read Excel file and access the data item?
> Thank you.
>
See R's Data Import/Export manual. At once, two solutions spring to
mind:
a) Export to csv or any other ASCII format and import in R.
b) RODBC
Uwe Ligges
__
On Wed, 3 Mar 2004 14:35:06 -0600
"Sivakumar Mohandass" <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its quality both o
Hi folks,
I guess it is a very premitive question.
If i have already created an image.
image(input)
and i want to add a line from (x0,y0) to (x1,y1) on that
image. how i can sketch a line superimposed on that.
Best regards
-m
__
[EMAIL PROTECTED]
Sivakumar Mohandass wrote:
>
> Dear all,
>
> A journal in which we wanted our manuscript published requires the
> figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
> files to these formats but it looses its quality both on the screen and
> on paper. Could some one please h
On 3 Mar 2004, Grace Conlon verbalised:
> In R, How to read Excel file and access the data item?
> Thank you.
The catdoc package (http://www.45.free.net/~vitus/ice/catdoc/) includes
a program xls2csv that converts xls to csv (comma separated value) files.
In a Unix-like system, I often do
blah
Presumably you have the data and R scripts that generated the jpegs. Just
substitute the calls to jpeg() (or bitmap()) to postscript(...,
onefile=FALSE) and you shall have the highest quality EPS.
HTH,
Andy
> From: Sivakumar Mohandass
>
> Dear all,
>
> A journal in which we wanted our manuscri
Dear R listers,
When using smooth.spline to interpolate data, results are generally
good. However, some cases produce totally unreasonable results.
The data are values of pressure, temperature, and salinity from a
probe that is lowered into the ocean, and the objective is to
interpolate temperat
Andy,
Thank you very much - that works and shows me how to go about working out
the range of possible values for the position co-ords.
Much appreciated.
Mark
> -Original Message-
> From: Liaw, Andy [mailto:[EMAIL PROTECTED]
> Sent: Thursday, March 04, 2004 2:07 AM
> To: 'Mark Van De Vyv
Hello
I am running R 1.8.1 on a Windows platform
I am attempting to fit an ordinal logistic regression model, using the
polr function, as described in Venables and Ripley. But when I try
model4 <- polr(ypsxcat~committed + as.factor(sex)
+ as.factor(drugusey) + anycsw + as.factor(sex)*committe
Dear all,
A journal in which we wanted our manuscript published requires the
figures as a tiff, eps or PowerPoint formated. I tried converting .jpeg
files to these formats but it looses its quality both on the screen and
on paper. Could some one please help.
Thanks in advance,
___
On Wed, 3 Mar 2004, Yongchao Ge wrote:
> I'd like to generate thousands of normal numbers from my C function using
> the C API functions provided R. I have two options:
>
> 1. double norm_rand(); (page 61 of R extension 1.8.1)
> 2. double rnorm(double mu, double sigma); (page 58 of R extension 1
XLSolutions Corporation ([1]www.xlsolutions-corp.com) is proud
to announce April 2004 2-day "R/S-plus Fundamentals and Programming
Techniques" in CANADA.
Toronto, Canada --> April 15-16, 2004
Reserve your seat now at the early bird rates! Payment due AFTER the
Frank E Harrell Jr <[EMAIL PROTECTED]> writes:
> Speaking of read.spss, which I have found handles labels and value labels
> very well, I am having a difficulty with character variables ending up as
> factors even when the number of levels equals the number of observations.
> I would be nice to h
Steven Lacey wrote:
> I want to create a dataframe where one of the columns does not hold
> individual elements, but a vector instead.
> ...
> However, if my dataframe does not already exist, then I cannot create a
> dataframe with vectors using the following syntax:
> data.frame(x1=list(c(5,6,7)
how to produce a Row Reduced Echelon Form for a matrix in R?
Aimin Yan
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Dear R People:
Here is a silly one:
Where is get.hist.quote, please?
Thanks,
Erin
mailto: [EMAIL PROTECTED]
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PLEASE do read the posting guide! http://www.R-project.o
> > I guess I'm not understanding the object that get.hist.quote makes. In
> > general, what are R facilities for discovering what a given object is?
>
> I suggest, you study first one of the beginners manuals of the R
> environment: http://cran.r-project.org/manuals.html
> Then you would easily se
If you wish to 'skip' (i.e. not interpolate) weekends in its, you could use
the following:
prices <- priceIts(instrument="ongc.ns")
plot(union(prices,newIts(start=start(prices),end=end(prices))),interp="none"
)
- Giles
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTE
In R, How to read Excel file and access the data item?
Thank you.
-
Yahoo! Search - Find what youre looking for faster.
[[alternative HTML version deleted]]
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Hi R People!
Are Tomas and Scott looking for partial correlation, please?
Do they mean something like "r_1.23", please?
Thanks,
Erin
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PLEASE do read the posting gui
Hi
I'd like to generate thousands of normal numbers from my C function using
the C API functions provided R. I have two options:
1. double norm_rand(); (page 61 of R extension 1.8.1)
2. double rnorm(double mu, double sigma); (page 58 of R extension 1.8.1)
If my understanding of R-exts is corre
I find it's just great to be able to say:
library(tseries)
x <- get.hist.quote(instrument="ongc.ns")
and it gets a full time-series of the stock price of the symbol
ongc.ns from Yahoo quote.
However, once my hopes have been raised by such beauty I get
disappointed when I do
plot(x)
and the annot
Torsten Hothorn <[EMAIL PROTECTED]> writes:
> Could anyone give me a a hint how I can convert 13264300800 to 2003/02/11
> again, please?
> ISOdate(1582,10,14) + 13264300800
[1] "2003-02-11 13:00:00 CET"
> ISOdate(1582,10,14) + 13142476800
[1] "1999-04-03 14:00:00 CEST"
[October 14, 1582 is Da
On Wed, 3 Mar 2004 11:21:12 -0500 (EST)
"Gabor Grothendieck" <[EMAIL PROTECTED]> wrote:
>
>
> I don't use SPSS but following through on your detective work
> can provide the likely answer.
>
> First note that both date numbers are evenly divisible by the number
> of seconds in a day, i.e. 24
On Wed, 3 Mar 2004, Peter Dalgaard wrote:
> Torsten Hothorn <[EMAIL PROTECTED]> writes:
>
> > Could anyone give me a a hint how I can convert 13264300800 to 2003/02/11
> > again, please?
>
> > ISOdate(1582,10,14) + 13264300800
> [1] "2003-02-11 13:00:00 CET"
> > ISOdate(1582,10,14) + 13142476
Torsten Hothorn wrote:
Hi,
I could not find any information on how `read.spss' deals with date
information. As an example, I created a file containing two variables,
one numeric (values = (1, 2)) and one of type "Datum" in SPSS (german
version with values "11.02.2003" and "03.04.1999" and I get i
Torsten Hothorn wrote:
I could not find any information on how `read.spss' deals with date
information. As an example, I created a file containing two variables,
one numeric (values = (1, 2)) and one of type "Datum" in SPSS (german
version with values "11.02.2003" and "03.04.1999" and I get in R:
S
Peter Dalgaard wrote:
"Anon." <[EMAIL PROTECTED]> writes:
I assume that this is because for much of the range, the integral is
basically zero.
The help page for integrate() says
When integrating over infinite intervals do so explicitly,
rather
than just using a large number as the endpoin
On Wed, Mar 03, 2004 at 03:25:14PM +0530, Ajay Shah wrote:
> I find it's just great to be able to say:
>
> library(tseries)
> x <- get.hist.quote(instrument="ongc.ns")
>
> and it gets a full time-series of the stock price of the symbol
> ongc.ns from Yahoo quote.
>
> However, once my hopes h
I don't use SPSS but following through on your detective work
can provide the likely answer.
First note that both date numbers are evenly divisible by the number
of seconds in a day, i.e. 24*60*60. This suggests that these numbers
are seconds since some origin.
Since we know "2003/02/11" cor
trellis.device(bg="white", color=F)
before your call to splom could make what you want but
take also a look at
?trellis.par.set
Stefano
> -Messaggio originale-
> Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> Inviato: mercoledì 3 marzo 2004 12.10
> A: r-help
> Oggetto: [R] Changin
On Wed, 3 Mar 2004, Heywood, Giles wrote:
> I get different results from match.call(), according to whether a function
> is dispatched via S3 or S4. Specifically, when I use S4 dispatch in the
> following example, the match.call() result is of length 1 less than I
> expect. I need to add an extr
On Wednesday 03 March 2004 05:09, [EMAIL PROTECTED] wrote:
> Context: Windows XP, R 1.8.1
>
> I'm studying Venables-Ripley "MASS" book and having a go at the many
> examples in library MASS. The code I'm checking (from script ch04.R) now
> is
>
> ..
> data(swiss)
> splom(~ swiss, aspect = "fil
You need to figure out the "user coordinates" for each panel, so you can put
the text within the range that's being plotted. Here's an example:
panel.myfitline <- function(x, y, digits=2, prefix="", cex.cor, ...)
{
usr <- par("usr")
res<-panel.smooth(x,y, col.smooth="blue", ...)
reg <
Interactive Plots with zoom options etc. can be performed using the
"iplots" library. It´s really very useful and can be downloaded from
http://stats.math.uni-augsburg.de/iPlots/index.shtml
Best regards
Chris
Barry Rowlingson wrote:
[EMAIL PROTECTED] wrote:
Hi, i don't understand how i cant zo
Sorry to ask a question so basic that it's almost silly ...
I have data which can be expressed in contingency table
form as
Factor1 Factor2 Counts1 Counts2
==
AU nAU1 nAU2
AV nAV1 nAV2
AW nAW1 nAW2
B
In Word, right click your picture, choose Format Picture,
select the Picture tab and under Color: choose Greyscale.
Depending on your printer, your print driver may also have
some options to this effect that you can access after choosing
File | Print.
There is undoubtedly a solution on the R s
The specific code given below is correct but, in general, using
POSIXt is not a good idea since it gives rise to subtle problems
with time zones.
You won't run into problems just converting character data
to POSIXlt, as the code below does, but most people who use
POSIXlt also use POSIXct and
Hi,
I could not find any information on how `read.spss' deals with date
information. As an example, I created a file containing two variables,
one numeric (values = (1, 2)) and one of type "Datum" in SPSS (german
version with values "11.02.2003" and "03.04.1999" and I get in R:
SPSSfile = url("h
Dear Ruben,
Yes, the iplots package works fine! You can download it from
http://stats.math.uni-augsburg.de/iPlots/index.shtml
Good luck!
Chris.
[EMAIL PROTECTED] wrote:
Hi, i dont understand ¿Graphics in R are interactives or not?, I hear the
the package iplots can do it (zoom, scaling etc),
Hi,
I want to create a dataframe where one of the columns does not hold
individual elements, but a vector instead.
For example, imagine the following dataframe without any vectors as elements
. . .
ex
X1 X2
1 5 15
2 7 12
3
On Wed, 2004-03-03 at 04:55, Spencer Graves wrote:
> Yes, but use sparingly, because any use of "<<-", "assign", etc.,
> from within a function to change something other than what appears in
> the standard return from a function call generates "spaghetti code" that
> is difficult to mainta
Your subject line is `partial autocorrelation', but that is not what you
describe in the body of the email.
Partial correlations are by definition between two variables, just like
correlations (but they can be partial-ed on other variables).
Please read the posting guide and the references ther
I get different results from match.call(), according to whether a function
is dispatched via S3 or S4. Specifically, when I use S4 dispatch in the
following example, the match.call() result is of length 1 less than I
expect. I need to add an extra comma to get the same results as in the S3
method
Here are three different ways (using x as defined in your
post):
with( month.day.year(x), day.of.week(month,day,year) )
do.call( "day.of.week", month.day.year(x) )
as.numeric(x-3)%%7 # uses fact that chron(3) is Sunday
---
Date: Mon, 1 Mar 2004 22:18:33 +0530
From: Ajay Shah <[EMAIL
Barry Rowlingson wrote:
[EMAIL PROTECTED] wrote:
Hi, i don't understand how i cant zoom in and zoom out a graphics
(plots)
exist a package for that? Thanks Ruben
I dont think you can do it quite like you can zoom in and out in a
program like 'photoshop'. All you can really do is redraw the pl
Hi, i dont understand ¿Graphics in R are interactives or not?, I hear the
the package iplots can do it (zoom, scaling etc), is true that?, Thanks Ruben
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[EMAIL PROTECTED] mailing list
https://www.stat.math.ethz.ch/mailman/listinfo/r-help
PLEASE do read
see plotCI in package gregmisc
---
Date: 03 Mar 2004 12:00:12 +0100
From: Peter Dalgaard <[EMAIL PROTECTED]>
To: Spencer Graves <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Subject: Re: [R] plot(x,y) with errors
[...]
I think there are special "plot with errorbars" in some of the "
Context: Windows XP, R 1.8.1
I'm studying Venables-Ripley "MASS" book and having a go at the many examples
in library MASS. The code I'm checking (from script ch04.R) now is
..
data(swiss)
splom(~ swiss, aspect = "fill",
panel = function(x, y, ...) {
panel.xyplot(x, y, ...); panel.lo
Here are a number of options:
plotOHLC(x) # in tseries package
or
require(chron)
x <- get.hist.quote( your.symbol, origin = chron(0) )
tt <- chron( time( x ) )
plot(tt, x[,"Close"])
or
# using tt and x just calculated
require(zoo)
plot(zoo(x,tt))
or
see Giles' post on his its package
thank you very much,
all the solutions work.
Regards,
Copex
Peter Dalgaard <[EMAIL PROTECTED]> wrote:
Spencer Graves writes:
> Does the following do what you want: > x<-c(1,2,3,4,5,6,7,8,9,10)
> > y<-c(1.3,2.5,4.6,5.3,5.9,6.7,7.4,8.5,9.4,10.4)
> > erry<-c(0.2,0.3,0.2,0.1,0.4,0.2,0.3,0.4,0.3,0.2)
Issue the command:
debug(eat.a.file)
and then run your eat.a.file call. This will step you through
the function showing where it goes wrong.
By the way,
- you don't need semicolons at the end of each line,
- the format you are specifying on chron is the default anyways
so you can leave ou
Spencer Graves <[EMAIL PROTECTED]> writes:
> Does the following do what you want: > x<-c(1,2,3,4,5,6,7,8,9,10)
> > y<-c(1.3,2.5,4.6,5.3,5.9,6.7,7.4,8.5,9.4,10.4)
> > erry<-c(0.2,0.3,0.2,0.1,0.4,0.2,0.3,0.4,0.3,0.2)
> >
> > plot(x, y)
> >
> > n <- length(x)
> > Y <- array(c(y-erry, y+e
If you use priceIts() in package 'its' (Irregular Time Series), you get
similar functionality, and the labelling etc in plot() recognizes the
calendar. You can also do further calendar-based extractions, etc.
- Giles
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECT
Dear R-list,
I try to calculate the 10- and 90-Percentile from grouped data.
Using 'lappy' it works fine but I have no success with 'gapply':
df <- data.frame(test = runif(100))
df <- df[order(df[,1]),]
z <- rep(1:10, each = 10)
lapply(split(df, z), function(x) quantile(x, probs=c(10,90)/100))
Yes, but use sparingly, because any use of "<<-", "assign", etc.,
from within a function to change something other than what appears in
the standard return from a function call generates "spaghetti code" that
is difficult to maintain. A month or a year from now, someone (the
developer or
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