Hi guys,
I got a problem when I was trying to use lattice to do some plot. Below is
one working example which can generate one curve for A and one curve for B
in each subplot. However, I would like to just show the points for B, not
connecting the dots. As for A, I still want a curve (dots are con
On Mar 13, 2011, at 1:28 AM, Dennis Murphy wrote:
Hi:
(1) The first argument to get.hist.quote() is instrument, not
instruments. I
concur with David that get.hist.quote()
takes a single character string as an argument.
(2) I tried running this with lapply() but got a download error on
Hi:
(1) The first argument to get.hist.quote() is instrument, not instruments. I
concur with David that get.hist.quote()
takes a single character string as an argument.
(2) I tried running this with lapply() but got a download error on the last
one:
getStockData("PBR-B")
trying URL '
http:
On Mar 12, 2011, at 11:35 PM, johnbeamer wrote:
Hello
I have a large series of data value -- effectivly say the point
across the
x-axis where a pitch crosses home plate. What I want to do is find
the % of
ground balls at various distances across home plate.
I therefore need to 'bin' the
On Mar 13, 2011, at 12:18 AM, algotr8der wrote:
Thanks David for the reply. I just tried the following with the same
result:
library(tseries)
tickers <- read.csv("testticker.txt", header=FALSE, sep=",")
tickers <- tickers[1]
V1
1 XOM
2 COP
3 PBR-A
4 FFIV
5SU
6 PBR-B
tickers
Hello
I have a large series of data value -- effectivly say the point across the
x-axis where a pitch crosses home plate. What I want to do is find the % of
ground balls at various distances across home plate.
I therefore need to 'bin' the two data sets I have - plate location for
ground balls an
Hi everyone .
I would like to integrate R with Microsoft Silverlight . I would like to
know whether we can connect R through any other software . If possible how ?
Could anyone help me with this
Thanks
--
Regards,
Vikram Selvaraju
Graduate Student
The University of Texas at Dallas
[[al
thank you！
--
View this message in context:
http://r.789695.n4.nabble.com/confirmatory-factor-analysis-in-R-tp3347618p3351230.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailma
Thanks David for the reply. I just tried the following with the same result:
>library(tseries)
>tickers <- read.csv("testticker.txt", header=FALSE, sep=",")
>tickers <- tickers[1]
V1
1 XOM
2 COP
3 PBR-A
4 FFIV
5SU
6 PBR-B
> tickers$V1 <- as.character(tickers$V1)
> tickers$V1
[1] "
If you are always using asp=1 then you can just do something like:
x <- rnorm(100, 1, 5)
y <- rnorm(100, 3, 4)
plot( x,y, asp=1)
r=2
nseg=360
x.cent <- 5
y.cent <- 7
xx <- x.cent + r*cos( seq(0,2*pi, length.out=nseg) )
yy <- y.cent + r*sin( seq(0,2*pi, length.out=nseg) )
lines(xx,yy, col='re
On Mar 12, 2011, at 7:57 PM, Mark Linderman wrote:
I am new to R and am sure this is simple, but I been unable to find a
solution.
I have 5 columns of data labeled "X", "Y", "A","B","C". I can easily
xyplot(Y ~ X | A) but I want the colors of the symbols to be based
upon the
values of B an
I am new to R and am sure this is simple, but I been unable to find a
solution.
I have 5 columns of data labeled "X", "Y", "A","B","C". I can easily
xyplot(Y ~ X | A) but I want the colors of the symbols to be based upon the
values of B and the shape of the symbols to be determined by C. There
Here is a working example that allows you to interactively label curves where
you want:
x <- c(200,210,230)
y1 <- c(0.1, 0.13, 0.1)
y2 <- c(0.11, 0.15,0.1)
y3 <- c(0.12,0.17,0.12)
df <- data.frame(x, y1,y2,y3)
lab <- c("0.1 mg/l","0.2 mg/l","0.3 mg/l")
plot(df$x, df$y1, type="l", col="blu
use the sqldf package:
> require(sqldf)
> a
time x
1 1.0 4
2 2.2 5
3 5.2 6
> b
time y
10 1
21 3
32 5
44 7
55 9
> sqldf("
+ select a.time, a.x, b.y
+ from a, b
+ where abs(a.time - b.time) < 0.5
+ ")
time x y
1 1.0 4 3
2 2.2 5 5
3 5.2 6 9
>
Sorry Alan but it's not my function. I just was looking for something similar
a while ago and googled to find The
ost whose url I sent you will have the name of the creator.
Actually I think I ended up just using symbols for my graph.
--- On Sat, 3/12/11, Allan Kamau wrote:
> From: Allan Kamau
On Mar 12, 2011, at 6:35 PM, derek wrote:
Problem solved:
require(graphics)
require(stats)
require(hmisc)
HBA=read.table("file",header =FALSE,sep = "", dec = ",",fileEncoding
= "",
encoding = "unknown", skip=60,nrows=210)
attach(HBA)
labcurve( list( One= list( V1,V2), Two= list( V3,V4),
Problem solved:
require(graphics)
require(stats)
require(hmisc)
HBA=read.table("file",header =FALSE,sep = "", dec = ",",fileEncoding = "",
encoding = "unknown", skip=60,nrows=210)
attach(HBA)
labcurve( list( One= list( V1,V2), Two= list( V3,V4),
Three=list( V5,V6), Four= list( V7,V8), Five= l
Please look at the labcurve function in the Hmisc package. Here is one line
from the documentation of ?labcurve
For automatic positioning of labels or keys, a curve is labeled at a point
that
is maximally separated from all of the other curves.
As you, the documentation uses the same words you
Hi all,
I don't know if anyone has any thoughts on this. I have been trying to move
from SAS Proc Mixed to R nlme and have an unusual result.
I have several subjects measured at four timepoints. I want to model the
within-subject correlation using an autoregressive structure. I've attached
the R
Thank you for reply. I think, in script you posted, I have to manually add
coordinates for label. I don't want do it manually. I would like to add
labels at the points, where the lines are maximally separated. (points of
maximum separation). Graphic representation of my data looks very similar to
i
On Mar 12, 2011, at 4:14 PM, flymer wrote:
Dear All,
Debuting in R, I'm facing a problem.
I have 2 vectors, say 'a' et 'b', and I'd like to merge them
according to
the proximity of their variable 'time'.
How to do to keep elements which satisfy (for example) 'a$time-b
$time<0.5'?
For exa
On Mar 12, 2011, at 3:08 PM, algotr8der wrote:
I am new to R so I apologize if my question is trivial. I have not
been able
to figure out whether what I want to do is even possible.
I have a data frame of stock ticker symbols which I store into R
space from
a txt file as follows:
tickers
Dear All,
Debuting in R, I'm facing a problem.
I have 2 vectors, say 'a' et 'b', and I'd like to merge them according to
the proximity of their variable 'time'.
How to do to keep elements which satisfy (for example) 'a$time-b$time<0.5'?
For example :
> a
time x
1 1.0 4
2 2.2 5
3 5.2 6
>
I am new to R so I apologize if my question is trivial. I have not been able
to figure out whether what I want to do is even possible.
I have a data frame of stock ticker symbols which I store into R space from
a txt file as follows:
tickers <- read.csv("stocks.txt", header=FALSE, sep=",")
ticke
There are instances when assignment
should be indicated by "<-" rather than "=".
One example is the function Sys.time().
sytem.time( x=1:1) # x is interpreted as an argument and is an
error here.
system.time( x<-1:1000) # times the creation of x with numbers 1 to 1
However, groupin
Hi:
Since you didn't provide any data, you're stuck with my example instead.
# Generate three sine curves on (0, 2*pi)
x <- seq(0, 2 * pi, length = 200)
df <- data.frame(x, y1 = sin(x), y2 = sin(2 * x), y3 = sin(4 * x))
# Create a small function to type (n), where n is a number to be input
mytex
On Sat, Mar 12, 2011 at 10:22 PM, David Winsemius
wrote:
>
> On Mar 12, 2011, at 10:56 AM, Allan Kamau wrote:
>
>>> --- On Sat, 3/12/11, Allan Kamau wrote:
>>>
From: Allan Kamau
Subject: [R] Drawing a circle on an existing graph.
To: r-help@r-project.org
Received: Saturday, M
On Mar 12, 2011, at 10:56 AM, Allan Kamau wrote:
--- On Sat, 3/12/11, Allan Kamau wrote:
From: Allan Kamau
Subject: [R] Drawing a circle on an existing graph.
To: r-help@r-project.org
Received: Saturday, March 12, 2011, 3:08 AM
I would like to draw a circle on a
graph I am plotting in R and
Hello R
I can not figure out, how to add text string or number to lines like on this
image:
http://r.789695.n4.nabble.com/file/n3350452/graph.png
I have some data series in text file.
My script:
require(graphics)
require(stats)
graf=read.table("file",header =FALSE,sep = "", dec = ",",fileEncodin
> --- On Sat, 3/12/11, Allan Kamau wrote:
>
>> From: Allan Kamau
>> Subject: [R] Drawing a circle on an existing graph.
>> To: r-help@r-project.org
>> Received: Saturday, March 12, 2011, 3:08 AM
>> I would like to draw a circle on a
>> graph I am plotting in R and use
>> the current plot's coordi
Stefan me.com> writes:
[snip]
> I'm just wondering if there is a equivalent to the fit function of
> gnuplot in R?
> If not, s there any good resource on fitting gaussians into a curve?
Try
?nls
?predict.nls
__
R-help@r-project.org mailing li
У Срд, 09/03/2011 у 23:29 -0800, Dennis Murphy піша:
> Hi:
>
> Here's one approach, although I imagine there are more efficient ways.
>
> # A function to strip spaces and return the first three non-blank elements
> of a string
> keyset <- function(x) substr(gsub(' ', '', x)[1], 1, 3)
Hello to ev
I would avoid eval(substitute(...)) in the function.
It makes the function hard to use in more general cases
(and hard to understand and subject to giving incorrect
answers in some cases).
Instead "quote" (or 'quote') the literal string that you pass
into the function. When you pass blue into the
The nls function does nonlinear least squares fits.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Stefan
>
So we'd had to debug either your code (which we do not have) or your
database server (or its connection).
Hence we cannot so very much without the code or reproducible examples.
Please also tell us the warnings you got when saying warnings().
Uwe Ligges
On 10.03.2011 16:51, Dr. Alireza Zol
Vlatka Matkovic Puljic gmail.com> writes:
>
> That was also my first thought.
> But I guess it has something to do with W and phihat
> (which I'm struggling to check
Again, it would help to post a reproducible example ...
hard to debug/diagnose by remote control. If you can't
possibly post
That was also my first thought.
But I guess it has something to do with W and phihat
(which I'm struggling to check?)
2011/3/12 Ben Bolker
> Vlatka Matkovic Puljic gmail.com> writes:
>
> >
> > >
> My first guess would have been that you had zeros and ones in
> your data, but you seem to h
Thank you so much, David. Your solution exactly suits my need.
formula() seems the key.
appreciate your help!
On Sat, Mar 12, 2011 at 10:22 AM, David Winsemius
wrote:
>
> On Mar 12, 2011, at 10:10 AM, Wensui Liu wrote:
>
>> Good morning, dear listers
>>
>> I am wondering how to do string evalua
Vlatka Matkovic Puljic gmail.com> writes:
>
> Maybe I should include data:
>
> results$cond
> [1] 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
[snip]
> [205] 0.950 0.960 0.980 0.980 0.999
>
> 2011/3/12 Vlatka Matkovic Puljic gmail.com>
>
> > I'm trying to do betar
On Mar 12, 2011, at 10:10 AM, Wensui Liu wrote:
Good morning, dear listers
I am wondering how to do string evaluation such that
model <- glm(Y ~ [STRING], data = mydata) where STRING <- "x1 + x2 +
x3"
It is very doable in other language such as SAS.
Also "very doable" in R. You need to
2011/3/11 Albyn Jones
> but presumably what you really want would be based on a joint confidence
> region for all the proportions.
>
I've had read "On Exact Methods for Testing Equality of Binomial
Proportions" by Akihito Matsuo, but still, this concept is for me unclear
and I got lost...
We h
Good morning, dear listers
I am wondering how to do string evaluation such that
model <- glm(Y ~ [STRING], data = mydata) where STRING <- "x1 + x2 + x3"
It is very doable in other language such as SAS.
Thank you so much for your insight!
__
R-help@r-
On Sat, Mar 12, 2011 at 3:08 AM, Allan Kamau wrote:
> I would like to draw a circle on a graph I am plotting in R and use
> the current plot's coordinate system.
> The most basic functionality I am looking for is the ability to plot a
> circle on an already existing graph by simply providing the x
Is this of any use? http://tolstoy.newcastle.edu.au/R/help/06/04/25821.html
--- On Sat, 3/12/11, Allan Kamau wrote:
> From: Allan Kamau
> Subject: [R] Drawing a circle on an existing graph.
> To: r-help@r-project.org
> Received: Saturday, March 12, 2011, 3:08 AM
> I would like to draw a circle
It may be a stupid question but have you loaded reshape?
library(reshape)
?melt
--- On Sat, 3/12/11, Deepika wrote:
> From: Deepika
> Subject: [R] how to use melt cast commands in R in window7
> To: r-help@r-project.org
> Received: Saturday, March 12, 2011, 3:45 AM
> Hi,
> I have installed R
Dear sir,
Thanks a lot for the solution.
It was such a simple solution, but people like me close their minds and don't
think of data.frame as a whole and keep on thinking about vector elements only.
I also almost got the solution when I tried
qq = unique(pp)
(qq <- sub(" .*","", qq)) but this
Hello R users,
I'm trying to do simulations for comparing cox and weibull
I have come across this problem:
Warning messages:
1: In survreg.fit(X, Y, weights, offset, init = init, controlvals = control,
:
Ran out of iterations and did not converge
2: In survreg.fit(X, Y, weights, offset, init
Please read the posting guide. "It is not working" doesn't give us any
clues. Tell us what you did and what happened.
Best,
Ista
On Sat, Mar 12, 2011 at 8:45 AM, Deepika wrote:
> Hi,
> I have installed R on my computer with windows 7 . I also installed reshape
> software, but I am not being able
On Sat, Mar 12, 2011 at 03:20:01AM -0800, Vincy Pyne wrote:
> Dear R helpers
>
> Suppose I have a data frame as given below
>
> mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8,
> 8, 8, 7, 7, 2, 2, 4))
>
[...]
>
> unique(mydat$x) will give me 1, 2, 5, 6? i.e. 4 valu
Hello,
On Mar 12, 2011, at 8:32 AM, Maas James Dr (MED) wrote:
Question 1:
I have a long list of variable names such as
first <- c("one","two","three")
and what I want to do is create a list of lists ... where the names
of each of overall lists components are "one","two", and "three".
Th
Question 1:
I have a long list of variable names such as
first <- c("one","two","three")
and what I want to do is create a list of lists ... where the names of each of
overall lists components are "one","two", and "three".
This is the same result as
second <- list(one=list(),two=list(),three
Hi:
The value of your title is a(n attached) data frame - perhaps you meant
'HMF' instead...
HTH,
Dennis
On Sat, Mar 12, 2011 at 2:41 AM, derek wrote:
> Hello R users, Im having this strange problem.
> http://r.789695.n4.nabble.com/file/n3350024/bad2.png
>
> I have txt file of this format (X.X
Solution:
main=HMF -> main="HMF"
--
View this message in context:
http://r.789695.n4.nabble.com/plot-generates-graph-with-coordinantes-written-over-it-tp3350024p3350163.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.o
Hi there,
I'm new to R, coming from gnuplot for the plotting facility...
I'm just wondering if there is a equivalent to the fit function of gnuplot in R?
If not, s there any good resource on fitting gaussians into a curve?
greetings,
Stefan
__
R-hel
Hello R users, Im having this strange problem.
http://r.789695.n4.nabble.com/file/n3350024/bad2.png
I have txt file of this format (X.X, Y.Y - random numbers, ...) with data
points:
nm A nm A nm A
X.X Y.Y X.X XX X.X
X.X Y.Y Y.Y Y.Y Y.Y
My script looks like this:
require(graphics)
require(stats)
I would like to draw a circle on a graph I am plotting in R and use
the current plot's coordinate system.
The most basic functionality I am looking for is the ability to plot a
circle on an already existing graph by simply providing the xy
coordinates for it's centre and the radius.
I am also look
Hi,
I have installed R on my computer with windows 7 . I also installed reshape
software, but I am not being able to work with melt cast commands . I have
chjecked the commands.It is not working.
Thankyou,
Deepika
[[alternative HTML version deleted]]
__
Thanks sir for your reply. Unfortunately I couldn't figure out the solution.
Vincy
--- On Sat, 3/12/11, Dennis Murphy wrote:
From: Dennis Murphy
Subject: Re: [R] Identifying unique pairs
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Saturday, March 12, 2011, 11:45 AM
Hi:
This problem
Maybe I should include data:
results$cond
[1] 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
[13] 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
[25] 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001 0.001
[37] 0.001 0.001 0.020 0.0
On Sat, 12 Mar 2011, Vlatka Matkovic Puljic wrote:
Dear R users,
I'm trying to do betareg on my dataset.
Dependent variable is not normally distributed and is proportion (of condom
use (0,1)).
But I'm having problems:
gyl<-betareg(cond ~ alcoh + drug, data=results)
Error in optim(par = start,
Dear R users,
I'm trying to do betareg on my dataset.
Dependent variable is not normally distributed and is proportion (of condom
use (0,1)).
But I'm having problems:
gyl<-betareg(cond ~ alcoh + drug, data=results)
Error in optim(par = start, fn = loglikfun, gr = gradfun, method = method, :
ini
Hi:
This problem came up the other day - see
http://stats.stackexchange.com/questions/7884/fast-ways-in-r-to-get-the-first-row-of-a-data-frame-grouped-by-an-identifier/7985#7985
Dennis
On Sat, Mar 12, 2011 at 3:20 AM, Vincy Pyne wrote:
> Dear R helpers
>
> Suppose I have a data frame as given
On 10 March 2011 02:07, Xiaobo Gu wrote:
>>
> set transactions ...[35 item(s), 8 transaction(s)] done [0.00s].
>>
>> That does not look right?
>
> I think it's because there are to few sample records, so all the rules
> are with 100% confidence
Sorry - I think you might have misunderstoo
Dear Dennis and Peter,
Thank you for the suggestions - they work very well with my example data.
I *think* I've got it up and running using my real data (with 2000 columns)
too, with:
sprintf("%s%04d", "X", c(1:2000))
then the stacking/unstacking operations.
Thanks again for the help
Best wish
On 2011-03-12 02:01, Dennis Murphy wrote:
Hi:
It would help if you named your variables such that alphanumeric ordering
doesn't disturb your variable ordering. Having been burned by this a few
times, I've learned the basics of sprintf() :) Here's your example
revisited, along with an alternative
Dear R helpers
Suppose I have a data frame as given below
mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8,
8, 8, 7, 7, 2, 2, 4))
mydat
x y
1 1 10
2 1 10
3 1 10
4 2 8
5 2 8
6 2 8
7 2
On 2011-03-11 17:10, jonbfish wrote:
Thanks again. I guess I have been fitting in splus gui and in R Commander, not
really calling it myself. I was thinking maybe there was a way to define it as
a class or something.
I will have to look at segments. Any thoughts if this might be easy to use wi
Hi:
It would help if you named your variables such that alphanumeric ordering
doesn't disturb your variable ordering. Having been burned by this a few
times, I've learned the basics of sprintf() :) Here's your example
revisited, along with an alternative stacking/unstacking display with
package re
Hi:
One answer is eval(substitute(...)): [Note: it's a good idea NOT to use the
same name on different objects - R is pretty good about distinguishing TEST
the function from TEST the data frame, but you could be playing with fire if
both objects have the same class but have different contents. Thi
Dear R users,
I'm having some problems with the stack() and unstack() functions, and
wondered if you could help.
I have a large data frame (400 rows x 2000 columns), which I need to reduce
to a single column of values (and therefore 80 rows), so that I can use
it in other operations (e.g.,
70 matches
Mail list logo
