[R] xyplot() groups scope issue
Hi all,
Tinkering with a wrapper for xyplot that will help me plot a bunch of
plots in a data analysis I'm doing and I ran into an odd error that
I'm guessing is a scope issue. Here's a very simple version of the code:
###
#load lattice
library(lattice)
#create the wrapper function
do.xyplot = function( plot.formula, data.frame, plot.groups){
print(plot.groups) #show me what the wrapper function thinks
'plot.groups' is
xyplot(
plot.formula
,data = data.frame
,groups = eval(plot.groups)
)
}
#create some data
mydata = as.data.frame(cbind( x = c(1:4), y = rep(1:2), z = rep(1:2,
each = 2) ))
#try to plot the data (fails)
do.xyplot(
plot.formula=formula(x~y)
,data.frame = mydata
,plot.groups = expression(z)
)
#after error message try again, this time declaring plot.groups as a
global variable (succeeds)
plot.groups = expression(z)
do.xyplot(
plot.formula=formula(x~y)
,data.frame = mydata
,plot.groups = expression(z)
)
#
I'm fine with declaring plot.groups before each call to do.xyplot,
but I'm curious if there's a simpler solution.
Mike
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://memetic.ca
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
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Re: [R] logistic regression
Maybe try making sure the data is numeric:
fac.to.num=function(x) as.numeric(as.character(x))
On 26-Jul-07, at 9:34 AM, Sullivan, Mary M wrote:
Greetings,
I am working on a logistic regression model in R and I am
struggling with the code, as it is a relatively new program for
me. In searching Google for 'logistic regression diagnostics' I
came Elizabeth Brown's Lecture 14 from her Winter 2004
Biostatistics 515 course (http://courses.washington.edu/b515/
l14.pdf) . I found most of the code to be very helpful, but I am
struggling with the lines on to calculate the observed and expected
values in the 10 groups created by the cut function. I get error
messages in trying to create the E and O matrices: R won't accept
assignment of fi1c==j and it won't calculate the sum.
I am wondering whether someone might be able to offer me some
assistance...my search of the archives was not fruitful.
Here is the code that I adapted from the lecture notes:
fit - fitted(glm.lyme)
fitc - cut(fit, br = c(0, quantile(fit, p = seq(.1, .9, .1)),1))
t-table(fitc)
fitc - cut(fit, br = c(0, quantile(fit, p = seq(.1, .9, .1)), 1),
labels = F)
t-table(fitc)
#Calculate observed and expected values in ea group
E - matrix(0, nrow=10, ncol = 2)
O - matrix(0, nrow=10, ncol=2)
for (j in 1:10) {
E[j, 2] = sum(fit[fitc==j])
E[j, 1] = sum((1- fit)[fitc==j])
O[j, 2] = sum(pcdata$lymdis[fitc==j])
O[j, 1] = sum((1-pcdata$lymdis)[fitc==j])
}
Here is the error message: Error in Summary.factor(..., na.rm =
na.rm) :
sum not meaningful for factors
I understand what it means; I just can't figure out how to get
around it or how to get the output printed in table form. Thank
you in advance for any assistance.
Mary Sullivan
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PLEASE do read the posting guide http://www.R-project.org/posting-
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and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://memetic.ca
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] using R for a reaction-time experiment
Psych grad student here, R user for 3 years. Using R for experiments is likely not advisable as it has no fine control of display timing (ex synching stimuli to the screen refresh, etc). On recommendation of colleagues I'm learning the Python language, which has a module called 'pygame' that is fantastic for psych experiments. Mike On 22-Jul-07, at 1:38 PM, Seth Roberts wrote: I want to use R to run a reaction-time experiment: Something appears on the screen, I respond by typing something (one keystroke), the system measures the speed of my response. R would be great for this if only I didn't have to hit Enter to enter that keystroke. I am doing such experiments now but they require two actions per trial: hit keystroke, hit Enter. Is there some way that R can be made to respond to a single keystroke (other than Enter)? -- View this message in context: http://www.nabble.com/using-R-for-a- reaction-time-experiment-tf4125643.html#a11732474 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (R) Using arguments for the empirical cumulative distribution function
?plot.ecdf() On 20-Jul-07, at 6:57 AM, squall44 wrote: Is there no one who can help me? :,( -- View this message in context: http://www.nabble.com/%28R%29-Using- arguments-for-the-empirical-cumulative-distribution-function- tf4111355.html#a11705448 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] linear interpolation of multiple random time series
Hi all,
Looking for tips on how I might more optimally solve this. I have
time series data (samples from a force sensor) that are not
guaranteed to be sampled at the same time values across trials. ex.
trial timex
1 1 1
1 5 4
1 7 9
1 12 20
2 1 0
2 3 5
2 9 10
2 13 14
2 19 22
2 24 32
Within each trial I'd like to use linear interpolation between each
successive time sample to fill in intermediary timepoints and x-
values, ex.
trial timex
1 1 1
1 2 1.75
1 3 2.5
1 4 3.25
1 5 4
1 6 6.5
1 7 9
1 8 11.2
1 9 13.4
1 10 15.6
1 11 17.8
1 12 20
2 1 0
2 2 2.5
2 3 5
2 4 5.83
2 5 6.67
2 6 7.5
2 7 8.33
2 8 9.17
2 9 10
2 10 11
2 11 12
2 12 13
2 13 14
2 14 15.3
2 15 16.7
2 16 18
2 17 19.3
2 18 20.7
2 19 22
2 20 24
2 21 26
2 22 28
2 23 30
2 24 32
The solution I've coded (below) involves going through the original
data frame line by line and is thus very slow (indeed, I had to
resort to writing to file as with a large data set I started running
into memory issues if I tried to create the new data frame in
memory). Any suggestions on a faster way to achieve what I'm trying
to do?
#assumes the first data frame above is stored as 'a'
arows = (length(a$x)-1)
write('', 'temp.txt')
for(i in 1:arows){
if(a$time[i+1] a$time[i]){
write.table(a[i,], 'temp.txt', row.names = F, col.names = F,
append
= T)
x1 = a$time[i]
x2 = a$time[i+1]
dx = x2-x1
if(dx != 1){
y1 = a$x[i]
y2 = a$x[i+1]
dy = y2-y1
slope = dy/dx
int = -slope*x1+y1
temp=a[i,]
for(j in (x1+1):(x2-1)){
temp$time = j
temp$x = slope*j+int
write.table(temp, 'temp.txt', row.names = F,
col.names = F,
append = T)
}
}
}else{
write.table(a[i,], 'temp.txt', row.names = F, col.names = F,
append
= T)
}
}
i=i+1
write.table(a[i,], 'temp.txt', row.names = F, col.names = F, append = T)
b=read.table('temp.txt',skip=1)
names(b)=names(a)
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Re: [R] linear interpolation of multiple random time series
Jim Holtman's solution works great, and will try the zoo method just for fun as well. Thanks to all of you :) Mike -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to read many files at one time?
On 14-Jul-07, at 6:16 PM, Zhang Jian wrote:
I want to load many files in the R. The names of the files are
Sim1.txt,
Sim2.txt, Sim3.txt, Sim4.txt, Sim5.txt and so on.
Can I read them at one time? What should I do? I can give the same
names in
R.
Thanks.
For example:
tst=paste(Sim,1:20,.txt,sep=) # the file names
tst
[1] Sim1.txt Sim2.txt Sim3.txt Sim4.txt Sim5.txt
Sim6.txt
[7] Sim7.txt Sim8.txt Sim9.txt Sim10.txt Sim11.txt
Sim12.txt
[13] Sim13.txt Sim14.txt Sim15.txt Sim16.txt Sim17.txt
Sim18.txt
[19] Sim19.txt Sim20.txt
data.name=paste(Sim,1:20,sep=) # the file names in R
data.name
[1] Sim1 Sim2 Sim3 Sim4 Sim5 Sim6 Sim7
Sim8 Sim9
[10] Sim10 Sim11 Sim12 Sim13 Sim14 Sim15 Sim16
Sim17 Sim18
[19] Sim19 Sim20
you could read each data file as a named element into a single list
data=list(NULL)
for(i in 1:length(tst)){
data[[i]]=read.table(tst[i])
names(data)[i]=data.name[i]
}
if you want to refer to Sim1, type
data$Sim1
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[R] Power calculation with measurement error
Hi all, Hopefully this will be quick, I'm looking for pointers to packages/ functions that would allow me to calculate the power of a t.test when the DV has measurement error. That is, I understand that, ceteris paribus, experiments using measure with more error (lower reliability) will have lower power. Mike -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] connecting to running process possible?
If you get the running process to write its results to file, then you could run a loop in R asking it to repeatedly read the file and plot the contents. It may work best if you plot to a file like a pdf. On 22-Jun-07, at 1:06 PM, Charles Cosse wrote: Hello, i'm trying to find a more modern system to reproduce the functionality that was available through the Histoscope program (from Fermilab). Namely, the capability of connecting to a running process and having plots update in realtime in response to new data. Is this possible with R? Thank you, Charles Cosse [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] surprising difference in log()
According to the description of floor(), the latter result is the correct one: 'floor takes a single numeric argument x and returns a numeric vector containing the largest integers *not greater than* the corresponding elements of x.' (emphasis added) floor(3) == 2 True On 26-Jun-07, at 11:09 AM, Fausto Galli wrote: Hello everybody My collegue and I noticed a strange behaviour of R on different platforms. It's a simple computation, but results are rather different. On Windows XP: floor(log(8,2)) [1] 3 which is what one should expect. Here's instead the result with Mac OS X (same version, 2.5.0 (2007-04-23)) floor(log(8,2)) [1] 2 Is it a bug in R or in the operating system? Anyway, it's quite a surprising one. _ Fausto Galli Institute of Finance University of Lugano Via G. Buffi 13 CH-6904 Lugano, Switzerland. +41 (0)58 666 4497 http://www.people.lu.unisi.ch/gallif __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Power calculation with measurement error
Thanks Greg, I've actually been programming precisely what you suggest since sending the request this morning (though your email was indeed helpful; I've never seen 'replicate()' and will see if it's faster than a loop). However, I was hoping that an analytic solution was extant and implemented somewhere. On 26-Jun-07, at 12:18 PM, Greg Snow wrote: I don't know of a current package that does this (others may), but if you know what you expect your data to look like you can simulate it and calculate power that way. Basically, write a function that will simulate data with the level of measurement error that you expect in the real data (or have the amount of measurement error passed in as a parameter so you can examine the effect of diffenent values). Then have the function compute the t test (or other test that you plan to do) and return the p-value from the test. Then you can simulate the process with a command like: out1 - replicate( 1000, myfunction(n=25, err=.1, diff=.5) ) And compute the power with: mean( out1 0.05 ) # or whatever alpha value you want. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Mike Lawrence Sent: Tuesday, June 26, 2007 5:13 AM To: Rhelp Subject: [R] Power calculation with measurement error Hi all, Hopefully this will be quick, I'm looking for pointers to packages/ functions that would allow me to calculate the power of a t.test when the DV has measurement error. That is, I understand that, ceteris paribus, experiments using measure with more error (lower reliability) will have lower power. Mike -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] surprising difference in log()
Sorry, my mistake. I should really leave looking at the help list until AFTER my morning coffee :P On 26-Jun-07, at 11:31 AM, Duncan Murdoch wrote: On 6/26/2007 10:20 AM, Mike Lawrence wrote: According to the description of floor(), the latter result is the correct one: 'floor takes a single numeric argument x and returns a numeric vector containing the largest integers *not greater than* the corresponding elements of x.' (emphasis added) floor(3) == 2 True 3 is not greater than 3, but it is greater than 2, so the result you quote above is wrong. You should see floor(3) [1] 3 floor(3) == 2 [1] FALSE Do you really see the result you posted? Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Power calculation with measurement error
On 26-Jun-07, at 8:12 AM, Mike Lawrence wrote: Hi all, Hopefully this will be quick, I'm looking for pointers to packages/ functions that would allow me to calculate the power of a t.test when the DV has measurement error. That is, I understand that, ceteris paribus, experiments using measure with more error (lower reliability) will have lower power. I came across a reference (http://memetic.ca/reliability.pdf) that provides a formula for calculating the noncentrality parameter for tests using imperfect measures (see Eq. 4), as well as a table of some resulting power estimates. However, while I have created a (very slow) monte carlo function that so far as I can tell matches their results, when I attempt to implement their analytic solution it's way off. Can anyone see what I'm doing incorrectly? n=100 r=.5 #reliability e=.5 #effect size delta=(sqrt(r*n)/2)*e power.t.test(n,delta,sig.level=.05,alternative='one.sided') Two-sample t test power calculation n = 100 delta = 1.767767 sd = 1 sig.level = 0.05 power = 1 alternative = one.sided NOTE: n is number in *each* group Meanwhile, their tables and my monte carlo method say that the power in that circumstance should be .7 Help? Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Power calculation with measurement error
On 26-Jun-07, at 2:36 PM, Mike Lawrence wrote:
On 26-Jun-07, at 8:12 AM, Mike Lawrence wrote:
Hi all,
Hopefully this will be quick, I'm looking for pointers to packages/
functions that would allow me to calculate the power of a t.test when
the DV has measurement error. That is, I understand that, ceteris
paribus, experiments using measure with more error (lower
reliability) will have lower power.
I came across a reference (http://memetic.ca/reliability.pdf) that
provides a formula for calculating the noncentrality parameter for
tests using imperfect measures (see Eq. 4), as well as a table of
some resulting power estimates. However, while I have created a (very
slow) monte carlo function that so far as I can tell matches their
results, when I attempt to implement their analytic solution it's way
off. Can anyone see what I'm doing incorrectly?
n=100
r=.5 #reliability
e=.5 #effect size
delta=(sqrt(r*n)/2)*e
power.t.test(n,delta,sig.level=.05,alternative='one.sided')
Two-sample t test power calculation
n = 100
delta = 1.767767
sd = 1
sig.level = 0.05
power = 1
alternative = one.sided
NOTE: n is number in *each* group
Meanwhile, their tables and my monte carlo method say that the power
in that circumstance should be .7
Found it; I was using power.t.test without being thorough in reading
its details. Sorry for the spam, and for anyone that's interested,
here's the final analytic solution:
#get power for a t.test, incorporating measurement error.
#n = total number of participants across your 2 groups
#r = estimated reliability of the measure used
#e = measured effect size
get.power=function(n,r,e,tails=2,alpha=.05){
d=(sqrt(r*n)/2)*e
a=1-ifelse(tails==2,alpha/2,alpha)
p=1-pt(qt(a,n-2),n-2,d)
return(p)
}
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Re: [R] how to create cumulative histogram from two independent variables?
?plot.ecdf plot.ecdf(rnorm(100),rexp(100)) On 20-Jun-07, at 5:02 PM, Jose Borreguero wrote: Hi all, I am extremely newbie to R. Can anybody jump-start me with any clues as to how do I get a cumulative histogram from two independent variables, cumhist(X,Y) ? -jose [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://memetic.ca Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: How to set degrees of freedom in cor.test?
You could calculate the confidence interval of the correlation at
your desired df: http://davidmlane.com/hyperstat/B8544.html
The below code takes as arguments the observed correlation, N, and
alpha, calculates the confidence interval and checks whether this
includes 0.
cor.test2=function(r,n,a=.05){
phi=function(x){
log((1+x)/(1-x))/2
}
inv.phi=function(x){
(exp(2*x)-1)/(exp(2*x)+1)
}
r.prime=phi(r)
err=qnorm(1-(a/2))/sqrt(n-3)
lims=c(inv.phi(r.prime-err),inv.phi(r.prime+err))
sig=ifelse(xor(all(0lims),all(0lims)),T,F)
return(sig)
}
On 14-Jun-07, at 5:40 AM, Florence Dufour wrote:
Hello,
I want to compute a correlation test but I do not want to use the
degrees of freedom that are calculated by default but I want to set a
particular number of degrees of freedom.
I looked in the manual, different other functions but I did not found
how to do it
Thanks in advance for your answers
Yours
Florence Dufour
PhD Student
AZTI Tecnalia - Spain
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://myweb.dal.ca/mc973993
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://myweb.dal.ca/mc973993
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] question about formula for lm
Yet another solution: with(X,lm(get(Ytext)~Xvar)) On 14-Jun-07, at 5:18 PM, Greg Snow wrote: Try: lm( formula( paste( Ytext, '~ Xvar' ) ), data=X) -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Pedro Mardones Sent: Thursday, June 14, 2007 1:14 PM To: R-help@stat.math.ethz.ch Subject: [R] question about formula for lm Dear all; Is there any way to make this to work?: .x-rnorm(50,10,3) .y-.x+rnorm(50,0,1) X-data.frame(.x,.y) colnames(X)-c(Xvar,Yvar) Ytext-Yvar lm(Ytext~Xvar,data=X) # doesn't run lm(Yvar~Xvar,data=X) # does run The main idea is to use Ytext as input in a function, so you just type Yvar and the model should fit Also, I need to avoid the expression X$Yvar~X$Xvar Thanks for any idea PM __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simultaneous computing
There's RMPI tutorial: http://ace.acadiau.ca/math/ACMMaC/Rmpi/index.html On 11-Jun-07, at 9:11 AM, Markus Schmidberger wrote: Hello, which possibilities are available in R for simultaneous or parallel computing? I only could find biopara (http://cran.r-project.org/src/contrib/Descriptions/biopara.html) Are there other possibilities? Are there special groups working on simultaneous computing with R? Thanks Markus -- Dipl.-Tech. Math. Markus Schmidberger Ludwig-Maximilians-Universität München IBE - Institut für medizinische Informationsverarbeitung, Biometrie und Epidemiologie __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregate to find majority level of a factor
This should do the trick. Also labels ties with NA.
a=as.data.frame(cbind(c(1,1,1,2,2,2,3,3,3,4,4),c
('big','big','small','big','small','small','small','small','small','big'
,'small')))
a$V2=factor(a$V2)
maj=function(x){
y=table(x)
z=which.max(y)
if(sum(y==max(y))==1){
return(names(y)[z])
}else{
return(NA)
}
}
aggregate(a$V2,list(a$V1),maj)
On 31-May-07, at 4:25 PM, Thompson, Jonathan wrote:
I want to use the aggregate function to summarize data by a factor (my
field plots), but I want the summary to be the majority level of
another
factor.
For example, given the dataframe:
Plot1 big
Plot1 big
Plot1 small
Plot2 big
Plot2 small
Plot2 small
Plot3 small
Plot3 small
Plot3 small
My desired result would be:
Plot1 big
Plot2 small
Plot3 small
I can't seem to find a scalar function that will give me the majority
level.
Thanks in advance,
Jonathan Thompson
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PLEASE do read the posting guide http://www.R-project.org/posting-
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and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Department of Psychology, Dalhousie University
Website: http://myweb.dal.ca/mc973993
Public calendar: http://icalx.com/public/informavore/Public
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
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Re: [R] test to compare significant correlation increase
Just a guess (please correct if I'm way off on this), but maybe you could look at the difference in r betwen 1 2 and see if the confidence interval (http://davidmlane.com/hyperstat/B8544.html) for this value given your sample size includes 0. On 30-May-07, at 11:24 AM, David Riano wrote: Hi! I am calculating correlation between two variables: 1. X versus Y 2. X versus Y(with a 3 steps lag) I would like to test if the correlation increase/decrease from 1 to 2 is significant or not. Is there any function in R to do this? any hints? Thanks for help :) David Riaño Center for Spatial Technologies and Remote Sensing (CSTARS) University of California 250-N, The Barn One Shields Avenue Davis, CA 95616-8527 USA 1-(517) 629-5499 http://www.cstars.ucdavis.edu/~driano/index.html http://www.cstars.ucdavis.edu __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] opinions please: text editors and reporting/Sweave?
One minor warning regarding LaTeX: I have encountered journals in the psychological field (specifically, journals of the psychonomic society) that refuse to accept articles prepared in LaTeX, even if they are submitted as PDF. I'm a big LaTeX fan myself, so I really can't comprehend this. On 30-May-07, at 6:07 PM, Greg Snow wrote: Tim, First, I personnally am a big fan of LaTeX, Emacs, and ESS and I think that in the long run you would benefit from learning all of them (probably start with Emacs, then ESS, then LaTeX once you already have a knowledge of Emacs and how it can help). Since you asked about the simplest way to go, you may want to look at the odfWeave package. This gives you the power of Sweave, but without having to learn LaTeX. It works with documents in openoffice (a free office suite similar to and mostly compatible with microsoft office (word)). Using this you can create your template using openoffice writer (or MS word, then convert using openoffice), run it through R/odfWeave, and have the result as another openoffice document that can then be converted to MS word or pdf. Personally, if I am doing something for myself, or in which the output format does not matter, then I use Sweave with LaTeX (using Emacs and ESS). But, often my results need to be sent to a client that will cut and past my results into an MS word document or power point presentation. Then I find it easier to use openoffice and odfWeave and have the end result be an MS word document that can be e-mailed to the client for them to mangle in ways they feel the need to. (there is also an HTML driver for Sweave I nthe the R2HTML package where you can have the template file resemble html and the final output is in html) Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Tim Howard Sent: Wednesday, May 30, 2007 2:43 PM To: r-help@stat.math.ethz.ch Subject: [R] opinions please: text editors and reporting/Sweave? dear all - I currently use Tinn-R as my text editor to work with code that I submit to R, with some output dumped to text files, some images dumped to pdf. (system: Windows 2K and XP, R 2.4.1 and R 2.5). We are using R for overnight runs to create large output data files for GIS, but then I need simple output reports for analysis results for each separate data set. Thus, I create many reports of the same style, but just based on different input data. I am recognizing that I need a better reporting system, so that I can create clean reports for each separate R run. This obviously means using Sweave and some implementation of LaTex, both of which are new to me. I've installed MikTex and successfully completed a demo or two for creating pdfs from raw LaTeX. It appears that if I want to ease my entry into the world of LaTeX, I might need to switch editors to something like Emacs (I read somewhere that Emacs helps with the TeX markup?). After quite a while wallowing at the Emacs site, I am finding that ESS is well integrated with R and might be the way to go. Aaaagh... I'm in way over my head! My questions: What, in your opinion, is the simplest way to integrate text and graphics reports into a single report such as a pdf file. If the answer to this is LaTeX and Sweave, is it difficult to use a text editor such as Tinn-R or would you strongly recommend I leave behind Tinn and move over to an editor that has more LaTeX help? In reading over Friedrich Leisch's Sweave User Manual (v 1.6.0) I am beginning to think I can do everything I need with my simple editor. Before spending many hours going down that path, I thought it prudent to ask the R community. It is likely I am misunderstanding some of the process here and any clarifications are welcome. Thank you in advance for any thoughts. Tim Howard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch
Re: [R] polygon error?
polygon expects to be given the vertices, so if you want the area under the curve you'll want to include vertices at density=0 z - pretty(c(-3,3), 100) ht - dnorm(z) plot(z,ht, type=l) zc - 1.645 ht-ht[zzc] z-z[zzc] ht-c(0,ht,0) z-c(z[1],z,z[length(z)]) polygon(z,ht,col='red') On 26-May-07, at 7:34 AM, LL wrote: Hi.. I'm not sure why polygon returns an area above the standard normal curve. z - pretty(c(-3,3), 100) ht - dnorm(z) data - data.frame(z=z, ht=ht) zc - 1.645 plot(data, type=l) lines(data) t - subset(data, zzc) polygon(t, col=red) Thanks, Lance [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculation of ratio distribution properties
MathCad code failed to attach last time. Here it is. On 25-May-07, at 2:24 PM, Mike Lawrence wrote: I came across this reference: http://www.informaworld.com/smpp/content? content=10.1080/03610920600683689 The authors sent me code (attached with permission) in MathCad to perform the calculations in which I'm interested. However, I do not have MathCad nor experience with its syntax, so I thought I'd send the code to the list to see if anyone with more experience with R and MathCad would be interested in making this code into a function or package of some sort Mike Begin forwarded message: From: Noyan Turkkan [EMAIL PROTECTED] Date: May 25, 2007 11:09:02 AM ADT To: Mike Lawrence [EMAIL PROTECTED] Subject: Rép. : R code for 'Density of the Ratio of Two Normal Random Variables'? Hi Mike I do not know if anyone coded my approach in R. However if you have acces to MathCad, I am including a MathCad file (also in PDF) which computes the density of the ratio of 2 dependent normal variables, its mean variance. If you do not have acces to MathCad, you will see that all the computations can be easily programmed in R, as I replaced Hypergeometric function by the Erf function. I am not very familiar with R but the erf function may be programmed as : (erf - function(x) 2*pnorm(x *sqrt(2)) - 1). Good luck. Noyan Turkkan, ing. Professeur titulaire directeur / Professor Head Dépt. de génie civil / Civil Eng. Dept. Faculté d'ingénierie / Faculty of Engineering Université de Moncton Moncton, N.B., Canada, E1A 3E9 Mike Lawrence [EMAIL PROTECTED] 05/25/07 9:20 am Hi Dr. Turkkan, I am working on a problem that necessitates the estimation of the mean and variance of the density of two dependent normal random variables and in my search for methods to achieve such estimation I came across your paper 'Density of the Ratio of Two Normal Random Variables' (2006). I'm not a statistics or math expert by any means, but I am quite familiar with the R programming language; do you happen to know whether anyone has coded your approach for R yet? Cheers, Mike -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein Ratio2depNV.pdf -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculation of ratio distribution properties
According to the paper I cited, there is controversy over the sufficiency of Hinkley's solution, hence their proposed more complete solution. On 25-May-07, at 2:45 PM, Lucke, Joseph F wrote: The exact ratio is given in On the Ratio of Two Correlated Normal Random Variables, D. V. Hinkley, Biometrika, Vol. 56, No. 3. (Dec., 1969), pp. 635-639. -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calculation of ratio distribution properties
Hi all, Looking to calculate the expected mean and variance of a ratio distribution where the source distributions are gaussian with known parameters and sample values are correlated. I see (from wikipedia: http://en.wikipedia.org/wiki/ Ratio_distribution#Gaussian_ratio_distribution) that this calculation is quite involved, so I'm hoping that someone has already coded a function to achieve this. Thanks, Mike -- Mike Lawrence Graduate Student, Department of Psychology, Dalhousie University Website: http://myweb.dal.ca/mc973993 Public calendar: http://icalx.com/public/informavore/Public The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Catenating factors.
It may be poor form, but whenever I deal with factors in this way I find as.character and as.numeric useful ex. #if your factor levels are letters z=factor(unique(c(as.character(x),as.character(y #if your factor levels are numbers z=factor(unique(c(as.numeric(as.character(x)),as.numeric(as.character (y) On 22-May-07, at 11:56 PM, Marc Schwartz wrote: On Wed, 2007-05-23 at 13:38 +1200, Rolf Turner wrote: I was recently asked by one of new colleagues how to combine, or catentate, or concatentate two factors. If x and y are factors, doing c(x,y) appears to coerce x and y to numeric mode before catenating them. So what does one do if one wants the result to be a factor whose levels are the union of the levels of x and y? I vaguely recall seeing this discussed in r- help, but I can't find anything on it there, nor in the FAQ. Am I missing something obvious? Rolf, I think that this thread from last November on R-devel may be helpful: http://tolstoy.newcastle.edu.au/R/e2/devel/06/11/1130.html HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- Mike Lawrence Graduate Student, Dalhousie University Department of Psychology http://myweb.dal.ca/mc973993 The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Parallel processes
You can find a decent tutorial on using RMPI here:
http://ace.acadiau.ca/math/ACMMaC/Rmpi/index.html
On 22-May-07, at 3:46 PM, Kuhn, Max wrote:
Erin,
There is a snow package (note the case) and also a few others.
Rlsf, is specific to grids/clusters that use the LSF queue system.
More generally, the nws package is more sophisticated and should
work on
most systems. Also, there is Rmpi (which I haven't used).
Max
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of hodgess
Sent: Tuesday, May 22, 2007 1:01 PM
To: [EMAIL PROTECTED]
Subject: [R] Parallel processes
Dear R People:
I was wondering if there were any packages for parallel
programming in
R.
According to the R-help, at one time there was a package called SNOW.
It doesn't seem
to exist anymore.
Any help would be much appreciated!
Sincerely,
Erin Hodgess
mailto: [EMAIL PROTECTED]
[[alternative(swapped) HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Dalhousie University Department of Psychology
http://myweb.dal.ca/mc973993
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequentially process a list
Try nested ifelse() statements to label the group
ex.
time$group=ifelse(time$time5,NA,
ifelse(time$time9,1,
ifelse(time$time13,NA,
ifelse(time$time17,2,NA)
)
)
)
Then use aggregate to find the max value.
ex.
time.max=aggregate(time$value,list(group=time$group),max)
On 21-May-07, at 6:39 AM, Robert wrote:
Hi dear R users,
I'm a R beginner and I have a basic question about sequential
treatments of lists.
I have a time based (i.e. events are consecutive) list of values of
a biological property.
Like :
time value
15
210
3 7
4 10
5 19
6 21
7 20
8 18
9 10
10 7
11 8
12 12
13 17
14 19
15 24
16 18
17 15
18 10
19 9
[...]
And I have to define a threshold and to attach each event to his
group, i.e. values upper the threshold.
Like, for a threshold value of 17
time value group
15 NA
210 NA
3 7 NA
4 10 NA
5 19 1
6 21 1
7 20 1
8 18 1
9 10 NA
10 7 NA
11 8 NA
12 12 NA
13 17 2
14 19 2
15 24 2
16 18 2
17 15 NA
18 10 NA
19 9 NA
[...]
The only solution that I have found is to do a sequentially read
and write :
for(i in 1:length(my_events_list))
{
[...]
}
But I very slow. Do you have another ideas ?
And after I need to extract maximum values for each group
Like :
groupmax_value
121
224
[...]
and for each event which is part of a group to know if is't a
ascending phase or no.
Yes, lot of questions !! Sorry, but I think that the solution may
be unique.
In advance, thank you a lot
regards
JS
__
___
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Dalhousie University Department of Psychology
http://myweb.dal.ca/mc973993
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequentially process a list
Sorry, obviously misread your help request. Jim's solution is what
you're looking for.
On 21-May-07, at 12:56 PM, Mike Lawrence wrote:
Try nested ifelse() statements to label the group
ex.
time$group=ifelse(time$time5,NA,
ifelse(time$time9,1,
ifelse(time$time13,NA,
ifelse(time$time17,2,NA)
)
)
)
Then use aggregate to find the max value.
ex.
time.max=aggregate(time$value,list(group=time$group),max)
On 21-May-07, at 6:39 AM, Robert wrote:
Hi dear R users,
I'm a R beginner and I have a basic question about sequential
treatments of lists.
I have a time based (i.e. events are consecutive) list of values of
a biological property.
Like :
time value
15
210
3 7
4 10
5 19
6 21
7 20
8 18
9 10
10 7
11 8
12 12
13 17
14 19
15 24
16 18
17 15
18 10
19 9
[...]
And I have to define a threshold and to attach each event to his
group, i.e. values upper the threshold.
Like, for a threshold value of 17
time value group
15 NA
210 NA
3 7 NA
4 10 NA
5 19 1
6 21 1
7 20 1
8 18 1
9 10 NA
10 7 NA
11 8 NA
12 12 NA
13 17 2
14 19 2
15 24 2
16 18 2
17 15 NA
18 10 NA
19 9 NA
[...]
The only solution that I have found is to do a sequentially read
and write :
for(i in 1:length(my_events_list))
{
[...]
}
But I very slow. Do you have another ideas ?
And after I need to extract maximum values for each group
Like :
groupmax_value
121
224
[...]
and for each event which is part of a group to know if is't a
ascending phase or no.
Yes, lot of questions !! Sorry, but I think that the solution may
be unique.
In advance, thank you a lot
regards
JS
_
_
___
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Dalhousie University Department of Psychology
http://myweb.dal.ca/mc973993
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Mike Lawrence
Graduate Student, Dalhousie University Department of Psychology
http://myweb.dal.ca/mc973993
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] How does glm(family='binomial') deal with perfect sucess?
Hi all, Trying to understand the logistic regression performed by glm (i.e. when family='binomial'), and I'm curious to know how it treats perfect success. That is, lets say I have the following summary data x=c(1,2,3,4,5,6) y=c(0,.04,.26,.76,.94,1) w=c(100,100,100,100,100,100) where x is y is the probability of success at each value of x, calculated across w observations. When I use glm my.glm.obj=glm(y~x,family='binomial',weights=w) the regression comes out fine, but if I try what I understand to be the equivalent lm procedure (i.e. fitting a straight line to the logit transformed y values): my.lm.obj=lm(qlogis(y)~x,weights=w) I get an error because, of course, logit(1) = log(1/0) = log(Inf) = Inf (similarly, logit(0) = log(0/1) = log(0) = -Inf). I'd be very interested to see how glm deals with these extremes. Cheers, Mike -- Mike Lawrence http://artsweb.uwaterloo.ca/~m4lawren The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to calculate area between ECDF and CDF?
Hi all,
I'm working with data to which I'm fitting three-parameter weibull
distributions (shape, scale shift). The data are of low sample sizes
(between 10 and 80 observations), so I'm reluctant to check my fits
using chi-square (also, I'd like to avoid bin choice issues). I'd use
the Kolmogorov-Smirnov test, but of course this is invalid when the
distribution parameters are estimated from the data.
So I'm tinkering with an alternative method (excuse my naiveté if this
is a bad idea, I'm a relative statistical novice) that calculates the
area of the difference between the ECDF of the data and the CDF of the
estimated function (somewhat like KS, which looks at the greatest
distance between these). My thought is to compare this observed area to
a distribution of simulated areas derived by monte carlo simulation
(draw N random samples from the estimated function, calculating area,
and repeat 1e5 times). If the observed area is greater than say 95% of
the simulated areas, then I'd reject the fit.
My problem is that I can't figure out how to efficiently calculate the
area between the ECDF and CDF functions. I can of course calculate the
integral of each easily, and if one were consistently larger than the
other simple subtraction of the integrals would yield the area between.
However, when the functions cross, as frequently occurs, the solution
seems much more complex. Any suggestions? Since as noted above I'll be
doing the area calculation 1e5 times or so per test, a computationally
frugal solution would be much appreciated!
Here's some code that I've been toying with:
#set up some true parameters
shape=2
scale=.5
shift=.3
n=10
#generate some observed data
obs=obs=rweibull(10,shape,scale)+shift
#lets say that the following are the estimated parameters from whatever
estimation process I'm using
est.shape=1.9
est.scale=.6
est.shift=.35
#Calculate area between ECDF and CDF of the function defined by the
#estimated parameters
# ???
#The following would work if the ECDF were consistently higher or lower
#than the CDF
#Get the CDF area between 0 and some large number (here, 10 is pretty
#large)
cdf.area=integrate(pweibull,0,10,shape=est.shape,scale=est.scale)
#Get the ECDF area.
#first get rid of the shift in obs
obs=obs-est.shift
#calculate area by multiplying cumulative proportions by distance
#between knots, then summing
#add knot at 10 to match cdf
k=c(knots(ecdf(obs)),10)
ecdf.area=vector(numeric,(n-1))
for(i in 1:n){
ecdf.area[i]=(k[i+1]-k[i])*(sum(obs=k[i])/n)
}
ecdf.area=sum(ecdf.area)
#again, subtraction of the areas works if the ecdf is consistently lower
#than the cdf
diff=cdf.area-ecdf.area
#or consistently higher than the cdf
diff=ecdf.area-cdf.area
#but how to calculate when the functions cross?
Cheers,
Mike
--
Mike Lawrence
http://artsweb.uwaterloo.ca/~m4lawren
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] nnet() fit criteria
Hi all, I'm using nnet() for non-linear regression as in Ch8.10 of MASS. I understand that nnet() by default optimizes least squares. I'm looking to have it instead optimize such that the mean error is zero (so that it is unbiased). Any suggestions on how this might be achieved? Cheers, Mike -- Mike Lawrence http://artsweb.uwaterloo.ca/~m4lawren The road to wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Translating R code + library into Fortran?
,plot=F)
if(length(temp$counts[temp$counts0])10){
ok=T
}
}
#train nn.scale with error checking
ok=F
while(ok==F){
nn2.scale=nnet(exp.scale~min+q1+med+mean+q3+max+samples,data=training.data,size=8,linout=T,rang=1e-08,maxit=500,trace=F)
cor.scale=predict(nn2.scale,training.data[,c(1:7)],type=raw)
temp=hist(cor.scale,plot=F)
if(length(temp$counts[temp$counts0])10){
ok=T
}
}
#train nn.shape with error checking
ok=F
while(ok==F){
nn2.shape=nnet(exp.shape~min+q1+med+mean+q3+max+samples,data=training.data,size=8,linout=T,rang=1e-08,maxit=500,trace=F)
cor.shape=predict(nn2.shape,training.data[,c(1:7)],type=raw)
temp=hist(cor.shape,plot=F)
if(length(temp$counts[temp$counts0])10){
ok=T
}
}
#estimation iteration 2
cor.shift=predict(nn2.shift,obs.stats,type=raw)
cor.scale=predict(nn2.scale,obs.stats,type=raw)
cor.shape=predict(nn2.shape,obs.stats,type=raw)
#record error
ind.shift.err[,i]=cor.shift-stadler$exp.shift
ind.scale.err[,i]=cor.scale-stadler$exp.scale
ind.shape.err[,i]=cor.shape-stadler$exp.shape
group.shift.err[i]=mean(cor.shift)-mean(stadler$exp.shift)
group.scale.err[i]=mean(cor.scale)-mean(stadler$exp.scale)
group.shape.err[i]=mean(cor.shape)-mean(stadler$exp.shape)
}
results=as.data.frame(rbind(cbind(sd(c(ind.shift.err[,1:162])),sd(c(ind.scale.err[,1:162])),sd(c(ind.shape.err[,1:162]))),cbind(sd(group.shift.err[1:162]),sd(group.scale.err[1:162]),sd(group.shape.err[1:162]
results
--
Mike Lawrence
http://arts.uwaterloo.ca/~m4lawren
The road to wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] error bars in lattice xyplot *with groups*
Hi all,
I'm trying to plot error bars in a lattice plot generated with xyplot. Deepayan
Sarkar has provided a very useful solution for simple circumstances
(https://stat.ethz.ch/pipermail/r-help/2005-October/081571.html), yet I am
having trouble getting it to work when the groups setting is enabled in
xyplot (i.e. multiple lines). To illustrate this, consider the singer data
generated by the above linked solution previously submitted:
#
library(lattice)
singer.split -
with(singer,
split(height, voice.part))
singer.ucl -
sapply(singer.split,
function(x) {
st - boxplot.stats(x)
c(st$stats[3], st$conf)
})
singer.ucl - as.data.frame(t(singer.ucl))
names(singer.ucl) - c(median, lower, upper)
singer.ucl$voice.part -
factor(rownames(singer.ucl),
levels = rownames(singer.ucl))
#now let's split up the voice.part factor into two factors,
singer.ucl$voice=factor(rep(c(1,2),4))
singer.ucl$range=factor(rep(c(Bass,Tenor,Alto,Soprano),each=2))
#here's Deepayan's previous solution, slightly modified to depict
# the dependent variable (median) and the error bars on the y-axis
# and the independent variable (voice.part) on the x-axis
prepanel.ci - function(x, y, ly, uy, subscripts, ...)
{
x - as.numeric(x)
ly - as.numeric(ly[subscripts])
uy - as.numeric(uy[subscripts])
list(ylim = range(y, uy, ly, finite = TRUE))
}
panel.ci - function(x, y, ly, uy, subscripts, pch = 16, ...)
{
x - as.numeric(x)
y - as.numeric(y)
ly - as.numeric(ly[subscripts])
uy - as.numeric(uy[subscripts])
panel.arrows(x, ly, x, uy, col = black,
length = 0.25, unit = native,
angle = 90, code = 3)
panel.xyplot(x, y, pch = pch, ...)
}
#this graph works
xyplot(median ~ voice.part,
data=singer.ucl,
ly = singer.ucl$lower,
uy = singer.ucl$upper,
prepanel = prepanel.ci,
panel = panel.ci,
type=b
)
#this one does not (it will plot, but will not seperate the groups)
xyplot(median ~ voice,
groups=range,
data=singer.ucl,
ly = singer.ucl$lower,
uy = singer.ucl$upper,
prepanel = prepanel.ci,
panel = panel.ci,
type=b
)
Any suggestions?
--
Mike Lawrence
[EMAIL PROTECTED]
The road to Wisdom? Well, it's plain and simple to express:
Err and err and err again, but less and less and less.
- Piet Hein
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Save output
I think this is what you're looking for: file_name_root = myfile for(i in 1:10) result- of_some_function out_file-paste(file_name_root, i, .txt, sep = ) write.table(result, file = out_file) } the above code produces 10 files (myfile1.txt to myfile10.txt), each containing whatever was calculated on that loop. If you have loops within loops and you want to write during the innermost loop, remember to take this into account like so: file_name_root = myfile for (i in 1:10) for(q in 1:10) result- of_some_function out_file-paste(file_name_root, i, q, .txt, sep = ) write.table(result, file = out_file) } } That way, when i iterates and when q resets to 1, you don't write over your previous output. Quoting Frank Schmid [EMAIL PROTECTED]: Dear R-Mastermind Within a while or a for-loop, is there a way that I can save to disk the results of the previous calculations at the end of each loop with filenames called file01.Rdata, file02.Rdata etc? So far, I have tried to write the outcome of each loop in a 3 dimensional array and saved it just once after the loop. Or is there another way so that I can keep the resulting matrix of every step in the loop? Thank you very much for your help Frank Schmid [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] What is Mandel's (Fitting) Test?
Hello everyone, A little background first: I have collected psychophysical data from 12 participants, and each participant's data is represented as a scatter plot (Percieved roughness versus Physical roughness). I would like to know whether, on average, this data is best fit by a linear function or by a quadratic function. (we have a priori reasons to expect a quadratic) Some of my colleagues have suggested the following methods of testing this: 1. For each participant, calculate the r-square values for linear and quadratic fits, z-transform the resulting values. Collect these z-transformed scores and then perform a dependent t-test across participants. If significant, then a quadratic fits better. 2. For each participant, calculate the amount of variance left over from the linear fit that is accounted for by the quadratic fit. Perform a one-sample t-test to see if this population of scores differs from zero 3. Same as #2, but z-transform before performing the t-test. However, I'm sure that these tests fail to take into account the fact that a quadratic function will generally have an advantage over a linear function simply by dint of having more terms to play with. So I've been looking for a test that takes this advantage into account and I came across something called the Mandel Test. It is available in the quantchem package, but the manual contains a very meagre description of it's details (assumptions, etc). Furthermore, besides biology/chemistry papers that reference it in passing, I've been able to find only one reference online that addresses it's use (http://www.econ.kuleuven.be/public/ndbae06/PDF-FILES/vanloco.doc), but even then it lacks specificity. So the question is, what is the Mandel Test? What are the assumptions and limitations of the test? Does it sound appropriate for my purposes (if not, how about the other tests suggested above)? How does it differ from the Lack-of-Fit test? Any help would be greatly appreciated. Cheers, Mike -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] xyplot auto.key issue
Hi Deepayan, Thanks for the reply, but when I enter the type = b code into the auto.key (see below) command I get the following message: Error in valid.pch(x$pch) : zero-length 'pch' Any suggestions? xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), xlab = Sample Size, ylab = Bias (ms), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ) ), # key settings auto.key = list ( type = b, lines = T, border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right ) ) Quoting Deepayan Sarkar [EMAIL PROTECTED]: On 7/30/05, Mike Lawrence [EMAIL PROTECTED] wrote: Quick correction: The lines lines = T, type = b in the par.settings section should not be there. They are remnants of my previous (failed) attempts at solving the problem. Below is the correct code: xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ) ), # key settings auto.key = list ( lines = T, size = 7, You seem to be missing a 'type=b' somewhere here. The type=b argument to xyplot is actually handled by the panel function. The key has type=l by default (see under 'key' in ?xyplot) and has to be changed explicitly. border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right, ) ) Quoting Mike Lawrence [EMAIL PROTECTED]: Hi all, I'm having a problem with the auto.key function in xyplot. I hate to bother the list like this and I'm positive I must be missing something very simple, yet I've spent the last day searching for a solution to no avail. Essentially, I want a key that contains entries in which the plot points are superimposed on a line of the same color as the points, like this: o--o--o Now, given the presence of the default divide command, I assume this is simple; indeed, I get the impression that this representation is supposed to be produced automatically. Yet I can't seem to get it to work! Now, I've incorporated various other tweaks to my xyplot function, so I'm wondering if these tweaks are somehow hindering my efforts. The function is pasted below; I am making a 3x3 plot, each panel contains 5 lines and it is these lines that I want represented in the key. See the comments for descriptions of the modifications. Any help would be greatly appreciated. Cheers, Mike xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c
Re: [R] xyplot auto.key issue
Hi again, Deepayan, I tried adding a pch = c(1,2,3,4,5) line in the auto key but the zero-length 'pch' error still occurs. Sundar, I tried your code using key instead of auto.key, and after tweaking it to fit my design (i.e. from x~y, groups = a to x~y | a, groups = b) it works perfectly! Thanks! Here's the code that finally worked (that is, the modifications to Sundar's code, which might be clearer for others to follow than my own): library(lattice) set.seed(1) z - expand.grid(x=1:10, b = LETTERS[1:5], a = LETTERS[10:18]) z$y - rnorm(nrow(z)) trellis.par.set(theme = col.whitebg()) par.line - trellis.par.get(superpose.line) par.symb - trellis.par.get(superpose.symbol) n - seq(nlevels(z$b)) my.key - list( space = right, border = TRUE, cex.title = 1.2, title = My Key, size = 7, lines = list( pch = par.symb$pch[n], lty = par.line$lty[n], col = par.line$col[n], type = b ), text = list( levels(z$b) ) ) xyplot( y ~ x | a, data = z, groups = b, pch = par.symb$pch[n], type = b, key = my.key ) Quoting Sundar Dorai-Raj [EMAIL PROTECTED]: Hi, Mike, Mike Lawrence wrote: Hi Deepayan, Thanks for the reply, but when I enter the type = b code into the auto.key (see below) command I get the following message: Error in valid.pch(x$pch) : zero-length 'pch' Any suggestions? Why not just ignore auto.key and use key? Personally, I use auto.key only when I want the defaults. If I want something more customized, then I use key. As in, library(lattice) set.seed(1) z - expand.grid(x=1:10, g = LETTERS[1:5]) z$y - rnorm(nrow(z)) trellis.par.set(theme = col.whitebg()) par.line - trellis.par.get(superpose.line) par.symb - trellis.par.get(superpose.symbol) n - seq(nlevels(z$g)) my.key - list(space = right, border = TRUE, cex.title = 1.2, title = My Key, size = 7, lines = list(pch = par.symb$pch[n], lty = par.line$lty[n], col = par.line$col[n], type = b), text = list(levels(z$g))) xyplot(y ~ x, z, groups = g, pch = par.symb$pch[n], type = b, key = my.key) xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), xlab = Sample Size, ylab = Bias (ms), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ) ), # key settings auto.key = list ( type = b, lines = T, border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right ) ) Quoting Deepayan Sarkar [EMAIL PROTECTED]: On 7/30/05, Mike Lawrence [EMAIL PROTECTED] wrote: Quick correction: The lines lines = T, type = b in the par.settings section should not be there. They are remnants of my previous (failed) attempts at solving the problem. Below is the correct code: xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ) ), # key settings auto.key = list ( lines = T, size = 7, You seem to be missing a 'type=b' somewhere here. The type=b argument to xyplot is actually handled by the panel function. The key has type=l by default (see under 'key' in ?xyplot) and has to be changed explicitly. border = T
Re: [R] How do you increase memeory?
memory.limit(size = x) where x is the desired memory limit in MB. Quoting Briggs, Meredith M [EMAIL PROTECTED]: Hello Function memory.size() =435109888. How do I increase it by, say 30%? Thanks Meredith [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] xyplot auto.key issue
Hi all, I'm having a problem with the auto.key function in xyplot. I hate to bother the list like this and I'm positive I must be missing something very simple, yet I've spent the last day searching for a solution to no avail. Essentially, I want a key that contains entries in which the plot points are superimposed on a line of the same color as the points, like this: o--o--o Now, given the presence of the default divide command, I assume this is simple; indeed, I get the impression that this representation is supposed to be produced automatically. Yet I can't seem to get it to work! Now, I've incorporated various other tweaks to my xyplot function, so I'm wondering if these tweaks are somehow hindering my efforts. The function is pasted below; I am making a 3x3 plot, each panel contains 5 lines and it is these lines that I want represented in the key. See the comments for descriptions of the modifications. Any help would be greatly appreciated. Cheers, Mike xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ), lines = T, type = b ), # key settings auto.key = list ( lines = T, size = 7, border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right, ) ) -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] xyplot auto.key issue
Quick correction: The lines lines = T, type = b in the par.settings section should not be there. They are remnants of my previous (failed) attempts at solving the problem. Below is the correct code: xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ) ), # key settings auto.key = list ( lines = T, size = 7, border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right, ) ) Quoting Mike Lawrence [EMAIL PROTECTED]: Hi all, I'm having a problem with the auto.key function in xyplot. I hate to bother the list like this and I'm positive I must be missing something very simple, yet I've spent the last day searching for a solution to no avail. Essentially, I want a key that contains entries in which the plot points are superimposed on a line of the same color as the points, like this: o--o--o Now, given the presence of the default divide command, I assume this is simple; indeed, I get the impression that this representation is supposed to be produced automatically. Yet I can't seem to get it to work! Now, I've incorporated various other tweaks to my xyplot function, so I'm wondering if these tweaks are somehow hindering my efforts. The function is pasted below; I am making a 3x3 plot, each panel contains 5 lines and it is these lines that I want represented in the key. See the comments for descriptions of the modifications. Any help would be greatly appreciated. Cheers, Mike xyplot( #basic settings bias ~ sample_size | measure, data = bias, groups = exp_tau, type = b, pch = c(1,2,3,4,5), #make strips transparent strip = function(bg, ...) strip.default(bg = 'transparent', ...), # tweak scales scales=list( x=list( at = c(20, 40, 60), tck = c(1,0), alternating = F ), y=list( at = c(-50, -25, 0, 25, 50), tck = c(1,0), alternating = F ) ), # tell key to match symbols to those used in the plot par.settings = list( superpose.symbol = list( cex = .8, pch = c(1,2,3,4,5) ), lines = T, type = b ), # key settings auto.key = list ( lines = T, size = 7, border = T, cex.title = 1.2, title = Expected Tau, text = c(30 ms, 80 ms, 130 ms, 180 ms, 230 ms), space = right, ) ) -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Mike Lawrence, BA(Hons) Research Assistant to Dr. Gail Eskes Dalhousie University QEII Health Sciences Centre (Psychiatry) [EMAIL PROTECTED] The road to Wisdom? Well, it's plain and simple to express: Err and err and err again, but less and less and less. - Piet Hein __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
