[R] Italicized title from index
Hi all! I've written a handy script that uses a for loop to allow me to generate a large number of figures and statistical outputs for a large dataset. I am using indexing to retrieve a species name for the title of my graphs- which worked fine. However, I need to italicize these species names. I originally used the paste function, and had no problems with indexing: *main=paste(Yield for , testsub[1,3], in management region , testsub[1,2])* The title looks like: *Yield for Gadus morhua in management region 4X5Y* I tried bquote from a related help thread, I tried to emulate it: *spp-testsub[1,3]* *region-testsub[1,2]* * * *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~ .(region))* Which doesn't seem to work at all, but when I try not putting anything after the italic: *main=bquote(Yield ~ for ~ italic(.(spp)))* I get: *Yield for 1* While *spp=Gadus morhua* I'm at wit's end, I tried to read about substitute, expression, and eval functions in the hopes I can figure it out, but I am lost! Thanks for any help! Cheers, -Jeremy N Undergraduate Researcher in Macroecology University of Ottawa Department of Biology Ad astra per alia porci! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Graph many points without hiding some
I have a very large dataset with three variables that I need to graph using
a scatterplot. However I find that the first variable gets masked by the
other two, so the graph looks entirely different depending on the order of
variables. Does anyone have any suggestions how to manage this?
This code is an illustration of what I am dealing with:
x - 1
plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
points(rnorm(x,mean=21),rnorm(x),col=2)
points(rnorm(x,mean=19),rnorm(x),col=3)
gives an entirely different looking graph to:
x - 1
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
despite being identical in all respects except for the order in which the
variables are plotted.
I have tried using pch=., however the colours are very difficult to
discern. I have experimented with a number of other symbols with no real
solution.
The only way that appears to work is to iterate the plot with a for loop,
and progressively add a few numbers from each variable, as below. However
although I can do this simply with random numbers as I have done here, this
is an extremely cumbersome method to use with real datasets.
plot(1,1,xlim=c(16,24),ylim=c(-4,4),col=white)
x - 100
for (i in 1:100) {
points(rnorm(x,mean=19),rnorm(x),col=3)
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
}
Is there some function in R that could solve this through automatically
iterating my data as above, using transparent symbols, or something else? Is
there some other way of solving this issue that I haven't thought of?
Thankyou,
Samuel Dennis
[[alternative HTML version deleted]]
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[R] Assign Names of columns in data.frame dinamically
Hello List.
I have many files of ECG, each one with 7 column and I need only the
second column and the name of each ECG. I am doing this but althought
I have various days trying this i haven't gotten, so I ask help with
this, if somebody cans help me I'll be so thankfully.
rm(list=ls())
# loadEcgFiles - function(Dir=.) {
Dir - .;
txtfiles - list.files(paste(Dir),'.txt$');
ecg = data.frame(ncol=1);
len = length(txtfiles);
for (i in 1:5 ) {
# i - 1;
filename = paste(projectDir, / ,
txtfiles[i],sep='');
sample = read.table(filename, nrow=75000);
sampleName = filename; sampleName=gsub(./,
,sampleName); sampleName=gsub(.txt, , sampleName);
temp - sample$V2;
names(temp) - sampleName;
ecg = cbind(ecg, temp);
#}
Thanks and sorry for my english.
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Re: [R] Graph many points without hiding some
On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis sjdenn...@gmail.com wrote:
I have a very large dataset with three variables that I need to graph using
a scatterplot. However I find that the first variable gets masked by the
other two, so the graph looks entirely different depending on the order of
variables. Does anyone have any suggestions how to manage this?
This code is an illustration of what I am dealing with:
x - 1
plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
points(rnorm(x,mean=21),rnorm(x),col=2)
points(rnorm(x,mean=19),rnorm(x),col=3)
gives an entirely different looking graph to:
x - 1
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
despite being identical in all respects except for the order in which the
variables are plotted.
I have tried using pch=., however the colours are very difficult to
discern. I have experimented with a number of other symbols with no real
solution.
The only way that appears to work is to iterate the plot with a for loop,
and progressively add a few numbers from each variable, as below. However
although I can do this simply with random numbers as I have done here, this
is an extremely cumbersome method to use with real datasets.
plot(1,1,xlim=c(16,24),ylim=c(-4,4),col=white)
x - 100
for (i in 1:100) {
points(rnorm(x,mean=19),rnorm(x),col=3)
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
}
Is there some function in R that could solve this through automatically
iterating my data as above, using transparent symbols, or something else? Is
there some other way of solving this issue that I haven't thought of?
Assume you are plotting variables y1, y2, y3 of the same length
against a common x, and you would like to assign colors say c(1,2,3).
You can automate the randomization of order as follows:
n = length(y1);
y = c(y1, y2, y3);
xx = rep(x, 3);
colors = rep(c(1,2,3), c(n, n, n));
order = sample(c(1:(3*n)));
plot(xx[order], y[order], col= colors[order])
I basically turn the y's into a single vector y with the corresponding
values of x stored in xx and the plotting colors, then randomize the
order using the sample function.
HTH,
Peter
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[R] ANCOVA for linear regressions without intercept
Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0 intercept). Is it possible to do ANCOVA for these regressions without intercepts? When I do summary(lm(body length ~ sex*head length)) this will include the intercepts as below Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept)-6.496971.68497 -3.856 0.000118 *** sexMale-9.393401.97760 -4.750 2.14e-06 *** sexUnknown -1.337912.35453 -0.568 0.569927 head_length 7.123070.05503 129.443 2e-16 *** sexMale:head_length 0.316310.06246 5.064 4.37e-07 *** sexUnknown:head_length 0.199370.07022 2.839 0.004556 ** --- Is there any way I can remove the intercepts so that I can simply compare the slopes with no intercept taken into account? Thanks for help in advance. Yusuke Fukuda __ R-help@r-project.org 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] Graph many points without hiding some
Hi:
I can think of a couple: (1) size reduction of the points; (2) alpha
transparency; (3) (1) + (2)
From your original plot in base graphics, I reduced cex to 0.2 and it didn't
look too bad:
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24), cex = 0.2)
points(rnorm(x,mean=20),rnorm(x),col=1, cex = 0.2)
points(rnorm(x,mean=21),rnorm(x),col=2, cex = 0.2)
AFAIK, base graphics doesn't have alpha transparency available, but the
ggplot2 package does. One approach is to adjust the alpha transparency on
default size points; another is to combine reduced point size with alpha
transparency. Here is your example rehashed for ggplot2.
require(ggplot2)
d - data.frame(x1 = rnorm(1, mean = 19), x2 = rnorm(1, mean = 20),
x3 = rnorm(1, mean = 21), x = rnorm(1))
# Basically stacking x1 - x3, creating two new vars named variable and value
dm - melt(d, id = 'x') # from reshape package, loads with ggplot2
# Alpha transparency is set to a low level with default point size,
# but the colors in the legend are muted by the level of transparency
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.05) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green'))
# A tradeoff is to reduce the point size and increase alpha a bit, but these
changes will
# also be reflected in the legend.
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.15, size = 1) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green'))
You may well find the legend to be useless for this example, so to get rid
of it,
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.15, size = 1) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green')) +
opts(legend.position = 'none')
The nice thing about the ggplot2 graph is that you can adjust the point size
and alpha transparency to your tastes. The default point size is 2 and the
default alpha = 1 (no transparency).
HTH,
Dennis
On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis sjdenn...@gmail.com wrote:
I have a very large dataset with three variables that I need to graph using
a scatterplot. However I find that the first variable gets masked by the
other two, so the graph looks entirely different depending on the order of
variables. Does anyone have any suggestions how to manage this?
This code is an illustration of what I am dealing with:
x - 1
plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
points(rnorm(x,mean=21),rnorm(x),col=2)
points(rnorm(x,mean=19),rnorm(x),col=3)
gives an entirely different looking graph to:
x - 1
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
despite being identical in all respects except for the order in which the
variables are plotted.
I have tried using pch=., however the colours are very difficult to
discern. I have experimented with a number of other symbols with no real
solution.
The only way that appears to work is to iterate the plot with a for loop,
and progressively add a few numbers from each variable, as below. However
although I can do this simply with random numbers as I have done here, this
is an extremely cumbersome method to use with real datasets.
plot(1,1,xlim=c(16,24),ylim=c(-4,4),col=white)
x - 100
for (i in 1:100) {
points(rnorm(x,mean=19),rnorm(x),col=3)
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
}
Is there some function in R that could solve this through automatically
iterating my data as above, using transparent symbols, or something else?
Is
there some other way of solving this issue that I haven't thought of?
Thankyou,
Samuel Dennis
[[alternative HTML version deleted]]
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] VECM with UNRESTRICTED TREND
Hello Greg, you include your trend as a (Nx1) matrix and use this for 'dumvar'. The matrix 'dumvar' is just added to the VECM as deterministic regressors and while you are referring to case 5, this is basically what you are after, if I am not mistaken. But we aware that this implies a quadratic trend for the levels. Best, Bernhard Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl] Gesendet: Mittwoch, 30. März 2011 20:50 An: Pfaff, Bernhard Dr.; r-help@r-project.org Betreff: Re: [R] VECM with UNRESTRICTED TREND Hello Bernhard, Thank You very much. Unfortunately I'm still not really sure how should I use dummy vars in this context... If I have a system of three variables (x, y, z), lag order = 2 and 1 cointegrating relation, what should I do? I mean, what kind of 'pattern' should be used to create those dummy variables, what should they represent and how many of them do I need? Many thanks in advance! Best, Greg 2011/3/30 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com Hello Greg, you can exploit the argument 'dumvar' for this. See ?ca.jo Best, Bernhard -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat Gesendet: Mittwoch, 30. März 2011 16:46 An: r-help@r-project.org Betreff: [R] VECM with UNRESTRICTED TREND Dear All, My question is: how can I estimate VECM system with unrestricted trend (aka case 5) option as a deterministic term? As far as I know, ca.jo in urca package allows for restricted trend only [vecm - ca.jo(data, type = trace/eigen, ecdet = trend, K = n, spec = transitory/longrun)]. Obviously, I don't have to do this in urca, so if another package gives the possibility, please let me know too! Thanks in advance! Greg [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * Confidentiality Note: The information contained in this message, and any attachments, may contain confidential and/or privileged material. It is intended solely for the person(s) or entity to which it is addressed. Any review, retransmission, dissemination, or taking of any action in reliance upon this information by persons or entities other than the intended recipient(s) is prohibited. If you received this in error, please contact the sender and delete the material from any computer. * [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Graph many points without hiding some
Hi.
You could also turn it into a 3D plot with some variation on the function
below:
plot4d-function(x,y,z, u, main=, xlab=, ylab=, zlab=, ulab=)
{
require(rgl)#may need to install this package first
#standard trick to get some intensity colors
uLim-range(u)
uLen-uLim[2] - uLim[1] + 1
colorlut-terrain.colors(uLen)
col-colorlut[u - uLim[1] + 1]
open3d()#Open new device
points3d(x=x, y=y, z=z, col=col)
aspect3d(x=1, y=1, z=1) #ensure bounding box is in cube-form
(scaling variables)
#note: if you want to flip an axis, use -1 in the statement above
axes3d() #Show axes
title3d(main = main, sub=paste(Green is low, ulab, , red is
high)
xlab = xlab, ylab = ylab, zlab = zlab)
}
HTH,
Nick Sabbe
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link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Peter Langfelder
Sent: donderdag 31 maart 2011 9:26
To: Samuel Dennis
Cc: R-help@r-project.org
Subject: Re: [R] Graph many points without hiding some
On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis sjdenn...@gmail.com wrote:
I have a very large dataset with three variables that I need to graph
using
a scatterplot. However I find that the first variable gets masked by the
other two, so the graph looks entirely different depending on the order of
variables. Does anyone have any suggestions how to manage this?
This code is an illustration of what I am dealing with:
x - 1
plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
points(rnorm(x,mean=21),rnorm(x),col=2)
points(rnorm(x,mean=19),rnorm(x),col=3)
gives an entirely different looking graph to:
x - 1
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
despite being identical in all respects except for the order in which the
variables are plotted.
I have tried using pch=., however the colours are very difficult to
discern. I have experimented with a number of other symbols with no real
solution.
The only way that appears to work is to iterate the plot with a for loop,
and progressively add a few numbers from each variable, as below. However
although I can do this simply with random numbers as I have done here,
this
is an extremely cumbersome method to use with real datasets.
plot(1,1,xlim=c(16,24),ylim=c(-4,4),col=white)
x - 100
for (i in 1:100) {
points(rnorm(x,mean=19),rnorm(x),col=3)
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
}
Is there some function in R that could solve this through automatically
iterating my data as above, using transparent symbols, or something else?
Is
there some other way of solving this issue that I haven't thought of?
Assume you are plotting variables y1, y2, y3 of the same length
against a common x, and you would like to assign colors say c(1,2,3).
You can automate the randomization of order as follows:
n = length(y1);
y = c(y1, y2, y3);
xx = rep(x, 3);
colors = rep(c(1,2,3), c(n, n, n));
order = sample(c(1:(3*n)));
plot(xx[order], y[order], col= colors[order])
I basically turn the y's into a single vector y with the corresponding
values of x stored in xx and the plotting colors, then randomize the
order using the sample function.
HTH,
Peter
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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.
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and provide commented, minimal, self-contained, reproducible code.
[R] nls.profile
Hello, I use nls.profile to compute confidence intervals of parameter estimates of a non-linear model. When computing the profiles, the model function produces an error for certain parameter combinations. Therefore nls fails and so does nls.profile. Is there a way to tell nls.profile to ignore errors and to proceed with the next parameter value? Cheers, Daniel. signature.asc Description: This is a digitally signed message part __ R-help@r-project.org 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] summing values by week - based on daily dates - but with somedates missing
Hi,
Yep, that was what it was doing. For a sum across week, try something like
get.week.flag - function(dd) {
## get weekday from the date in dd and code it as Monday = 1, Tuesday = 2 etc
idd =
factor(weekdays(dd),levels=c(Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday))
## convert to numeric
ndd = as.numeric(idd)
## flag entries where weekday code gets less (this will flag changes in week)
wflag = c(FALSE,(ndd[-length(idd)] ndd[-1]))
## cumulative sum to get the week flag
cumsum(wflag) + 1
}
to get a week flag (this is assuming that your data is sorted by date, if not
you'll have to sort it first). If you want the week to start on a different
day, just change the ordering of the weekdays in the levels statement.
data.frame(date=myframe$date,day=weekdays(myframe$date),week=get.week.flag(myframe$date))
seems to indicate that the function is doing what it should, so you can then
amend the previous code to use get.week.flag instead of weekdays, as in
sum.by.week - function(ff) {
by.day - split(ff$value,get.week.flag(ff$dates))
lapply(by.day,sum)
}
by.grp - split(myframe,myframe$group)
lapply(by.grp,sum.by.week)
Martyn
-Original Message-
From: Dimitri Liakhovitski [mailto:dimitri.liakhovit...@gmail.com]
Sent: 30 March 2011 18:03
To: Martyn Byng
Cc: r-help
Subject: Re: [R] summing values by week - based on daily dates - but with
somedates missing
Thank you, Martyn.
But it looks like this way we are getting sums by day - i.e., across
all Mondays, all Tuesdays, etc.
Maybe I did not explain well, sorry! The desired output would contain
sums for each WHOLE week - across all days that comprise that week -
Monday through Sunday.
Makes sense?
Dimitri
On Wed, Mar 30, 2011 at 12:53 PM, Martyn Byng martyn.b...@nag.co.uk wrote:
Hi,
How about something like:
sum.by.day - function(ff) {
by.day - split(ff$value,weekdays(ff$dates))
lapply(by.day,sum)
}
by.grp - split(myframe,myframe$group)
lapply(by.grp,sum.by.day)
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Dimitri Liakhovitski
Sent: 30 March 2011 15:23
To: r-help
Subject: [R] summing values by week - based on daily dates - but with
somedates missing
Dear everybody,
I have the following challenge. I have a data set with 2 subgroups,
dates (days), and corresponding values (see example code below).
Within each subgroup: I need to aggregate (sum) the values by week -
for weeks that start on a Monday (for example, 2008-12-29 was a
Monday).
I find it difficult because I have missing dates in my data - so that
sometimes I don't even have the date for some Mondays. So, I can't
write a proper loop.
I want my output to look something like this:
group dates value
group.1 2008-12-29 3.0937
group.1 2009-01-05 3.8833
group.1 2009-01-12 1.362
...
group.2 2008-12-29 2.250
group.2 2009-01-05 1.4057
group.2 2009-01-12 3.4411
...
Thanks a lot for your suggestions! The code is below:
Dimitri
### Creating example data set:
mydates-rep(seq(as.Date(2008-12-29), length = 43, by = day),2)
myfactor-c(rep(group.1,43),rep(group.2,43))
set.seed(123)
myvalues-runif(86,0,1)
myframe-data.frame(dates=mydates,group=myfactor,value=myvalues)
(myframe)
dim(myframe)
## Removing same rows (dates) unsystematically:
set.seed(123)
removed.group1-sample(1:43,size=11,replace=F)
set.seed(456)
removed.group2-sample(44:86,size=11,replace=F)
to.remove-c(removed.group1,removed.group2);length(to.remove)
to.remove-to.remove[order(to.remove)]
myframe-myframe[-to.remove,]
(myframe)
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
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[R] statistical question
Dear List! I want to compare medians of non normal distributed data. Is it possible and usefull to calculate 95% confidence intervals for medians? And if so - how can this be achieved in R? Thanks a lot! Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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] dfsane arguments
Hi there,
I'm trying to solve 2 nonlinear equations in 2 unknowns using the BB
package.
The first part of my program solves 3 ODEs using the deSolve package. This
part works. The output is used as parameter values in the functions I need
to solve.
The second part is to solve 2 equations in 2 unknowns. This does not work. I
get the error message unexpected end of input. So what inputs am I missing
here? As I understand it the arguments I have excluded from dfsane(), such
as control, are set to default?
parameters - c(K_vv= 0.0047,
K_rv=-0.0268,
K_rr=0.3384,
theta_v=107.4039,
theta_r =5.68,
Sigma_rv=0.0436,
Sigma_rr=0.1145,
lambda_v=0,
lambda_r=-0.0764
)
state - c(b_1 = 0,
b_2 = 0,
a = 0)
Kristian - function(t, state, parameters){
with(as.list(c(state, parameters)),{
db_1 =
-((K_vv+lambda_v)*b_1+(K_rv+Sigma_rv*lambda_v+Sigma_rr*lambda_r)*b_2+0.5*(b_1)^2+Sigma_rv*b_1*b_2+0.5*((Sigma_rv)^2+(Sigma_rr)^2)*(b_2)^2
)
db_2 = -K_rr*b_2+1
da = K_vv*theta_v*b_1+(K_rv*theta_v+K_rr*theta_r)*b_2
})
}
times - seq(0, 10, by = 0.5)
library(deSolve)
out - ode(y = state, times = times, func = Kristian, parms = parameters)
# constructing output as a matrix
outmat - as.matrix(out)
library(BB) #loading BB package
Bo - function(x, s){
f - rep(NA, length(x))
f[1] - (1-exp(outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2]))/
( exp(-outmat[1,4]-outmat[1,3]*x[1]-outmat[1,2]*x[2])
+ exp(-outmat[2,4]-outmat[2,3]*x[1]-outmat[2,2]*x[2])
+ exp(-outmat[3,4]-outmat[3,3]*x[1]-outmat[3,2]*x[2])
+ exp(-outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2]))-s[1]
f[2] -(1-exp(-outmat[20,4]-outmat(20,3)*x[1]-outmat[20,2]*x[2]))/
( exp(-outmat[1,4]-outmat[1,3]*x[1]-outmat[1,2]*x[2])
+ exp(-outmat[2,4]-outmat[2,3]*x[1]-outmat[2,2]*x[2])
+ exp(-outmat[3,4]-outmat[3,3]*x[1]-outmat[3,2]*x[2])
+ exp(-outmat[4,4]-outmat[4,3]*x[1]-outmat[4,2]*x[2])
+ exp(-outmat[5,4]-outmat[5,3]*x[1]-outmat[5,2]*x[2])
+ exp(-outmat[5,4]-outmat[5,3]*x[1]-outmat[5,2]*x[2])
+ exp(-outmat[6,4]-outmat[6,3]*x[1]-outmat[6,2]*x[2])
+ exp(-outmat[7,4]-outmat[7,3]*x[1]-outmat[7,2]*x[2])
+ exp(-outmat[8,4]-outmat[8,3]*x[1]-outmat[8,2]*x[2])
+ exp(-outmat[9,4]-outmat[9,3]*x[1]-outmat[9,2]*x[2])
+ exp(-outmat[10,4]-outmat[10,3]*x[1]-outmat[10,2]*x[2])
+ exp(-outmat[11,4]-outmat[11,3]*x[1]-outmat[11,2]*x[2])
+ exp(-outmat[12,4]-outmat[12,3]*x[1]-outmat[12,2]*x[2])
+ exp(-outmat[13,4]-outmat[13,3]*x[1]-outmat[13,2]*x[2])
+ exp(-outmat[14,4]-outmat[14,3]*x[1]-outmat[14,2]*x[2])
+ exp(-outmat[15,4]-outmat[15,3]*x[1]-outmat[15,2]*x[2])
+ exp(-outmat[16,4]-outmat[16,3]*x[1]-outmat[16,2]*x[2])
+ exp(-outmat[17,4]-outmat[17,3]*x[1]-outmat[17,2]*x[2])
+ exp(-outmat[18,4]-outmat[18,3]*x[1]-outmat[18,2]*x[2])
+ exp(-outmat[19,4]-outmat[19,3]*x[1]-outmat[19,2]*x[2])
+ exp(-outmat[20,4]-outmat[20,3]*x[1]-outmat[20,2]*x[2])) -s[2]
f
s - c(0.03, 0.045)
p-c(0.5, 0.5)
ans - dfsane(par=p, fn=Bo, s=s)
ans$par
Any help will be much appreciated!
Thank you,
Kristian Lind
[[alternative HTML version deleted]]
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Re: [R] statistical question
The default rank test in the quantreg package would look like this summary(rq(y ~ d, tau = .5)) where d is a factor variable indicating which sample the elements of y belonged to. Summary returns a confidence interval for the coef of the factor variable -- if this interval excludes zero at the chosen alpha level then the difference in medians is significant. Roger Koenker rkoen...@illinois.edu On Mar 31, 2011, at 4:15 AM, Anna Lee wrote: Dear List! I want to compare medians of non normal distributed data. Is it possible and usefull to calculate 95% confidence intervals for medians? And if so - how can this be achieved in R? Thanks a lot! Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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@r-project.org 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] Simple lattice question
DeaR ComRades,
require(lattice)
data - data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')
How do I put a legend for the grouping variable in the empty upper-right panel?
TIA
Rubén
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
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Re: [R] how about a p- operator?
In addition to %...% operators one can define := (I haven't seen this
possibility documented anywhere but it's used in a package) which
seems to have different precedence.
`:=`-`%-%` # the %-% by John Fox
a
[1] -1
a := 1000 + 1
[1] 1001
a
[1] 1001
a %-% 1000 + 1
[1] 1001
a
[1] 1000
Regards,
Kenn
Kenn Konstabel
National Institute for Health Development
Hiiu 42
Tallinn
On Thu, Mar 31, 2011 at 2:50 AM, William Dunlap wdun...@tibco.com wrote:
The %...% operators are not a panacea.
they have the same precedence as `*`
and `/` (I think) so you get things like:
x %-% 10 - 8 # %-% has higher precedence than -
[1] 2
x # not what you thought it would be
[1] 10
x %-% 10 ^3 # but lower than ^
[1] 1000
x # this is what you expected
[1] 1000
It isn't that hard to write a package with your
own parser in it. Just have it generate the
call tree from your input text and call eval()
on it.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of John Fox
Sent: Wednesday, March 30, 2011 4:34 PM
To: 'Carl Witthoft'
Cc: r-help@r-project.org
Subject: Re: [R] how about a p- operator?
Dear Carl,
I think that the following does what you want:
`%-%` - function(e1, e2){
+ e1 - deparse(substitute(e1))
+ env - parent.frame()
+ assign(e1, e2, envir=env)
+ e2
+ }
x %-% 10
[1] 10
x
[1] 10
But, as has been pointed out, it's probably easier just to
parenthesize the
usual assignment command.
Regards,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]
On Behalf Of Carl Witthoft
Sent: March-30-11 7:00 PM
To: r-help@r-project.org
Subject: [R] how about a p- operator?
I was cursing Matlab again today (what else is new) because
the default
action for every Matlab command is to spew the result to
the console, and
one must remember to put that darn ; at the end of every line.
So I just wondered: was there ever a discussion as to
providing some
modified version of the - and - operators in R to do
the reverse?
That is, since R does not print the values of a command
to the console,
what if there were an operator such that
newobject p- somefunction()
would do the same as
print(newobject - somefunction())
Any thoughts?
Carl
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
[R] That dreaded floating point trap
Hi,
I had a piece of code which looped over a decimal vector like this:
for( i in where ){
thisdata - subset(herde, herde$mlr = i)
# do stuff with thisdata..
}
'where' is a vector like seq(-1, 1, by=0.1)
My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different if
'where' was seq(-1, 1, by=0.1) than when 'where' was seq(-0.8, 1, by=0.1)
It went away after I changed the first line to:
thisdata - subset(herde, herde$mlr = round(i, digits=1))
This is that floating point trap the R inferno pdf talked about,
right? That file talked about the problem, but didn't offer a solution.
Similar things happened when I created a table() from a vector with
values in seq(-1, 1, by=0.1)
Do I really have to round every float at every occurence from now on, or
is there another solution? I only found all.equal() and identical(), but
I want to subset for observations with a value /greater/ than something.
Thanks in advance,
Alex
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Re: [R] VECM with UNRESTRICTED TREND
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = trend, K = 2, spec =
longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1 as well? And doesn't
it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
matrix 'dumvar' is just added to the VECM as deterministic regressors and
while you are referring to case 5, this is basically what you are after, if
I am not mistaken. But we aware that this implies a quadratic trend for the
levels.
Best,
Bernhard
--
*Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
*Gesendet:* Mittwoch, 30. März 2011 20:50
*An:* Pfaff, Bernhard Dr.; r-help@r-project.org
*Betreff:* Re: [R] VECM with UNRESTRICTED TREND
Hello Bernhard,
Thank You very much. Unfortunately I'm still not really sure how should I
use dummy vars in this context...
If I have a system of three variables (x, y, z), lag order = 2 and 1
cointegrating relation, what should I do? I mean, what kind of 'pattern'
should be used to create those dummy variables, what should they represent
and how many of them do I need?
Many thanks in advance!
Best,
Greg
2011/3/30 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you can exploit the argument 'dumvar' for this. See ?ca.jo
Best,
Bernhard
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
Gesendet: Mittwoch, 30. März 2011 16:46
An: r-help@r-project.org
Betreff: [R] VECM with UNRESTRICTED TREND
Dear All,
My question is:
how can I estimate VECM system with unrestricted trend (aka
case 5) option as a deterministic term?
As far as I know, ca.jo in urca package allows for restricted trend
only [vecm
- ca.jo(data, type = trace/eigen, ecdet = trend, K =
n, spec = transitory/longrun)].
Obviously, I don't have to do this in urca, so if another
package gives the possibility, please let me know too!
Thanks in advance!
Greg
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Re: [R] That dreaded floating point trap
On 31-Mar-11 11:24:01, Alexander Engelhardt wrote:
Hi,
I had a piece of code which looped over a decimal vector like this:
for( i in where ){
thisdata - subset(herde, herde$mlr = i)
# do stuff with thisdata..
}
'where' is a vector like seq(-1, 1, by=0.1)
My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different
if 'where' was seq(-1, 1, by=0.1) than when 'where' was
, when you wan
It went away after I changed the first line to:
thisdata - subset(herde, herde$mlr = round(i, digits=1))
This is that floating point trap the R inferno pdf talked about,
right? That file talked about the problem, but didn't offer a solution.
Similar things happened when I created a table() from a vector with
values in seq(-1, 1, by=0.1)
Do I really have to round every float at every occurence from now on,
or is there another solution? I only found all.equal() and identical(),
but I want to subset for observations with a value /greater/ than
something.
Thanks in advance,
Alex
A very straightforward way to avoid this problem is to construct the
sequence by multiplying a sequence of integers by an approriate
constant. E.g. for your first example:
for( i in where ){
thisdata - subset(herde, herde$mlr = i)
# do stuff with thisdata..
}
'where' is a vector like 0.1*((-10):10)
[ instead of seq(-1, 1, by=0.1) ]
and then, when you want to change to seq(-0.8, 1, by=0.1),
use instead 0.1*(-80,10).
Hoping this helps,
Ted.
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 31-Mar-11 Time: 12:52:35
-- XFMail --
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Re: [R] how about a p- operator?
Dear Ken and Bill,
Thanks for the comments. I obviously didn't think about the precedence issue
and was unaware that one could define :=.
Best,
John
-Original Message-
From: Kenn Konstabel [mailto:lebats...@gmail.com]
Sent: March-31-11 7:12 AM
To: William Dunlap
Cc: John Fox; Carl Witthoft; r-help@r-project.org
Subject: Re: [R] how about a p- operator?
In addition to %...% operators one can define := (I haven't seen this
possibility documented anywhere but it's used in a package) which seems to
have different precedence.
`:=`-`%-%` # the %-% by John Fox
a
[1] -1
a := 1000 + 1
[1] 1001
a
[1] 1001
a %-% 1000 + 1
[1] 1001
a
[1] 1000
Regards,
Kenn
Kenn Konstabel
National Institute for Health Development Hiiu 42 Tallinn
On Thu, Mar 31, 2011 at 2:50 AM, William Dunlap wdun...@tibco.com wrote:
The %...% operators are not a panacea.
they have the same precedence as `*`
and `/` (I think) so you get things like:
x %-% 10 - 8 # %-% has higher precedence than -
[1] 2
x # not what you thought it would be
[1] 10
x %-% 10 ^3 # but lower than ^
[1] 1000
x # this is what you expected
[1] 1000
It isn't that hard to write a package with your own parser in it.
Just have it generate the call tree from your input text and call
eval() on it.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of John Fox
Sent: Wednesday, March 30, 2011 4:34 PM
To: 'Carl Witthoft'
Cc: r-help@r-project.org
Subject: Re: [R] how about a p- operator?
Dear Carl,
I think that the following does what you want:
`%-%` - function(e1, e2){
+ e1 - deparse(substitute(e1))
+ env - parent.frame()
+ assign(e1, e2, envir=env)
+ e2
+ }
x %-% 10
[1] 10
x
[1] 10
But, as has been pointed out, it's probably easier just to
parenthesize the usual assignment command.
Regards,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]
On Behalf Of Carl Witthoft
Sent: March-30-11 7:00 PM
To: r-help@r-project.org
Subject: [R] how about a p- operator?
I was cursing Matlab again today (what else is new) because
the default
action for every Matlab command is to spew the result to
the console, and
one must remember to put that darn ; at the end of every line.
So I just wondered: was there ever a discussion as to
providing some
modified version of the - and - operators in R to do
the reverse?
That is, since R does not print the values of a command
to the console,
what if there were an operator such that
newobject p- somefunction()
would do the same as
print(newobject - somefunction())
Any thoughts?
Carl
__
R-help@r-project.org 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.
__
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PLEASE do read the posting guide
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__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Italicized title from index
Try this: plot(1, main = bquote(Yield for ~ italic(.(spp)) ~ in management region ~ .(region))) On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman jnewm...@uottawa.ca wrote: Hi all! I've written a handy script that uses a for loop to allow me to generate a large number of figures and statistical outputs for a large dataset. I am using indexing to retrieve a species name for the title of my graphs- which worked fine. However, I need to italicize these species names. I originally used the paste function, and had no problems with indexing: *main=paste(Yield for , testsub[1,3], in management region , testsub[1,2])* The title looks like: *Yield for Gadus morhua in management region 4X5Y* I tried bquote from a related help thread, I tried to emulate it: *spp-testsub[1,3]* *region-testsub[1,2]* * * *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~ .(region))* Which doesn't seem to work at all, but when I try not putting anything after the italic: *main=bquote(Yield ~ for ~ italic(.(spp)))* I get: *Yield for 1* While *spp=Gadus morhua* I'm at wit's end, I tried to read about substitute, expression, and eval functions in the hopes I can figure it out, but I am lost! Thanks for any help! Cheers, -Jeremy N Undergraduate Researcher in Macroecology University of Ottawa Department of Biology Ad astra per alia porci! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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] Italicized title from index
Better: plot(1, main = bquote(Yield for ~ italic(.(as.character(spp))) ~ in management region ~ .(region))) You have a factor, so you need convert it to character On Thu, Mar 31, 2011 at 8:59 AM, Henrique Dallazuanna www...@gmail.com wrote: Try this: plot(1, main = bquote(Yield for ~ italic(.(spp)) ~ in management region ~ .(region))) On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman jnewm...@uottawa.ca wrote: Hi all! I've written a handy script that uses a for loop to allow me to generate a large number of figures and statistical outputs for a large dataset. I am using indexing to retrieve a species name for the title of my graphs- which worked fine. However, I need to italicize these species names. I originally used the paste function, and had no problems with indexing: *main=paste(Yield for , testsub[1,3], in management region , testsub[1,2])* The title looks like: *Yield for Gadus morhua in management region 4X5Y* I tried bquote from a related help thread, I tried to emulate it: *spp-testsub[1,3]* *region-testsub[1,2]* * * *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~ .(region))* Which doesn't seem to work at all, but when I try not putting anything after the italic: *main=bquote(Yield ~ for ~ italic(.(spp)))* I get: *Yield for 1* While *spp=Gadus morhua* I'm at wit's end, I tried to read about substitute, expression, and eval functions in the hopes I can figure it out, but I am lost! Thanks for any help! Cheers, -Jeremy N Undergraduate Researcher in Macroecology University of Ottawa Department of Biology Ad astra per alia porci! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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] That dreaded floating point trap
A very straightforward way to avoid this problem is to construct the
sequence by multiplying a sequence of integers by an approriate
constant. E.g. for your first example:
for( i in where ){
thisdata- subset(herde, herde$mlr= i)
# do stuff with thisdata..
}
'where' is a vector like 0.1*((-10):10)
[ instead of seq(-1, 1, by=0.1) ]
and then, when you want to change to seq(-0.8, 1, by=0.1),
use instead 0.1*(-80,10).
Hi,
this helps, thank you.
But if this code is in a function, and some user supplies a vector, I
will still have to round it in the function, I guess.
It's weird how 0.1 is different from round(0.1, digits=1) , but I don't
want to read that 90 page long floating point paper which was referenced
somewhere :)
Thanks,
Alex
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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] VECM with UNRESTRICTED TREND
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea,
but still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
creating the trend variable.
Best,
Bernhard
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = trend, K = 2, spec =
longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
drop = FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1 as well? And
doesn't it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and use this for
'dumvar'. The matrix 'dumvar' is just added to the VECM as deterministic
regressors and while you are referring to case 5, this is basically what you
are after, if I am not mistaken. But we aware that this implies a quadratic
trend for the levels.
Best,
Bernhard
Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
Gesendet: Mittwoch, 30. März 2011 20:50
An: Pfaff, Bernhard Dr.; r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED TREND
Hello Bernhard,
Thank You very much. Unfortunately I'm still not really
sure how should I use dummy vars in this context...
If I have a system of three variables (x, y, z), lag
order = 2 and 1 cointegrating relation, what should I do? I mean, what kind of
'pattern' should be used to create those dummy variables, what should they
represent and how many of them do I need?
Many thanks in advance!
Best,
Greg
2011/3/30 Pfaff, Bernhard Dr.
bernhard_pf...@fra.invesco.com
Hello Greg,
you can exploit the argument 'dumvar' for this.
See ?ca.jo
Best,
Bernhard
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Im
Auftrag von Grzegorz Konat
Gesendet: Mittwoch, 30. März 2011 16:46
An: r-help@r-project.org
Betreff: [R] VECM with UNRESTRICTED TREND
Dear All,
My question is:
how can I estimate VECM system with
unrestricted trend (aka
case 5) option as a deterministic term?
As far as I know, ca.jo in urca package
allows for restricted trend
only [vecm
- ca.jo(data, type = trace/eigen, ecdet
= trend, K =
n, spec = transitory/longrun)].
Obviously, I don't have to do this in urca,
so if another
package gives the possibility, please let me
know too!
Thanks in advance!
Greg
[[alternative HTML version deleted]]
__
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Re: [R] VECM with UNRESTRICTED TREND
'time' was a trend variable from my.data set. Equivalent to the output of
the command 'matrix' you just gave me.
So now I did:
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
mat1 - matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar=mat1)
and the output is:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = trace, ecdet = const, K = 2, spec = longrun, :
No column names in 'dumvar', using prefix 'exo' instead.
What do I do wrong?
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
creating the trend variable.
Best,
Bernhard
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = trend, K = 2, spec =
longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
= FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1 as well? And
doesn't it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
matrix 'dumvar' is just added to the VECM as deterministic regressors and
while you are referring to case 5, this is basically what you are after, if
I am not mistaken. But we aware that this implies a quadratic trend for the
levels.
Best,
Bernhard
--
*Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
*Gesendet:* Mittwoch, 30. März 2011 20:50
*An:* Pfaff, Bernhard Dr.; r-help@r-project.org
*Betreff:* Re: [R] VECM with UNRESTRICTED TREND
Hello Bernhard,
Thank You very much. Unfortunately I'm still not really sure how should I
use dummy vars in this context...
If I have a system of three variables (x, y, z), lag order = 2 and 1
cointegrating relation, what should I do? I mean, what kind of 'pattern'
should be used to create those dummy variables, what should they represent
and how many of them do I need?
Many thanks in advance!
Best,
Greg
2011/3/30 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you can exploit the argument 'dumvar' for this. See ?ca.jo
Best,
Bernhard
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
Gesendet: Mittwoch, 30. März 2011 16:46
An: r-help@r-project.org
Betreff: [R] VECM with UNRESTRICTED TREND
Dear All,
My question is:
how can I estimate VECM system with unrestricted trend (aka
case 5) option as a deterministic term?
As far as I know, ca.jo in urca package allows for restricted trend
only [vecm
- ca.jo(data, type = trace/eigen, ecdet = trend, K =
n, spec = transitory/longrun)].
Obviously, I don't have to do this in urca, so if another
package gives the possibility, please let me know too!
Thanks in advance!
Greg
[[alternative HTML version deleted]]
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*
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] That dreaded floating point trap
On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt this helps, thank you. But if this code is in a function, and some user supplies a vector, I will still have to round it in the function, I guess. It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want to read that 90 page long floating point paper which was referenced somewhere :) Or you could try the much shorter R FAQ 7.31. Turns out it isn't weird at all, if you are a computer. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org 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] fonts in mosaic
If you are doing multiple plots like this, there is no way I know of to specify the fonts for labeling *once* for all such plots. However, you can do something like this to save typing and keep things consistent: my.largs - list( gp_labels = gpar(fontsize = 12, fontfamily = calibri), gp_varnames = gpar(fontsize = 16, fontfamily = calibri)) mosaic(UCBAdmissions, labeling_args=my.largs) mosaic(UCBAdmissions, labeling_args=my.largs, shade=TRUE) etc. On 3/30/2011 2:42 PM, Henrique Dallazuanna wrote: Try this: windowsFonts(calibri = windowsFont(Calibri)) mosaic(UCBAdmissions, labeling_args = list( gp_labels = gpar(fontsize = 12, fontfamily = calibri), gp_varnames = gpar(fontsize = 16, fontfamily = calibri) )) On Wed, Mar 30, 2011 at 3:25 PM, Erich Neuwirth erich.neuwi...@univie.ac.at wrote: Achim I simply want to replace the font R uses on mosaic (whatever it is) by a font of my choice (say Calibri or Arial) because I need to embed the R charts in a PowerPoint presentation and want the fonts to match. And I want the most simple way of accomplishing this. I worked my way through the strucplot vignette, but I could not extract enough information there. Is there some information about the proper font names to use in R? Personally, I simply change the size of the device I'm plotting on. When I plot on a large device, the fonts will be relatively smaller, and vice versa. This is what I do when including graphics in PDF files (papers, slides, reports, etc.). For fine control, you can set the arguments of the labeling function employed. ?strucplot shows that the default is ?labeling_border which has several arguments. For example you can set the graphical parameters of the labels (gp_labels) or the graphical parameters of the variable names (gp_varnames). Both arguments take ?gpar lists (grid graphical parameters). For example you may do __ R-help@r-project.org 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. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org 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] VECM with UNRESTRICTED TREND
Well, without further information, I do not know, but try the following
library(urca)
example(ca.jo)
trend - matrix(1:nrow(sjf), ncol = 1)
colnames(trend) - trd
ca.jo(sjf, type = trace, ecdet = const, K = 2, spec = longrun, dumvar =
trend)
Best,
Bernhard
Von: Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
Gesendet: Donnerstag, 31. März 2011 14:40
An: Pfaff, Bernhard Dr.; r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED TREND
'time' was a trend variable from my.data set. Equivalent to the output
of the command 'matrix' you just gave me.
So now I did:
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
mat1 - matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar=mat1)
and the output is:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
drop = FALSE] :
non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, :
No column names in 'dumvar', using prefix 'exo' instead.
What do I do wrong?
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Bernhard,
thank You so much one again! Now I (more or less)
understand the idea, but still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
What is 'time'? Just employ matrix(seq(1:nrow(dat1)),
ncol = 1) for creating the trend variable.
Best,
Bernhard
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet =
trend, K = 2, spec = longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) -
lag + 1L), , drop = FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1
as well? And doesn't it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr.
bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and
use this for 'dumvar'. The matrix 'dumvar' is just added to the VECM as
deterministic regressors and while you are referring to case 5, this is
basically what you are after, if I am not mistaken. But we aware that this
implies a quadratic trend for the levels.
Best,
Bernhard
Von: Grzegorz Konat
[mailto:grzegorz.ko...@ibrkk.pl]
Gesendet: Mittwoch, 30. März 2011 20:50
An: Pfaff, Bernhard Dr.;
r-help@r-project.org
Betreff: Re: [R] VECM with UNRESTRICTED
TREND
Hello Bernhard,
Thank You very much. Unfortunately I'm
still not really sure how should I use dummy vars in this context...
If I have a system of three variables
(x, y, z), lag order = 2 and 1 cointegrating relation, what should I do? I
mean, what kind of 'pattern' should be used to create those dummy variables,
what should they represent and how many of them do I need?
Many thanks in advance!
Best,
Greg
Re: [R] That dreaded floating point trap
Am 31.03.2011 14:41, schrieb Sarah Goslee: On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt this helps, thank you. But if this code is in a function, and some user supplies a vector, I will still have to round it in the function, I guess. It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want to read that 90 page long floating point paper which was referenced somewhere :) Or you could try the much shorter R FAQ 7.31. Turns out it isn't weird at all, if you are a computer. You're a computer! :) But yes.. the FAQ entry was where I found all.equal and the referenced 90-page-paper. But I didn't find out how to do a subset with 'somevector 0.4'. I think I'll have to round the numbers every time now.. or use some other not-so-pretty workaround like 'somevector 0.4 - 0.05' for 0.1-binned data. __ R-help@r-project.org 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] Lists of tables and conditional statements
Thanks Henrique - that worked like a charm - I had tried lots of other combinations before seeing your reply - wished I had asked sooner! Alan -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Wednesday, March 30, 2011 10:09 PM To: Herbert, Alan G Cc: r-help@r-project.org Help Subject: Re: [R] Lists of tables and conditional statements On Mar 30, 2011, at 7:27 PM, Henrique Dallazuanna wrote: Try this: lapply(l, function(x)x[x[,'Sum'] == 3,]) If this is the right answer, you should send a solved message. The dput extract was incomplete. -- David. On Wed, Mar 30, 2011 at 7:38 PM, Herbert, Alan G aherb...@bu.edu wrote: Hi R-users, I have a list containing numeric tables of differing row length. I want to make a new list that contains only rows from tables with a Sum greater than 3, plus the names of each table. I was wondering whether there is an elegant way to do this using apply of related functions as this list has many thousands of such tables. Here is an example of the list $AACS POOL INFO pool1 pool2 pool6 pool7 pool8 pool.all Sum 12:125561133:novel 0 0 0 0 10 1 12:125570904:novel 0 0 0 0 10 1 12:125571014:novel 0 1 0 0 00 1 12:125571038:novel 0 0 0 1 00 1 12:125575996:novel 0 0 0 1 00 1 12:125591844:rs2297478 1 0 1 0 01 3 12:125599114:novel 0 0 0 1 00 1 12:125612668:novel 0 0 0 0 10 1 12:125612839:rs900411 1 0 1 0 11 4 12:125626650:novel 0 0 0 0 10 1 12:125626737:novel 0 0 0 1 00 1 $AADAC POOL INFO pool1 pool2 pool5 pool6 pool7 pool8 pool.all Sum 3:151542411:novel 0 0 0 0 1 0 1 2 3:151542412:novel 0 0 0 0 1 0 1 2 3:151542643:novel 0 1 0 0 0 0 0 1 3:151545322:rs2410836 0 1 0 0 0 0 1 2 3:151545323:rs62272918 0 1 0 0 0 0 1 2 3:151545509:novel 0 0 1 0 0 0 1 2 3:151545601:rs1803155 1 1 1 1 1 1 1 7 3:151545721:novel 0 0 1 0 0 0 0 1 3:151545802:novel 0 0 0 0 1 0 0 1 3:151545824:novel 0 1 0 0 0 0 0 1 Thanks for your help [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org 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] That dreaded floating point trap
n Thu, Mar 31, 2011 at 3:56 PM, Alexander Engelhardt a...@chaotic-neutral.de wrote: Am 31.03.2011 14:41, schrieb Sarah Goslee: On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt this helps, thank you. But if this code is in a function, and some user supplies a vector, I will still have to round it in the function, I guess. It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want to read that 90 page long floating point paper which was referenced somewhere :) Or you could try the much shorter R FAQ 7.31. Turns out it isn't weird at all, if you are a computer. You're a computer! :) But yes.. the FAQ entry was where I found all.equal and the referenced 90-page-paper. But I didn't find out how to do a subset with 'somevector 0.4'. I think I'll have to round the numbers every time now.. or use some other not-so-pretty workaround like 'somevector 0.4 - 0.05' for 0.1-binned data. You could define your own %% and %% to do rounding and comparison in one step. Kenn __ R-help@r-project.org 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] VECM with UNRESTRICTED TREND
The code you gave me works fine with Finland, but the same for my data -
does not!
I do:
library(urca)
data(my.data)
dat1 - my.data[, c(dY, X, dM)]
trend - matrix(1:nrow(dat1), ncol = 1)
colnames(trend) - trd
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar = trend)
and the result is again:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
non-numeric argument to binary operator
I attach my dataset in xls format. If you have 5 minutes and wish to check
it out, I'd be extremely grateful!
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Well, without further information, I do not know, but try the following
library(urca)
example(ca.jo)
trend - matrix(1:nrow(sjf), ncol = 1)
colnames(trend) - trd
ca.jo(sjf, type = trace, ecdet = const, K = 2, spec = longrun,
dumvar = trend)
Best,
Bernhard
--
*Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
*Gesendet:* Donnerstag, 31. März 2011 14:40
*An:* Pfaff, Bernhard Dr.; r-help@r-project.org
*Betreff:* Re: [R] VECM with UNRESTRICTED TREND
'time' was a trend variable from my.data set. Equivalent to the output of
the command 'matrix' you just gave me.
So now I did:
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
mat1 - matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar=mat1)
and the output is:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
= FALSE] :
non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = trace, ecdet = const, K = 2, spec = longrun,
:
No column names in 'dumvar', using prefix 'exo' instead.
What do I do wrong?
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
creating the trend variable.
Best,
Bernhard
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = trend, K = 2, spec =
longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
drop = FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1 as well? And
doesn't it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
matrix 'dumvar' is just added to the VECM as deterministic regressors and
while you are referring to case 5, this is basically what you are after, if
I am not mistaken. But we aware that this implies a quadratic trend for the
levels.
Best,
Bernhard
--
*Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
*Gesendet:* Mittwoch, 30. März 2011 20:50
*An:* Pfaff, Bernhard Dr.; r-help@r-project.org
*Betreff:* Re: [R] VECM with UNRESTRICTED TREND
Hello Bernhard,
Thank You very much. Unfortunately I'm still not really sure how should I
use dummy vars in this context...
If I have a system of three variables (x, y, z), lag order = 2 and 1
cointegrating relation, what should I do? I mean, what kind of 'pattern'
should be used to create those dummy variables, what should they represent
and how many of them do I need?
Many thanks in advance!
Best,
Greg
2011/3/30 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you can exploit the argument 'dumvar' for this. See ?ca.jo
Best,
Bernhard
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Im Auftrag von Grzegorz Konat
Gesendet: Mittwoch, 30. März 2011 16:46
An: r-help@r-project.org
Betreff: [R] VECM with UNRESTRICTED TREND
Dear All,
My question is:
how can I estimate VECM system with unrestricted trend (aka
case 5) option as a deterministic term?
As far as I know, ca.jo in urca package allows for restricted trend
only [vecm
- ca.jo(data, type = trace/eigen, ecdet = trend, K =
n, spec = transitory/longrun)].
Obviously, I don't have to do this in urca, so if another
package gives the possibility, please let me know too!
Thanks in advance!
Greg
[[alternative HTML version deleted]]
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Re: [R] That dreaded floating point trap
Dear Alexander, Instead of testing 'somevector 0.4', test 'abs(somevector - 0.4) some.small.number' Best regards, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek team Biometrie Kwaliteitszorg Gaverstraat 4 9500 Geraardsbergen Belgium Research Institute for Nature and Forest team Biometrics Quality Assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Alexander Engelhardt Verzonden: donderdag 31 maart 2011 14:57 Aan: r-help@r-project.org Onderwerp: Re: [R] That dreaded floating point trap Am 31.03.2011 14:41, schrieb Sarah Goslee: On Thu, Mar 31, 2011 at 8:14 AM, Alexander Engelhardt this helps, thank you. But if this code is in a function, and some user supplies a vector, I will still have to round it in the function, I guess. It's weird how 0.1 is different from round(0.1, digits=1) , but I don't want to read that 90 page long floating point paper which was referenced somewhere :) Or you could try the much shorter R FAQ 7.31. Turns out it isn't weird at all, if you are a computer. You're a computer! :) But yes.. the FAQ entry was where I found all.equal and the referenced 90-page-paper. But I didn't find out how to do a subset with 'somevector 0.4'. I think I'll have to round the numbers every time now.. or use some other not-so-pretty workaround like 'somevector 0.4 - 0.05' for 0.1-binned data. __ R-help@r-project.org 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@r-project.org 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 do t-test in r for difference of mean
I am trying to do t-test to test whether the mean of one one column of the data frame is greater then another. please help me out. -- Arkajyoti Jana M. Phil/ 2nd semester Centre for Economic Studies and planning School of Social Sciences Jawaharlal Nehru University New Delhi-67 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] summing values by week - based on daily dates - but with somedates missing
Thank you so much, everyone, for your help.
Extremely valuable suggestions and extremely valuable learnings!
Dimitri
On Thu, Mar 31, 2011 at 5:03 AM, Martyn Byng martyn.b...@nag.co.uk wrote:
Hi,
Yep, that was what it was doing. For a sum across week, try something like
get.week.flag - function(dd) {
## get weekday from the date in dd and code it as Monday = 1, Tuesday = 2 etc
idd =
factor(weekdays(dd),levels=c(Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday))
## convert to numeric
ndd = as.numeric(idd)
## flag entries where weekday code gets less (this will flag changes in week)
wflag = c(FALSE,(ndd[-length(idd)] ndd[-1]))
## cumulative sum to get the week flag
cumsum(wflag) + 1
}
to get a week flag (this is assuming that your data is sorted by date, if not
you'll have to sort it first). If you want the week to start on a different
day, just change the ordering of the weekdays in the levels statement.
data.frame(date=myframe$date,day=weekdays(myframe$date),week=get.week.flag(myframe$date))
seems to indicate that the function is doing what it should, so you can then
amend the previous code to use get.week.flag instead of weekdays, as in
sum.by.week - function(ff) {
by.day - split(ff$value,get.week.flag(ff$dates))
lapply(by.day,sum)
}
by.grp - split(myframe,myframe$group)
lapply(by.grp,sum.by.week)
Martyn
-Original Message-
From: Dimitri Liakhovitski [mailto:dimitri.liakhovit...@gmail.com]
Sent: 30 March 2011 18:03
To: Martyn Byng
Cc: r-help
Subject: Re: [R] summing values by week - based on daily dates - but with
somedates missing
Thank you, Martyn.
But it looks like this way we are getting sums by day - i.e., across
all Mondays, all Tuesdays, etc.
Maybe I did not explain well, sorry! The desired output would contain
sums for each WHOLE week - across all days that comprise that week -
Monday through Sunday.
Makes sense?
Dimitri
On Wed, Mar 30, 2011 at 12:53 PM, Martyn Byng martyn.b...@nag.co.uk wrote:
Hi,
How about something like:
sum.by.day - function(ff) {
by.day - split(ff$value,weekdays(ff$dates))
lapply(by.day,sum)
}
by.grp - split(myframe,myframe$group)
lapply(by.grp,sum.by.day)
Martyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Dimitri Liakhovitski
Sent: 30 March 2011 15:23
To: r-help
Subject: [R] summing values by week - based on daily dates - but with
somedates missing
Dear everybody,
I have the following challenge. I have a data set with 2 subgroups,
dates (days), and corresponding values (see example code below).
Within each subgroup: I need to aggregate (sum) the values by week -
for weeks that start on a Monday (for example, 2008-12-29 was a
Monday).
I find it difficult because I have missing dates in my data - so that
sometimes I don't even have the date for some Mondays. So, I can't
write a proper loop.
I want my output to look something like this:
group dates value
group.1 2008-12-29 3.0937
group.1 2009-01-05 3.8833
group.1 2009-01-12 1.362
...
group.2 2008-12-29 2.250
group.2 2009-01-05 1.4057
group.2 2009-01-12 3.4411
...
Thanks a lot for your suggestions! The code is below:
Dimitri
### Creating example data set:
mydates-rep(seq(as.Date(2008-12-29), length = 43, by = day),2)
myfactor-c(rep(group.1,43),rep(group.2,43))
set.seed(123)
myvalues-runif(86,0,1)
myframe-data.frame(dates=mydates,group=myfactor,value=myvalues)
(myframe)
dim(myframe)
## Removing same rows (dates) unsystematically:
set.seed(123)
removed.group1-sample(1:43,size=11,replace=F)
set.seed(456)
removed.group2-sample(44:86,size=11,replace=F)
to.remove-c(removed.group1,removed.group2);length(to.remove)
to.remove-to.remove[order(to.remove)]
myframe-myframe[-to.remove,]
(myframe)
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
R-help@r-project.org 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.
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This
Re: [R] how about a p- operator?
On 11-03-30 7:00 PM, Carl Witthoft wrote: I was cursing Matlab again today (what else is new) because the default action for every Matlab command is to spew the result to the console, and one must remember to put that darn ; at the end of every line. So I just wondered: was there ever a discussion as to providing some modified version of the - and - operators in R to do the reverse? That is, since R does not print the values of a command to the console, what if there were an operator such that newobjectp- somefunction() would do the same as print(newobject- somefunction()) Any thoughts? Others have given alternatives. Just some comments on this particular proposal: We already have a problem that x-3 is slightly ambiguous: does it mean x - 3 # yes or x -3 # no I think the compound operator p- would just repeat this problem. What does xp-3 mean? If the amount of typing in print(x - 3) is a problem, that's a user interface issue, so it should be addressed at the user interface level, not by changing the language. A front-end could make it easy to transform x - 3 into the longer expression (or an equivalent), or just make it easy to examine the .Last.value variable. Duncan Murdoch __ R-help@r-project.org 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] That dreaded floating point trap
On 11-03-31 7:24 AM, Alexander Engelhardt wrote:
Hi,
I had a piece of code which looped over a decimal vector like this:
for( i in where ){
thisdata- subset(herde, herde$mlr= i)
# do stuff with thisdata..
}
'where' is a vector like seq(-1, 1, by=0.1)
The solution to this problem is to take steps by representable numbers,
not by numbers like 0.1 that can't be represented exactly. For example,
seq(-1, 1, by=0.25) has exact entries, because fractions with small
powers of 2 in the denominator are all exactly representable. (Small
depends on the numerator, but for fractions between 0 and 1 it's about
52, so not really so small.)
Duncan Murdoch
My problem was: 'nrow(thisdata)' in loop repetition 0.4 was different if
'where' was seq(-1, 1, by=0.1) than when 'where' was seq(-0.8, 1, by=0.1)
It went away after I changed the first line to:
thisdata- subset(herde, herde$mlr= round(i, digits=1))
This is that floating point trap the R inferno pdf talked about,
right? That file talked about the problem, but didn't offer a solution.
Similar things happened when I created a table() from a vector with
values in seq(-1, 1, by=0.1)
Do I really have to round every float at every occurence from now on, or
is there another solution? I only found all.equal() and identical(), but
I want to subset for observations with a value /greater/ than something.
Thanks in advance,
Alex
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Re: [R] Simple lattice question
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')
How do I put a legend for the grouping variable in the empty upper-right panel?
See the help page for xyplot for an example using the iris data.
You just need to add something like
auto.key = list(x = .6, y = .7, corner = c(0, 0))
to your lattice call. See the 'key' entry on the ?xyplot page.
Peter Ehlers
TIA
Rubén
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
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[R] another statistical question
Dear List! I have a unverse (basic population) which is not normally distributed. Now from this universe I take some subsets. Each subset is normally distributed within itself. I now want to compare the subsets and see if they differ significantly. So what is my assumption - normal distributed data and therefore comparison of means, or non normal distributed data and therefore comparison of medians? Cheers, Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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] Simple lattice question
Thanks Peters!
Just a few minor glitches now:
require(lattice)
data - data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0), points=TRUE,
lines=TRUE, pch=1:4, lty=1:4, type='b',
text=list(lab = as.character(unique(data$z)
I have an extra column of symbols on the legend,
and,
would like to add a title to the legend. Such as 'main' for plots.
Any suggestions?
Rubén
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
-Mensaje original-
De: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Enviado el: jueves, 31 de marzo de 2011 15:41
Para: Rubén Roa
CC: r-help@r-project.org
Asunto: Re: [R] Simple lattice question
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data-
data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='bl
ack',type='b')
How do I put a legend for the grouping variable in the
empty upper-right panel?
See the help page for xyplot for an example using the iris data.
You just need to add something like
auto.key = list(x = .6, y = .7, corner = c(0, 0))
to your lattice call. See the 'key' entry on the ?xyplot page.
Peter Ehlers
TIA
Rubén
__
__
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
__
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PLEASE do read the posting guide
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Re: [R] another statistical question
Am 31.03.2011 15:46, schrieb Anna Lee: Dear List! I have a unverse (basic population) which is not normally distributed. Now from this universe I take some subsets. Each subset is normally distributed within itself. I now want to compare the subsets and see if they differ significantly. So what is my assumption - normal distributed data and therefore comparison of means, or non normal distributed data and therefore comparison of medians? If you want to do an ANOVA, the assumption is normality /within/ the groups. That is, Y given X=x is normal distributed. Also, you want the same variance within each group (group = your subset = factor value). Short answer: Means. (I am only 95% certain) __ R-help@r-project.org 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] choosing best 'match' for given factor
Folks,
I have a 'matching' matrix between variables A, X, L, O:
a - structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c(A, X, L, O), c(A, X, L, O)))
a
A X L O
A 1.00 0.41 0.58 0.75
X 0.41 1.00 0.60 0.86
L 0.58 0.75 1.00 0.83
O 0.60 0.86 0.83 1.00
And I have a search vector of variables
v - c(X, O)
I want to write a function bestMatch(searchvector, matchMat) such that for each
variable in searchvector, I get the variable that it has the highest match to -
but searching only among variables to the left of it in the 'matching' matrix,
and not matching with any variable in searchvector itself.
So in the above example, although X has the highest match (0.86) with O, I
can't choose O as it's to the right of X (and also because O is in the
searchvector v already); I'll have to choose A.
For O, I will choose L, the variable it's best matched with - as it can't
match X already in the search vector.
My function bestMatch(v, a) will then return c(A, L)
My matrix a is quite large, and I have a long list of search vectors v, so I
need an efficient method.
I wrote this:
bestMatch - function(searchvector, matchMat) {
sapply(searchvector, function(cc) {
y - matchMat[!(rownames(matchMat) %in%
searchvector) (index(rownames(matchMat)) match(cc, rownames(matchMat))),
cc, drop = FALSE];
rownames(y)[which.max(y)]
})
}
Any advice?
Thanks,
Murali
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[R] error in recode.defalt ....object '.data' not found
Dear colleagues, working with the data frame below, trying to reverse two variables I the error message below. i searched through the help list but could not find any postings which could help me solve the situation. I tried attaching and detaching the data frame to no avail. Yours, Simon Kiss *DATA FRAME 'data.frame': 1569 obs. of 9 variables: $ equal : num 3 4 3 2 3 4 2 3 2 2 ... $ disc : num 3 2 3 3 2 2 3 3 3 3 ... $ family: num 3 2 2 2 3 2 2 1 2 1 ... $ special : num 3 3 4 4 3 3 4 4 3 4 ... $ immigrants: num 3 8 3 8 3 3 4 1 1 2 ... $ wedlock : num 3 3 3 3 3 2 2 8 2 3 ... $ crime : num 3 2 2 1 2 3 1 8 2 1 ... $ breakdown : num 3 3 3 2 2 4 8 2 2 4 ... $ nonwhites : num 2 4 3 3 2 2 3 4 3 3 ... *RECODE social$nonwhites-recode(social$nonwhites, 1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA) *ERROR Error in recode.default(social$nonwhites, 1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA) : object '.data' not found * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org 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] error in recode.defalt ....object '.data' not found
Dear Simon, Your intention appears to be to call the recode() function in the car package. That function is not generic, and consequently has no default method, leading me to believe that you've called a recode() function in some other package. I hope this helps, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Thu, 31 Mar 2011 11:01:42 -0400 Simon Kiss sjk...@gmail.com wrote: Dear colleagues, working with the data frame below, trying to reverse two variables I the error message below. i searched through the help list but could not find any postings which could help me solve the situation. I tried attaching and detaching the data frame to no avail. Yours, Simon Kiss *DATA FRAME 'data.frame': 1569 obs. of 9 variables: $ equal : num 3 4 3 2 3 4 2 3 2 2 ... $ disc : num 3 2 3 3 2 2 3 3 3 3 ... $ family: num 3 2 2 2 3 2 2 1 2 1 ... $ special : num 3 3 4 4 3 3 4 4 3 4 ... $ immigrants: num 3 8 3 8 3 3 4 1 1 2 ... $ wedlock : num 3 3 3 3 3 2 2 8 2 3 ... $ crime : num 3 2 2 1 2 3 1 8 2 1 ... $ breakdown : num 3 3 3 2 2 4 8 2 2 4 ... $ nonwhites : num 2 4 3 3 2 2 3 4 3 3 ... *RECODE social$nonwhites-recode(social$nonwhites, 1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA) *ERROR Error in recode.default(social$nonwhites, 1=4; 2=3; 3=2; 4=1; 8=NA; -9=NA) : object '.data' not found * Simon J. Kiss, PhD Assistant Professor, Wilfrid Laurier University 73 George Street Brantford, Ontario, Canada N3T 2C9 Cell: +1 519 761 7606 __ R-help@r-project.org 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@r-project.org 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] AEM package
I'm trying to do AEM eigenfunctions, but I have a different number of points
into each reservoir (e.g.: reservoir 1-4 points, reservoir 2-6 points,
reservoir 3-2 points). Can someone help me?Besides this in the Legendre's
book exemple in the part:length.edge-vector(length=nrow(edges.b))for (i in
1:nrow(edges.b)){length.edge[i]-D1.mat[edges.b[i,1], edges.b[i,2]]}There is
some problem with the lenght argument.Thanks
Fernanda Melo Carneiro contato: (62) 3521-1480 e 8121-7374www.ecoevol.ufg.br
Laboratório de Ecologia Teórica e Síntese (UFG)
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Re: [R] fonts in mosaic
The easiest way of changing the font used for labels by the windows graphics device (opened by a call to windows()) seems to be the following: Let us assume we want to use the font Consolas for all labels: windowsFonts(myfont=Consolas) par(family=myfont) If one later on wants to change the default font for graphics, one has to run both commands again. Just issueing a new windowsFonts command will not change the font used by the graphics device. __ R-help@r-project.org 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] ANCOVA for linear regressions without intercept
If you haven't already received an answer, a careful reading of ?formula will provide it. -- Bert On Wed, Mar 30, 2011 at 11:42 PM, Yusuke Fukuda yusuke.fuk...@nt.gov.auwrote: Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0 intercept). Is it possible to do ANCOVA for these regressions without intercepts? When I do summary(lm(body length ~ sex*head length)) this will include the intercepts as below Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept)-6.496971.68497 -3.856 0.000118 *** sexMale-9.393401.97760 -4.750 2.14e-06 *** sexUnknown -1.337912.35453 -0.568 0.569927 head_length 7.123070.05503 129.443 2e-16 *** sexMale:head_length 0.316310.06246 5.064 4.37e-07 *** sexUnknown:head_length 0.199370.07022 2.839 0.004556 ** --- Is there any way I can remove the intercepts so that I can simply compare the slopes with no intercept taken into account? Thanks for help in advance. Yusuke Fukuda __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to do t-test in r for difference of mean
Hi Arkajyoti, Please learn to help yourself. Start with help.search(t-test) Also read one of the many introductory R tutorials on the web. Best, Ista On Thu, Mar 31, 2011 at 9:07 AM, arkajyoti jana arkajyo...@gmail.com wrote: I am trying to do t-test to test whether the mean of one one column of the data frame is greater then another. please help me out. -- Arkajyoti Jana M. Phil/ 2nd semester Centre for Economic Studies and planning School of Social Sciences Jawaharlal Nehru University New Delhi-67 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org 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] choosing best 'match' for given factor
Hi Murali.
I haven't compared, but this is what I would do:
bestMatch-function(searchVector, matchMat)
{
searchRow-unique(sort(match(searchVector, colnames(matchMat #if
you're sure, you could drop unique
cat(Original row indices:)
print(searchRow)
matchMat-matchMat[, -searchRow, drop=FALSE] #avoid duplicates
altogether
cat(Corrected Matrix:\n)
print(matchMat)
correctedRows-searchRow - seq_along(searchRow) + 1 #works because
of the sort above
cat(Corrected row indices:)
print(correctedRows)
sapply(correctedRows, function(cr){
lookWhere-matchMat[cr, seq(cr-1)]
cat(Will now look into:\n)
print(lookWhere)
cc-which.max(lookWhere)
cat(Max at position, cc, \n)
colnames(matchMat)[cc]
})
}
I don't think there's that much difference. Depending on specific sizes, it
may be more or less costly to first shrink the search matrix like I do. And
similarly depending, I may be better still if you remove the rows that
you're not interested in as well (some more but similar index trickery
required then.
HTH,
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of murali.me...@avivainvestors.com
Sent: donderdag 31 maart 2011 16:46
To: r-help@r-project.org
Subject: [R] choosing best 'match' for given factor
Folks,
I have a 'matching' matrix between variables A, X, L, O:
a - structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c(A, X, L, O), c(A, X, L, O)))
a
A X L O
A 1.00 0.41 0.58 0.75
X 0.41 1.00 0.60 0.86
L 0.58 0.75 1.00 0.83
O 0.60 0.86 0.83 1.00
And I have a search vector of variables
v - c(X, O)
I want to write a function bestMatch(searchvector, matchMat) such that for
each variable in searchvector, I get the variable that it has the highest
match to - but searching only among variables to the left of it in the
'matching' matrix, and not matching with any variable in searchvector
itself.
So in the above example, although X has the highest match (0.86) with O,
I can't choose O as it's to the right of X (and also because O is in the
searchvector v already); I'll have to choose A.
For O, I will choose L, the variable it's best matched with - as it
can't match X already in the search vector.
My function bestMatch(v, a) will then return c(A, L)
My matrix a is quite large, and I have a long list of search vectors v, so I
need an efficient method.
I wrote this:
bestMatch - function(searchvector, matchMat) {
sapply(searchvector, function(cc) {
y - matchMat[!(rownames(matchMat) %in%
searchvector) (index(rownames(matchMat)) match(cc, rownames(matchMat))),
cc, drop = FALSE];
rownames(y)[which.max(y)]
})
}
Any advice?
Thanks,
Murali
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Re: [R] another statistical question
On 2011-03-31 07:44, Alexander Engelhardt wrote: Am 31.03.2011 15:46, schrieb Anna Lee: Dear List! I have a unverse (basic population) which is not normally distributed. Now from this universe I take some subsets. Each subset is normally distributed within itself. I now want to compare the subsets and see if they differ significantly. So what is my assumption - normal distributed data and therefore comparison of means, or non normal distributed data and therefore comparison of medians? If you want to do an ANOVA, the assumption is normality /within/ the groups. That is, Y given X=x is normal distributed. Also, you want the same variance within each group (group = your subset = factor value). The equal-variances assumption is not strictly required. R has the multiple-groups equivalent of the two-sample t-test. See ?oneway.test. Peter Ehlers Short answer: Means. (I am only 95% certain) __ R-help@r-project.org 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@r-project.org 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 put line linking two plots
Here is a way to do it using just base graphics: layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE)) plot(runif(10), type='b', ylim=c(0,1)) x.tmp - grconvertX(4, to='ndc') y.tmp - grconvertY(0.9, to='ndc') plot(runif(20), type='l', ylim=c(0,1)) par(xpd=NA) segments( 10, 1, grconvertX(x.tmp, from='ndc'), grconvertY(y.tmp, from='ndc'), col='red' ) plot(runif(20), type='l') -- 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 Paul Murrell Sent: Wednesday, March 30, 2011 4:29 PM To: Mario Valle Cc: R-help@r-project.org Subject: Re: [R] How to put line linking two plots Hi On 30/03/2011 10:54 p.m., Mario Valle wrote: Hello! Suppose I have three charts like below. The top chart is a general overview and the bottom charts are related so some point of this chart. To make clear this relationship I want to draw a line between (4,0.9) in the top chart and (10,1) in the bottom-left one. Currently I add it manually using Inkscape on the resulting pdf file. Is it possible to add it inside R? Should I switch to other charting packages? You'll have your work cut out using traditional graphics, but this is doable in grid-based graphics. For example, ... library(grid) library(lattice) set.seed(123) print(xyplot(runif(10)~1:10, type=b), position=c(0, .5, 1, 1), prefix=top, more=TRUE) print(xyplot(runif(20)~1:20, type=l), position=c(0, 0, .5, .5), prefix=left, more=TRUE) print(xyplot(runif(20)~1:20, type=l), position=c(.5, 0, 1, .5), prefix=right) trellis.focus(panel, 1, 1, prefix=top) grid.move.to(unit(4, native), unit(.9, native)) trellis.unfocus() trellis.focus(panel, 1, 1, prefix=left, clip.off=TRUE) grid.line.to(unit(10, native), unit(1, native)) trellis.unfocus() Paul Thanks for the advice! mario set.seed(123) pdf(test.pdf, width=14, height=7) layout(matrix(c(1,1,2,3), 2, 2, byrow = TRUE)) plot(runif(10), type='b') plot(runif(20), type='l') plot(runif(20), type='l') dev.off() R 2.12.2 on Windows 7 (32bits) -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 p...@stat.auckland.ac.nz http://www.stat.auckland.ac.nz/~paul/ __ R-help@r-project.org 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@r-project.org 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] Italicized title from index
Hi, Thank you *very* much. I'll remember that in the future! Cheers, -Jeremy N Undergraduate Researcher in Macroecology University of Ottawa Department of Biology Ad astra per alia porci! On Thu, Mar 31, 2011 at 8:01 AM, Henrique Dallazuanna www...@gmail.comwrote: Better: plot(1, main = bquote(Yield for ~ italic(.(as.character(spp))) ~ in management region ~ .(region))) You have a factor, so you need convert it to character On Thu, Mar 31, 2011 at 8:59 AM, Henrique Dallazuanna www...@gmail.com wrote: Try this: plot(1, main = bquote(Yield for ~ italic(.(spp)) ~ in management region ~ .(region))) On Thu, Mar 31, 2011 at 2:35 AM, Jeremy Newman jnewm...@uottawa.ca wrote: Hi all! I've written a handy script that uses a for loop to allow me to generate a large number of figures and statistical outputs for a large dataset. I am using indexing to retrieve a species name for the title of my graphs- which worked fine. However, I need to italicize these species names. I originally used the paste function, and had no problems with indexing: *main=paste(Yield for , testsub[1,3], in management region , testsub[1,2])* The title looks like: *Yield for Gadus morhua in management region 4X5Y* I tried bquote from a related help thread, I tried to emulate it: *spp-testsub[1,3]* *region-testsub[1,2]* * * *main=bquote(Yield ~ for ~ italic(.(spp)) ~ in ~ management ~ region ~ .(region))* Which doesn't seem to work at all, but when I try not putting anything after the italic: *main=bquote(Yield ~ for ~ italic(.(spp)))* I get: *Yield for 1* While *spp=Gadus morhua* I'm at wit's end, I tried to read about substitute, expression, and eval functions in the hopes I can figure it out, but I am lost! Thanks for any help! Cheers, -Jeremy N Undergraduate Researcher in Macroecology University of Ottawa Department of Biology Ad astra per alia porci! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Graph many points without hiding some
Just a note, Base graphics does support transparency as long as the device
plotting to supports it.
--
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 Dennis Murphy
Sent: Thursday, March 31, 2011 1:36 AM
To: Samuel Dennis
Cc: R-help@r-project.org
Subject: Re: [R] Graph many points without hiding some
Hi:
I can think of a couple: (1) size reduction of the points; (2) alpha
transparency; (3) (1) + (2)
From your original plot in base graphics, I reduced cex to 0.2 and it
didn't
look too bad:
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24), cex = 0.2)
points(rnorm(x,mean=20),rnorm(x),col=1, cex = 0.2)
points(rnorm(x,mean=21),rnorm(x),col=2, cex = 0.2)
AFAIK, base graphics doesn't have alpha transparency available, but the
ggplot2 package does. One approach is to adjust the alpha transparency
on
default size points; another is to combine reduced point size with
alpha
transparency. Here is your example rehashed for ggplot2.
require(ggplot2)
d - data.frame(x1 = rnorm(1, mean = 19), x2 = rnorm(1, mean =
20),
x3 = rnorm(1, mean = 21), x = rnorm(1))
# Basically stacking x1 - x3, creating two new vars named variable and
value
dm - melt(d, id = 'x') # from reshape package, loads with ggplot2
# Alpha transparency is set to a low level with default point size,
# but the colors in the legend are muted by the level of transparency
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.05) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green'))
# A tradeoff is to reduce the point size and increase alpha a bit, but
these
changes will
# also be reflected in the legend.
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.15, size = 1) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green'))
You may well find the legend to be useless for this example, so to get
rid
of it,
ggplot(dm, aes(x = x, y = value, colour = variable)) + theme_bw() +
geom_point(alpha = 0.15, size = 1) +
scale_colour_manual(values = c('x1' = 'black',
'x2' = 'red', 'x3' = 'green')) +
opts(legend.position = 'none')
The nice thing about the ggplot2 graph is that you can adjust the point
size
and alpha transparency to your tastes. The default point size is 2 and
the
default alpha = 1 (no transparency).
HTH,
Dennis
On Wed, Mar 30, 2011 at 10:04 PM, Samuel Dennis sjdenn...@gmail.com
wrote:
I have a very large dataset with three variables that I need to graph
using
a scatterplot. However I find that the first variable gets masked by
the
other two, so the graph looks entirely different depending on the
order of
variables. Does anyone have any suggestions how to manage this?
This code is an illustration of what I am dealing with:
x - 1
plot(rnorm(x,mean=20),rnorm(x),col=1,xlim=c(16,24))
points(rnorm(x,mean=21),rnorm(x),col=2)
points(rnorm(x,mean=19),rnorm(x),col=3)
gives an entirely different looking graph to:
x - 1
plot(rnorm(x,mean=19),rnorm(x),col=3,xlim=c(16,24))
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
despite being identical in all respects except for the order in which
the
variables are plotted.
I have tried using pch=., however the colours are very difficult to
discern. I have experimented with a number of other symbols with no
real
solution.
The only way that appears to work is to iterate the plot with a for
loop,
and progressively add a few numbers from each variable, as below.
However
although I can do this simply with random numbers as I have done
here, this
is an extremely cumbersome method to use with real datasets.
plot(1,1,xlim=c(16,24),ylim=c(-4,4),col=white)
x - 100
for (i in 1:100) {
points(rnorm(x,mean=19),rnorm(x),col=3)
points(rnorm(x,mean=20),rnorm(x),col=1)
points(rnorm(x,mean=21),rnorm(x),col=2)
}
Is there some function in R that could solve this through
automatically
iterating my data as above, using transparent symbols, or something
else?
Is
there some other way of solving this issue that I haven't thought of?
Thankyou,
Samuel Dennis
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Re: [R] Simple lattice question
On 2011-03-31 06:58, Rubén Roa wrote:
Thanks Peters!
Just a few minor glitches now:
require(lattice)
data- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0), points=TRUE,
lines=TRUE, pch=1:4, lty=1:4, type='b',
text=list(lab = as.character(unique(data$z)
I have an extra column of symbols on the legend,
and,
would like to add a title to the legend. Such as 'main' for plots.
Any suggestions?
for key(), make 'lines' into a list:
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),
pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0),
title=title here, cex.title=.9, lines.title=3,
lines=list(pch=1:4, lty=1:4, type='b'),
text=list(lab = as.character(unique(data$z)
Peter Ehlers
Rubén
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
-Mensaje original-
De: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Enviado el: jueves, 31 de marzo de 2011 15:41
Para: Rubén Roa
CC: r-help@r-project.org
Asunto: Re: [R] Simple lattice question
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data-
data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|SP,data=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='bl
ack',type='b')
How do I put a legend for the grouping variable in the
empty upper-right panel?
See the help page for xyplot for an example using the iris data.
You just need to add something like
auto.key = list(x = .6, y = .7, corner = c(0, 0))
to your lattice call. See the 'key' entry on the ?xyplot page.
Peter Ehlers
TIA
Rubén
__
__
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
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Re: [R] choosing best 'match' for given factor
Folks:
I think the following may be somewhat faster, as it avoids sorting:
bmat - function(mx,vec)
{
nm - colnames(mx)
ivec - match(vec,nm)
sapply(ivec,function(k){
if(k==1)NA else {
lookat - setdiff(seq_len(k-1),ivec) ## only those to left and not
in search vector ##
nm[lookat[which.max(mx[lookat,k] )]]
}
}
)
}
-- Bert
On Thu, Mar 31, 2011 at 8:30 AM, Nick Sabbe nick.sa...@ugent.be wrote:
Hi Murali.
I haven't compared, but this is what I would do:
bestMatch-function(searchVector, matchMat)
{
searchRow-unique(sort(match(searchVector, colnames(matchMat #if
you're sure, you could drop unique
cat(Original row indices:)
print(searchRow)
matchMat-matchMat[, -searchRow, drop=FALSE] #avoid duplicates
altogether
cat(Corrected Matrix:\n)
print(matchMat)
correctedRows-searchRow - seq_along(searchRow) + 1 #works because
of the sort above
cat(Corrected row indices:)
print(correctedRows)
sapply(correctedRows, function(cr){
lookWhere-matchMat[cr, seq(cr-1)]
cat(Will now look into:\n)
print(lookWhere)
cc-which.max(lookWhere)
cat(Max at position, cc, \n)
colnames(matchMat)[cc]
})
}
I don't think there's that much difference. Depending on specific sizes, it
may be more or less costly to first shrink the search matrix like I do. And
similarly depending, I may be better still if you remove the rows that
you're not interested in as well (some more but similar index trickery
required then.
HTH,
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of murali.me...@avivainvestors.com
Sent: donderdag 31 maart 2011 16:46
To: r-help@r-project.org
Subject: [R] choosing best 'match' for given factor
Folks,
I have a 'matching' matrix between variables A, X, L, O:
a - structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c(A, X, L, O), c(A, X, L, O)))
a
A X L O
A 1.00 0.41 0.58 0.75
X 0.41 1.00 0.60 0.86
L 0.58 0.75 1.00 0.83
O 0.60 0.86 0.83 1.00
And I have a search vector of variables
v - c(X, O)
I want to write a function bestMatch(searchvector, matchMat) such that for
each variable in searchvector, I get the variable that it has the highest
match to - but searching only among variables to the left of it in the
'matching' matrix, and not matching with any variable in searchvector
itself.
So in the above example, although X has the highest match (0.86) with O,
I can't choose O as it's to the right of X (and also because O is in the
searchvector v already); I'll have to choose A.
For O, I will choose L, the variable it's best matched with - as it
can't match X already in the search vector.
My function bestMatch(v, a) will then return c(A, L)
My matrix a is quite large, and I have a long list of search vectors v, so I
need an efficient method.
I wrote this:
bestMatch - function(searchvector, matchMat) {
sapply(searchvector, function(cc) {
y - matchMat[!(rownames(matchMat) %in%
searchvector) (index(rownames(matchMat)) match(cc, rownames(matchMat))),
cc, drop = FALSE];
rownames(y)[which.max(y)]
})
}
Any advice?
Thanks,
Murali
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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.
--
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.
-- Maimonides (1135-1204)
Bert Gunter
Genentech Nonclinical Biostatistics
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and provide commented, minimal, self-contained, reproducible code.
[R] Sequential multiple regression
Hello, In the past I have tended to reside more in the ANOVA camp but am trying to become more familiar with regression techniques in R. I would like to get the F change from a model as I take away factors: SO... mod1-lm(y~x1+x2+x3)...mod2-lm(y~x1,x2)...mod3-lm(y~x1) I can do this by hand by running several models in R and taking the MSr1/MSe1, MSr2/MSe2... This is slow and I know there's a better way. In SPSS (which I no longer use) I could easily obtain these results (F-change) as documented by Professor Andy Fields: http://www.statisticshell.com/multireg.pdf You can see the F changes for his two IV model yielding 2 F changes. Maybe it's the language I'm using (sequential multiple regression) that yields me poor results in searching the archives and Rseek. The results tend to be around hierarchal regression (I'm not familiar with this terminology being in the ANOVA camp). When I look at the hier.part package and run the examples it doesn't seem to give me the F change I'm looking for. The step function in the base program reduces the model but takes away the non sig. IV's (which is a great approach but I'm really after those F changes). As is the usually the case I'm sure R does this simply and beautifully, I'm just not experienced with the statistical vocabulary and techniques around regression to find what I'm looking for. F-change values with R: Any help would be appreciated. Thank you in advance, Tyler [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sequential multiple regression
?drop1 -- Bert On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker tyler_rin...@hotmail.com wrote: Hello, In the past I have tended to reside more in the ANOVA camp but am trying to become more familiar with regression techniques in R. I would like to get the F change from a model as I take away factors: SO... mod1-lm(y~x1+x2+x3)...mod2-lm(y~x1,x2)...mod3-lm(y~x1) I can do this by hand by running several models in R and taking the MSr1/MSe1, MSr2/MSe2... This is slow and I know there's a better way. In SPSS (which I no longer use) I could easily obtain these results (F-change) as documented by Professor Andy Fields: http://www.statisticshell.com/multireg.pdf You can see the F changes for his two IV model yielding 2 F changes. Maybe it's the language I'm using (sequential multiple regression) that yields me poor results in searching the archives and Rseek. The results tend to be around hierarchal regression (I'm not familiar with this terminology being in the ANOVA camp). When I look at the hier.part package and run the examples it doesn't seem to give me the F change I'm looking for. The step function in the base program reduces the model but takes away the non sig. IV's (which is a great approach but I'm really after those F changes). As is the usually the case I'm sure R does this simply and beautifully, I'm just not experienced with the statistical vocabulary and techniques around regression to find what I'm looking for. F-change values with R: Any help would be appreciated. Thank you in advance, Tyler [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org 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] statistical question
You could do bootstrapping for confidence intervals or permutation tests for just comparing the medians. There are other tools as well, just be sure that you understand what you are testing and what assumptions are being made in whatever method you choose. -- 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 Anna Lee Sent: Thursday, March 31, 2011 3:15 AM To: r-help@r-project.org Subject: [R] statistical question Dear List! I want to compare medians of non normal distributed data. Is it possible and usefull to calculate 95% confidence intervals for medians? And if so - how can this be achieved in R? Thanks a lot! Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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@r-project.org 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] choosing best 'match' for given factor
Try this:
bestMatch - function(search, match) {
colnames(match)[pmax(apply(match[,search], 2, which.max) - 1, 1)]
}
On Thu, Mar 31, 2011 at 11:46 AM, murali.me...@avivainvestors.com wrote:
Folks,
I have a 'matching' matrix between variables A, X, L, O:
a - structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c(A, X, L, O), c(A, X, L, O)))
a
A X L O
A 1.00 0.41 0.58 0.75
X 0.41 1.00 0.60 0.86
L 0.58 0.75 1.00 0.83
O 0.60 0.86 0.83 1.00
And I have a search vector of variables
v - c(X, O)
I want to write a function bestMatch(searchvector, matchMat) such that for
each variable in searchvector, I get the variable that it has the highest
match to - but searching only among variables to the left of it in the
'matching' matrix, and not matching with any variable in searchvector itself.
So in the above example, although X has the highest match (0.86) with O,
I can't choose O as it's to the right of X (and also because O is in the
searchvector v already); I'll have to choose A.
For O, I will choose L, the variable it's best matched with - as it can't
match X already in the search vector.
My function bestMatch(v, a) will then return c(A, L)
My matrix a is quite large, and I have a long list of search vectors v, so I
need an efficient method.
I wrote this:
bestMatch - function(searchvector, matchMat) {
sapply(searchvector, function(cc) {
y - matchMat[!(rownames(matchMat) %in%
searchvector) (index(rownames(matchMat)) match(cc, rownames(matchMat))),
cc, drop = FALSE];
rownames(y)[which.max(y)]
})
}
Any advice?
Thanks,
Murali
__
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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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] Sequential multiple regression
Bert and anyone else with info, First, Bert thank you for your quick reply. drop1 gives the results as a type II anova. Is there a way to make drop1 give you type I anova (the args don't appear to have a way to do so)? Another package/function perhaps? Tyler Date: Thu, 31 Mar 2011 09:32:02 -0700 Subject: Re: [R] Sequential multiple regression From: gunter.ber...@gene.com To: tyler_rin...@hotmail.com CC: r-help@r-project.org ?drop1 -- Bert On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker tyler_rin...@hotmail.com wrote: Hello, In the past I have tended to reside more in the ANOVA camp but am trying to become more familiar with regression techniques in R. I would like to get the F change from a model as I take away factors: SO... mod1-lm(y~x1+x2+x3)...mod2-lm(y~x1,x2)...mod3-lm(y~x1) I can do this by hand by running several models in R and taking the MSr1/MSe1, MSr2/MSe2... This is slow and I know there's a better way. In SPSS (which I no longer use) I could easily obtain these results (F-change) as documented by Professor Andy Fields: http://www.statisticshell.com/multireg.pdf You can see the F changes for his two IV model yielding 2 F changes. Maybe it's the language I'm using (sequential multiple regression) that yields me poor results in searching the archives and Rseek. The results tend to be around hierarchal regression (I'm not familiar with this terminology being in the ANOVA camp). When I look at the hier.part package and run the examples it doesn't seem to give me the F change I'm looking for. The step function in the base program reduces the model but takes away the non sig. IV's (which is a great approach but I'm really after those F changes). As is the usually the case I'm sure R does this simply and beautifully, I'm just not experienced with the statistical vocabulary and techniques around regression to find what I'm looking for. F-change values with R: Any help would be appreciated. Thank you in advance, Tyler [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nls.profile
On Thu, Mar 31, 2011 at 4:02 AM, Daniel Kaschek daniel.kasc...@physik.uni-freiburg.de wrote: Hello, I use nls.profile to compute confidence intervals of parameter estimates of a non-linear model. When computing the profiles, the model function produces an error for certain parameter combinations. Therefore nls fails and so does nls.profile. Is there a way to tell nls.profile to ignore errors and to proceed with the next parameter value? Not easily. What I tend to do is to adjust (increase) the alphamax parameter (which is a misnomer, it should have been alphamin but I got confused) until the profile can succeed. __ R-help@r-project.org 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] one question about bioconductor
dear lady and gentalmen: i am gaoshan from kansas university. i used such coding to deal with gel data data - ReadAffy() Warning messages: 1: In file(out, wt) : cannot open file 'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd3f0b2e85': No such file or directory 2: In file(out, wt) : cannot open file 'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd1d7427a' eset - rma(data) Error in gzfile(file, r) : cannot open the connection In addition: Warning message: In gzfile(file, r) : cannot open compressed file 'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\file5f6a6cb5', probable reason 'No such file or directory' can you help me to fix it? thank u very much gao shan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to update R?
Hi, My R version is old, and I would like to update it. How can I update R and keep all the libraries? Thanks Best, Yunfei Li -- Research Assistant Department of Statistics School of Molecular Biosciences Biotechnology Life Sciences Building 427 Washington State University Pullman, WA 99164-7520 Phone: 509-339-5096 http://www.wsu.edu/~ye_lab/people.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] one question about bioconductor
On Thu, Mar 31, 2011 at 09:32:06AM -0700, wang peter wrote: dear lady and gentalmen: i am gaoshan from kansas university. i used such coding to deal with gel data data - ReadAffy() Warning messages: 1: In file(out, wt) : cannot open file 'C:\Users\gaoshan\AppData\Local\Temp\RtmpvsyXOV\Rhttpd3f0b2e85': No such file or directory As the message says: there is something wrong with the path. In order to get more helpful replies, you should show the actual code you used and also give a hint about the spoecific packages you were using. E.g. ReadAffy most certainly requires at least a filename which seems to be missing from your comamnd above. In addition, I recommend to post your question on the bioconductor mailing list. cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan Maximus-von-Imhof-Forum 3 85354 Freising, Germany http://webclu.bio.wzw.tum.de/~pagel/ __ R-help@r-project.org 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] Cluster analysis, factor variables, large data set
Dear R helpers, I have a large data set with 36 variables and about 50.000 cases. The variabels represent labour market status during 36 months, there are 8 different variable values (e.g. Full-time Employment, Student,...) Only cases with at least one change in labour market status is included in the data set. To analyse sub sets of the data, I have used daisy in the cluster-package to create a distance matrix and then used pam (or pamk in the fpc-package), to get a k-medoids cluster-solution. Now I want to analyse the whole set. clara is said to cope with large data sets, but the first step in the cluster analysis, the creation of the distance matrix must be done by another function since clara only works with numeric data. Is there an alternative to the daisy - clara route that does not require as much RAM? What functions would you recommend for a cluster analysis of this kind of data on large data set? regards, Hans Ekbrand __ R-help@r-project.org 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] read password-protected files
Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to update R?
This question has been asked by many people already. The easiest way is: 1) install the new version 2) copy all or the libraries that you installed later from the library folder of older version to the new version 3) uninstall the old version 4) do a library update in the new version Done! ...Tao From: Li, Yunfei yunfei...@wsu.edu To: r-help@r-project.org Sent: Thu, March 31, 2011 10:16:23 AM Subject: [R] How to update R? Hi, My R version is old, and I would like to update it. How can I update R and keep all the libraries? Thanks Best, Yunfei Li -- Research Assistant Department of Statistics School of Molecular Biosciences Biotechnology Life Sciences Building 427 Washington State University Pullman, WA 99164-7520 Phone: 509-339-5096 http://www.wsu.edu/~ye_lab/people.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to update R?
Hi Yunfei, It really depends to some extent on the version of R (and your OS) you were using. Here is a link to a blog post that discusses some potential strategies: http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-the-r-code-to-do-it-on-windows/ There are also a number of messages if you search the R-help archive. However, old version of libraries may not be portable to the most recent R, so you may need to update R and update all your libraries too. Cheers, Josh On Thu, Mar 31, 2011 at 10:16 AM, Li, Yunfei yunfei...@wsu.edu wrote: Hi, My R version is old, and I would like to update it. How can I update R and keep all the libraries? Thanks Best, Yunfei Li -- Research Assistant Department of Statistics School of Molecular Biosciences Biotechnology Life Sciences Building 427 Washington State University Pullman, WA 99164-7520 Phone: 509-339-5096 http://www.wsu.edu/~ye_lab/people.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org 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] Cluster analysis, factor variables, large data set
Dear Hans, clara doesn't require a distance matrix as input (and therefore doesn't require you to run daisy), it will work with the raw data matrix using Euclidean distances implicitly. I can't tell you whether Euclidean distances are appropriate in this situation (this depends on the interpretation and variables and particularly on how they are scaled), but they may be fine at least after some transformation and standardisation of your variables. Hope this helps, Christian On Thu, 31 Mar 2011, Hans Ekbrand wrote: Dear R helpers, I have a large data set with 36 variables and about 50.000 cases. The variabels represent labour market status during 36 months, there are 8 different variable values (e.g. Full-time Employment, Student,...) Only cases with at least one change in labour market status is included in the data set. To analyse sub sets of the data, I have used daisy in the cluster-package to create a distance matrix and then used pam (or pamk in the fpc-package), to get a k-medoids cluster-solution. Now I want to analyse the whole set. clara is said to cope with large data sets, but the first step in the cluster analysis, the creation of the distance matrix must be done by another function since clara only works with numeric data. Is there an alternative to the daisy - clara route that does not require as much RAM? What functions would you recommend for a cluster analysis of this kind of data on large data set? regards, Hans Ekbrand __ R-help@r-project.org 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. *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche __ R-help@r-project.org 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 update R?
Forgot to mention that my way of updating is for Windows only. ...Tao From: Joshua Wiley jwiley.ps...@gmail.com To: Li, Yunfei yunfei...@wsu.edu Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:05:29 AM Subject: Re: [R] How to update R? Hi Yunfei, It really depends to some extent on the version of R (and your OS) you were using. Here is a link to a blog post that discusses some potential strategies: http://www.r-statistics.com/2010/04/changing-your-r-upgrading-strategy-and-the-r-code-to-do-it-on-windows/ There are also a number of messages if you search the R-help archive. However, old version of libraries may not be portable to the most recent R, so you may need to update R and update all your libraries too. Cheers, Josh On Thu, Mar 31, 2011 at 10:16 AM, Li, Yunfei yunfei...@wsu.edu wrote: Hi, My R version is old, and I would like to update it. How can I update R and keep all the libraries? Thanks Best, Yunfei Li -- - Research Assistant Department of Statistics School of Molecular Biosciences Biotechnology Life Sciences Building 427 Washington State University Pullman, WA 99164-7520 Phone: 509-339-5096 http://www.wsu.edu/~ye_lab/people.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org 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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read password-protected files
Hi Tao, You will need to give us more details. CSV files are pure text. How are they password protected? Are they Excel files? In a password protected compressed format? Encrypted? Best Regards, Josh On Thu, Mar 31, 2011 at 11:00 AM, Shi, Tao shida...@yahoo.com wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org 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] read password-protected files
On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org 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] read password-protected files
On Thu, Mar 31, 2011 at 11:00:48AM -0700, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is a way to read them in in R without manually removing the password protection for each file? I doubt that there is such a thing as a password protected csv file. They are just text files, after all. So I guess you have something else. How or what did the presumed pasword protection? cu Philipp -- Dr. Philipp Pagel Lehrstuhl für Genomorientierte Bioinformatik Technische Universität München Wissenschaftszentrum Weihenstephan Maximus-von-Imhof-Forum 3 85354 Freising, Germany http://webclu.bio.wzw.tum.de/~pagel/ __ R-help@r-project.org 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] read password-protected files
Hi Josh and Prof. Ripley, Thanks for the quick replies! Sorry about the HTML email. It's the default for my yahoo account and forgot to switch. The question was asked by a colleague of mine. After double-checking, yes, they are actually Excel files. Any idea on how to approach this? Thanks! ...Tao From: Prof Brian Ripley rip...@stats.ox.ac.uk Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:15:35 AM Subject: Re: [R] read password-protected files On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? [[elided Yahoo spam]] ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org 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] read password-protected files
Hi Tao, There are a number of R packages to work with reading/writing Excel files (e.g., xlsReadWrite). If Excel is on the system in question, as long as you have opened the file in Excel, you should be able to use RODBC with RDCOMClient or rcom, but I am not sure those will work if it is password protected unless it is currently unlocked in Excel. HTH, Josh On Thu, Mar 31, 2011 at 11:33 AM, Shi, Tao shida...@yahoo.com wrote: Hi Josh and Prof. Ripley, Thanks for the quick replies! Sorry about the HTML email. It's the default for my yahoo account and forgot to switch. The question was asked by a colleague of mine. After double-checking, yes, they are actually Excel files. Any idea on how to approach this? Thanks! ...Tao From: Prof Brian Ripley rip...@stats.ox.ac.uk To: Shi, Tao shida...@yahoo.com Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:15:35 AM Subject: Re: [R] read password-protected files On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 __ R-help@r-project.org 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] read password-protected files
Thanks, Josh. I use xlsReadWrite routinely. It would be nice if it has a password option. ...Tao - Original Message From: Joshua Wiley jwiley.ps...@gmail.com To: Shi, Tao shida...@yahoo.com Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:50:27 AM Subject: Re: [R] read password-protected files Hi Tao, There are a number of R packages to work with reading/writing Excel files (e.g., xlsReadWrite). If Excel is on the system in question, as long as you have opened the file in Excel, you should be able to use RODBC with RDCOMClient or rcom, but I am not sure those will work if it is password protected unless it is currently unlocked in Excel. HTH, Josh On Thu, Mar 31, 2011 at 11:33 AM, Shi, Tao shida...@yahoo.com wrote: Hi Josh and Prof. Ripley, Thanks for the quick replies! Sorry about the HTML email. It's the default for my yahoo account and forgot to switch. The question was asked by a colleague of mine. After double-checking, yes, they are actually Excel files. Any idea on how to approach this? Thanks! ...Tao From: Prof Brian Ripley rip...@stats.ox.ac.uk To: Shi, Tao shida...@yahoo.com Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:15:35 AM Subject: Re: [R] read password-protected files On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? Thank you very much! ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 __ R-help@r-project.org 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] read password-protected files
Using RDCOMCliente: library(RDCOMClient) eApp - COMCreate(Excel.Application) wk - eApp$Workbooks()$Open(Filename=your_file,Password=your_password) tf - tempfile() wk$Sheets(1)$SaveAs(tf, 3) DF - read.table(sprintf(%s.txt, tf), header = TRUE, sep = \t) On Thu, Mar 31, 2011 at 3:33 PM, Shi, Tao shida...@yahoo.com wrote: Hi Josh and Prof. Ripley, Thanks for the quick replies! Sorry about the HTML email. It's the default for my yahoo account and forgot to switch. The question was asked by a colleague of mine. After double-checking, yes, they are actually Excel files. Any idea on how to approach this? Thanks! ...Tao From: Prof Brian Ripley rip...@stats.ox.ac.uk Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:15:35 AM Subject: Re: [R] read password-protected files On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? [[elided Yahoo spam]] ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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] Cluster analysis, factor variables, large data set
On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote: Dear Hans, clara doesn't require a distance matrix as input (and therefore doesn't require you to run daisy), it will work with the raw data matrix using Euclidean distances implicitly. I can't tell you whether Euclidean distances are appropriate in this situation (this depends on the interpretation and variables and particularly on how they are scaled), but they may be fine at least after some transformation and standardisation of your variables. The variables are unordered factors, stored as integers 1:9, where 1 means Full-time employment 2 means Part-time employment 3 means Student 4 means Full-time self-employee ... Does euclidean distances make sense on unordered factors coded as integers? __ R-help@r-project.org 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] transparent grays?
Is there a grayscale equivalent to alpha levels in rgb? Example: I have the following to make red transparent circles overlap with previously plotted blue symbols. symbols(x=sites$long,y=sites$lat,circles=log(sites$prop.nem +1),add=T,inches=F,bg=rgb(red=1,green=0,blue=0, alpha=0.5),fg=rgb(red=1,green=0,blue=0, alpha=0.5)) I'm having a hard time coming up with a grayscale equivalent. Thanks! __ R-help@r-project.org 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] Cluster analysis, factor variables, large data set
On Thu, Mar 31, 2011 at 08:48:02PM +0200, Hans Ekbrand wrote: On Thu, Mar 31, 2011 at 07:06:31PM +0100, Christian Hennig wrote: Dear Hans, clara doesn't require a distance matrix as input (and therefore doesn't require you to run daisy), it will work with the raw data matrix using Euclidean distances implicitly. I can't tell you whether Euclidean distances are appropriate in this situation (this depends on the interpretation and variables and particularly on how they are scaled), but they may be fine at least after some transformation and standardisation of your variables. The variables are unordered factors, stored as integers 1:9, where 1 means Full-time employment 2 means Part-time employment 3 means Student 4 means Full-time self-employee ... Does euclidean distances make sense on unordered factors coded as integers? To be clear, here is an extract my.df.full[900:910, 16:19] PL210F.first.year PL210G.first.year PL210H.first.year PL210I.first.year 900 2 2 1 2 901 1 1 1 1 902 1 1 1 1 903 2 2 2 2 904 1 1 1 1 905 2 2 2 2 906 7 8 2 7 907 5 5 5 5 908 1 1 1 1 909 1 1 1 1 910 1 1 1 1 class(my.df.full[,16]) [1] integer __ R-help@r-project.org 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] Cluster analysis, factor variables, large data set
On Thu, Mar 31, 2011 at 11:48 AM, Hans Ekbrand h...@sociologi.cjb.net wrote:
The variables are unordered factors, stored as integers 1:9, where
1 means Full-time employment
2 means Part-time employment
3 means Student
4 means Full-time self-employee
...
Does euclidean distances make sense on unordered factors coded as
integers?
It probably doesn't. You said you have some 36 observations for each
case, correct? You can turn these 36 observations into a vector of
length 36 * 9 on which Euclidean distance will make some sense, namely
k changes will produce a distance of sqrt(2*k). For each observation
with value p (p between 1 and 9), create a vector r = c(0,0,1,0,...0)
where the entry 1 is in the p-th component. Hence, if values p1 and p2
are the same, euclidean distance between r1 and r2 is zero; if they
are not the same, Euclidan distance is sqrt(2).
Here's some possible R code:
transform = function(obsVector, maxVal)
{
templateMat = matrix(0, maxVal, maxVal);
diag(templateMat) = 1;
return(as.vector(templateMat[, obsVector]));
}
set.seed(10)
n = 4;
m = 5;
max = 4;
data = matrix(sample(c(1:max), n*m, replace = TRUE), m, n);
data
[,1] [,2] [,3] [,4]
[1,]3312
[2,]1332
[3,]3324
[4,]1242
[5,]4141
trafoData = apply(data, 2, transform, maxVal = max);
trafoData
[,1] [,2] [,3] [,4]
[1,]0010
[2,]0001
[3,]1100
[4,]0000
[5,]1000
[6,]0001
[7,]0110
[8,]0000
[9,]0000
[10,]0010
[11,]1100
[12,]0001
[13,]1000
[14,]0101
[15,]0000
[16,]0010
[17,]0101
[18,]0000
[19,]0000
[20,]1010
The code assumes that cases are in columns and observations in rows of
data. Examine data and trafoData to see how the transformation works.
Once you have the transformed data, simply apply your favorite
clustering method that uses Euclidean distance.
HTH,
Peter
__
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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.
__
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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] rank of Matrix
Dear list, Can anyone tell me how to obtain the rank of a sparse Matrix, for example from package Matrix (class dgCMatrix)? Here is an example of QR decomposition of a sparse matrix (from the sparseQR class help). library(Matrix) data(KNex) mm - KNex$mm str(mmQR - qr(mm)) Similarly, using the functions/classes from the relatively new MatrixModels package: library(MatrixModels) str(trial - data.frame(counts=c(18,17,15,20,10,20,25,13,12), outcome=gl(3,1,9,labels=LETTERS[1:3]), treatment=gl(3,3,labels=letters[1:3]))) glmS - glm4(counts ~ 0+outcome + treatment, poisson, trial, verbose = TRUE, sparse = TRUE) str(glmS) str(X - glmS@pred@X) str(QR - qr(X)) I'm not very familiar with matrix decomposition, but is the 'p' slot from a sparseQR object the same as $pivot from base qr? Additionally, how do I obtain the rank of the input matrix? qr from the base package produces an object with several components, two of which are rank, and pivot, is there an equivalent way of getting at these values/vectors for sparse matrices? Basically, I am trying to get at the variance-covariance matrix in order to compute t-values for the coefficients of a (sparse) GLM (i.e. a summary function for class glpModel). Am I going about this the right way? Are they perhaps more efficient ways of doing this when working with sparse matrices. Is there a preferred way of calculating t-statistics? Note that 'rankMatrix' from package Matrix appears to densify the matrix before use, so this is not a viable option. I have asked a similar question before [1], though I am now trying to 'answer' it myself by digging in and actually coding a bit more myself, however, as you can see I'm still a bit stuck :-( Thanks for any suggestions, Carson [1] http://www.mail-archive.com/r-help@r-project.org/msg130145.html -- Carson J. Q. Farmer ISSP Doctoral Fellow National Centre for Geocomputation National University of Ireland, Maynooth, http://www.carsonfarmer.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] transparent grays?
Try this: with(trees, symbols(Height, Volume, circles=Girth/24, inches=FALSE, fg = gray30, bg = sapply(apply(rbind(col2rgb(gray(seq(0,1, l = 10)))/255, alpha = 0.5), 2, as.list), do.call, what = rgb))) On Thu, Mar 31, 2011 at 2:52 PM, Frostygoat frostyg...@gmail.com wrote: Is there a grayscale equivalent to alpha levels in rgb? Example: I have the following to make red transparent circles overlap with previously plotted blue symbols. symbols(x=sites$long,y=sites$lat,circles=log(sites$prop.nem +1),add=T,inches=F,bg=rgb(red=1,green=0,blue=0, alpha=0.5),fg=rgb(red=1,green=0,blue=0, alpha=0.5)) I'm having a hard time coming up with a grayscale equivalent. Thanks! __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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] generate random numbers
Hey List, does anyone know how I can generate a vector of random numbers from a given distribution? Something like rnorm just for non normal distributions??? Thanks a lot! Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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] Sequential multiple regression
See http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf ?update.lm -- Bert On Thu, Mar 31, 2011 at 10:27 AM, Tyler Rinker tyler_rin...@hotmail.com wrote: Bert and anyone else with info, First, Bert thank you for your quick reply. drop1 gives the results as a type II anova. Is there a way to make drop1 give you type I anova (the args don't appear to have a way to do so)? Another package/function perhaps? Tyler Date: Thu, 31 Mar 2011 09:32:02 -0700 Subject: Re: [R] Sequential multiple regression From: gunter.ber...@gene.com To: tyler_rin...@hotmail.com CC: r-help@r-project.org ?drop1 -- Bert On Thu, Mar 31, 2011 at 9:24 AM, Tyler Rinker tyler_rin...@hotmail.com wrote: Hello, In the past I have tended to reside more in the ANOVA camp but am trying to become more familiar with regression techniques in R. I would like to get the F change from a model as I take away factors: SO... mod1-lm(y~x1+x2+x3)...mod2-lm(y~x1,x2)...mod3-lm(y~x1) I can do this by hand by running several models in R and taking the MSr1/MSe1, MSr2/MSe2... This is slow and I know there's a better way. In SPSS (which I no longer use) I could easily obtain these results (F-change) as documented by Professor Andy Fields: http://www.statisticshell.com/multireg.pdf You can see the F changes for his two IV model yielding 2 F changes. Maybe it's the language I'm using (sequential multiple regression) that yields me poor results in searching the archives and Rseek. The results tend to be around hierarchal regression (I'm not familiar with this terminology being in the ANOVA camp). When I look at the hier.part package and run the examples it doesn't seem to give me the F change I'm looking for. The step function in the base program reduces the model but takes away the non sig. IV's (which is a great approach but I'm really after those F changes). As is the usually the case I'm sure R does this simply and beautifully, I'm just not experienced with the statistical vocabulary and techniques around regression to find what I'm looking for. F-change values with R: Any help would be appreciated. Thank you in advance, Tyler [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics -- Men by nature long to get on to the ultimate truths, and will often be impatient with elementary studies or fight shy of them. If it were possible to reach the ultimate truths without the elementary studies usually prefixed to them, these would not be preparatory studies but superfluous diversions. -- Maimonides (1135-1204) Bert Gunter Genentech Nonclinical Biostatistics __ R-help@r-project.org 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] generate random numbers
There are several non-normal distributions built in that you could just use their generationg functions. There are packages (most with dist somewhere in the name) that allow for functions on standard distributions including combining together distributions. If your distribution is not one of the above, but has a well defined inverse cumulative distribution then you can just generate random uniforms (runif) and put them into the inverse. If none of the above works then there are still many tools available such as rejection sampling or metropolis-hastings sampling. The logspline package can approximate a distribution from a dataset and generate values from the approximation. We would need to know more about the actual distribution that you are interested in to be of more specific help. -- 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 Anna Lee Sent: Thursday, March 31, 2011 1:24 PM To: r-help@r-project.org Subject: [R] generate random numbers Hey List, does anyone know how I can generate a vector of random numbers from a given distribution? Something like rnorm just for non normal distributions??? Thanks a lot! Anna -- Der Inhalt dieser E-Mail ist vertraulich. Sollte Ihnen die E-Mail irrtümlich zugesandt worden sein, bitte ich Sie, mich unverzüglich zu benachrichtigen und die E-Mail zu löschen. This e-mail is confidential. If you have received it in error, please notify me immediately and delete it from your system. __ R-help@r-project.org 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@r-project.org 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] generate random numbers
On 31-Mar-11 19:23:33, Anna Lee wrote: Hey List, does anyone know how I can generate a vector of random numbers from a given distribution? Something like rnorm just for non normal distributions??? Thanks a lot! Anna SUppose we give your distribution the name Dist. The generic approach would start by defining a function for the inverse of its cumulative distribution. Call this qDist. Then qDist(runif(1000)) would generate 1000 values from the distribution Dist. As a ready-made example, qnorm is the inverse of pnorm, the cumulative distribution function of the Normal distribution. Then qnorm(runif(1000)) would act just like rnorm(1000), though the sequence of values would be different (a different algorithm) -- and also rnorm() would be more efficient (being specially written). Depending on what your desired distribution is, you may find that an rDist has already been written for it. There are many distributions already in R for which the family of functions dDist, pDist, qDist and rDist are provided. For more specific advice, please give us information about the specific distribution you want to sample from! Ted. E-Mail: (Ted Harding) ted.hard...@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 31-Mar-11 Time: 20:50:52 -- XFMail -- __ R-help@r-project.org 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] fit.mult.impute() in Hmisc
I tried multiple imputation with aregImpute() and
fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1.
The warning message below suggests that summary(f) of
fit.mult.impute() would only use the last imputed data set.
Thus, the whole imputation process is ignored.
Not using a Design fitting function; summary(fit)
will use standard errors, t, P from last imputation only.
Use vcov(fit) to get the correct covariance matrix,
sqrt(diag(vcov(fit))) to get s.e.
But the standard errors in summary(f) agree with the values
from sqrt(diag(vcov(f))) to the 4th decimal point. It would
seem that summary(f) actually adjusts for multiple
imputation?
Does summary(f) in Hmisc 3.8-3 actually adjust for MI?
If it does not adjust for MI, then how do I get the
MI-adjusted coefficients and standard errors?
I can't seem to find answers in the documentations, including
rereading section 8.10 of the Harrell (2001) book Googling
located a thread in R-help back in 2003, which seemed dated.
Many thanks in advance for the help,
Yuelin.
http://idecide.mskcc.org
---
library(Hmisc)
Loading required package: survival
Loading required package: splines
data(kyphosis, package = rpart)
kp - lapply(kyphosis, function(x)
+ { is.na(x) - sample(1:length(x), size = 10); x })
kp - data.frame(kp)
kp$kyp - kp$Kyphosis == present
set.seed(7)
imp - aregImpute( ~ kyp + Age + Start + Number, dat = kp, n.impute = 10,
+ type = pmm, match = closest)
Iteration 13
f - fit.mult.impute(kyp ~ Age + Start + Number, fitter=glm, xtrans=imp,
+ family = binomial, data = kp)
Variance Inflation Factors Due to Imputation:
(Intercept) Age Start Number
1.061.281.171.12
Rate of Missing Information:
(Intercept) Age Start Number
0.060.220.140.10
d.f. for t-distribution for Tests of Single Coefficients:
(Intercept) Age Start Number
2533.47 193.45 435.79 830.08
The following fit components were averaged over the 10 model fits:
fitted.values linear.predictors
Warning message:
In fit.mult.impute(kyp ~ Age + Start + Number, fitter = glm, xtrans = imp, :
Not using a Design fitting function; summary(fit) will use
standard errors, t, P from last imputation only. Use vcov(fit) to get the
correct covariance matrix, sqrt(diag(vcov(fit))) to get s.e.
f
Call: fitter(formula = formula, family = binomial, data = completed.data)
Coefficients:
(Intercept) AgeStart Number
-3.6971 0.0118 -0.1979 0.6937
Degrees of Freedom: 80 Total (i.e. Null); 77 Residual
Null Deviance: 80.5
Residual Deviance: 58 AIC: 66
sqrt(diag(vcov(f)))
(Intercept) Age Start Number
1.5444782 0.0063984 0.0652068 0.2454408
-0.1979/0.0652068
[1] -3.0350
summary(f)
Call:
fitter(formula = formula, family = binomial, data = completed.data)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.240 -0.618 -0.288 -0.109 2.409
Coefficients:
Estimate Std. Error z value Pr(|z|)
(Intercept) -3.6971 1.5445 -2.39 0.0167
Age 0.0118 0.00641.85 0.0649
Start-0.1979 0.0652 -3.03 0.0024
Number0.6937 0.24542.83 0.0047
(Dispersion parameter for binomial family taken to be 1)
Null deviance: 80.508 on 80 degrees of freedom
Residual deviance: 57.965 on 77 degrees of freedom
AIC: 65.97
Number of Fisher Scoring iterations: 5
=
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Re: [R] Simple lattice question
On Mar 31, 2011, at 8:41 AM, Peter Ehlers wrote:
On 2011-03-31 03:39, Rubén Roa wrote:
DeaR ComRades,
require(lattice)
data- data.frame(SP=sort(rep(as.factor(c('A','B','C','D','E')),12)),
x=rpois(60,10),
y=rep(c(rep(0,4),rep(10,4),rep(20,4)),5),
z=rep(1:4,15))
xyplot(x~y|
SP
,data
=data,groups=z,layout=c(2,3),pch=1:4,lty=1:4,col='black',type='b')
How do I put a legend for the grouping variable in the empty upper-
right panel?
See the help page for xyplot for an example using the iris data.
You just need to add something like
auto.key = list(x = .6, y = .7, corner = c(0, 0))
to your lattice call. See the 'key' entry on the ?xyplot page.
On my machine the default color scheme is TRUE, so the auto.key comes
out with colored circles. This incantation gets the key to match the
plotting symbols and lines:
xyplot(x~y|SP, data=data,groups=z, layout=c(2,3),
par.settings=simpleTheme(pch=1:4,lty=1:4,col='black'), type=b,
auto.key = list(x = .6, y = .7, lines=TRUE, corner = c(0, 0)) )
Peter Ehlers
TIA
Rubén
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] Effects - plot the marginal effect
Hello, I try to plot the marginal effect by using package effects (example of the graph i want to get is in the attached picture). All variables are continuous. Here is regression function, results and error effect function gives: mreg01 = lm(a90$enep1 ~ a90$enpres + a90$proximity1 + (a90$enpres * a90$proximity1), data=a90) summary(mreg01) Call: lm(formula = a90$enep1 ~ a90$enpres + a90$proximity1 + (a90$enpres * a90$proximity1), data = a90) Residuals: Min 1Q Median 3Q Max -2.3173 -1.3349 -0.5713 0.8938 8.1084 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 4.2273 0.3090 13.683 2e-16 *** a90$enpres 0.4225 0.2319 1.822 0.072250 . a90$proximity1 -3.8797 1.0984 -3.532 0.000696 *** a90$enpres:a90$proximity1 0.8953 0.4101 2.183 0.032025 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 2.029 on 78 degrees of freedom Multiple R-squared: 0.2128, Adjusted R-squared: 0.1826 F-statistic: 7.031 on 3 and 78 DF, p-value: 0.0003029 plot(effect(a90$enpres:a90$proximity1, mreg01))Warning messages:1: In a90$enpres:a90$proximity1 : numerical expression has 82 elements: only the first used2: In a90$enpres:a90$proximity1 : numerical expression has 82 elements: only the first used3: In analyze.model(term, mod, xlevels, default.levels) : 0 does not appear in the modelError in plot(effect(a90$enpres:a90$proximity1, mreg01)) : error in evaluating the argument 'x' in selecting a method for function 'plot' Thanks in advance. Tomas attachment: graph.png__ R-help@r-project.org 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] nlme for exact binomial model
Hi I am conducting a meta-analysis on proportions and I'm interested in using the exact binomial likeliood for the within-study effects rather than the approximate approaches and I don't want to use the quasi-likelihood approach in R's glmmPQL. Does anyone know how I can do this using nlme? I can do it in SAS nlmixed but I would rather use R -this is my SAS code: proc nlmixed data=work.temp ; parms mup=2 vp= 08; /*initial values*/ rawprop=1/(1+exp(-truep)); /*unkown true p logit proportion*/ model numberofisolates~binomial(sumisolatestudy, rawprop) ; random truep~normal(mup,vp) subject=studyid; run; Thanks B. Aletta Nonyane, PhD Assistant Scientist Department of International Health Johns Hopkins Bloomberg School of Public Health 615 N. Wolfe Street Baltimore, MD, 21205 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] VECM with UNRESTRICTED TREND
Dear Pfaff,
Would that be possible to fit a Time varying VECM using urca?
Yours,
Renoir
On Thursday, March 31, 2011, Grzegorz Konat grzegorz.ko...@ibrkk.pl wrote:
The code you gave me works fine with Finland, but the same for my data -
does not!
I do:
library(urca)
data(my.data)
dat1 - my.data[, c(dY, X, dM)]
trend - matrix(1:nrow(dat1), ncol = 1)
colnames(trend) - trd
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar = trend)
and the result is again:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop =
FALSE] :
non-numeric argument to binary operator
I attach my dataset in xls format. If you have 5 minutes and wish to check
it out, I'd be extremely grateful!
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Well, without further information, I do not know, but try the following
library(urca)
example(ca.jo)
trend - matrix(1:nrow(sjf), ncol = 1)
colnames(trend) - trd
ca.jo(sjf, type = trace, ecdet = const, K = 2, spec = longrun,
dumvar = trend)
Best,
Bernhard
--
*Von:* Grzegorz Konat [mailto:grzegorz.ko...@ibrkk.pl]
*Gesendet:* Donnerstag, 31. März 2011 14:40
*An:* Pfaff, Bernhard Dr.; r-help@r-project.org
*Betreff:* Re: [R] VECM with UNRESTRICTED TREND
'time' was a trend variable from my.data set. Equivalent to the output of
the command 'matrix' you just gave me.
So now I did:
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
mat1 - matrix(seq(1:nrow(dat1)), ncol = 1)
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = const, K = 2, spec =
longrun, dumvar=mat1)
and the output is:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), , drop
= FALSE] :
non-numeric argument to binary operator
In addition: Warning message:
In ca.jo(dat1, type = trace, ecdet = const, K = 2, spec = longrun,
:
No column names in 'dumvar', using prefix 'exo' instead.
What do I do wrong?
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Bernhard,
thank You so much one again! Now I (more or less) understand the idea, but
still have problem with its practical application.
I do (somewhat following example 8.1 in your textbook):
library(urca)
data(my.data)
names(my.data)
attach(my.data)
dat1 - my.data[, c(dY, X, dM)]
dat2 - cbind(time)
What is 'time'? Just employ matrix(seq(1:nrow(dat1)), ncol = 1) for
creating the trend variable.
Best,
Bernhard
args('ca.jo')
yxm.vecm - ca.jo(dat1, type = trace, ecdet = trend, K = 2, spec =
longrun, dumvar=dat2)
The above code produces following output:
Error in r[i1, , drop = FALSE] - r[-nrow(r):-(nrow(r) - lag + 1L), ,
drop = FALSE] :
non-numeric argument to binary operator
What does that mean? Should I use cbind command to dat1 as well? And
doesn't it transform the series into series of integer numbers?
Thank you once again (especially for your patience).
Best,
Greg
2011/3/31 Pfaff, Bernhard Dr. bernhard_pf...@fra.invesco.com
Hello Greg,
you include your trend as a (Nx1) matrix and use this for 'dumvar'. The
matrix 'dumvar' is just added to the VECM as deterministic regressors and
while you are referring to case 5, this is basically what you are after, if
I am not mistaken. But we aware that this implies a quadratic trend for the
levels
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[R] Create Variable names dynamically
Hi,
I want to create variable names from within my code, but can't find any
documentation for this.
An example is probably the best way to illustrate. I am reading data in from a
file, doing a bunch of stuff, and want to generate variables with my output.
(I could make a list of lists and name all the elements, but I really want
separate variables.)
#
#This is just a dummy example, please excuse any shortcuts...
data - read.table(file, )
animals - (data[,animal])
animals
cat, dog, horse # Not known what these are before I read the data file
# do a bunch of stuff
mean_cat - abc
var_cat - dfd
mean_dog - 123
var_dog - 453
etc..
##
I thought of trying to use the paste() function to create the variable name,
but that doesn't work:
for( animal in animals){
paste(mean, animal _) - 123
}
Any ideas???
Thanks
--
Noah Silverman
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Re: [R] Create Variable names dynamically
Take a look in ?assign
On Thu, Mar 31, 2011 at 5:42 PM, Noah Silverman n...@smartmediacorp.com wrote:
Hi,
I want to create variable names from within my code, but can't find any
documentation for this.
An example is probably the best way to illustrate. I am reading data in from
a file, doing a bunch of stuff, and want to generate variables with my
output. (I could make a list of lists and name all the elements, but I
really want separate variables.)
#
#This is just a dummy example, please excuse any shortcuts...
data - read.table(file, )
animals - (data[,animal])
animals
cat, dog, horse # Not known what these are before I read the data file
# do a bunch of stuff
mean_cat - abc
var_cat - dfd
mean_dog - 123
var_dog - 453
etc..
##
I thought of trying to use the paste() function to create the variable name,
but that doesn't work:
for( animal in animals){
paste(mean, animal _) - 123
}
Any ideas???
Thanks
--
Noah Silverman
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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Re: [R] read password-protected files
Thanks, Henrique! I'll try that. ...Tao - Original Message From: Henrique Dallazuanna www...@gmail.com To: Shi, Tao shida...@yahoo.com Cc: Prof Brian Ripley rip...@stats.ox.ac.uk; r-help@r-project.org Sent: Thu, March 31, 2011 12:00:45 PM Subject: Re: [R] read password-protected files Using RDCOMCliente: library(RDCOMClient) eApp - COMCreate(Excel.Application) wk - eApp$Workbooks()$Open(Filename=your_file,Password=your_password) tf - tempfile() wk$Sheets(1)$SaveAs(tf, 3) DF - read.table(sprintf(%s.txt, tf), header = TRUE, sep = \t) On Thu, Mar 31, 2011 at 3:33 PM, Shi, Tao shida...@yahoo.com wrote: Hi Josh and Prof. Ripley, Thanks for the quick replies! Sorry about the HTML email. It's the default for my yahoo account and forgot to switch. The question was asked by a colleague of mine. After double-checking, yes, they are actually Excel files. Any idea on how to approach this? Thanks! ...Tao From: Prof Brian Ripley rip...@stats.ox.ac.uk Cc: r-help@r-project.org Sent: Thu, March 31, 2011 11:15:35 AM Subject: Re: [R] read password-protected files On Thu, 31 Mar 2011, Shi, Tao wrote: Hi list, I have a bunch of .csv files that are password-protected. I wonder if there is There is nothing in the CSV standard about password protection. I very much doubt that these actually CSV files. So please follow the posting guide [*] a way to read them in in R without manually removing the password protection for each file? [[elided Yahoo spam]] ...Tao [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [*] as above: and as you sent HTML you are either full of contempt for the helpeRs or failed to do your homework. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 __ R-help@r-project.org 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org 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] Create Variable names dynamically
Perfect.
Thank You
On Mar 31, 2011, at 1:47 PM, Mark Leeds wrote:
hi noah: assign(thing you paste, 123, envir=whatever) should work I think.
On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman n...@smartmediacorp.com
wrote:
Hi,
I want to create variable names from within my code, but can't find any
documentation for this.
An example is probably the best way to illustrate. I am reading data in from
a file, doing a bunch of stuff, and want to generate variables with my
output. (I could make a list of lists and name all the elements, but I
really want separate variables.)
#
#This is just a dummy example, please excuse any shortcuts...
data - read.table(file, )
animals - (data[,animal])
animals
cat, dog, horse # Not known what these are before I read the data
file
# do a bunch of stuff
mean_cat - abc
var_cat - dfd
mean_dog - 123
var_dog - 453
etc..
##
I thought of trying to use the paste() function to create the variable name,
but that doesn't work:
for( animal in animals){
paste(mean, animal _) - 123
}
Any ideas???
Thanks
--
Noah Silverman
__
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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.
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and provide commented, minimal, self-contained, reproducible code.
[R] Is it possible to run unit tests after a package installation?
Dear all, I'm deep into Chambers' SoDA and R-exts.html but I can't find all answers. The thing is that I would like to run my unit tests right after a package installation. That is while the command R CMD check package_name runs all files named package_name/tests/*.R (so unit tests can be placed there) I wonder if it is possible to run them on some destination/target machine after the package has been installed. My goal is to enforce reliable results from calls to the package functions independently on the machine the package is being installed. The only way I can come up with (and is not even close to a package post-install check) is to place them in the directory where examples can be show using the example() command. But package_name/tests/Examples does not look as the place it should be (or at least it does not work for me for the time being). One of the difficulties I foresee also is how to state that my unit test framework of choice (be RUnit or testthat depending on my mood) is a dependency but not something that must be loaded before my package does. That is, it must be installed in the system but it should not be loaded. Any hints or comments on the above will be highly appreciated. Thanks in advance. jcb! __ R-help@r-project.org 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] generate random numbers
On 01/04/11 08:50, Ted Harding wrote: On 31-Mar-11 19:23:33, Anna Lee wrote: Hey List, does anyone know how I can generate a vector of random numbers from a given distribution? Something like rnorm just for non normal distributions??? Thanks a lot! Anna SUppose we give your distribution the name Dist. The generic approach would start by defining a function for the inverse of its cumulative distribution. Call this qDist. Then qDist(runif(1000)) would generate 1000 values from the distribution Dist. As a ready-made example, qnorm is the inverse of pnorm, the cumulative distribution function of the Normal distribution. Then qnorm(runif(1000)) would act just like rnorm(1000), though the sequence of values would be different (a different algorithm) -- and also rnorm() would be more efficient (being specially written). Depending on what your desired distribution is, you may find that an rDist has already been written for it. There are many distributions already in R for which the family of functions dDist, pDist, qDist and rDist are provided. For more specific advice, please give us information about the specific distribution you want to sample from! Ted. I can point to one general implementation which might be helpful, and even the function names are the same. In the version of DistributionUtils on R-Forge you will find functions pDist and qDist which should give the distribution function and quantile function of any continuous unimodal distribution. Provisos: there may be problems with distributions with very heavy tails, and generally the routines could be slow. David Scott -- _ David Scott Department of Statistics The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 923 5055, or +64 9 373 7599 ext 85055 Email: d.sc...@auckland.ac.nz, Fax: +64 9 373 7018 Director of Consulting, Department of Statistics __ R-help@r-project.org 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 Model with curve fitting parameter?
I have a model Q=K*A*(R^r)*(S^s) A, R, and S are data I have and K is a curve fitting parameter. I have linearized as log(Q)=log(K)+log(A)+r*log(R)+s*log(S) I have taken the log of the data that I have and this is the model formula without the K part lm(Q~offset(A)+R+S, data=x) What is the formula that I should use? Thanks for all of your help. I can provide a subset of data if necessary. -- Stephen Sefick | Auburn University | | Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849 | |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org 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] Create Variable names dynamically
The best thing to do is to understand how to use 'list' for this
purpose. Much easier to handle the information.
On Thu, Mar 31, 2011 at 4:42 PM, Noah Silverman n...@smartmediacorp.com wrote:
Hi,
I want to create variable names from within my code, but can't find any
documentation for this.
An example is probably the best way to illustrate. I am reading data in from
a file, doing a bunch of stuff, and want to generate variables with my
output. (I could make a list of lists and name all the elements, but I
really want separate variables.)
#
#This is just a dummy example, please excuse any shortcuts...
data - read.table(file, )
animals - (data[,animal])
animals
cat, dog, horse # Not known what these are before I read the data file
# do a bunch of stuff
mean_cat - abc
var_cat - dfd
mean_dog - 123
var_dog - 453
etc..
##
I thought of trying to use the paste() function to create the variable name,
but that doesn't work:
for( animal in animals){
paste(mean, animal _) - 123
}
Any ideas???
Thanks
--
Noah Silverman
__
R-help@r-project.org 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.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
__
R-help@r-project.org 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] regular expression
Hi, I am stuck on this: how to specify a match pattern that means not to include abc? I tried: grep(^(abc), hello, value=T) should return hello. while grep(^(abc), hello abcd foo, value=T) should return character(0). But both returned character(0). Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
