[R] new line between '}' and 'else' in function body
Hi list members,
?else tells us
In particular, you should not have a newline between '}' and
'else' to avoid a syntax error in entering a 'if ... else'
construct at the keyboard or via 'source'.
but there's no syntax error when you break the line between } and
else in a function, e.g.
f = function(x) {
if (x) {
1
} # a new line here!
else {
2
}
}
f(TRUE)
[1] 1
f(FALSE)
[1] 2
Seems strange...
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
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[R] extract tables as data.frames from HTML source
Hi,
I wonder whether there is any convenient function (or package) to
extract tables from a HTML page? e.g. from
http://www.google.com/finance/historical?q=SHE:002251
I know we can readLines('URL'), gsub('td...', '...', source), ...
and at last get the numbers; I'm writing to ask whether someone has
already contributed a more general function (with the package XML or
other packages). Thanks!
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
__
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.
Re: [R] new line between '}' and 'else' in function body
Hi,
That's because the parser knows how to deal with that stuff. However,
when you type the same if/else at the command line, it will be parsed
line by line, and the evaluator will not wait for the else to evaluate
the if. Try to copy and paste your if/else to the command line.
Romain
Yihui Xie wrote:
Hi list members,
?else tells us
In particular, you should not have a newline between '}' and
'else' to avoid a syntax error in entering a 'if ... else'
construct at the keyboard or via 'source'.
but there's no syntax error when you break the line between } and
else in a function, e.g.
f = function(x) {
if (x) {
1
} # a new line here!
else {
2
}
}
f(TRUE)
[1] 1
f(FALSE)
[1] 2
Seems strange...
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
__
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[R] Survey Package with Binary Data (no Standard Errors reported)
Hi, I'm trying to get standard errors for some of the variables in my data frame. One of the questions on my survey is whether faculty coordinate across curriculum to include Arts Education as subject matter. All the responses are coded in zeros and ones obviously. For some of the other variables I have a 2 for those that responded with Don't Know. I'm getting NA for mean and standard deviations from svymean. Am I doing something wrong of can the survey package not handle this type of data? Here's my code. survey - svydesign(id=~1, data=General, strata=~Grade.Level) Warning message: In svydesign.default(id = ~1, data = General, strata = ~Grade.Level) : No weights or probabilities supplied, assuming equal probability summary(survey) Stratified Independent Sampling design (with replacement) svydesign(id = ~1, data = General, strata = ~Grade.Level) Probabilities: Min. 1st Qu. MedianMean 3rd Qu.Max. 1 1 1 1 1 1 Stratum Sizes: Elementary High Middle obs 312 236156 design.PSU312 236156 actual.PSU312 236156 Data variables: [1] Grade.Level Curriculum [3] Field.Trips Residencies [5] PTA.Support Community.Open.Performances [7] Visual.Arts.Attendance Literary.Arts.Attendance [9] Arts.Organization.Membership Arts.Essential svymean(~Curriculum, survey) mean SE Curriculum NA NA ??? PJ __ 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] Intraday financial returns
Hello, I would like to create a function that computes intraday returns of a financial asset on a calendar time basis, without making any loop. For instance, I want to get price returns every 60 seconds. The main problem is that the times series of prices is irregularly spaced in time. I have looked in the zoo or its classes but have not found any answer to my problem. Thanks in advance. Nicolas Huth [[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] Add missing values/timestamps
Thanks a lot! The way with zoo worked perfect. Here is the code I've used finally: data.input01 -read.csv(./1_15min.txt, header = TRUE, sep = ;, quote=\, dec=,, fill = TRUE, comment.char=) data.input02 -read.csv(./2_15min.txt, header = TRUE, sep = ;, quote=\, dec=,, fill = TRUE, comment.char=) data.input03 -read.csv(./3_15min.txt, header = TRUE, sep = ;, quote=\, dec=,, fill = TRUE, comment.char=) data.troughput01 - rbind(data.input01,data.input02,data.input03) data.test - seq(as.POSIXct(2006-01-01 00:00:00),as.POSIXct(2006-12-31 23:45:00),900) data.troughput02 - as.zoo(data.troughput01$V2) index(data.troughput02) - as.POSIXct(data.troughput01$V1) data.output01 - merge(data.troughput04,zoo(,data.test)) data.output02 - na.approx(data.output01,na.rm = FALSE) Gabor Grothendieck wrote: Try this where we read in a zoo series and then merge it with a zero width regularly spaced series to create the result. Lines - V1,V2 2008-10-14 08:45:00,94411.08 2008-10-14 08:50:00,90745.45 2008-10-14 08:55:00,82963.35 2008-10-14 09:00:00,75684.38 2008-10-14 09:05:00,78931.82 2008-10-14 09:20:00,74580.11 2008-10-14 09:25:00,69666.48 2008-10-14 09:30:00,77794.89 library(zoo) z - read.zoo(textConnection(Lines), header = TRUE, sep = ,, tz = ) tt - seq(time(z)[1], time(z)[length(z)], 5*60) merge(z, zoo(, tt)) The last statement's output will be: merge(z, zoo(, tt)) 2008-10-14 08:45:00 2008-10-14 08:50:00 2008-10-14 08:55:00 2008-10-14 09:00:00 94411.0890745.4582963.35 75684.38 2008-10-14 09:05:00 2008-10-14 09:10:00 2008-10-14 09:15:00 2008-10-14 09:20:00 78931.82 NA NA 74580.11 2008-10-14 09:25:00 2008-10-14 09:30:00 69666.4877794.89 See the 3 zoo vignettes for use of zoo series and R News 4/1 (for date/time info). On Mon, Mar 30, 2009 at 10:38 AM, j.k kat...@gmx.at wrote: Hello alltogheter, I have the following problem and maybe someone can help me with it. I have a list of values with times. They look like that: V1 V2 1 2008-10-14 08:45:00 94411.08 2 2008-10-14 08:50:00 90745.45 3 2008-10-14 08:55:00 82963.35 4 2008-10-14 09:00:00 75684.38 5 2008-10-14 09:05:00 78931.82 6 2008-10-14 09:20:00 74580.11 7 2008-10-14 09:25:00 69666.48 8 2008-10-14 09:30:00 77794.89 I have these data combined from different series of measurements. As you can see the problem is that between these series are gaps which I want to fill. The format of the time is POSIXct Are there any suggestions how I can fill these missing times and afterwards interpolate/predict their values? Thanks in advance Johannes -- View this message in context: http://www.nabble.com/Add-missing-values-timestamps-tp22784737p22784737.html Sent from the R help mailing list archive at Nabble.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. __ 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. -- View this message in context: http://www.nabble.com/Add-missing-values-timestamps-tp22784737p22863325.html Sent from the R help mailing list archive at Nabble.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] extract tables as data.frames from HTML source
Yihui Xie wrote: I wonder whether there is any convenient function (or package) to extract tables from a HTML page? e.g. from http://www.google.com/finance/historical?q=SHE:002251 Try a search on R (I prefer markmail search) http://r-project.markmail.org/search/?q=extract%20html Dieter -- View this message in context: http://www.nabble.com/extract-tables-as-data.frames-from-HTML-source-tp22862641p22863816.html Sent from the R help mailing list archive at Nabble.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] how to sort and plot data?
Hem wrote: user_id website time 20google0930 21yahoo0935 20facebook1000 25facebook1015 61google0940 ... My problem is how to sort the data? So that, I can get information about one user_id viewed how many website perday? Maybe you were looking at the wrong item, because what you want is not sorting, but a table. Check the documentation of table or ftable. Dieter -- View this message in context: http://www.nabble.com/how-to-sort-and-plot-data--tp22861661p22863918.html Sent from the R help mailing list archive at Nabble.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] finding best fitting model
Benedikt Niesterok wrote: Is there a function in r to find the best fitting model for a set of data? I would like to know if my data are related exponentially,linearly or if there is a logarithmic correlation between my x and y values. There is no one-stop method to do this. I would suggest that you try the alternatives by doing the transformation (implicit in lm or explicit on data) and checking the residuals of the fit. Check documentation of plot.lm for an example or get the chapter in MASS on the subject. Dieter -- View this message in context: http://www.nabble.com/finding-best-fitting-model-tp22852224p22864012.html Sent from the R help mailing list archive at Nabble.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.
[R] chaining consensus in cluster analyses
Dear R-helpers, conducting different community ecology analyses my main aim is to find groupings in the data and geographical borders between communities and to prove them statistically. So after conducting a global test (Mantel) I am running a NMS and cluster analyses. These are followed by other tests. However, I have two questions concerning the cluster analyses which I was not able to solve so far: 1) I would like to calculate % chaining for specific clusters. How can I do that in R? 2) I would like to calculate several clusters with different methods and then merge them in a consensus cluster, similar to a consensus tree used in genetic analyses. Is this possible and if yes how? Any help or hint is greatly appreciated! with kind regards Johannes -- Project Coordinator BIOTA West Amphibians Museum of Natural History Dep. of Research (Herpetology) Invalidenstrasse 43 D-10115 Berlin Tel: +49 (0)30 2093 8708 Fax: +49 (0)30 2093 8565 http://www.biota-africa.org http://community-ecology.biozentrum.uni-wuerzburg.de __ 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 sort and plot data?
Hi,
There is definitely a more elegant way of doing this which I don't know
about (without a for loop), but try this:
mat - matrix(NA, nrow = max(user_id), ncol = 2)
mat[,1] - 1:max(user_id) # 1st column of matrix is the user ID
for (i in 1:max(user_id)){
temp1 - subset(data, user_id = i)
temp2 - unique(temp1$website)
mat[2,i] - length(temp2)
}
The matrix will give you user id and number of sites visited, provided user
id ranges from 1 to the number of users. There must be a way to do this
using table, but I cant figure it out.
Cheers,
Umesh
On Fri, Apr 3, 2009 at 1:42 PM, Dieter Menne
dieter.me...@menne-biomed.dewrote:
Hem wrote:
user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google0940
...
My problem is how to sort the data? So that, I can get information about
one
user_id viewed how many website perday?
Maybe you were looking at the wrong item, because what you want is not
sorting, but a table.
Check the documentation of table or ftable.
Dieter
--
View this message in context:
http://www.nabble.com/how-to-sort-and-plot-data--tp22861661p22863918.html
Sent from the R help mailing list archive at Nabble.com.
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[[alternative HTML version deleted]]
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[R] DierckxSpline fitting with different sets of y-values in one time
Dear R users, I have a question about the Package DierckxSpline. I have tried to find the answer by myself but it didn't worked out. I wondered if Dierckxspline can use different sets of y values in one time to fit a line with knot. I have different sets of Y values representing the same thing for different voxels (in an fmri image). I have already fitted the data in different graphs and I know now that the plots are comparable for the different data sets (so for the different voxels) but I wanted to include all the information in one plot, If it's possible. So I want to know the best fitting line for the whole cluster of voxels except for one voxel. If there is no possibility to do this I tought it would be an option to take the mean of the different y-values and use those values but I don't know if this is mathematical right to do. I hope someone can help me with this. Thank you very much, Elisabeth Jonckers GIfMI Ghent University Hospital Belgium [[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] [R} seasonal differencing
Hi, -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Joseph Magagnoli Sent: Thursday, April 02, 2009 5:36 PM To: r-help@r-project.org Subject: [R] [R} seasonal differencing Hi all, I was wondering how to construct a seasonal differenced time series variable. I used the following code to construct a 12 span seasonal difference seasonal-diff(V2, lag=12, differences=1) is this correct? thank you in advance joe I think this is fine. Just playing around (see below), I obtained what I would expect. :-) V2 - rep(x=1:12, times=5) seasonal - diff(V2, lag=12, differences=1) V2 [1] 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 [26] 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 [51] 3 4 5 6 7 8 9 10 11 12 seasonal [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [39] 0 0 0 0 0 0 0 0 0 0 Hope this helps, Roland -- This mail has been sent through the MPI for Demographic Research. Should you receive a mail that is apparently from a MPI user without this text displayed, then the address has most likely been faked. If you are uncertain about the validity of this message, please check the mail header or ask your system administrator for assistance. __ 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] new line between '}' and 'else' in function body
Thanks, Romain! So I think, for consistency, the following result
deparse(parse(text = '
+ f = function(x) {
+if (x) {
+1
+} # a new line here!
+else {
+2
+}
+ }
+ ')
+ )
[1] structure(expression(f = function(x) { if (x) {
[3] 1 }
[5] else { 2
[7] } }), srcfile =
environment)
should be
[1] structure(expression(f = function(x) { if (x) {
[3] 1 } else {
[5] 2 }
[7] }), srcfile = environment)
instead.
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
On Fri, Apr 3, 2009 at 2:48 PM, Romain Francois
romain.franc...@dbmail.com wrote:
Hi,
That's because the parser knows how to deal with that stuff. However, when
you type the same if/else at the command line, it will be parsed line by
line, and the evaluator will not wait for the else to evaluate the if. Try
to copy and paste your if/else to the command line.
Romain
Yihui Xie wrote:
Hi list members,
?else tells us
In particular, you should not have a newline between '}' and
'else' to avoid a syntax error in entering a 'if ... else'
construct at the keyboard or via 'source'.
but there's no syntax error when you break the line between } and
else in a function, e.g.
f = function(x) {
if (x) {
1
} # a new line here!
else {
2
}
}
f(TRUE)
[1] 1
f(FALSE)
[1] 2
Seems strange...
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
--
Romain Francois
Independent R Consultant
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Odp: finding best fitting model
Hi
I adopted an idea from Chemical Engineering where one guy used set of
functions to check which can fit data well and used Excel for it. So I
tried if I can do it in R.
It is very rough and needs data to be a two column data frame with x in
the first column together with list of formulas.
You can use it like
nls.find(modely1, data)
and you can try to give it some starting values but it can arrive to
results without them. You can also set start=T, which will use default
method for nls. The result is list of model parameters ordered by residual
sum of residual sqares. It does not help you to think about your data, it
does not use any type of diagnostics.
But you can use it with some care.
regards
Petr
Below is three set of models and the function itself.
modely1 - list(
y~1/(a-x),
y~x^2-a*x,
y~a^x,
y~1-a^x,
y~a^(1/x),
y~x^a,
y~1-1/(x^a),
y~a^(1-log(x)),
y~1-exp(-a*x),
y~a*x^2-log(x),
y~a/cosh(x),
y~a*cosh(x/a))
modely2 - list(
y~a*x+b,
y~1/(a+b*x),
y~a/(b+x),
y~a+b*log(x),
y~a*exp(b*x),
y~a*exp(b/x),
y~a*(1-exp(b*x)),
y~(a/(a-b))*(exp(-b*x)-exp(-a*x)),
y~1/(1+exp(-a*(x-b))),
y~a*x^b,
y~a*(1+x)^b,
y~(a+x)^b,
y~(a-x)^b,
y~1-1/(1+a*x)^b,
y~a*x^(b*x),
y~a*x^(b/x),
y~x/(a+b*x-(a+b)*x^2),
y~a*cos(x-b),
y~a/(sin(x-b))
)
modely3 - list(
y~a+b*x+c*x^2,
y~1/(a+b*x+c*x^2),
y~a+b/x+c/(x^2),
y~a+b*log(x-c),
y~a+b*log(1+c^x),
y~a+b*x^c,
y~a*(x-b)^c,
y~a*(x/b)^c,
y~a*b^x+x^c,
y~a*x^(b*x-c),
y~(a*x^c)/(1+b*x^c),
y~(a*x^c)/(b^c+x^c),
y~a+b*exp(c*x),
y~a-b*exp(-c*x),
y~a*(1-b*exp(c*x)),
y~a*(1-exp(-b*x))^c,
y~a+(b*(exp(c*x)-1))/c,
y~exp(a+b*x+c*x^2),
y~a/(1+exp(-b*(x-c))),
y~a*sin((pi*(x-b))/c)
)
nls.find - function(formula, data, start=NULL, ...){
ll - length(formula)
result2 - vector(list, ll)
result1 - rep(NA, ll)
names(data) - c(x,y)
for(i in 1:ll) {
if (is.null(start)) {
n - length(all.vars(formula[[i]]))-2
start - as.list(rep(mean(data[,2], na.rm=T),n))
names(start) - letters[1:n]
fit-try(nls(formula[[i]], data, start=start, ...))
} else {
if (is.logical(start)) {
fit-try(nls(formula[[i]], data, ...))
} else {
fit-try(nls(formula[[i]], data, start, ...)) }
}
if( class(fit)==try-error) result1[i] - NA else result1[i] -
sum(resid(fit)^2, na.rm=T)
if( class(fit)==try-error) result2[[i]] - NA else result2[[i]] -
coef(fit)
}
ooo-order(result1)
result - mapply(c, sq.resid = result1, result2, SIMPLIFY=FALSE)
names(result) - paste(1:ll, as.character(formula), sep=/)
result[ooo]
}
Petr Pikal
petr.pi...@precheza.cz
724008364, 581252140, 581252257
r-help-boun...@r-project.org napsal dne 02.04.2009 18:40:43:
Hello,
Is there a function in r to find the best fitting model for a set of
data?
I would like to know if my data are related exponentially,linearly or if
there
is a logarithmic correlation between my x and y values.
To get a better imagination I've added the graphics at the end of this
mail
as an attachment.
Thanks, Ben
--
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Help pasting string as object name
I have a data frame containing monthly observations of the 'density' of each US
state, recorded in variables named density.AL, density.AK, density.AZ,
and so on for all 50 states. The data frame (called d) also contains a variable
called Date which is encoded as a string in the format Jan-09, Feb-09,
etc.
I also have a vector st.list-c(AL, AK, AZ,...) of length 50.
I would like a new dataframe with st.list in one column and the value of
d$density for that state when d$Date==Feb-09 in another column.
How can I do this?
Here is what I have tried:
names - rep(d$density,length(st.list))
templist - as.vector(mapply(paste, names, st.list ,sep=.))
d.2-data.frame()
for (i in 1:length(templist)) {
d.2$density[i] - subset(parse(file=,templist[i]),d$Date==Feb-09)
i-i+1 } ### hangs!
Thanks for any help!
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[R] plyr and table question
Dear all, I'm puzzled by the following example inspired by a recent question on R-help, cc - textConnection(user_id website time 20google0930 21yahoo0935 20facebook1000 25facebook1015 61google0940) d - read.table(cc, head=T) ; close(cc) table(d$user_id) # count the occurrences # now I'd like to include these results in the original data.frame, ddply(d, .(website), transform, count = table(user_id)) # why two new columns? I just can't understand how this is different from, ddply(d, .(website), transform, count = sum(user_id)) Many thanks, baptiste __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] new line between '}' and 'else' in function body
you can always wrap the whole if/else statement into innocent braces or
parentheses, as in
y = {
if (x) 1
else 2 }
y = (
if (x) 1
else 2 )
it doesn't have to be a function, and there is no need for the
assignment either -- you just need to tell the parser that the input
hasn't ended. that's a matter of taste, but i find
{ if (x) 1
else 2 }
more readable than something like
if (x) { 1
} else 2
vQ
Yihui Xie wrote:
Thanks, Romain! So I think, for consistency, the following result
deparse(parse(text = '
+ f = function(x) {
+if (x) {
+1
+} # a new line here!
+else {
+2
+}
+ }
+ ')
+ )
[1] structure(expression(f = function(x) { if (x) {
[3] 1 }
[5] else { 2
[7] } }), srcfile =
environment)
should be
[1] structure(expression(f = function(x) { if (x) {
[3] 1 } else {
[5] 2 }
[7] }), srcfile = environment)
instead.
Regards,
Yihui
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Re: [R] Help pasting string as object name
Hi,
Is this what you want?
d - data.frame(density.AL = seq(1, 10),
density.AK = seq(1, 10), # many others...
Date=letters[1:10]) # dummy example
library(reshape)
melt(subset(d, Date == b), id=Date)
BTW, I spotted a few awkward things in your code,
st - c(AL, AK)
vars - paste(d$density, st, sep=.) # easier than mapply etc.
more importantly, in the for loop you should not be incrementing i
manually (as in a while loop), it's already taken care of by the for{}
construct.
HTH,
baptiste
On 3 Apr 2009, at 10:40, Rob Denniker wrote:
I have a data frame containing monthly observations of the 'density'
of each US state, recorded in variables named density.AL,
density.AK, density.AZ, and so on for all 50 states. The data
frame (called d) also contains a variable called Date which is
encoded as a string in the format Jan-09, Feb-09, etc.
I also have a vector st.list-c(AL, AK, AZ,...) of length 50.
I would like a new dataframe with st.list in one column and the
value of d$density for that state when d$Date==Feb-09 in another
column.
How can I do this?
Here is what I have tried:
names - rep(d$density,length(st.list))
templist - as.vector(mapply(paste, names, st.list ,sep=.))
d.2-data.frame()
for (i in 1:length(templist)) {
d.2$density[i] - subset(parse(file=,templist[i]),d$Date==Feb-09)
i-i+1 } ### hangs!
Thanks for any help!
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_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] Functions Accessing Databases
Hi,
There seems to be something wrong with your function getinfo.
Try first to substitute your dbGetQuery by a print statement, so that you
can see what is going on:
getinfo=function(t){print(the SQL query)}
allinfo=sapply(c(1985:2007),getinfo)
If these results look good, you can change your function back to
dbGetQuery(con, )
If not, show your results, and we will look again
Bart
Bob Roberts-2 wrote:
Hello,
I'm accessing a MySQL database from inside R and had a problem with a
function. In the database, there is data from years 1985 to 2007 that I
would like to retrieve. I can easily get the data from one year by the
following:
info1985=dbGetQuery(con, statement='the SQL query')
Inside the statement, I have a column that is set to the desired year
(e.g. table.column=1985) through a WHERE clause, but when I write a
function like the following:
getinfo=function(t){dbGetQuery(con,statement='the SQL query')}
and instead of using a specific year, I substitute in t (e.g.
table.column=t) in a WHERE clause
When I go to do
allinfo=sapply(c(1985:2007),getinfo)
I get the following error message: RS-DBI driver: (could not run
statement: Unknown column 't' in 'where clause')
Is there anyway around this error? Or is not possible to substitute in
years with a function and sapply through SQL?
Thanks so much.
[[alternative HTML version deleted]]
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[R] course in ecological data analysis
Dear all, For my PhD study I'm looking for relevant courses/workshops (short term) in ecological data anlysis with R in Europe. After 2 days searching I'm convinced that google is probably not the right medium to find this information. If anyone can help me I will be most grateful. Best regards - J. Capelle [[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] factorise variables in a data.frame
Dear list,
I often need to convert several variables from numeric or integer into
factors (before plotting, for instance), as in the following example,
d - data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 -
within(d, {
x = factor(x)
y = factor(y)
})
str(d)
str(d2)
I'd like to write a function factorise() which takes a data.frame and
a vector of variable names, and returns the original data.frame with
the desired variables converted to factor,
factorise - function(d, f)
***ply(d, f, factor) # some apply function
also, perhaps a defactorise() function doing the reverse operation
with as.numeric.
I played with the plyr package and the base apply family for a while
but can't find any concise construct.
Best regards,
baptiste
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] extract tables as data.frames from HTML source
If the question is specific to getting stock data from
google finance then check out getSymbols.google
in the quantmod package.
Also note that there exists an r-sig-finance list for questions
pertaining to R and finance.
On Fri, Apr 3, 2009 at 2:18 AM, Yihui Xie xieyi...@gmail.com wrote:
Hi,
I wonder whether there is any convenient function (or package) to
extract tables from a HTML page? e.g. from
http://www.google.com/finance/historical?q=SHE:002251
I know we can readLines('URL'), gsub('td...', '...', source), ...
and at last get the numbers; I'm writing to ask whether someone has
already contributed a more general function (with the package XML or
other packages). Thanks!
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
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[R] Trouble extracting graphic results from a bootstrap
Hi,
I'm trying to extract a histogram over the results from a bootstrap. However
I keep receiving the error message Error in hist.default(boot.lrtest$ll,
breaks = scott) : 'x' must be numeric.
The bootstrap I'm running looks like:
boot.test - function(data, indeces, maxit=20) {
+ y1 - fit1+e1[indeces]
+ mod1 - glm(y1 ~ X1-1, maxit=maxit)
+ y2 - fit2+e2[indeces]
+ mod2 - glm(y2~1, maxit=maxit)
+ ll - 2*(logLik(mod1)-logLik(mod2))
+ ll
+ }
boot.lrtest - boot(data=M1, statistic=boot.test, R=2000, maxit=100);
hist(boot.lrtest$ll, breaks=scott) # This results in the error message
stated above
So my question is: what am I doing wrong?
//Anders
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Re: [R] factorise variables in a data.frame
baptiste auguie wrote:
Dear list,
I often need to convert several variables from numeric or integer into
factors (before plotting, for instance), as in the following example,
d - data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 -
within(d, {
x = factor(x)
y = factor(y)
})
str(d)
str(d2)
I'd like to write a function factorise() which takes a data.frame and
a vector of variable names, and returns the original data.frame with
the desired variables converted to factor,
would this not be good enough:
# dummy data
data = data.frame(x=1:10, y=1:10)
# a factorizer
factorize = function(data, columns=names(data)) {
data[columns] = lapply(data[columns], as.factor)
data }
sapply(factorize(data, 'x'), is)
# $x factor ...
# $y integer ...
lapply(factorize(data), is)
# $x factor ...
# $y factor ...
factorise - function(d, f)
***ply(d, f, factor) # some apply function
also, perhaps a defactorise() function doing the reverse operation
with as.numeric.
then, perhaps,
# an izer
ize = function(data, columns=names(data), izer=as.factor) {
data[columns] = lapply(data[columns], izer)
data }
ize(data, 'x', as.logical)
or even
ize = function(izer)
function(data, columns=names(data)) {
data[columns] = lapply(data[columns], izer)
data }
logicalize = ize(as.logical)
characterize = ize(as.character)
lapply(logicalize(data), is)
# $x logical ...
# $y logical ...
lapply(characterize(data, 'x'), is)
# $x character ...
# $y integer ...
etc.
vQ
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[R] Schoenfeld Residuals
Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs produced 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf My data (nearma) has a lot of rem.Remtime entries which are equal i.e large amounts of tied data. If I remove the entries where this is the case from the dataset I get the results I want! Please can someone explain why removing paients with tied remission time has such an effect on the code and also how to remedy the problem without removing patients? Thank you very much, Laura. __ 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] factorise variables in a data.frame
Excellent!
I felt it was fairly trivial but i can be quite dense on Friday
mornings.
I really like the generalisation.
Many thanks,
baptiste
On 3 Apr 2009, at 12:11, Wacek Kusnierczyk wrote:
baptiste auguie wrote:
Dear list,
I often need to convert several variables from numeric or integer
into
factors (before plotting, for instance), as in the following example,
d - data.frame(
x = seq(1, 10),
y = seq(1, 10),
z = rnorm(10),
a = letters[1:10])
d2 -
within(d, {
x = factor(x)
y = factor(y)
})
str(d)
str(d2)
I'd like to write a function factorise() which takes a data.frame and
a vector of variable names, and returns the original data.frame with
the desired variables converted to factor,
would this not be good enough:
# dummy data
data = data.frame(x=1:10, y=1:10)
# a factorizer
factorize = function(data, columns=names(data)) {
data[columns] = lapply(data[columns], as.factor)
data }
sapply(factorize(data, 'x'), is)
# $x factor ...
# $y integer ...
lapply(factorize(data), is)
# $x factor ...
# $y factor ...
factorise - function(d, f)
***ply(d, f, factor) # some apply function
also, perhaps a defactorise() function doing the reverse operation
with as.numeric.
then, perhaps,
# an izer
ize = function(data, columns=names(data), izer=as.factor) {
data[columns] = lapply(data[columns], izer)
data }
ize(data, 'x', as.logical)
or even
ize = function(izer)
function(data, columns=names(data)) {
data[columns] = lapply(data[columns], izer)
data }
logicalize = ize(as.logical)
characterize = ize(as.character)
lapply(logicalize(data), is)
# $x logical ...
# $y logical ...
lapply(characterize(data, 'x'), is)
# $x character ...
# $y integer ...
etc.
vQ
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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.
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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[R] Curve fitting,FDA for biological data
Dear all, Another newbie just got attracted to this mailing list. I am a biologist currently working my way through R, had sort play around with python earlier this year. I have some data exhibiting periodicity ** my data consists of peaks and valleys, with peaks arising due to the presence of a repetitive structural unit,** with x being a reference grid (position along a chromosome) and y being strength of signal (this y signal fluctuates to give rise to the peaks and valleys). ie presence of a structural unit along a chromosome gives rise to a peak in my data. I would like to use a curve fitting algorithm (I guess something like a fourier analysis and/or splines). Due to the nature of the data I would like to look for periodicities at different scales (along the x grid). So say 2-4 different splines/curves are probably enough to describe the 40,000 occourences of the repetitive structural unit in my data, while say 4-6 of these units could exhibit certain patterns in the way they group together. I assume in my case, I can consider my x axis (position) to be equivalent to a time x axis as in signal processing. I considered using the FDA package (silverman and ramsey I think). Does anyone have an ideas if this is the right way to go or suggestions etc PS I have highlighted in the attached gif with red, the occourence of the repetitive signal (differences in the wavelength for example could be important but not more than 4 would be required to fit all data), and in yellow a hypothetical occourence of a periodicity in a different scale Thanks a lot Dr Triantafyllos Gkikopoulos The University of Dundee is a registered Scottish charity, No: SC015096 __ 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] plyr and table question
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie ba...@exeter.ac.uk wrote: Dear all, I'm puzzled by the following example inspired by a recent question on R-help, cc - textConnection(user_id website time 20 google 0930 21 yahoo 0935 20 facebook 1000 25 facebook 1015 61 google 0940) d - read.table(cc, head=T) ; close(cc) table(d$user_id) # count the occurrences # now I'd like to include these results in the original data.frame, ddply(d, .(website), transform, count = table(user_id)) # why two new columns? Because ddply expects a data frame as output from your aggregation function. When the output isn't a data frame, it calls as.data.frame, which in this case produces a data frame with two columns. Hadley -- http://had.co.nz/ __ 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] Schoenfeld Residuals
I am not sure that ties are the only reason. If I create a few ties in the ovarian dataset that Therneau and Lumley provide, all I get are some warnings: ovarian[4:5, 1] - mean(ovarian[4:5, 1]) ovarian[6:8, 1] - mean(ovarian[6:8, 1]) fit - coxph( Surv(futime, fustat) ~ age + rx, ovarian) temp- cox.zph(fit) plot(temp) Warning messages: 1: In approx(xx, xtime, seq(min(xx), max(xx), length.out = 17)[2 * : collapsing to unique 'x' values 2: In approx(xtime, xx, temp) : collapsing to unique 'x' values The error message you get is requesting a finite ylim. Have you considered acceding with that request? Alternative: Assuming the number of tied survival times is modest, have you tried jitter-ing the rem.Remtime variable a few times to see it the results are stable? If the number of ties is large, then you need to review Thernaeu Gramsch section 3.3 -- David Winsemius On Apr 3, 2009, at 7:57 AM, Laura Bonnett wrote: Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs produced 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf My data (nearma) has a lot of rem.Remtime entries which are equal i.e large amounts of tied data. If I remove the entries where this is the case from the dataset I get the results I want! Please can someone explain why removing paients with tied remission time has such an effect on the code and also how to remedy the problem without removing patients? Thank you very much, Laura. __ 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] Geometric Brownian Motion Process with Jumps
Hi,
I have been using maxLik to do some MLE of Geometric Brownian Motion Process
and everything has been going fine, but know I have tried to do it with jumps.
I have create a vector of jumps and then added this into my log-likelihood
equation, know I am getting a message:
NA in the initial gradient
My codes is hear
#
n-length(combinedlr)
j-c(1,2,3,4,5,6,7,8,9,10)
gbmploglik-function(param){
mu-param[1]
sigma-param[2]
lamda-param[3]
nu-param[4]
gama-param[5]
logLikVal- - n*lamda - .5*n*log(2*pi) + sum(log(sum(for(j in
1:10)(cat((lamda^j/factorial(j))*(1/((sigma^2+j*gama^2)^.5)*exp( -
(combinedlr-mu-j*nu)^2/2*(sigma^2+j*gama^2
logLikVal
}
rescbj- maxLik(gbmploglik, grad = NULL, hess = NULL, start=c(0,1,1,1,1),
method = Newton-Raphson)
summary(rescbj)
#
I am also was wondering if anyone know if there was a package that dealt with
Geometric Brownian Motion Process augmented with jumps. Then I could just put
that into my code and might resolve the issue.
Any suggest as to how to resolved this issue, are greatly apprecaited.
Yours truly,
JP
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[R] Linear model, finding the slope
Hi for some data I working on I am merely plotting time against temperature for a variable named filmclip. So for example, I have volunteers who watched various film clips and have used infared camera to monitor the temperature on their face at every second of the clip. The variable names I have used are Normalised ( for the temperature) and Frame (for the time in seconds). So I have fitted a linear model model-lm(Normalised~Frame,data=All,subset=((Subject==1)(Filmclip==Whatever) and coef(model) gives me an intercept value and a value for the slope. Now what I want to do is find out if the slope is significant or not. So far I just have values such as 0.02211 for example and have no idea if this is to be interpreted as significant or not. Sorry if I haven't been clear but any advice on how to find out what values are significant would be greatly appreciated. -- View this message in context: http://www.nabble.com/Linear-model%2C-finding-the-slope-tp22865254p22865254.html Sent from the R help mailing list archive at Nabble.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.
[R] Titles on lattice colorkey
Dear R-ers, I'm not sure if this is a missing feature, a support request, or stupidity on my part, but nevertheless, its a question. Is it possible to add titles to colorkey legends? As far as I can tell, there is a command to do it for normal key legends, but not for colorkeys. eg it works for a normal key, created through auto.key xyplot(decrease ~ treatment, OrchardSprays, groups = rowpos, type = a, auto.key = list(space = right, points = FALSE, lines = TRUE,title=Key title)) but there is no comparable command for a colorkey x - seq(pi/4, 5 * pi, length = 100) y - seq(pi/4, 5 * pi, length = 100) r - as.vector(sqrt(outer(x^2, y^2, +))) grid - expand.grid(x=x, y=y) grid$z - cos(r^2) * exp(-r/(pi^3)) levelplot(z~x*y, grid, cuts = 50, scales=list(log=e), xlab=, ylab=, main=Weird Function, sub=with log scales, region = TRUE, colorkey = list(space=right,title=Doesn't work)) Cheers, Mark __ 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] plyr and table question
baptiste auguie-2 wrote: ddply(d, .(website), transform, count = table(user_id)) # why two new columns? Try this to see why: as.data.frame(table(d$user_id)) This works more like you expect: ddply(d, .(website), transform, count = unclass(table(user_id))) - Tom -- View this message in context: http://www.nabble.com/plyr-and-table-question-tp22865174p22868047.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to do this the R way
Hi. I am sure there is a better way in R to do this then using a loop but I
am new to it and not sure what to do. I think it might be something about
using a function as an argument but not sure.
I have a 1 x 2000 vector TS2 which has entries from the set {x: x is in Z
and 0x8} (where Z is the set of Integers).
Then I also have a 5050 x 7 matrix called 'perm' whose entries are also from
the set {x: x is in Z and 0x8}
I want to construct the following transformation of TS2 which will still be
a vector of size 1 x 2000 and which I will call 'newTS2' such that:
newTS2 =(perm[100.TS2[1]]. perm[100.TS2[2]]. perm[100.TS2[3]]. ... ,
perm[100.TS2[2000]])
Is there a nice way to do this without a loop?
Thank you.
--
View this message in context:
http://www.nabble.com/how-to-do-this-%22the-R-way%22-tp22864354p22864354.html
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Re: [R] plyr and table question
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie ba...@exeter.ac.uk wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1 20 google 930 2
2 20 facebook 1000 2
3 21 yahoo 935 1
4 25 facebook 1015 1
5 61 google 940 1
Have I missed a built-in function to obtain this result?
ddply(d, .(user_id), transform, count = nrow)
?
Hadley
--
http://had.co.nz/
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Re: [R] Linear model, finding the slope
Melissa2k9 wrote: Hi for some data I working on I am merely plotting time against temperature for a variable named filmclip. So for example, I have volunteers who watched various film clips and have used infared camera to monitor the temperature on their face at every second of the clip. The variable names I have used are Normalised ( for the temperature) and Frame (for the time in seconds). So I have fitted a linear model model-lm(Normalised~Frame,data=All,subset=((Subject==1)(Filmclip==Whatever) and coef(model) gives me an intercept value and a value for the slope. Now what I want to do is find out if the slope is significant or not. So far I just have values such as 0.02211 for example and have no idea if this is to be interpreted as significant or not. Sorry if I haven't been clear but any advice on how to find out what values are significant would be greatly appreciated. See ?lm and ?summary.lm and try the examples. Uwe Ligges __ 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] Convert factor to double?
Hi! I'm reading a tab-seperated CVS file with: test1 - read.table(data.txt, header=TRUE) It's in the following format: Date_Time qK qL vL vP ... 0 30 22 110 88 ... ... (BTW: It seems to me R shifts the column descriptions by one.) Anyway, I would like to Fourier-transform one column. So I say: fft(test1$vP) Error in levels(x)[x] : invalid subscript type 'complex' I guess this is some typing error but I don't know which function I have to use for conversion. Some more Info about the data: test1$vP[1:10] [1] 110 108 116 118 114 120 117 111 95 118 166 Levels: - 0 1 10 100 101 102 103 104 105 106 107 108 109 11 110 111 ... 99 class( test1$vPkw) [1] factor I would appreciate any hints! Thanks in advance! Best, Philip -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plyr and table question
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1 20 google 930 2
2 20 facebook 1000 2
3 21yahoo 935 1
4 25 facebook 1015 1
5 61 google 940 1
Have I missed a built-in function to obtain this result?
Thanks,
baptiste
On 3 Apr 2009, at 14:16, hadley wickham wrote:
On Fri, Apr 3, 2009 at 4:43 AM, baptiste auguie ba...@exeter.ac.uk
wrote:
Dear all,
I'm puzzled by the following example inspired by a recent question on
R-help,
cc - textConnection(user_id website time
20google0930
21yahoo0935
20facebook1000
25facebook1015
61google0940)
d - read.table(cc, head=T) ; close(cc)
table(d$user_id) # count the occurrences
# now I'd like to include these results in the original data.frame,
ddply(d, .(website), transform, count = table(user_id)) # why two new
columns?
Because ddply expects a data frame as output from your aggregation
function. When the output isn't a data frame, it calls as.data.frame,
which in this case produces a data frame with two columns.
Hadley
--
http://had.co.nz/
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] lme between-group and within-group covariance
I will try to make this more precise. In the lme() function, the correlation argument allows the user to specify a within-group correlation structure, i.e. the structure of the Lambda matrix using the mixed model notation in Pineiro and Bates. What I want to do is specify a distinct structure for the Psi matrix (same notation), that is, a correlation structure for the random effects. If lme() doesn't allow for this, is there any other function that I could use? MUHC-Research wrote: Dear R users, I would be interested in using the lme() function to fit a linear mixed model to a longitudinal dataset. I know this function allows for the specification of a within-group covariance structure. However, does it allow for the explicit specification of a between-group covariance structure? Being able to specify both separately would be very important in the context of my project since, as might be expected, they have different implications/interpretations. For instance, the mixed procedure in SAS allows users to specify the two structures separately by adding a value for the type argument after the RANDOM statement and the REPEATED statement. My question is thus if we can do the same with lme(). I thank you most sincerely for your help. -- View this message in context: http://www.nabble.com/lme-between-group-and-within-group-covariance-tp22834748p22868945.html Sent from the R help mailing list archive at Nabble.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] Plotting multiple ablines
I've really been on a roll this week; the formula for the lines that I
presented was completely wrong.
But I'm glad I learned about mapply. I used this:
mapply(abline,
(converge$kY + tan((90-converge$kT) * pi / 180)*(-converge$kX)),
tan((90-converge$kT) * pi / 180))
Tom!
On Thu, Apr 2, 2009 at 8:29 AM, r...@quantide.com r...@quantide.com wrote:
May be:
plot(c(-1, 1) , c(-1, 1), type = n)
n = 4
a = rep(0, n)
b = 1:n/n
fun = function(i, a, b, col = 1 , ...) {
abline(a[i], b[i], col = col[i], ...)
}
lapply(1:n, fun, a=a, b=b, col = 1:n)
Andrea
Thomas Levine wrote:
I really want to do this:
abline(
a=tan(-kT*pi/180),
b=kY-tan(-kT*pi/180)*kX
)
where kX,kY and kT are vectors of equal length. But I can't do that
with abline unless I use a loop, and I haven't figured out the least
unelegant way of writing the loop yet. So is there a way to do this
without a loop?
Or if I am to resort to the loop, what's the best way of doing it
considering that I have some missing data? Here's the mess that I
wrote.
converge - na.omit(data.frame(kX,kY,kT))
for (z in (length(converge$kT)))
{abline(
a=tan(converge$kT[z]*pi/180),
b=converge$kY[z]-tan(-converge$kT[z]*converge$kX[z]*pi/180)
)}
I think the missing data are causing the problem; this happens when I run:
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
'a' and 'b' must be finite
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[[alternative HTML version deleted]]
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Re: [R] plyr and table question
of course!
Thanks,
baptiste
On 3 Apr 2009, at 14:48, hadley wickham wrote:
On Fri, Apr 3, 2009 at 8:43 AM, baptiste auguie ba...@exeter.ac.uk
wrote:
That makes sense, so I can do something like,
count - function(x){
as.integer(unclass(table(x)))
}
count(d$user_id)
ddply(d, .(user_id), transform, count = count(user_id))
user_id website time count
1 20 google 930 2
2 20 facebook 1000 2
3 21yahoo 935 1
4 25 facebook 1015 1
5 61 google 940 1
Have I missed a built-in function to obtain this result?
ddply(d, .(user_id), transform, count = nrow)
?
Hadley
--
http://had.co.nz/
_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
__
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.
Re: [R] Schoenfeld Residuals
Thank you for your comments. I have about 200 out of 2000 tied data points which makes the situation more complicated! I'll have at look at the book section you referred to. With regards to making the ylim finite, I'm not sure how I can go about that given that I don't understand why it isn't already! Thank you for your help, Laura 2009/4/3 David Winsemius dwinsem...@comcast.net: I am not sure that ties are the only reason. If I create a few ties in the ovarian dataset that Therneau and Lumley provide, all I get are some warnings: ovarian[4:5, 1] - mean(ovarian[4:5, 1]) ovarian[6:8, 1] - mean(ovarian[6:8, 1]) fit - coxph( Surv(futime, fustat) ~ age + rx, ovarian) temp- cox.zph(fit) plot(temp) Warning messages: 1: In approx(xx, xtime, seq(min(xx), max(xx), length.out = 17)[2 * : collapsing to unique 'x' values 2: In approx(xtime, xx, temp) : collapsing to unique 'x' values The error message you get is requesting a finite ylim. Have you considered acceding with that request? Alternative: Assuming the number of tied survival times is modest, have you tried jitter-ing the rem.Remtime variable a few times to see it the results are stable? If the number of ties is large, then you need to review Thernaeu Gramsch section 3.3 -- David Winsemius On Apr 3, 2009, at 7:57 AM, Laura Bonnett wrote: Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs produced 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf My data (nearma) has a lot of rem.Remtime entries which are equal i.e large amounts of tied data. If I remove the entries where this is the case from the dataset I get the results I want! Please can someone explain why removing paients with tied remission time has such an effect on the code and also how to remedy the problem without removing patients? Thank you very much, Laura. __ 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] Geometric Brownian Motion Process with Jumps
Hi,
The error message is clear in that the gradient cannot be evaluated at your
starting value for the parameters.
Is your likelihood a smooth function of parameters? If so, then provide a
different starting value. If it is not smooth, then you may have to use a
method that does not depend on gradients, such as Nelder-Mead.
Can you provide a reproducible example, which would help us dig deeper into
your problem?
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: rvarad...@jhmi.edu
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John-Paul Taylor
Sent: Friday, April 03, 2009 8:30 AM
To: r-help@r-project.org
Subject: Re: [R] Geometric Brownian Motion Process with Jumps
Hi,
I have been using maxLik to do some MLE of Geometric Brownian Motion Process
and everything has been going fine, but know I have tried to do it with
jumps. I have create a vector of jumps and then added this into my
log-likelihood equation, know I am getting a message:
NA in the initial gradient
My codes is hear
#
n-length(combinedlr)
j-c(1,2,3,4,5,6,7,8,9,10)
gbmploglik-function(param){
mu-param[1]
sigma-param[2]
lamda-param[3]
nu-param[4]
gama-param[5]
logLikVal- - n*lamda - .5*n*log(2*pi) + sum(log(sum(for(j in
1:10)(cat((lamda^j/factorial(j))*(1/((sigma^2+j*gama^2)^.5)*exp( -
(combinedlr-mu-j*nu)^2/2*(sigma^2+j*gama^2
logLikVal
}
rescbj- maxLik(gbmploglik, grad = NULL, hess = NULL, start=c(0,1,1,1,1),
method = Newton-Raphson)
summary(rescbj)
#
I am also was wondering if anyone know if there was a package that dealt
with Geometric Brownian Motion Process augmented with jumps. Then I could
just put that into my code and might resolve the issue.
Any suggest as to how to resolved this issue, are greatly apprecaited.
Yours truly,
JP
__
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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] [R-pkgs] New R graphics package targeting HTML 5 canvas element
Hi, canvas is a new R package implementing a graphics device that emits javascript code conforming to the HTML 5 CanvasRenderingContext2D interface. Available on CRAN soon, but you can get it here immediately: http://www.rforge.net/canvas If you have access to a beta web browser like Firefox 3.1 or later, you can see example plots here: http://www.rforge.net/canvas/plots/index.html Clicking on one of the plots will take you to a comparator page where the plot is rendered twice, once with the canvas package and once with the png(type='cairo') function. The implementation is almost complete, however there are some parts that need work like font metrics and dotted/dashed lines. If anyone is interested in helping flesh out the remaining bits, I'd be much obliged. Best, Jeff ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Geometric Brownian Motion Process with Jumps
Hi,
The error message is clear in that the gradient cannot be evaluated at your
starting value for the parameters.
Is your likelihood a smooth function of parameters? If so, then provide a
different starting value. If it is not smooth, then you may have to use a
method that does not depend on gradients, such as Nelder-Mead.
Can you provide a reproducible example, which would help us dig deeper into
your problem?
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email: rvarad...@jhmi.edu
Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John-Paul Taylor
Sent: Friday, April 03, 2009 8:30 AM
To: r-help@r-project.org
Subject: Re: [R] Geometric Brownian Motion Process with Jumps
Hi,
I have been using maxLik to do some MLE of Geometric Brownian Motion Process
and everything has been going fine, but know I have tried to do it with
jumps. I have create a vector of jumps and then added this into my
log-likelihood equation, know I am getting a message:
NA in the initial gradient
My codes is hear
#
n-length(combinedlr)
j-c(1,2,3,4,5,6,7,8,9,10)
gbmploglik-function(param){
mu-param[1]
sigma-param[2]
lamda-param[3]
nu-param[4]
gama-param[5]
logLikVal- - n*lamda - .5*n*log(2*pi) + sum(log(sum(for(j in
1:10)(cat((lamda^j/factorial(j))*(1/((sigma^2+j*gama^2)^.5)*exp( -
(combinedlr-mu-j*nu)^2/2*(sigma^2+j*gama^2
logLikVal
}
rescbj- maxLik(gbmploglik, grad = NULL, hess = NULL, start=c(0,1,1,1,1),
method = Newton-Raphson)
summary(rescbj)
#
I am also was wondering if anyone know if there was a package that dealt
with Geometric Brownian Motion Process augmented with jumps. Then I could
just put that into my code and might resolve the issue.
Any suggest as to how to resolved this issue, are greatly apprecaited.
Yours truly,
JP
__
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.
__
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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] Convert factor to double?
p.si...@gmx.net wrote: Hi! I'm reading a tab-seperated CVS file with: test1 - read.table(data.txt, header=TRUE) It's in the following format: Date_Time qK qL vL vP ... 0 30 22 110 88 ... ... (BTW: It seems to me R shifts the column descriptions by one.) No, look into your data file first. Anyway, I would like to Fourier-transform one column. So I say: fft(test1$vP) Error in levels(x)[x] : invalid subscript type 'complex' I guess this is some typing error but I don't know which function I have to use for conversion. Some more Info about the data: test1$vP[1:10] [1] 110 108 116 118 114 120 117 111 95 118 166 Levels: - 0 1 10 100 101 102 103 104 105 106 107 108 109 11 110 111 ... 99 Obviously there is some - sign in your data file. You might want to fix that prior to importing the data. You cannot convert easily to numeric with artifacts like the string - in your data. Uwe Ligges class( test1$vPkw) [1] factor I would appreciate any hints! Thanks in advance! Best, Philip -- __ 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] (no subject)
Dear R users, I am trying to do exact matching on a large dataset (500.000 obs), about equal size of treatment and controll group, with replacement: As for the moment I use the Match function of the Matching library. I match on 2 covariates and all observations in the treatment group have at least one exact counterpart in the controllgroup. Now I want to introduce observation weights. I set ties=FALSE, as I want exactly one by one matching: Is there a way which makes that I draw randomly from the individuals in the controllgroup which have the same values of covariates as the individual in the treatmentgroup, setting the probabilities to be drawn proportional to the weights of the individual in the CT? E.g. I have three individuals which all have the same value for the covariates as the one observation I want to find a partner for, and the first of the three individuals has a very large weight: Now when drawing randomly among those three I want the probability that the first one is dr! awn to be very large. I'd really appreciate any suggestions: the weights option does not do the job, this seems to work only if setting ties=TRUE Thanks Dirk -- __ 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 this the R way
onyourmark wrote:
Hi. I am sure there is a better way in R to do this then using a loop but I
am new to it and not sure what to do. I think it might be something about
using a function as an argument but not sure.
I have a 1 x 2000 vector TS2 which has entries from the set {x: x is in Z
and 0x8} (where Z is the set of Integers).
Then I also have a 5050 x 7 matrix called 'perm' whose entries are also from
the set {x: x is in Z and 0x8}
I want to construct the following transformation of TS2 which will still be
a vector of size 1 x 2000 and which I will call 'newTS2' such that:
newTS2 =(perm[100.TS2[1]]. perm[100.TS2[2]]. perm[100.TS2[3]]. ... ,
Perhaps you can explain what the last line is supposed to do? This is
invalid R code and also not a known mathematical notation for me
Uwe Ligges
perm[100.TS2[2000]])
Is there a nice way to do this without a loop?
Thank you.
__
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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] for loop for extracting linear model info
Melissa2k9 wrote:
Hi,
I have written a for loop as such:
model-lm(Normalised~Frame,data=All,subset=((Subject==1)(Filmclip==Strand)))
summary(model)
###
#To extract just the Adjusted R squared
###
rsq-summary(model)[[9]]
###
#To extract just the slope
###
slope-summary(model)[[4]][[2]]
###
#To extract only the p value from the t test for slope
###
pvalue-summary(model)[[4]][[8]]
data-data.frame(slope,pvalue,rsq)
###
#To extract this info for all films
for (i in c(1:8,10:20,22:29))
{
model_1-lm(Normalised~Frame,data=All,subset=((Subject==i)(Filmclip==Strand)))
summary(model_1)
slope-summary(model_1)[[4]][[2]]
pvalue-summary(model_1)[[4]][[8]]
rsq-summary(model_1)[[9]]
data2-data.frame(slope,pvalue,rsq)
data2-rbind(data,data2)
}
I want this to run for all i but so far I am only getting two entries in my
data frame, one for the first subject, and another.
You are overwriting the old data2 with a new one that consists of those
two in each iteration of the loop ...
Uwe Ligges
Does anyone know where I am going wrong in my code so I can have this data
for all subjects 1-8,10-20, and 22-29.
Thanks
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Re: [R] course in ecological data analysis
Try r-sig-ecol...@r-project.org bests miltinho brazil-toronto On Fri, Apr 3, 2009 at 6:27 AM, Capelle, Jacob jacob.cape...@wur.nl wrote: Dear all, For my PhD study I'm looking for relevant courses/workshops (short term) in ecological data anlysis with R in Europe. After 2 days searching I'm convinced that google is probably not the right medium to find this information. If anyone can help me I will be most grateful. Best regards - J. Capelle [[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.htmlhttp://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.
[R] embed?
I have a question on the function 'embed'. I ran the example x - 1:10 embed(x, dimension=3) This gives the output: [,1] [,2] [,3] [1,]321 [2,]432 [3,]543 [4,]654 [5,]765 [6,]876 [7,]987 [8,] 1098 I don't quite understand the output and why it is useful. First, there are only 8 rows down from 10 and the first element starts with 3. Of course I can think of explanations as to what is occuring but I cannot see how this is useful. I am sure it has application as i see this command used in much of the source but I just cannot see it now. The documentation states: Each row of the resulting matrix consists of sequences x[t], x[t-1], ..., x[t-dimension+1], where t is the original index of x. If x is a matrix, i.e., x contains more than one variable, then x[t] consists of the tth observation on each variable. This explanation doesn't seem to account for the dimension argument. Thank you for your comments. Kevin __ 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] DierckxSpline fitting with different sets of y-values in one time
Dear Elisabeth: Have you tried it? I have not, but I suspect the answer is no. What problem are you trying to solve? You might get more useful suggestions from this list if you provide commented, minimal, self-contained, reproducible code describing your problem and what you've tried to solve it, as suggested in the posting guide www.R-project.org/posting-guide.html. With fmri images, I might look first at the fields package, because it is explicitly designed for curve, surface and function fitting with an emphasis on splines, spatial data and spatial statistics. The major methods include cubic, robust, and thin plate splines, multivariate Kriging and Kriging for large data sets. I have not used it, but it sounds to me like you might want thin plate splines. Paul Dierckx (1993) Curve and Surface Fitting with Splines (Oxford Science Publications) discusses thin plate splines, and Dierckx wrote Fortran to fit them. However, the DierckxSpline package does not currently connect to those capabilities. If univariate splines would do, I might start with the fda package. The theory behind that is documented in two books by Ramsay and Silverman. Hope this helps. Spencer Graves Jonckers Elisabeth wrote: Dear R users, I have a question about the Package DierckxSpline. I have tried to find the answer by myself but it didn't worked out. I wondered if Dierckxspline can use different sets of y values in one time to fit a line with knot. I have different sets of Y values representing the same thing for different voxels (in an fmri image). I have already fitted the data in different graphs and I know now that the plots are comparable for the different data sets (so for the different voxels) but I wanted to include all the information in one plot, If it's possible. So I want to know the best fitting line for the whole cluster of voxels except for one voxel. If there is no possibility to do this I tought it would be an option to take the mean of the different y-values and use those values but I don't know if this is mathematical right to do. I hope someone can help me with this. Thank you very much, Elisabeth Jonckers GIfMI Ghent University Hospital Belgium [[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-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] embed?
Its lets you perform rolling summaries using apply: apply(embed(1:10, 3), 1, mean) [1] 2 3 4 5 6 7 8 9 Note that 2 is the mean of 1:3, 3 is the mean of 2:4, ..., 9 is the mean of 8:10. On Fri, Apr 3, 2009 at 11:04 AM, rkevinbur...@charter.net wrote: I have a question on the function 'embed'. I ran the example x - 1:10 embed(x, dimension=3) This gives the output: [,1] [,2] [,3] [1,] 3 2 1 [2,] 4 3 2 [3,] 5 4 3 [4,] 6 5 4 [5,] 7 6 5 [6,] 8 7 6 [7,] 9 8 7 [8,] 10 9 8 I don't quite understand the output and why it is useful. First, there are only 8 rows down from 10 and the first element starts with 3. Of course I can think of explanations as to what is occuring but I cannot see how this is useful. I am sure it has application as i see this command used in much of the source but I just cannot see it now. The documentation states: Each row of the resulting matrix consists of sequences x[t], x[t-1], ..., x[t-dimension+1], where t is the original index of x. If x is a matrix, i.e., x contains more than one variable, then x[t] consists of the tth observation on each variable. This explanation doesn't seem to account for the dimension argument. Thank you for your comments. Kevin __ 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] embed?
Kevin, The documentation is quite clear. What embedding does is that it takes a scalar time series, x[t], and embeds it in a higher-dimensional space of dimension, dimension. The entries in the matrix you see are the indices of the time-series. So, for example, if dimension = 2, you embed your time-series on a 2-Dim space: (x, y), where the points are: (x[2], x[1]), (x[3], x[2]), ..., (x[N], x[N-1]). embed(x, dimension=2) [,1] [,2] [1,]21 [2,]32 [3,]43 [4,]54 [5,]65 [6,]76 [7,]87 [8,]98 [9,] 109 This is allso known as Ruelle-Takens embedding in non-linear dynamical systems, where this device is helpful in detecting the existence of a low-dimensional attractor of the time-series. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: rvarad...@jhmi.edu Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of rkevinbur...@charter.net Sent: Friday, April 03, 2009 11:05 AM To: r-help@r-project.org Subject: [R] embed? I have a question on the function 'embed'. I ran the example x - 1:10 embed(x, dimension=3) This gives the output: [,1] [,2] [,3] [1,]321 [2,]432 [3,]543 [4,]654 [5,]765 [6,]876 [7,]987 [8,] 1098 I don't quite understand the output and why it is useful. First, there are only 8 rows down from 10 and the first element starts with 3. Of course I can think of explanations as to what is occuring but I cannot see how this is useful. I am sure it has application as i see this command used in much of the source but I just cannot see it now. The documentation states: Each row of the resulting matrix consists of sequences x[t], x[t-1], ..., x[t-dimension+1], where t is the original index of x. If x is a matrix, i.e., x contains more than one variable, then x[t] consists of the tth observation on each variable. This explanation doesn't seem to account for the dimension argument. Thank you for your comments. Kevin __ 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] course in ecological data analysis
At 05:27 AM 4/3/2009, Capelle, Jacob wrote: Dear all, For my PhD study I'm looking for relevant courses/workshops (short term) in ecological data anlysis with R in Europe. After 2 days searching I'm convinced that google is probably not the right medium to find this information. If anyone can help me I will be most grateful. Best regards - J. Capelle Try http://www.statistics.com/ Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: r...@lcfltd.com Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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] Curve fitting,FDA for biological data
Here is the gif that didn't come through earlier http://www.nabble.com/file/p22870832/signal.gif signal.gif -- View this message in context: http://www.nabble.com/Curve-fitting%2CFDA-for-biological-data-tp22868069p22870832.html Sent from the R help mailing list archive at Nabble.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] embed?
rkevinbur...@charter.net wrote: I have a question on the function 'embed'. I ran the example x - 1:10 embed(x, dimension=3) This gives the output: [,1] [,2] [,3] [1,]321 [2,]432 [3,]543 [4,]654 [5,]765 [6,]876 [7,]987 [8,] 1098 I don't quite understand the output and why it is useful. First, there are only 8 rows down from 10 and the first element starts with 3. Of course I can think of explanations as to what is occuring but I cannot see how this is useful. I am sure it has application as i see this command used in much of the source but I just cannot see it now. The documentation states: Each row of the resulting matrix consists of sequences x[t], x[t-1], ..., x[t-dimension+1], where t is the original index of x. If x is a matrix, i.e., x contains more than one variable, then x[t] consists of the tth observation on each variable. This explanation doesn't seem to account for the dimension argument. following this 'explanation', the first row consists of values x[t], x[t-1], ... x[t-3+1], that is, x[t], x[t-1], x[t-2]. how does t, the original index of x, relate to positions in the matrix? does it correspond to the row number, or the column number? it can't be the former, because then the first row would include x[1], x[0], x[-1] -- nonsense. it can't be the latter, because the first row would include x[1], x[1], x[1] (nonsense), and so all other rows (nonsense). for a vector, say x, the output, say matrix, contains values calculated as follows: m[i,j] = x[i + dimension - j], with i = 1, ..., length(x)-dimension+1 and j = 1, ..., dimension so that you have a rolling window over the vector, with row indices corresponding to the start of the window, and column indices corresponding to the position within the window. arguably, the authors could have done their homework better. vQ __ 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] Curve fitting,FDA for biological data
What is your end goal? If it is to try and account for the variability of the timeseries you may want to look at ?spectrum If it is to model the periodicity... Stephen Sefick On Fri, Apr 3, 2009 at 11:30 AM, trias t.gkikopou...@dundee.ac.uk wrote: Here is the gif that didn't come through earlier http://www.nabble.com/file/p22870832/signal.gif signal.gif -- View this message in context: http://www.nabble.com/Curve-fitting%2CFDA-for-biological-data-tp22868069p22870832.html Sent from the R help mailing list archive at Nabble.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. -- Stephen Sefick 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 __ 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] embed?
Hi kevin: one use ( there are probably many others ) is for use inside a vector autoregression model where the RHS is lags of the independent variable. So, x_t is the 3rd column, x_t-1 is the second, x_t-2 is the third etc. On Apr 3, 2009, rkevinbur...@charter.net wrote: I have a question on the function 'embed'. I ran the example x - 1:10 embed(x, dimension=3) This gives the output: [,1] [,2] [,3] [1,] 3 2 1 [2,] 4 3 2 [3,] 5 4 3 [4,] 6 5 4 [5,] 7 6 5 [6,] 8 7 6 [7,] 9 8 7 [8,] 10 9 8 I don't quite understand the output and why it is useful. First, there are only 8 rows down from 10 and the first element starts with 3. Of course I can think of explanations as to what is occuring but I cannot see how this is useful. I am sure it has application as i see this command used in much of the source but I just cannot see it now. The documentation states: Each row of the resulting matrix consists of sequences x[t], x[t-1], ..., x[t-dimension+1], where t is the original index of x. If x is a matrix, i.e., x contains more than one variable, then x[t] consists of the tth observation on each variable. This explanation doesn't seem to account for the dimension argument. Thank you for your comments. Kevin __ [1]r-h...@r-project.org mailing list [2]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [3]http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. References 1. mailto:R-help@r-project.org 2. https://stat.ethz.ch/mailman/listinfo/r-help 3. http://www.R-project.org/posting-guide.html __ 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] Help pasting string as object name
What a very useful package! Thanks for pointing out its existence.
Sadly ?melt is basically useless, but I did find the following quasi-vignette
by the author of the reshape package to be quite useful.
http://www.jstatsoft.org/v21/i12/paper
Cheers.
-Original Message-
From: ba...@exeter.ac.uk
Sent: Fri, 3 Apr 2009 11:02:31 +0100
To: bearmarketsr...@inbox.com
Subject: Re: [R] Help pasting string as object name
Hi,
Is this what you want?
d - data.frame(density.AL = seq(1, 10),
density.AK = seq(1, 10), # many others...
Date=letters[1:10]) # dummy example
library(reshape)
melt(subset(d, Date == b), id=Date)
BTW, I spotted a few awkward things in your code,
st - c(AL, AK)
vars - paste(d$density, st, sep=.) # easier than mapply etc.
more importantly, in the for loop you should not be incrementing i
manually (as in a while loop), it's already taken care of by the for{}
construct.
HTH,
baptiste
On 3 Apr 2009, at 10:40, Rob Denniker wrote:
I have a data frame containing monthly observations of the 'density'
of each US state, recorded in variables named density.AL,
density.AK, density.AZ, and so on for all 50 states. The data
frame (called d) also contains a variable called Date which is
encoded as a string in the format Jan-09, Feb-09, etc.
I also have a vector st.list-c(AL, AK, AZ,...) of length 50.
I would like a new dataframe with st.list in one column and the
value of d$density for that state when d$Date==Feb-09 in another
column.
How can I do this?
Here is what I have tried:
names - rep(d$density,length(st.list))
templist - as.vector(mapply(paste, names, st.list ,sep=.))
d.2-data.frame()
for (i in 1:length(templist)) {
d.2$density[i] - subset(parse(file=,templist[i]),d$Date==Feb-09)
i-i+1 } ### hangs!
Thanks for any help!
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_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] Fit unequal variance model in R
Feng Jingyu wrote: I used gls and it still does not provide me different estimates of variance for each treatment group. Did I do anything wrong? lm3-gls(GSI~treatment,data=z,weights=varIdent(form=~treatment),method=ML) try weights = varIdent(form~1|treatment) See the example in library/nlme/scripts/ch05.r ,fm1Orth.gls Dieter -- View this message in context: http://www.nabble.com/Fit-unequal-variance-model-in-R-tp22829549p22873236.html Sent from the R help mailing list archive at Nabble.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] Constrined dependent optimization.
I have decided to use this SANN approach to my problem but to keep the run time
reasonable instead of 20,000 variables I will randomly sample this space to get
the number of variables under 100. But I want to do this a number of times. Is
there someone who could help me set up WINBUGS to repeat this optimization N
times each time randomly picking 100 of the possible 20,000.
Comments?
Kevin
Paul Smith phh...@gmail.com wrote:
Apparently, the convergence is faster if one uses this new swap function:
swapfun - function(x,N=100) {
loc - c(sample(1:(N/2),size=1,replace=FALSE),sample((N/2):100,1))
tmp - x[loc[1]]
x[loc[1]] - x[loc[2]]
x[loc[2]] - tmp
x
}
It seems that within 20 millions of iterations, one gets the exact
optimal solution, which does not take too long.
Paul
On Mon, Mar 30, 2009 at 5:11 PM, Paul Smith phh...@gmail.com wrote:
Optim with SANN also solves your example:
---
f - function(x) sum(c(1:50,50:1)*x)
swapfun - function(x,N=100) {
loc - sample(N,size=2,replace=FALSE)
tmp - x[loc[1]]
x[loc[1]] - x[loc[2]]
x[loc[2]] - tmp
x
}
N - 100
opt1 -
optim(fn=f,par=sample(1:N,N),gr=swapfun,method=SANN,control=list(maxit=5,fnscale=-1,trace=10))
opt1$par
opt1$value
---
We need to specify a large number of iterations to get the optimal
solution. The objective function at the optimum is 170425, and one
gets a close value with optim and SANN.
Paul
On Mon, Mar 30, 2009 at 2:22 PM, Hans W. Borchers
hwborch...@googlemail.com wrote:
Image you want to minimize the following linear function
f - function(x) sum( c(1:50, 50:1) * x / (50*51) )
on the set of all permutations of the numbers 1,..., 100.
I wonder how will you do that with lpSolve? I would simply order
the coefficients and then sort the numbers 1,...,100 accordingly.
I am also wondering how optim with SANN could be applied here.
As this is a problem in the area of discrete optimization resp.
constraint programming, I propose to use an appropriate program
here such as the free software Bprolog. I would be interested to
learn what others propose.
Of course, if we don't know anything about the function f then
it amounts to an exhaustive search on the 100! permutations --
probably not a feasible job.
Regards, Hans Werner
Paul Smith wrote:
On Sun, Mar 29, 2009 at 9:45 PM, rkevinbur...@charter.net wrote:
I have an optimization question that I was hoping to get some suggestions
on how best to go about sovling it. I would think there is probably a
package that addresses this problem.
This is an ordering optimzation problem. Best to describe it with a
simple example. Say I have 100 bins each with a ball in it numbered
from 1 to 100. Each bin can only hold one ball. This optimization is that
I have a function 'f' that this array of bins and returns a number. The
number returned from f(1,2,3,4) would return a different number from
that of f(2,1,3,4). The optimization is finding the optimum order of
these balls so as to produce a minimum value from 'f'.I cannot use the
regular 'optim' algorithms because a) the values are discrete, and b) the
values are dependent ie. when the variable representing the bin
location is changed (in this example a new ball is put there) the
existing ball will need to be moved to another bin (probably swapping
positions), and c) each variable is constrained, in the example above
the only allowable values are integers from 1-100. So the problem becomes
finding the optimum order of the balls.
Any suggestions?
If your function f is linear, then you can use lpSolve.
Paul
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[R] Conversions From standard to metric units
I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. Is there a package in R for this already? If not I believe that I am going to write some of the most often used in function form. My question is should I include this in my StreamMetabolism package. It is not along the same theme lines, but could loosely fit. The reason that I ask is that I don't want to clutter CRAN with a small package containing some conversion functions because I am to lazy to source them into R every time that I use them, but I also don't want the StreamMetabolism package to turn into StephenMisc Fuctions. Thoughts, comments, or suggestions would be appreciated. -- Stephen Sefick 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 __ 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] Trouble extracting graphic results from a bootstrap
On Apr 3, 2009, at 7:10 AM, Anders Bjorn wrote:
Hi,
I'm trying to extract a histogram over the results from a bootstrap.
However
I keep receiving the error message Error in hist.default(boot.lrtest
$ll,
breaks = scott) : 'x' must be numeric.
The bootstrap I'm running looks like:
boot.test - function(data, indeces, maxit=20) {
+ y1 - fit1+e1[indeces]
+ mod1 - glm(y1 ~ X1-1, maxit=maxit)
+ y2 - fit2+e2[indeces]
+ mod2 - glm(y2~1, maxit=maxit)
+ ll - 2*(logLik(mod1)-logLik(mod2))
+ ll
+ }
boot.lrtest - boot(data=M1, statistic=boot.test, R=2000, maxit=100);
hist(boot.lrtest$ll, breaks=scott) # This results in the error
message
stated above
So my question is: what am I doing wrong?
At least two things, possibly 3 or 4:
1) not specifying from which package the function boot comes from,
and ..
2) not offering a reproducible example. and possibly ..
3) assuming that boot.lrtest$ll will have any meaning, with the
corollary..
4) not looking at boot.lrtest$ll with the str function.
Regarding 3), when I look at the documentation for boot in package
boot, I see no component of the value returned from a boot call that
has the name ll. The name ll will be thrown away after the
function call and you need to refer to components of the boot object
by the names that the developers used.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Updating a data frame
To my earlier question about updating a dataframe, and certainty that this
has been solved several times before, Dr. Winsemius suggests (Thanks!):
I am sure this is not the most elegant method, but it will work.
new - merge(nn,uu, by = c(a,b), all.x=T)
new$y - with( new, (ifelse(!is.na(x.y), y.y, y.x) ) )
new$x - with( new, (ifelse(!is.na(x.y), x.y, x.x) ) )
new[ , c(a, b, x, y)]
a b x y
1 1 1 100.000 -100.
2 1 2 0.76826360.28821953
3 1 3 0.62744360.08373154
4 2 1 0.27503190.55738251
5 2 2 200.000 -100.
6 2 3 0.19193330.88803516
7 3 1 0.36065590.72215405
8 3 2 0.93962690.10943878
9 3 3 300.000 -100.
10 4 1 400.000 -100.
11 4 2 0.56332170.63063714
12 4 3 0.37129220.40779830
That was one of the methods I've considered, but the with() function
cleans it up considerably. I'll have to get more familiar with that.
I've also tried a merge()-less update that seems to work. Still not quite
as transparent as an SQL UPDATE or SAS MERGE step. Can anyone suggest an
improvement or alternate to either method?
nn - expand.grid('a'=1:4, 'b'=1:3)
nn$x - runif(nrow(nn))
nn$y - runif(nrow(nn))
uu - rbind(data.frame('a'=1, 'b'=1, 'x'=100, 'y'=-100)
,data.frame('a'=2, 'b'=2, 'x'=200, 'y'=-100)
,data.frame('a'=4, 'b'=1, 'x'=400, 'y'=-100)
,data.frame('a'=3, 'b'=3, 'x'=300, 'y'=-100)
)
# This works, but it's sensitive to the order of the updates in uu.
corr - nn
corr[paste(corr$a, corr$b) %in% paste(uu$a, uu$b),]$x - uu$x
corr[paste(corr$a, corr$b) %in% paste(uu$a, uu$b),]$y - uu$y
# try updates in a different order, and it puts the updates in the wrong
rows
uu-uu[c(2,4,3,1),]
corr - nn
corr[paste(corr$a, corr$b) %in% paste(uu$a, uu$b),]$x - uu$x
corr[paste(corr$a, corr$b) %in% paste(uu$a, uu$b),]$y - uu$y
# ordering the dataframes first seems to work well.
ord.uu - uu[order(uu$a, uu$b),]
corr - nn[order(nn$a, nn$b),]
corr[paste(corr$a, corr$b) %in% paste(ord.uu$a, ord.uu$b),]$x - ord.uu$x
corr[paste(corr$a, corr$b) %in% paste(ord.uu$a, ord.uu$b),]$y - ord.uu$y
Enjoy the days,
cur
--
Curt Seeliger, Data Ranger
Raytheon Information Services - Contractor to ORD
seeliger.c...@epa.gov
541/754-4638
[[alternative HTML version deleted]]
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Re: [R] Conversions From standard to metric units
stephen sefick wrote: I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. Is there a package in R for this already? If not I believe that I am going to write some of the most often used in function form. My question is should I include this in my StreamMetabolism package. It is not along the same theme lines, but could loosely fit. The reason that I ask is that I don't want to clutter CRAN with a small package containing some conversion functions because I am to lazy to source them into R every time that I use them, but I also don't want the StreamMetabolism package to turn into StephenMisc Fuctions. Thoughts, comments, or suggestions would be appreciated. RSiteSearch turns up the MedUnits package; maybe your conversions are already there, or would fit within that package. 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.
[R] Removing backslashes from data
I am trying to check for backslashes in data, then remove them when I find them, but am having a difficult time figuring out the best way to do it. I know the backslash is the escape character in R, and I should be able to use 'gsub' to accomplish this, but I all I seem to be getting are errors. For example: If entry is: Hello\World I want: HelloWorld There are several entries with backslashes, and I need to find them and delete them. All help is welcome, thank you. ___ Drew Conway Ph.D. Student Department of Politics, New York University agc...@nyu.edu http://homepages.nyu.edu/~agc282 [[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] Fit unequal variance model in R
Feng Jingyu wrote: Thanks a lot. The problem is solved. It took me a while to understand the output from the R. With little calculation, I am able to match results from R to SAS. To conserve you sanity, don't try it. They will be different. Dieter -- View this message in context: http://www.nabble.com/Fit-unequal-variance-model-in-R-tp22829549p22873997.html Sent from the R help mailing list archive at Nabble.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] Conversions From standard to metric units
I had a similar need for conversions in optics. I put together several functions and data on r-forge, where it does not clutter CRAN but can still be shared conveniently with others. baptiste On 3 Apr 2009, at 19:27, Duncan Murdoch wrote: stephen sefick wrote: I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. Is there a package in R for this already? If not I believe that I am going to write some of the most often used in function form. My question is should I include this in my StreamMetabolism package. It is not along the same theme lines, but could loosely fit. The reason that I ask is that I don't want to clutter CRAN with a small package containing some conversion functions because I am to lazy to source them into R every time that I use them, but I also don't want the StreamMetabolism package to turn into StephenMisc Fuctions. Thoughts, comments, or suggestions would be appreciated. RSiteSearch turns up the MedUnits package; maybe your conversions are already there, or would fit within that package. 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. _ Baptiste Auguié School of Physics University of Exeter Stocker Road, Exeter, Devon, EX4 4QL, UK Phone: +44 1392 264187 http://newton.ex.ac.uk/research/emag __ 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] data.frame to array?
I have a list of data.frames str(bins) List of 19217 $ 100026:'data.frame': 1 obs. of 6 variables: ..$ Sku : chr 100026 ..$ Bin : chr T149C ..$ Count: int 108 ..$ X: int 20 ..$ Y: int 149 ..$ Z: chr 3 $ 100030:'data.frame': 1 obs. of 6 variables: ... As you can see one 'column' is Count. This list seems to contain 19217 data.frames. I would like to create an array of 19217 integers which hold the values of the Count column. I have tried the obvious (to me): bins[[1:3]]$Count But that returns NULL instead of an array of length 3 that I was expecting. Interestingly bins[[1]]$Count returns the first Count in the list of data frames. How do I get all of the Counts? Thank you. Kevin __ 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] Working with and without formula in function
I am attempting to write a function that is flexible enough to respond to the
user providing a formula (with a data= argument) or not (similar to
plot(x,y) versus plot(y~x,data=data)). I have found a method to work with
this in a simple case but am having trouble determining how to find a
variable from within the data= argument that is not part of the formula.
The following illustrates my problem ...
# an illustrative function -- I know that plotrix::thigmophobe.labels() does
what this function does
myplot - function(x,y=NULL,data=NULL,label=NULL) {
if (class(x)==formula) {
mf - model.frame(x,data=data)
x - mf[,2]
y - mf[,1]
}
if (is.null(y)) stop(Y-axis variable is missing)
plot(x,y)
if (!is.null(label)) text(x,y,label)
}
# dummy data
df -
data.frame(x=runif(10),y=runif(10),grp=factor(rep(c(Yes,No),each=5)))
# both calls work as expected
with(df,myplot(x,y))
myplot(y~x,data=df)
# only first works as I would hope
with(df,myplot(x,y,label=grp))
myplot(y~x,data=df,label=grp)
# this works but is clumsy
myplot(y~x,data=df,label=df$grp)
Any help with how to make this function recognize the grp variable in df
without having to type df$grp when supplying it to the label= argument
would be greatly appreciated. Thank you in advance.
--
View this message in context:
http://www.nabble.com/Working-with-and-without-formula-in-function-tp22874005p22874005.html
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and provide commented, minimal, self-contained, reproducible code.
[R] dendrogram rect.hclust() not working?
I have tried to use rect.hclust() to draw a rectangle around a set of
leaves, but am running into trouble.
The rect.hclust() is drawing two rects instead of one, and of the
wrong size:
scoreClusterObj - hclust(scoreDistanceObj, method=clustMethod)
order - scoreClusterObj$order
orderedLabels - rep(0, length(order))
for (orderIndex in 1:length(order)) {
# this puts a name to the permutation of leaves, done by hclust()
orderedLabels[orderIndex] - classes[order[orderIndex]]
}
scoreDendrogramObj - as.dendrogram(scoreClusterObj)
coloredLeafScoreDendrogramObj - dendrapply(scoreDendrogramObj,
markColoredLeaves)
scoreDendrogramPlot - plot(coloredLeafScoreDendrogramObj,
horiz=FALSE, axes=FALSE)
significantClustersInScoreDendrogramObj -
dendrapply(coloredLeafScoreDendrogramObj, markSignificantClusters)
I have the local functions markColoredLeaves() -- which changes the
colors of certain leaves, and it works fine -- and another function
called markSignificantClusters(), in which I try to draw a
rect.hclust() if a condition is met (i.e. a cluster is statistically
significant):
markSignificantClusters - function (n) {
if (!is.leaf(n)) {
a - attributes(n)
leafList - unlist(dendrapply(n, listLabels))
if (nodesContainCertainLeaves) {
ma - match(leafList, orderedLabels)
print (paste (min-ma, min(ma), max-ma, max(ma), sep=
))
r - rect.hclust(scoreClusterObj, h = a$height, which = c(min(ma),
max(ma)), border = 2)
print (r)
quit()
}
}
}
For testing, I have a call to quit() the script after the first
qualifying node has a rect drawn around it.
So I run this script, and when I look at the runtime log output (from
the print() statements), it finds the correct, qualifying node
containing the following two items:
[1] clusters
[1] +v_stat3_01 +v_stat1_01
These two leaves are located at positions 5 and 6 of the tree. This is
correct output from the statistical test. So I should only get one
rect drawn, of width 2, containing leaves 5 and 6.
Also, the ma variable is returning the correct leaf range (between 5
and 6, inclusively), so I know I'm passing the correct leaf range to
the rect.hclust() function.
But in my graphical output, I get two rects at positions 6 and 7:
http://www.flickr.com/photos/alexreynolds/3409263765/sizes/o/
This doesn't seem to be an offset issue, for two reasons:
1. I am getting two rects, not one. Looking at the output from the
print(r) statement, I see why two rects are drawn:
[[1]]
+v_stat1_01
273
[[2]]
+v_e2f1_q6_01
326
2. When I re-run the script, I can occasionally get different cluster
results. In one case, where I should get one rect of size 3
(containing 3 leaves), instead I get two rects containing 2 and 4
leaves, resp. separated by several other clusters.
Worse, if I plot the dendrogram horizontally, the rects are drawn of
completely wrong dimensions:
scoreDendrogramPlot - plot(coloredLeafScoreDendrogramObj, horiz=TRUE,
axes=FALSE)
yields:
http://www.flickr.com/photos/alexreynolds/3410126060/sizes/o/
Is there a way to use rect.hclust() that works reliably (in both
orientations, or even in one orientation)?
Alternatively, is there a way to modify the thickness and color of the
edges that draw down from a significant node?
I tried adding this to my markSignificantClusters() function, within
the if (nodesContainCertainLeaves) block, to modify the edgePar
settings of a qualifying node, to no effect:
n - dendrapply(n, function(e) { attr(e, edgePar) - list(lty=3,
col=red); e })
If you got this far through this message, thanks. :) I would be
grateful for any advice.
Thanks,
Alex
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Re: [R] Removing backslashes from data
Below works but it has two backslashes in the word. maybe someone can explain why the 4 and 2 works but 2 1 doesn't ? thanks. gsub(,,Hello\\World,perl=TRUE) On Apr 3, 2009, Andrew Conway agc...@nyu.edu wrote: I am trying to check for backslashes in data, then remove them when I find them, but am having a difficult time figuring out the best way to do it. I know the backslash is the escape character in R, and I should be able to use 'gsub' to accomplish this, but I all I seem to be getting are errors. For example: If entry is: Hello\World I want: HelloWorld There are several entries with backslashes, and I need to find them and delete them. All help is welcome, thank you. ___ Drew Conway Ph.D. Student Department of Politics, New York University [1]agc...@nyu.edu [2]http://homepages.nyu.edu/~agc282 [[alternative HTML version deleted]] __ [3]r-h...@r-project.org mailing list [4]https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide [5]http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. References 1. mailto:agc...@nyu.edu 2. http://homepages.nyu.edu/~agc282 3. mailto:R-help@r-project.org 4. https://stat.ethz.ch/mailman/listinfo/r-help 5. http://www.R-project.org/posting-guide.html __ 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] Removing backslashes from data
On 4/3/2009 2:33 PM, Andrew Conway wrote: I am trying to check for backslashes in data, then remove them when I find them, but am having a difficult time figuring out the best way to do it. I know the backslash is the escape character in R, and I should be able to use 'gsub' to accomplish this, but I all I seem to be getting are errors. For example: If entry is: Hello\World I want: HelloWorld There are several entries with backslashes, and I need to find them and delete them. All help is welcome, thank you. This gets a little confusing because print() automatically doubles backslashes. To see what's really going on, you need to use cat. Then the following can be seen to work: input - \Hello\\World\ cat(input, \n) output - gsub(, , input) cat(output, \n) Duncan Murdoch ___ Drew Conway Ph.D. Student Department of Politics, New York University agc...@nyu.edu http://homepages.nyu.edu/~agc282 [[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-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] Fit unequal variance model in R
Feng Jingyu wrote: Hi For my purpose, I need to match variance estimates for each group from R and SAS. They do match now. Consider yourself a lucky man! Dieter -- View this message in context: http://www.nabble.com/Fit-unequal-variance-model-in-R-tp22829549p22874048.html Sent from the R help mailing list archive at Nabble.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.
[R] R. A. Fisher Test for value of periodorgam
Is there in some pacakage this test? i mean that input will be only the value of periodogram, and the function will say that are significant. or is there only the way to calcule Fishers statistics W =Vi/V1 + · · · + Vm, for i =1 m and test it step by step ? thanks -- View this message in context: http://www.nabble.com/R.-A.-Fisher-Test-for-value-of-periodorgam-tp22874210p22874210.html Sent from the R help mailing list archive at Nabble.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.
[R] Weighted principal components analysis?
Hello R-ers, I'm trying to do a weighted principal components analysis. I couldn't find any such option with princomp or prcomp. Does anyone know of a package or way to do this? More specifically, the observations I'm working with are averages from populations of varying sizes. I thus need to weight the observations by sample size. Ideally I could apply these weights at the cell level (i.e., allowing sample size to vary within observations across variables), but even applying them just to the observations would get me most of the way there. I'm using R v2.8.1 on Windows XP. I've searched Help and the R site and had no luck. Thanks for any help you can provide. Cheers, Alan Cohen Centre for Global Health Research Toronto, Ontario __ 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] data.frame to array?
On Fri, 2009-04-03 at 11:45 -0700, rkevinbur...@charter.net wrote: I have a list of data.frames str(bins) List of 19217 $ 100026:'data.frame': 1 obs. of 6 variables: ..$ Sku : chr 100026 ..$ Bin : chr T149C ..$ Count: int 108 ..$ X: int 20 ..$ Y: int 149 ..$ Z: chr 3 $ 100030:'data.frame': 1 obs. of 6 variables: ... As you can see one 'column' is Count. This list seems to contain 19217 data.frames. I would like to create an array of 19217 integers which hold the values of the Count column. I have tried the obvious (to me): bins[[1:3]]$Count sapply(bins, `[[`, Count) will get you this as a vector. HTH G But that returns NULL instead of an array of length 3 that I was expecting. Interestingly bins[[1]]$Count returns the first Count in the list of data frames. How do I get all of the Counts? Thank you. Kevin __ 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. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% 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.
[R] compute effect sizes (partial eta)
hi, is there a function that calculates and prints out the partial etas of every independent variable in a linear model? thanks for any help! __ 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] data.frame to array?
I do not think that the form [[1:3]] is legit. ltest - list( a, b, c, d) ltest[[1:3]] Error in ltest[[1:3]] : recursive indexing failed at level 2 You might try with single brackets: ltest[1:3] [[1]] [1] a [[2]] [1] b [[3]] [1] c -- David Winsemius On Apr 3, 2009, at 2:45 PM, rkevinbur...@charter.net wrote: I have a list of data.frames str(bins) List of 19217 $ 100026:'data.frame': 1 obs. of 6 variables: ..$ Sku : chr 100026 ..$ Bin : chr T149C ..$ Count: int 108 ..$ X: int 20 ..$ Y: int 149 ..$ Z: chr 3 $ 100030:'data.frame': 1 obs. of 6 variables: ... As you can see one 'column' is Count. This list seems to contain 19217 data.frames. I would like to create an array of 19217 integers which hold the values of the Count column. I have tried the obvious (to me): bins[[1:3]]$Count But that returns NULL instead of an array of length 3 that I was expecting. Interestingly bins[[1]]$Count returns the first Count in the list of data frames. How do I get all of the Counts? Thank you. Kevin __ 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] data.frame to array?
On Fri, Apr 3, 2009 at 1:45 PM, rkevinbur...@charter.net wrote: I have a list of data.frames str(bins) List of 19217 $ 100026:'data.frame': 1 obs. of 6 variables: ..$ Sku : chr 100026 ..$ Bin : chr T149C ..$ Count: int 108 ..$ X : int 20 ..$ Y : int 149 ..$ Z : chr 3 $ 100030:'data.frame': 1 obs. of 6 variables: ... As you can see one 'column' is Count. This list seems to contain 19217 data.frames. I would like to create an array of 19217 integers which hold the values of the Count column. I have tried the obvious (to me): bins[[1:3]]$Count But that returns NULL instead of an array of length 3 that I was expecting. Interestingly bins[[1]]$Count returns the first Count in the list of data frames. How do I get all of the Counts? Why not turn your list of data frames into a single data frame? bindf - do.call(rbind, bins) bindf$Count Hadley -- http://had.co.nz/ __ 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] US county map question
Hi R-help: I'm just an old guy and new to this list... But have been using R for years now. I want to make a map of counties in the US with shaded colors that depend on the level of variable Y that I want to map. I have the US county and state fips codes and the Y variable. How do I do this? Please reply to p...@cdc.gov Thanks, Phil Smith p...@cdc.gov [[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] data.frame to array?
David Winsemius wrote: I do not think that the form [[1:3]] is legit. sure it is. ltest - list( a, b, c, d) ltest[[1:3]] Error in ltest[[1:3]] : recursive indexing failed at level 2 read the error message: *recursive* indexing failed. that's because ltest[[1]] has only one element while you wanted its second element (and that element's third element). ltest = list(list(2, as.list(1:3))) ltest[[1:3]] # 3 is just fine. vQ __ 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] Discriminant Analysis - Obtaining Classification Functions
Hello! I need some help with the linear discriminant analysis in R. I have some plant samples (divided into several groups) on which I measured a few quantitative characteristics. Now, I need to infer some classification rules usable for identifying new samples. I have used the function lda from the MASS library in a usual fashion: lda.1 - lda(groups~char1+char2+char3, data=xxx) I'd like to obtain the classification functions for the particular groups, with the aid of which I could classify unknown samples. I know I can use predict.lda to classify such samples, but I need to obtain some equations into which I could simply put the measured values of an unknown sample manually and the result would predict which group the sample most probably belongs to (like in eg. STATISTICA). I haven't found out how to extract these functions from the lda output. Could somebody give me some advice? Thank you in advance, Pavel Kur __ 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] Changing the y-axis units of a lattice histogram
Hello, I need to multiply the number of counts in the y-axis of a lattice histogram by a constant factor, such that the plot would represent a different type of variable plotted in the y-axis, not counts. Can this be done? Thank you, Judith __ 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] Changing the y-axis units of a lattice histogram
Try: library(lattice) histogram( ~ height | voice.part, data = singer, type = c, scales = list(y = list(at = seq(0, 20, 5), labels = seq(0, 200, 50 HTH, --sundar On Fri, Apr 3, 2009 at 2:01 PM, Judith Flores jur...@yahoo.com wrote: Hello, I need to multiply the number of counts in the y-axis of a lattice histogram by a constant factor, such that the plot would represent a different type of variable plotted in the y-axis, not counts. Can this be done? Thank you, Judith __ 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] Multiple use of par()
Hi all,
I created a plot function which used par(mfcol=c(2,1)) so that I could
have two plots together using just one command.
For exampe:
plot.foo - function(data){
par(mfcol=c(2,1))
hist(data)
plot(data)
}
Later I wanted to show 4 of these foo objects in the same picture. So
I used par(mfcol=c(2,2)) again at the beginning of the code like:
par(mfcol=c(2,2))
plot(foo.1)
plot(foo.2)
plot(foo.3)
plot(foo.4)
but this time the par() command inside of the functions seem to be
overwriting the par() command at the very begining. Can anyone please
give me some advise on dealing with this? I guess that I may either
need to change the way I plot foo, e.g. using some function rather
than par(), or use some parameters at the beginning. Thank you very
much,
Best wishes,
--
彭河森 Hesen Peng
http://hesen.peng.googlepages.com/
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Re: [R] Weighted principal components analysis?
see the row.w argument of the function dudi.pca in the ade4 package. Cheers. Alan Cohen wrote: Hello R-ers, I'm trying to do a weighted principal components analysis. I couldn't find any such option with princomp or prcomp. Does anyone know of a package or way to do this? More specifically, the observations I'm working with are averages from populations of varying sizes. I thus need to weight the observations by sample size. Ideally I could apply these weights at the cell level (i.e., allowing sample size to vary within observations across variables), but even applying them just to the observations would get me most of the way there. I'm using R v2.8.1 on Windows XP. I've searched Help and the R site and had no luck. Thanks for any help you can provide. Cheers, Alan Cohen Centre for Global Health Research Toronto, Ontario __ 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. -- Stéphane DRAY (d...@biomserv.univ-lyon1.fr ) Laboratoire BBE-CNRS-UMR-5558, Univ. C. Bernard - Lyon I 43, Bd du 11 Novembre 1918, 69622 Villeurbanne Cedex, France Tel: 33 4 72 43 27 57 Fax: 33 4 72 43 13 88 http://pbil.univ-lyon1.fr/members/dray/ __ 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] Linear model, finding the slope
I'm not sure what you are doing when you Normalize. Would you explain? To see if the slope is significant, look at the model summary, in your example: summary(model) Charles Annis, P.E. charles.an...@statisticalengineering.com phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Melissa2k9 Sent: Friday, April 03, 2009 5:51 AM To: r-help@r-project.org Subject: [R] Linear model, finding the slope Hi for some data I working on I am merely plotting time against temperature for a variable named filmclip. So for example, I have volunteers who watched various film clips and have used infared camera to monitor the temperature on their face at every second of the clip. The variable names I have used are Normalised ( for the temperature) and Frame (for the time in seconds). So I have fitted a linear model model-lm(Normalised~Frame,data=All,subset=((Subject==1)(Filmclip==Whateve r) and coef(model) gives me an intercept value and a value for the slope. Now what I want to do is find out if the slope is significant or not. So far I just have values such as 0.02211 for example and have no idea if this is to be interpreted as significant or not. Sorry if I haven't been clear but any advice on how to find out what values are significant would be greatly appreciated. -- View this message in context: http://www.nabble.com/Linear-model%2C-finding-the-slope-tp22865254p22865254. html Sent from the R help mailing list archive at Nabble.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. __ 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] Discriminant Analysis - Obtaining Classification Functions
reauire(MASS) ; ?predict.lda should enlighten you. Glancing at VR4 might be a bit more illuminating... HTH Emmanuel Charpentier Le vendredi 03 avril 2009 à 22:29 +0200, Pavel Kúr a écrit : Hello! I need some help with the linear discriminant analysis in R. I have some plant samples (divided into several groups) on which I measured a few quantitative characteristics. Now, I need to infer some classification rules usable for identifying new samples. I have used the function lda from the MASS library in a usual fashion: lda.1 - lda(groups~char1+char2+char3, data=xxx) I'd like to obtain the classification functions for the particular groups, with the aid of which I could classify unknown samples. I know I can use predict.lda to classify such samples, but I need to obtain some equations into which I could simply put the measured values of an unknown sample manually and the result would predict which group the sample most probably belongs to (like in eg. STATISTICA). I haven't found out how to extract these functions from the lda output. Could somebody give me some advice? Thank you in advance, Pavel Kur __ 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] [OT ?] rant (was : Re: Conversions From standard to metric units)
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit : I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. rant US/Imperial units are *not* standard units. The former metric system is now called Système International (International System) for a reason, which is *not* gallocentrism of a few 6e7 frogs, but rather laziness of about 5.6e9 losers who refuse to load their memories with meaningless conversion factors... /rant Emmanuel Charpentier who has served his time with pounds per cubic feet, furlongs per fortnight, BTU and other figments of British/American sadistic imagination, thank you very much... /rant # Again, didn't work the first time... Is there a package in R for this already? If not I believe that I am going to write some of the most often used in function form. My question is should I include this in my StreamMetabolism package. It is not along the same theme lines, but could loosely fit. The reason that I ask is that I don't want to clutter CRAN with a small package containing some conversion functions because I am to lazy to source them into R every time that I use them, but I also don't want the StreamMetabolism package to turn into StephenMisc Fuctions. Thoughts, comments, or suggestions would be appreciated. __ 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] [OT ?] rant (was : Re: Conversions From standard to metric units)
On 4/04/2009, at 10:37 AM, Emmanuel Charpentier wrote: Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit : I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. rant US/Imperial units are *not* standard units. The former metric system is now called Système International (International System) for a reason, which is *not* gallocentrism of a few 6e7 frogs, but rather laziness of about 5.6e9 losers who refuse to load their memories with meaningless conversion factors... /rant snip Right on, Red Freak!!! cheers, Rolf ## Attention: This e-mail message is privileged and confidential. If you are not the intended recipient please delete the message and notify the sender. Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshal www.marshalsoftware.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] [OT ?] rant (was : Re: Conversions From standard to metric units)
On 03/04/2009 5:37 PM, Emmanuel Charpentier wrote: Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit : I am starting to use R for almost any sort of calculation that I need. I am a biologist that works in the states, and there is often a need to convert from standard units to metric units. rant US/Imperial units are *not* standard units. But they are fun: you should see the arguments you can have about whether imperial fluid ounces are the same volume as US fluid ounces. (They're not: US ounces are bigger. But not big enough so that their gallons catch up!) Duncan Murdoch The former metric system is now called Système International (International System) for a reason, which is *not* gallocentrism of a few 6e7 frogs, but rather laziness of about 5.6e9 losers who refuse to load their memories with meaningless conversion factors... /rant Emmanuel Charpentier who has served his time with pounds per cubic feet, furlongs per fortnight, BTU and other figments of British/American sadistic imagination, thank you very much... /rant # Again, didn't work the first time... Is there a package in R for this already? If not I believe that I am going to write some of the most often used in function form. My question is should I include this in my StreamMetabolism package. It is not along the same theme lines, but could loosely fit. The reason that I ask is that I don't want to clutter CRAN with a small package containing some conversion functions because I am to lazy to source them into R every time that I use them, but I also don't want the StreamMetabolism package to turn into StephenMisc Fuctions. Thoughts, comments, or suggestions would be appreciated. __ 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] [OT ?] rant (was : Re: Conversions From standard to metricunits)
For science yes. For pleasure I'll still take a pint instead of 570ml!
Murray
- Original Message -
From: Rolf Turner r.tur...@auckland.ac.nz
To: Emmanuel Charpentier charp...@bacbuc.dyndns.org
Cc: r-h...@stat.math.ethz.ch
Sent: Friday, April 03, 2009 6:18 PM
Subject: Re: [R] [OT ?] rant (was : Re: Conversions From standard to
metricunits)
On 4/04/2009, at 10:37 AM, Emmanuel Charpentier wrote:
Le vendredi 03 avril 2009 à 14:17 -0400, stephen sefick a écrit :
I am starting to use R for almost any sort of calculation that I need.
I am a biologist that works in the states, and there is often a need
to convert from standard units to metric units.
rant
US/Imperial units are *not* standard units. The former metric system
is now called Système International (International System) for a
reason, which is *not* gallocentrism of a few 6e7 frogs, but rather
laziness of about 5.6e9 losers who refuse to load their memories with
meaningless conversion factors...
/rant
snip
Right on, Red Freak!!!
cheers,
Rolf
##
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[R] error in trmesh (alphahull package)
Hello R community, I have cross-posted with r-sig-geo as this issue could fall under either interest group I believe. I just came accross the alphahull package and am very pleased I may not need to use CGAL anymore for this purpose. However, I am having a problem computing alpha shapes with my point data, and it seems to have to do with the spatial configuration of my points (which form long, skinny, contiguous clusters). If I use rnorm to generate the same sample size with the same mean and SD, I can compute the alpha shape without trouble (which is why I think it might have to do with the spatial arrangement of my points). See below: # first here's a quick sample of my data (UTM coordinates) xcoords[1:5] [1] 670080.2 670080.2 670080.2 670080.2 670080.2 ycoords[1:5] [1] 5005501 5005499 5005498 5005497 5005495 xcoords[1:5] #try the ashape routine with error alpha.shape-ashape(xcoords,ycoords,15) Error in tri.mesh(X) : error in trmesh #get statistics to generate a similar dataset to test against length(xcoords) [1] 26257 length(ycoords) [1] 26257 mean(xcoords) [1] 670462.4 mean(ycoords) [1] 5005382 sd(xcoords) [1] 149.3114 sd(ycoords) [1] 181.5950 #generate the test data xtest-rnorm(26257,670462.4,149.3) ytest-rnorm(26257,5005382,181.60) # try ashape routine with success alpha.shape-ashape(xtest,ytest,15) class(alpha.shape) [1] ashape Thanks for any insight into this! Murray ps I am able to compute the alpha shapes for this same dataset without problem using CGAL but I find it a pain to work with __ 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] [OT ?] rant (was : Re: Conversions From standard to metricunits)
G'day Murray, On Fri, 3 Apr 2009 20:01:30 -0400 Murray Cooper myrm...@earthlink.net wrote: For science yes. For pleasure I'll still take a pint instead of 570ml! And you might experience a disappointment then if you order it in the US where the pint is apparently 450ml; at least, I won several bets on whether a pint is less or more than half a litre against American visitors The stake being, of course, the next round of drinks. :) And, if memory serves correctly, here in Singapore I once ordered a pint of draught beer and was served half a litre... Confused me completely... That's when I thought that the metric system is so much safer until I ordered my next beer, according to the menu a bottled beer of 400ml, it came in a 333ml bottle... Cheers, Berwin === Full address = Berwin A TurlachTel.: +65 6515 4416 (secr) Dept of Statistics and Applied Probability+65 6515 6650 (self) Faculty of Science FAX : +65 6872 3919 National University of Singapore 6 Science Drive 2, Blk S16, Level 7 e-mail: sta...@nus.edu.sg Singapore 117546http://www.stat.nus.edu.sg/~statba __ 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.
