Re: [R] lm and time series.
That is the thing. As a new comer to 'R' I don't understand how to write a
formula when all I have is a time series. I don't know how to express the
independent and dependent variables in a formula when the object is a time
series. So please just solve this simple example and I will extrapolate from
there.
Say the units of the time series is days and the value at each point is the
response. If I wanted to fit a straiight line through the following time series:
y - 4:7
t - ts(y)
So this is saying to me something like 4 units were sold on the first day, 5 on
the second, 6 on the third, and 7 on the fourth.
So given the time series t I want to find the slope and inercept:
y = m*x + b
with x in days and the respoinse would be the number of units sold. I need to
find 'm' and 'b'. If all I have is t (the time series above) then what would be
the formula, and for that matter the arguments to lm to give the desired result?
fit - lm(???)
Thank you.
Kevin
stephen sefick [EMAIL PROTECTED] wrote:
So you want time as the independent variable? Let's say that the
units of y in your first example were seconds- couldn't you just use a
regular lm and say that the units were seconds, minutes, or what ever?
I am probably out of my league here, but I am just not understanding
what it is that you want. a time series is just a series of data
points indexed by time. Arima maybe, or some other cool times series
modeling approach- wavelet, spectral density- for frequency domain
type things... What are you trying to accomplish?
On Fri, Sep 5, 2008 at 5:47 PM, [EMAIL PROTECTED] wrote:
I want to fit a function to time series. If I had:
x - 1:4
y - 1:4
lm(y~x)
This would fit a simple line to the four points. But if it is represented
as a time series
x - 1:4
t - ts(x)
lm()
So I have a time series in the object t. How do I write a formula for lm?
What do I put in the formula for x and y when I only have t (the time
series).
Kevin
stephen sefick [EMAIL PROTECTED] wrote:
what do you want to do?
On Fri, Sep 5, 2008 at 3:22 PM, [EMAIL PROTECTED] wrote:
I am sorry but I looked at ?lm and could not see any guidance on
writting a formula. If I have two arrays or a data set then I know how
to do that (y ~ x) but for a time series I am not sure how to write y or
x.
Thank you.
Kevin
Gabor Grothendieck [EMAIL PROTECTED] wrote:
The Time Series section in ?lm should be self explanatory. If you are
using
diff's and lag's then look at the dyn package.
On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote:
I did a ?lm and it said basically to be careful when using lm and a
time series. But my question is probably more to do with my
inexperience that anything. If I have a time series object 'ti' how
do I write the formula? The response is the value at any particular
time and the time is basically the index of the time series. But I
don't know how to put that into a formula.
Thank you.
Kevin
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--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
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
--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
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
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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] annotating objects in workspace
See ?comment On Fri, 5 Sep 2008, Alexy Khrabrov wrote: Is there a way to associate descriptions with the objects in the workspace, and later retrieve them to know what the object was created for? Thanks, Alexy __ 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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lm and time series.
On Fri, 5 Sep 2008, [EMAIL PROTECTED] wrote: That is the thing. As a new comer to 'R' I don't understand how to write a formula when all I have is a time series. I don't know how to express the independent and dependent variables in a formula when the object is a time series. So please just solve this simple example and I will extrapolate from there. Say the units of the time series is days and the value at each point is the response. If I wanted to fit a straiight line through the following time series: y - 4:7 t - ts(y) So this is saying to me something like 4 units were sold on the first day, 5 on the second, 6 on the third, and 7 on the fourth. So given the time series t I want to find the slope and inercept: y = m*x + b with x in days and the respoinse would be the number of units sold. I need to find 'm' and 'b'. If all I have is t (the time series above) then what would be the formula, and for that matter the arguments to lm to give the desired result? fit - lm(???) y - ts(4:7) lm(y ~ time(y)) And as previously pointed out to you: To preserve the time series properties, look at ?lm, the dynlm and dyn packages. Z Thank you. Kevin stephen sefick [EMAIL PROTECTED] wrote: So you want time as the independent variable? Let's say that the units of y in your first example were seconds- couldn't you just use a regular lm and say that the units were seconds, minutes, or what ever? I am probably out of my league here, but I am just not understanding what it is that you want. a time series is just a series of data points indexed by time. Arima maybe, or some other cool times series modeling approach- wavelet, spectral density- for frequency domain type things... What are you trying to accomplish? On Fri, Sep 5, 2008 at 5:47 PM, [EMAIL PROTECTED] wrote: I want to fit a function to time series. If I had: x - 1:4 y - 1:4 lm(y~x) This would fit a simple line to the four points. But if it is represented as a time series x - 1:4 t - ts(x) lm() So I have a time series in the object t. How do I write a formula for lm? What do I put in the formula for x and y when I only have t (the time series). Kevin stephen sefick [EMAIL PROTECTED] wrote: what do you want to do? On Fri, Sep 5, 2008 at 3:22 PM, [EMAIL PROTECTED] wrote: I am sorry but I looked at ?lm and could not see any guidance on writting a formula. If I have two arrays or a data set then I know how to do that (y ~ x) but for a time series I am not sure how to write y or x. Thank you. Kevin Gabor Grothendieck [EMAIL PROTECTED] wrote: The Time Series section in ?lm should be self explanatory. If you are using diff's and lag's then look at the dyn package. On Fri, Sep 5, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote: I did a ?lm and it said basically to be careful when using lm and a time series. But my question is probably more to do with my inexperience that anything. If I have a time series object 'ti' how do I write the formula? The response is the value at any particular time and the time is basically the index of the time series. But I don't know how to put that into a formula. 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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 -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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. __ 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] request
hi I am expecting the mails that will flow in R- help after subscribing this but could not. Can you provide me the right form. Thanks. [[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] request: most repeated sequnce
Dear R community Hope every one be in best of his/her health. I have a situation in which there are s-sectors. Each sector is further divided into r-rows and c-columns. All it makes an array having dimension (r,c,s). e.g. x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4, 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0) x=array(x,dim=c(3,6,5)) x , , 1 [,1] [,2] [,3] [,4] [,5] [,6] [1,]100000 [2,]100000 [3,]100000 , , 2 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123000 [2,]123000 [3,]120000 , , 3 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]134000 , , 4 [,1] [,2] [,3] [,4] [,5] [,6] [1,]123400 [2,]123400 [3,]123400 , , 5 [,1] [,2] [,3] [,4] [,5] [,6] [1,]120000 [2,]223400 [3,]223400 I want to get the most repeated sequence (row-wise) of values in each sector. e.g. in sector 1 i.e. , , 1 the most repeated sequence is 1 (ignoring zeros). In , , 2 the most repeated sequence is 1 2 3. Similarly in last sector i.e. , , 5 such sequence is 2 2 3 4. Any body can help to solve this problem. Thanks best regards Muhammad Azam [[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] Hopefully an easy error bar question
Hi im trying to add error bars to my barplots, there very basic, i have a few
grapghs where the y variable is different but on all the X variable is Age
(Adult and Juvenile) however this is split into two levels so i have males and
females, so my graph basically has four bars on it.
I know how to add eror bars for instance when there is only one level eg lookng
at the diffrence between male and female lizards and tree height and have used
the code:
error.bars-function(yv,z,nn) {
xv-
barplot(yv,ylim=c(0,(max(yv)+max(z))),names=nn,ylab=deparse(substitute(yv)
))
g=(max(xv)-min(xv))/50
for (i in 1:length(xv)) {
lines(c(xv[i],xv[i]),c(yv[i]+z[i],yv[i]-z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]+z[i], yv[i]+z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]-z[i], yv[i]-z[i]))
}}
Have then worked out the standard errors and added that using:
se-c()
and then added the final bit of code:
labels-as.character(levels(Sex2))
ybar-as.vector(tapply(Height2,Sex2,mean))
error.bars(ybar,se,labels)
So i was wondering, in order to get error bars on my barplot with two levels do
i need to change the last bit of code. For example im looking at mean body
condition (y) against age (x) and sex (level), i tried to change the code to
the following but it didnt work, any suggestions?
labels-as.character(levels(list(sex,Age)))
ybar-as.vector(tapply(ConditionIndex,sex,Age,mean))
error.bars(ybar,se,labels)
[[alternative HTML version deleted]]
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Re: [R] Plotting the complex fft in 3D?
Just another remakr on this thread. I you have time series and think about its fourier transform (EE language) then you should know that the statistical language of that is spectral analysis or maybe frequency domain time-series analysis and the R function to consider should definitely be spectrum() which is a wrapper (among others) to spec.pgram() -- which calls fft() -- for computing the so-called periodogram. If you learn more about the topic, you will learn that in almost all cases you'd consider a *smoothed* version of the periodogram, etc etc (because the so-called *raw* periodgram is *in*consistent as an estimate fo the underlying true spectrum). The Time-Series chapter/section of the MASS book is very helpful here, IIRC. Martin Maechler, ETH Zurich OB == Oliver Bandel [EMAIL PROTECTED] on Thu, 4 Sep 2008 23:16:07 +0200 writes: OB Zitat von Duncan Murdoch [EMAIL PROTECTED]: On 04/09/2008 4:44 PM, Oliver Bandel wrote: Zitat von Duncan Murdoch [EMAIL PROTECTED]: Oliver Bandel wrote: Hello, OB [...] plot3d doesn't support that directly, but you could plot with type='n', then use segments3d to add the lines. BTW: how to change the perspective? I did not found an angle-parameter for the plot3d()-function. Just grab it with your mouse and drag. OB Wow, coool! :-) OB Well, rgl I think gl stands for OpenGl. OB Fine. :-) Alternatively, play3d(spin3d()) will spin it, or par3d(userMatrix=rotationMatrix(...)) for a fixed setting. OB Ok, some thinsg to play with. OB Thank you. OB Ciao, OB Oliver OB __ OB R-help@r-project.org mailing list OB https://stat.ethz.ch/mailman/listinfo/r-help OB PLEASE do read the posting guide http://www.R-project.org/posting-guide.html OB 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] [R-pkgs] New caret packages
New major versions of the caret packages (caret 3.37, caretLSF 1.23 and caretNWS 0.23) have been uploaded to CRAN. caret is a package for building and evaluating a wide variety of predictive models. There are functions for pre-processing, tuning models using resampling, visualizing the results, calculating performance and estimating variable importance. caretNWS and caretLSF are two parallel processing versions that can reduce the training time when multiple compute nodes are available. The project is now hosted on R-Forge. The homepage is http://caret.r-forge.r-project.org/ The package currently includes model tuning/resampling for the following models: lm, single trees (C4.5, rpart, ctree, logistic model trees), mars (via earth), boosted models (ada, gbm, blackboost, glmboost, gamboost, logitboost), bagged models (trees, earth, fda), randomforests (randomforest and cforest), rule-based models (Ripper and M5 prime), discriminant models (lda, fda, rda, ssda, slda), kernel methods (lssvm, ksvm, rvm, gausspr), nnet, nnet with initial pca step, multinom, pls, plsda, gpls, nearest shrunken centroids, the lasso, the elastic net, supervised pca, knn, lvq and NaiveBayes. Recent changes include: - Estimation of class probabilities from PLS discriminant analysis using Bayes rule (in addition to softmax) - Added predict.train and predit.list - More lattice plots to visualize resampling results (xyplot, stripplot, densitplot, histogram) - User-specified performance metrics for resampling - User-specified algorithms for determining the optimal tuning parameters (instead of highest/lowest) - A CHANGES files now exists to track the specifics of the version changes Max ___ R-packages mailing list [EMAIL PROTECTED] 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.
[R] new to R
hi im starting with R.have no idea to start...plz help -- View this message in context: http://www.nabble.com/new-to-R-tp19342903p19342903.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] new to R
On Sat, Sep 6, 2008 at 5:10 AM, sudeshna [EMAIL PROTECTED] wrote: hi im starting with R.have no idea to start...plz help Search the Internet for online tutorials and/or read an introductory book (search for them, e.g., on Amazon.Com). Good luck, Paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] new to R
sudeshna wrote: hi im starting with R.have no idea to start...plz help Hi sudeshna, There are several beginner's guides on the CRAN website. Go to: http://cran.r-project.org and select Contributed (second last option on the left). Jim __ 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] Hopefully an easy error bar question
Brown, Heidi wrote:
Hi im trying to add error bars to my barplots, there very basic, i have a few
grapghs where the y variable is different but on all the X variable is Age
(Adult and Juvenile) however this is split into two levels so i have males and
females, so my graph basically has four bars on it.
I know how to add eror bars for instance when there is only one level eg lookng at the diffrence between male and female lizards and tree height and have used the code:
error.bars-function(yv,z,nn) {
xv-
barplot(yv,ylim=c(0,(max(yv)+max(z))),names=nn,ylab=deparse(substitute(yv)
))
g=(max(xv)-min(xv))/50
for (i in 1:length(xv)) {
lines(c(xv[i],xv[i]),c(yv[i]+z[i],yv[i]-z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]+z[i], yv[i]+z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]-z[i], yv[i]-z[i]))
}}
Have then worked out the standard errors and added that using:
se-c()
and then added the final bit of code:
labels-as.character(levels(Sex2))
ybar-as.vector(tapply(Height2,Sex2,mean))
error.bars(ybar,se,labels)
So i was wondering, in order to get error bars on my barplot with two levels do
i need to change the last bit of code. For example im looking at mean body
condition (y) against age (x) and sex (level), i tried to change the code to
the following but it didnt work, any suggestions?
labels-as.character(levels(list(sex,Age)))
ybar-as.vector(tapply(ConditionIndex,sex,Age,mean))
error.bars(ybar,se,labels)
Hi Heidi,
I think you may be having trouble with the x positions of the error
bars. barplot returns these positions, so if you prepend xpos- to the
barplot call, it will give you the positions for your error bars.
Jim
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Re: [R] Plotting the complex fft in 3D?
Hello Martin, Zitat von Martin Maechler [EMAIL PROTECTED]: Just another remakr on this thread. I you have time series and think about its fourier transform (EE language) then you should know that the statistical language of that is spectral analysis or maybe frequency domain time-series analysis and the R function to consider should definitely be [...] Yes, I'm coming from the EE world, but I also know the other terms. The term spectral analysis is also used in EE, maybe not that often. Also frequency domain time-series analysis is used in EE, but maybe only at the university, later at the job, short terms spectral analysis or fft are used. (But it may differ from country to country.) I tried spectrum now on my example data, and it looks quite different to the result of fft(). It looks very close to what one gets as output from a spectrum analyzer (measurement harware). So it's quite nice to use this. :-) spectrum() which is a wrapper (among others) to spec.pgram() -- which calls fft() -- for computing the so-called periodogram. If you learn more about the topic, you will learn that in almost all cases you'd consider a *smoothed* version of the periodogram, etc etc (because the so-called *raw* periodgram is *in*consistent as an estimate fo the underlying true spectrum). Well, what do you mean with inconcistent? And why is spectrum() better? Do you talk about things like windowing for becoming more appropriate results? Even if the output from spectrum() looks more like what I know from measurement hardware, it might not always be better. Can yo explain, why better using this? The FFT only creates coefficients for certain seperated frequencies. It depends on number of samlpes how accurate the result is. And if the samples aren't an integer multiple of the frequency in the measured signal, this creates errors in the results. Possibly this is, what you are talking about? Why is spectrum() better? Would be nice to have an explanation, on how it's results are created, so that I can understand, when which kind of analysis is better. For the non-EE analysis, why is there fft() used and not spectrum()? For what kind of analysis is what function better? The Time-Series chapter/section of the MASS book is very helpful here, IIRC. MASS book? Do you mean the documentation of the MASS package? Well, so much packages R is very powerful and provides a lot of things. But where to start? Any idea? I started with some introductional papers and got a good overview. But a more systematical approach might be better. Can you recommend some things to read? I also have seen that there is one german book about programming in R. I will look at it in more detail, when the new edition will be published (sept/oct). But maybe you can recommend other readings as well? Thanks, Oliver __ 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] restricted bootstrap
On Thu, 4 Sep 2008, Grant Gillis wrote: Hello Professor Ripely, Sorry for not being clear. I posted after a long day of struggling. Also my toy distance matrix should have been symmetrical. Simply put I have spatially autocorrelated data collected from many points. I would like to do a linear regression on these data. To deal with the autocrrelation I want to resample a subset of my data with replacement but I need to restrict subsets such that no two locations where data was collected are closer than Xm apart (further apart than the autocrrelation in the data). That is impossible. Resampling with replacement will give duplicated locations (with a very high probability) and those have distance zero. If you want a subsample (necessarily without replacement) you have a hard-core point process on a discrete set. It's possible that the MCMC methods we used for Strauss processes can be made to work in that case, but it is also possible that the state space is reducible and so more elaborate algorithms are needed. I do think it would be much easier to take autocorrelation into account in your linear model fit. There are many ways to do that, e.g. MASS::lm.gls, and in fact uless the correlations are very high OLS is likely to be quite efficient (but you need to use e.g. a sandwich estimator to get reliable standard errors). Thanks for having a look at this for me. I will look up the hard-core spatial point process. Grant 2008/9/4 Prof Brian Ripley [EMAIL PROTECTED] I see nothing here to do with the 'bootstrap', which is sampling with replacement. Do you know what you mean exactly by 'randomly sample'? In general the way to so this is to sample randomly (uniformly, whatever) and reject samples that do not meet your restriction. For some restrictions there are more efficient algorithms, but I don't understand yours. (What are the 'rows'? Do you want to sample rows in space or xy locations? How come 'dist' is not symmetric?) For some restrictions, an MCMC sampling scheme is needed, the hard-core spatial point process being a related example. On Wed, 3 Sep 2008, Grant Gillis wrote: Hello List, I am not sure that I have the correct terminology here (restricted bootstrap) which may be hampering my archive searches. I have quite a large spatially autocorrelated data set. I have xy coordinates and the corresponding pairwise distance matrix (metres) for each row. I would like to randomly sample some number of rows but restricting samples such that the distance between them is larger than the threshold of autocorrelation. I have been been unsuccessfully trying to link the 'sample' function to values in the distance matrix. My end goal is to randomly sample M thousand rows of data N thousand times calculating linear regression coefficients for each sample but am stuck on taking the initial sample. I believe I can figure out the rest. Example Question I would like to radomly sample 3 rows further but withe the restriction that they are greater than 100m apart example data: main data: y- c(1, 2, 9, 5, 6) x-c( 1, 3, 5, 7, 9) z-c(2, 4, 6, 8, 10) a-c(3, 9, 6, 4 ,4) maindata-cbind(y, x, z, a) y x x a [1,] 1 1 1 3 [2,] 2 3 3 9 [3,] 9 5 5 6 [4,] 5 7 7 4 [5,] 6 9 9 4 distance matrix: row1-c(0, 123, 567, 89) row2-c(98, 0, 345, 543) row3-c(765, 90, 0, 987) row4-c(654, 8, 99, 0) dist-rbind(row1, row2, row3, row4) [,1] [,2] [,3] [,4] row10 123 567 89 row2 980 345 543 row3 765 900 987 row4 6548 990 Thanks for all of the help in the past and now Cheers Grant [[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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/http://www.stats.ox.ac.uk/%7Eripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list
[R] Help use try function with boot
Hi R users, Is is possible for me to use the try function with boot? I would to do the bootstraping with a nonlinear model(it works well when R 1000). But it does not work very well (when R is large) thus I try to use try to resolve. I put the try function in two cases: case1: put the try in front of the boot c1.try-try(boot(c1data, statistic = c1.fun, R=3999),silent=T) c1.try [1] Error in nls(formula = density ~ nmf(time, alpha, delta, psi, tau, gamma), : \n Convergence failure: false convergence (8)\n attr(,class) [1] try-error case2: put the try in front of the nls c1.nmf-try(nls(density~nmf(time, alpha, delta, psi, tau, gamma), +algorithm=port,data=c1, +lower=c(alpha=0.1, delta=0, psi=0.2, tau=1, gamma=-5), +upper=c(alpha=0.6, delta=0.1, psi=3, tau=7, gamma=2), +start=c(alpha=0.35, delta=0, psi=0.99, tau=4.5, gamma=-1.2)),silent=T) c1.try-boot(c1data, statistic = c1.fun, R=3999) Error in nls(formula = density ~ nmf(time, alpha, delta, psi, tau, gamma), : Convergence failure: iteration limit reached without convergence Any suggestion will be helpful. Many thanks in advance Chunhao __ 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] Hopefully an easy error bar question
Brown, Heidi wrote:
Hi im trying to add error bars to my barplots, there very basic, i have a few
grapghs where the y variable is different but on all the X variable is Age
(Adult and Juvenile) however this is split into two levels so i have males and
females, so my graph basically has four bars on it.
I know how to add eror bars for instance when there is only one level eg lookng at the diffrence between male and female lizards and tree height and have used the code:
error.bars-function(yv,z,nn) {
xv-
barplot(yv,ylim=c(0,(max(yv)+max(z))),names=nn,ylab=deparse(substitute(yv)
))
g=(max(xv)-min(xv))/50
for (i in 1:length(xv)) {
lines(c(xv[i],xv[i]),c(yv[i]+z[i],yv[i]-z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]+z[i], yv[i]+z[i]))
lines(c(xv[i]-g,xv[i]+g),c(yv[i]-z[i], yv[i]-z[i]))
}}
Have then worked out the standard errors and added that using:
se-c()
and then added the final bit of code:
labels-as.character(levels(Sex2))
ybar-as.vector(tapply(Height2,Sex2,mean))
error.bars(ybar,se,labels)
So i was wondering, in order to get error bars on my barplot with two levels do
i need to change the last bit of code. For example im looking at mean body
condition (y) against age (x) and sex (level), i tried to change the code to
the following but it didnt work, any suggestions?
labels-as.character(levels(list(sex,Age)))
ybar-as.vector(tapply(ConditionIndex,sex,Age,mean))
error.bars(ybar,se,labels)
See http://biostat.mc.vanderbilt.edu/DynamitePlots for many reasons not
to use dynamite plots.
Frank
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Test for equality of complicatedly related average correlations
Dear R-Users, I am currently looking for a way to test the equality of two correlations that are related in a very special way. Let me describe the situation with an example. - There are 100 respondents, and there are 2 points in time, t=1 and t=2. - For each of the respondents and at each of the time points, I have information on 10 X-variables and on 10 Y-variables. - Based on this information, I calculate two correlations for each respondent: cor(X[t=1],X[t=2]) and cor(Y[t=1],Y[t=2]), with X and Y being the vectors of the corresponding 10 variables. - Now I get the average correlations over the whole sample using Fishers Z-transformation, i.e. I have mean(cor(X[t=1],X[t=2])) and mean(cor(X[t=1],X[t=2])) and want to know if the mean correlations are significantly different! I haven't found any test that deals with exactly my situation. Therefore, I simply apply a paired t-test based on the individual z-correlations. From my point of view this should be ok, because of the z's normality. However, I am unsure if there is a better way to test the hypothesis that I am interested in? I'd be grateful for any comment or hint. Thank you very much, Ralph - Ralph Wirth University Erlangen-Nuremberg, Chair of Statistics GfK Group, Department of Methods and Product Development -- View this message in context: http://www.nabble.com/Test-for-equality-of-complicatedly-related-average-correlations-tp19346312p19346312.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] loop
I have to calculate a formula that gives me a ten components vector. I want
to see how the components behave at varying the variable i.
But when i run the following function:
bbayes-c()
for(i in 1:100) {
+ xx-t(X1_10)%*%X1_10
+ xxmeno1-solve(xx)
+ V-xxmeno1*i
+ Vmeno1-solve(V)
+ tx-t(X1_10)
+ prpar-solve(Vmeno1+xx)
+ snpar-tx%*%y
+ bbayes-prpar%*%snpar
+ }
it oly gives me the vector calculated with the last value of the sequence,
in this case 100.
Could you please suggest how to print all the 100 vectors calculated for i
in 1:100.
Thank you in advance.
Davide Crapis
--
View this message in context:
http://www.nabble.com/loop-tp19346683p19346683.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] request
you are subscribed to the r-help list. On Sat, Sep 6, 2008 at 4:48 AM, Ram Kumar Basnet [EMAIL PROTECTED] wrote: hi I am expecting the mails that will flow in R- help after subscribing this but could not. Can you provide me the right form. Thanks. [[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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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] Articles about comparision between R and others softwares
Hi Ricardo, You can search for comparisons by entering the packages that interest you at: http://finzi.psych.upenn.edu/search.html Michael Mitchell wrote an interesting comparison of SAS, SPSS, Stata and R at: http://www.ats.ucla.edu/stat/technicalreports/ That report says little about R, but Patrick Burns' excellent rejoinder to that report fills in much of the missing R material. It is at that link too. The accuracy of various stat packages, including R, is in: Keeling, Kellie B. and Pavur, Robert J. A comparative study of the reliability of nine statistical software packages. 8, May 1, 2007, Computational Statistics Data Analysis, Vol. 51, pp. 3811-3831. I've got an 80-page comparison of many features of R to SAS and SPSS at: http://RforSASandSPSSusers.com That document focuses more on the language basics and data management rather than statistics. My book by the same name adds graphics and basic statistics to the mix. That should finally be printed in a few weeks. Cheers, Bob = Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting Center U of TN Office of Information Technology 200 Stokely Management Center, Knoxville, TN 37996-0520 Voice: (865) 974-5230 FAX: (865) 974-4810 Email: [EMAIL PROTECTED] Web: http://oit.utk.edu/scc, News: http://listserv.utk.edu/archives/statnews.html = -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of ricardo13 Sent: Friday, September 05, 2008 3:46 PM To: r-help@r-project.org Subject: [R] Articles about comparision between R and others softwares Hi Do you know some articles, papers, something than tell about comparision between R and others softwares statisticals. Thank You Ricardo -- View this message in context: http://www.nabble.com/Articles-about- comparision-between-R-and-others-softwares-tp19338210p19338210.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] Plotting the complex fft in 3D?
Oliver Bandel wrote: Hello Martin, Zitat von Martin Maechler [EMAIL PROTECTED]: Just another remakr on this thread. I you have time series and think about its fourier transform (EE language) then you should know that the statistical language of that is spectral analysis or maybe frequency domain time-series analysis and the R function to consider should definitely be [...] Yes, I'm coming from the EE world, but I also know the other terms. The term spectral analysis is also used in EE, maybe not that often. Also frequency domain time-series analysis is used in EE, but maybe only at the university, later at the job, short terms spectral analysis or fft are used. (But it may differ from country to country.) I tried spectrum now on my example data, and it looks quite different to the result of fft(). It looks very close to what one gets as output from a spectrum analyzer (measurement harware). So it's quite nice to use this. :-) spectrum() which is a wrapper (among others) to spec.pgram() -- which calls fft() -- for computing the so-called periodogram. If you learn more about the topic, you will learn that in almost all cases you'd consider a *smoothed* version of the periodogram, etc etc (because the so-called *raw* periodgram is *in*consistent as an estimate fo the underlying true spectrum). Well, what do you mean with inconcistent? And why is spectrum() better? Do you talk about things like windowing for becoming more appropriate results? Even if the output from spectrum() looks more like what I know from measurement hardware, it might not always be better. Can yo explain, why better using this? The FFT only creates coefficients for certain seperated frequencies. It depends on number of samlpes how accurate the result is. And if the samples aren't an integer multiple of the frequency in the measured signal, this creates errors in the results. Possibly this is, what you are talking about? Why is spectrum() better? Would be nice to have an explanation, on how it's results are created, so that I can understand, when which kind of analysis is better. For the non-EE analysis, why is there fft() used and not spectrum()? For what kind of analysis is what function better? spectrum() and spec.pgram() use fft() to calculate results. But some information is thrown away, some information is merged appropriately, and maybe tapering and padding is applied. Well, it really depends on your tasks which functions to use. I found myself frequently using fft() and doing the rest manually, because spectrum() is sometimes too intelligent for me ... The Time-Series chapter/section of the MASS book is very helpful here, IIRC. MASS book? Do you mean the documentation of the MASS package? No, the other way round: the Springer book Modern Applied Statistics with S, 4th edition, 2002, by Venables and Ripley has some package (MASS) as supplementary material. Best wishes, Uwe Ligges Well, so much packages R is very powerful and provides a lot of things. But where to start? Any idea? I started with some introductional papers and got a good overview. But a more systematical approach might be better. Can you recommend some things to read? I also have seen that there is one german book about programming in R. I will look at it in more detail, when the new edition will be published (sept/oct). But maybe you can recommend other readings as well? Thanks, Oliver __ 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] request
ss == stephen sefick [EMAIL PROTECTED]
on Sat, 6 Sep 2008 09:40:50 -0400 writes:
ss you are subscribed to the r-help list.
Huh??? How would *you* know that?
The list of subscribers to R-help is not at all public,
and the fact that Ram Kumar's posting went through may well be
the result of the fact that we have a kind group of R-help
moderators who spend a fraction of their valuable time to let
people occasionally post even if unsubscribed.
Search for volunteer on
https://stat.ethz.ch/mailman/listinfo/r-help
{please, Ram, also do read the above web page carefully ..}
Regards,
Martin Maechler,
ETH Zurich (provider of all the @r-project.org mailing lists)
ss On Sat, Sep 6, 2008 at 4:48 AM, Ram Kumar Basnet [EMAIL PROTECTED]
wrote:
hi
I am expecting the mails that will flow in R- help after subscribing
this but could not. Can you provide me the right form.
Thanks.
[[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.
ss --
ss Stephen Sefick
ss Research Scientist
ss Southeastern Natural Sciences Academy
ss Let's not spend our time and resources thinking about things that are
ss so little or so large that all they really do for us is puff us up and
ss make us feel like gods. We are mammals, and have not exhausted the
ss annoying little problems of being mammals.
ss -K. Mullis
ss __
ss R-help@r-project.org mailing list
ss https://stat.ethz.ch/mailman/listinfo/r-help
ss PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
ss 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] Hopefully an easy error bar question
FEH == Frank E Harrell [EMAIL PROTECTED]
on Sat, 06 Sep 2008 07:19:33 -0500 writes:
FEH Brown, Heidi wrote:
Hi im trying to add error bars to my barplots, there very basic, i have
a few grapghs where the y variable is different but on all the X variable is
Age (Adult and Juvenile) however this is split into two levels so i have males
and females, so my graph basically has four bars on it.
I know how to add eror bars for instance when there is only one level eg
lookng at the diffrence between male and female lizards and tree height and
have used the code:
error.bars-function(yv,z,nn) {
xv-
barplot(yv,ylim=c(0,(max(yv)+max(z))),names=nn,ylab=deparse(substitute(yv)
))
[ omitting the abomination ]
Have then worked out the standard errors and added that using:
se-c()
and then added the final bit of code:
labels-as.character(levels(Sex2))
ybar-as.vector(tapply(Height2,Sex2,mean))
error.bars(ybar,se,labels)
So i was wondering, in order to get error bars on my barplot with two
levels do i need to change the last bit of code. For example im looking at mean
body condition (y) against age (x) and sex (level), i tried to change the code
to the following but it didnt work, any suggestions?
labels-as.character(levels(list(sex,Age)))
ybar-as.vector(tapply(ConditionIndex,sex,Age,mean))
error.bars(ybar,se,labels)
FEH See http://biostat.mc.vanderbilt.edu/DynamitePlots for many reasons
not
FEH to use dynamite plots.
Ah! Very good! Let's hope this eventually leads to progress
i.e., to much less use of these..
Thank you, Frank!
Martin Maechler, ETH Zurich
FEH --
FEH Frank E Harrell Jr Professor and Chair School of Medicine
FEH Department of Biostatistics Vanderbilt University
FEH __
FEH R-help@r-project.org mailing list
FEH https://stat.ethz.ch/mailman/listinfo/r-help
FEH PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
FEH and provide commented, minimal, self-contained, reproducible code.
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Re: [R] request
I apologize, my naive understanding that if an email was sent to the
list then the sender is subscribed. However, my hypothesis is
falseifiable.
On Sat, Sep 6, 2008 at 10:33 AM, Martin Maechler
[EMAIL PROTECTED] wrote:
ss == stephen sefick [EMAIL PROTECTED]
on Sat, 6 Sep 2008 09:40:50 -0400 writes:
ss you are subscribed to the r-help list.
Huh??? How would *you* know that?
The list of subscribers to R-help is not at all public,
and the fact that Ram Kumar's posting went through may well be
the result of the fact that we have a kind group of R-help
moderators who spend a fraction of their valuable time to let
people occasionally post even if unsubscribed.
Search for volunteer on
https://stat.ethz.ch/mailman/listinfo/r-help
{please, Ram, also do read the above web page carefully ..}
Regards,
Martin Maechler,
ETH Zurich (provider of all the @r-project.org mailing lists)
ss On Sat, Sep 6, 2008 at 4:48 AM, Ram Kumar Basnet [EMAIL PROTECTED]
wrote:
hi
I am expecting the mails that will flow in R- help after subscribing
this but could not. Can you provide me the right form.
Thanks.
[[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.
ss --
ss Stephen Sefick
ss Research Scientist
ss Southeastern Natural Sciences Academy
ss Let's not spend our time and resources thinking about things that are
ss so little or so large that all they really do for us is puff us up and
ss make us feel like gods. We are mammals, and have not exhausted the
ss annoying little problems of being mammals.
ss -K. Mullis
ss __
ss R-help@r-project.org mailing list
ss https://stat.ethz.ch/mailman/listinfo/r-help
ss PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
ss and provide commented, minimal, self-contained, reproducible code.
--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
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] new to R
On 9/6/08, sudeshna [EMAIL PROTECTED] wrote: hi im starting with R.have no idea to start...plz help -- http://www.math.ilstu.edu/dhkim/Rstuff/Rtutor.html http://www.statmethods.net/index.html http://rforsasandspssusers.com/ Rcmdr rattle http://www.sciviews.org/_rgui/ http://zoonek2.free.fr/UNIX/48_R/all.html http://cran.r-project.org/other-docs.html http://www.r-project.org/doc/bib/R-publications.html http://cran.r-project.org/manuals.html Also search this Mailing-List archives. You'll find many pointers lying around. Good luck, Liviu __ 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] request: most repeated sequnce
Dear R community
Initially i thought my problem has been solved but one thing which i found e.g.
if
1. All the elements of a sector are zero e.g
, , 7
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000000000 0
[2,]000000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
2. Majority of the rows consist of zeros e.g.
, , 5
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]440000000 0
[2,]440000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
Actually
zeros are not my values. I get values and fill the remaining parts with
zeros like x=array(0,dim=c(3,6,5)). Now according to first strategy
000000000 0 are most repeated
sequence of rows in both of above cases. But i don't want to consider
cases where all elements are zeros and interested to get 44
0000000 0 or just 4 4 in case 2.
Thanks and best regards
Muhammad Azam
- Original Message
From: jim holtman [EMAIL PROTECTED]
To: Muhammad Azam [EMAIL PROTECTED]
Cc: R Help r-help@r-project.org; R-help request [EMAIL PROTECTED]
Sent: Saturday, September 6, 2008 2:39:19 PM
Subject: Re: [R] request: most repeated sequnce
Here is a start. You can delete the zeros:
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
+ 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
x=array(x,dim=c(3,6,5))
apply(x,3,function(.mat){
+ rows - table(apply(.mat,1,function(z){
+ paste(z,collapse=' ')
+ }))
+ names(rows[which.max(rows)])
+ })
[1] 1 0 0 0 0 0 1 2 3 0 0 0 1 2 3 4 0 0 1 2 3 4 0 0 2 2 3 4 0 0
On Sat, Sep 6, 2008 at 4:54 AM, Muhammad Azam [EMAIL PROTECTED] wrote:
Dear R community
Hope every one be in best of his/her health. I have a situation in which
there are s-sectors. Each sector is further divided into r-rows and
c-columns. All it makes an array having dimension (r,c,s). e.g.
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
x=array(x,dim=c(3,6,5))
x
, , 1
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]100000
[2,]100000
[3,]100000
, , 2
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123000
[2,]123000
[3,]120000
, , 3
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123400
[2,]123400
[3,]134000
, , 4
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123400
[2,]123400
[3,]123400
, , 5
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]120000
[2,]223400
[3,]223400
I want to get the most repeated sequence (row-wise) of values in each sector.
e.g. in sector 1 i.e. , , 1
the most repeated sequence is 1 (ignoring zeros). In , , 2 the most repeated
sequence is 1 2 3. Similarly in last sector i.e.
, , 5 such sequence is 2 2 3 4. Any body can help to solve this problem.
Thanks
best regards
Muhammad Azam
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] subsetting a data frame
Hello How can I change the function to get the rows with the sum (x+y+z) = 10? Thank you very much Joseph - Original Message From: Marc Schwartz [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Wednesday, September 3, 2008 3:24:58 PM Subject: Re: [R] subsetting a data frame on 09/03/2008 05:06 PM joseph wrote: I have a data frame that looks like this: V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 where V3 is in the form x:y:z Can someone show me how to subset the rows where the values of x, y and z = 10: V1 V2 V3 df1:2:1 cd1:0:9 Thanks Joseph How about this: DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)), ] V1 V2V3 2 d f 1:2:1 3 c d 1:0:9 Basically, use strsplit() to break apart 'V3': strsplit(as.character(DF$V3), :) [[1]] [1] 0 1 12 [[2]] [1] 1 2 1 [[3]] [1] 1 0 9 The use sapply() to crawl the list, converting the elements to numerics and do the value comparison: sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)) [1] FALSE TRUE TRUE The above then returns the logical vector to subset the rows of 'DF'. HTH, Marc Schwartz [[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] Mode value
Hello everyone, I would like to know if there is any function to calculate the mode value, or I have to build one to do it. Thanks so much Carlos __ 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] Mode value
look here: http://www.nabble.com/how-to-calculate-the-mode-of-a-continuous-variable-td19214243.html#a19214243 On Sat, Sep 6, 2008 at 1:24 PM, Carlos Morales [EMAIL PROTECTED] wrote: Hello everyone, I would like to know if there is any function to calculate the mode value, or I have to build one to do it. Thanks so much Carlos __ 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 Research Scientist Southeastern Natural Sciences Academy 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] Mode value
Hi Carolos,
I know that it is not a elegant soluction, but may work. Almost for integer
values. Take care with float values.
modevalue-function(x)
{
x.freq-data.frame(table(x))
x.freq.max-max(x.freq$Freq)
x.freq.selected-subset(x.freq, x.freq$Freq==x.freq.max)
return(c(unlist(x.freq.selected$x)))
}
x-sample(x=0:10, size=200, replace=T)
modevalue(x)
On Sat, Sep 6, 2008 at 2:24 PM, Carlos Morales
[EMAIL PROTECTED]wrote:
Hello everyone,
I would like to know if there is any function to calculate the mode value,
or I have to build one to do it.
Thanks so much
Carlos
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http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
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Re: [R] Hopefully an easy error bar question
Martin Maechler maechler at stat.math.ethz.ch writes: FEH == Frank E Harrell f.harrell at vanderbilt.edu on Sat, 06 Sep 2008 07:19:33 -0500 writes: FEH See http://biostat.mc.vanderbilt.edu/DynamitePlots for many reasons not FEH to use dynamite plots. Ah! Very good! Let's hope this eventually leads to progress i.e., to much less use of these.. Thank you, Frank! Martin Maechler, ETH Zurich FEH -- Yes, but: http://emdbolker.wikidot.com/blog:dynamite (repeats Frank's criticisms, but also raises the point that boxplots show the range of the data, but not much about the statistical inferences on the data ...) comments welcome. Ben Bolker __ 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] subsetting a data frame
Hi Jorge I got the rows where V3 looks like this 10:10:10; Ithe sum here is 30 and not 10. I want the rows where the sum is 10 for exaple 5:5:0 and 2:2:6 thanks Joseph - Original Message From: Jorge Ivan Velez [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Sent: Saturday, September 6, 2008 10:43:09 AM Subject: Re: [R] subsetting a data frame Dear Joseph, Try DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) == 10)), ] HTH, Jorge On Sat, Sep 6, 2008 at 1:24 PM, joseph [EMAIL PROTECTED] wrote: Hello How can I change the function to get the rows with the sum (x+y+z) = 10? Thank you very much Joseph - Original Message From: Marc Schwartz [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Wednesday, September 3, 2008 3:24:58 PM Subject: Re: [R] subsetting a data frame on 09/03/2008 05:06 PM joseph wrote: I have a data frame that looks like this: V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 where V3 is in the form x:y:z Can someone show me how to subset the rows where the values of x, y and z = 10: V1 V2 V3 df1:2:1 cd1:0:9 Thanks Joseph How about this: DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)), ] V1 V2V3 2 d f 1:2:1 3 c d 1:0:9 Basically, use strsplit() to break apart 'V3': strsplit(as.character(DF$V3), :) [[1]] [1] 0 1 12 [[2]] [1] 1 2 1 [[3]] [1] 1 0 9 The use sapply() to crawl the list, converting the elements to numerics and do the value comparison: sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)) [1] FALSE TRUE TRUE The above then returns the logical vector to subset the rows of 'DF'. HTH, Marc Schwartz [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mode value
On 09/06/08 17:24, Carlos Morales wrote: Hello everyone, I would like to know if there is any function to calculate the mode value, or I have to build one to do it. I just did this the other day. Funny you should ask. It finds the mode of each row of a matrix called Pbest. Pbest.p - apply(Pbest,1,function(x) x[which.max(as.vector(table(x)))][1]) So to get the mode of a vector v1, the following should work (but I haven't tried it): v1[which.max(as.vector(table(v1)))][1] -- Jonathan Baron, Professor of Psychology, University of Pennsylvania Home page: http://www.sas.upenn.edu/~baron __ 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 use try function with boot
Hi Jinsong,
I try to put the try in the c1.fun but it still does not work. Do
you mean this?
c1.fun-try(function(data,i){
+ d-data
+ d$density-d$fitted+d$res[i]
+ coef(update(c1.nmf,data=d))
+ } ,silent=T)
c1.try-boot(c1data, statistic = c1.fun, R=5000)
Error in nls(formula = density ~ nmf(time, alpha, delta, psi, tau, gamma), :
Convergence failure: false convergence (8)
Thanks
Chunhao
Quoting Jinsong Zhao [EMAIL PROTECTED]:
[EMAIL PROTECTED] wrote:
Hi R users,
Is is possible for me to use the try function with boot? I would to
do the bootstraping with a nonlinear model(it works well when R
1000). But it does not work very well (when R is large) thus I try
to use try to resolve. I put the try function in two cases:
case1: put the try in front of the boot
c1.try-try(boot(c1data, statistic = c1.fun, R=3999),silent=T)
c1.try
[1] Error in nls(formula = density ~ nmf(time, alpha, delta, psi,
tau, gamma), : \n Convergence failure: false convergence (8)\n
attr(,class)
[1] try-error
case2: put the try in front of the nls
c1.nmf-try(nls(density~nmf(time, alpha, delta, psi, tau, gamma),
+algorithm=port,data=c1,
+lower=c(alpha=0.1, delta=0, psi=0.2, tau=1, gamma=-5),
+upper=c(alpha=0.6, delta=0.1, psi=3, tau=7, gamma=2),
+start=c(alpha=0.35, delta=0, psi=0.99, tau=4.5,
gamma=-1.2)),silent=T)
c1.try-boot(c1data, statistic = c1.fun, R=3999)
Error in nls(formula = density ~ nmf(time, alpha, delta, psi, tau,
gamma), :
Convergence failure: iteration limit reached without convergence
Any suggestion will be helpful.
Many thanks in advance
Chunhao
put try() in your statistic function. in your case, it may be inside
the c1.fun function.
HTH,
Jinsong
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Re: [R] subsetting a data frame
#something like this? V1 - c(1:10) V2 - c(0,5,7,8,1,6,5,13,7,0) V3 - c(9,5,6,8,1,7,5,33,88,0) z - cbind(V1,V2,V3) row.sums - rowSums(z) d - cbind(z, row.sums) subset(d, row.sums==10) On Sat, Sep 6, 2008 at 2:25 PM, joseph [EMAIL PROTECTED] wrote: Hi Jorge I got the rows where V3 looks like this 10:10:10; Ithe sum here is 30 and not 10. I want the rows where the sum is 10 for exaple 5:5:0 and 2:2:6 thanks Joseph - Original Message From: Jorge Ivan Velez [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Sent: Saturday, September 6, 2008 10:43:09 AM Subject: Re: [R] subsetting a data frame Dear Joseph, Try DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) == 10)), ] HTH, Jorge On Sat, Sep 6, 2008 at 1:24 PM, joseph [EMAIL PROTECTED] wrote: Hello How can I change the function to get the rows with the sum (x+y+z) = 10? Thank you very much Joseph - Original Message From: Marc Schwartz [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Wednesday, September 3, 2008 3:24:58 PM Subject: Re: [R] subsetting a data frame on 09/03/2008 05:06 PM joseph wrote: I have a data frame that looks like this: V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 where V3 is in the form x:y:z Can someone show me how to subset the rows where the values of x, y and z = 10: V1 V2 V3 df1:2:1 cd1:0:9 Thanks Joseph How about this: DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)), ] V1 V2V3 2 d f 1:2:1 3 c d 1:0:9 Basically, use strsplit() to break apart 'V3': strsplit(as.character(DF$V3), :) [[1]] [1] 0 1 12 [[2]] [1] 1 2 1 [[3]] [1] 1 0 9 The use sapply() to crawl the list, converting the elements to numerics and do the value comparison: sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)) [1] FALSE TRUE TRUE The above then returns the logical vector to subset the rows of 'DF'. HTH, Marc Schwartz [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] subsetting a data frame
Sorry I didn't read the problem carefully. On Sat, Sep 6, 2008 at 2:53 PM, Jorge Ivan Velez [EMAIL PROTECTED] wrote: Hi Joseph, Try this: # Data set DF=read.table(textConnection(V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 be2:2:6 fc5:5:0),header=TRUE) closeAllConnections() target=10 DF[sapply(strsplit(as.character(DF$V3), :), function(x) sum(as.numeric(x))== target), ] HTH, Jorge On Sat, Sep 6, 2008 at 2:25 PM, joseph [EMAIL PROTECTED] wrote: Hi Jorge I got the rows where V3 looks like this 10:10:10; Ithe sum here is 30 and not 10. I want the rows where the sum is 10 for exaple 5:5:0 and 2:2:6 thanks Joseph - Original Message From: Jorge Ivan Velez [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Sent: Saturday, September 6, 2008 10:43:09 AM Subject: Re: [R] subsetting a data frame Dear Joseph, Try DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) == 10)), ] HTH, Jorge On Sat, Sep 6, 2008 at 1:24 PM, joseph [EMAIL PROTECTED] wrote: Hello How can I change the function to get the rows with the sum (x+y+z) = 10? Thank you very much Joseph - Original Message From: Marc Schwartz [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Wednesday, September 3, 2008 3:24:58 PM Subject: Re: [R] subsetting a data frame on 09/03/2008 05:06 PM joseph wrote: I have a data frame that looks like this: V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 where V3 is in the form x:y:z Can someone show me how to subset the rows where the values of x, y and z = 10: V1 V2 V3 df1:2:1 cd1:0:9 Thanks Joseph How about this: DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)), ] V1 V2V3 2 d f 1:2:1 3 c d 1:0:9 Basically, use strsplit() to break apart 'V3': strsplit(as.character(DF$V3), :) [[1]] [1] 0 1 12 [[2]] [1] 1 2 1 [[3]] [1] 1 0 9 The use sapply() to crawl the list, converting the elements to numerics and do the value comparison: sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)) [1] FALSE TRUE TRUE The above then returns the logical vector to subset the rows of 'DF'. HTH, Marc Schwartz [[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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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.
[R] how to address last and all but last column in dataframe
Dear R-colleagues, another question from a newbie: I am creating a lot of simple pivot-charts from my raw data using the reshape-package. In these charts we have medical doctors judging videos in the columns and the videos they judge in the rows. Simple example of chart/data.frame input with two categories 1/0: video 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 1010 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I recently learned, that I can easily create a confusion matrix out of this data using the following commands: pairs-data.frame(pred=factor(unlist(input[2:21])),ref=factor(input[,22])) pred-pairs$pred ref - pairs$ref library (caret) confusionMatrix(pred, ref, positive=1) - where column 21 is the reference/goldstandard. My problem is now, that I analyse data.frames with an unknown count of columns. So to get rid of the first and last column for the pred variable and to select the last column for the ref variable, I have to look at the data.frame before doing the above commands to set the proper column numbers. It would be very comfortable, if I could address the last column not by number (where I have to count beforehand) but by a variable last column. Probably there is a more easy solution for this problem using the names of the columns as well: the reference is always number 21 the first column is always called video. So I tried: attach(input) pairs-data.frame(pred=factor(unlist(input[[,-c(video,21)]])),ref=factor(input[[21]])) which does not work unfortunately :-(. I'd be very happy in case someone could help me out, cause I am really tired of counting - there are a lot of tables to analyse... Cheers and greetings from Munich, Felix __ 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] I don't know how to run a r-code written in emacs
Hi, I just installed R, I'm work in UBUNTU and I don't have idea about how to run a r-code written in emacs into the shell. Well I am in a shell, and obviously I can run simple commands over there, Must I compile the program? if yes, How must I do that? what is the extension? I really appreciate your help -- View this message in context: http://www.nabble.com/I-don%27t-know-how-to-run-a-r-code-written-in-emacs-tp19348030p19348030.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to address last and all but last column in dataframe
Not sure where your input came from. It's not in a format I would have expected of an R object and the first line is not in a form that would be particularly easy to read into a valid R object. Numbers are no legitimate object names. It's also not clear what you want to do with the duplicated line numbers at the beginning. Your question implies that you do not consider them part of the data. In the future a worked example along the lines of that constructed by Jorge Ivan Velez in a recent answer to another question might increase chances of a prompt reply with tested code: # Data set DF=read.table(textConnection(V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 be2:2:6 fc5:5:0),header=TRUE) closeAllConnections() The length of a dataframe is the number of columns. ?length Dataframes can be referenced using the extract operation e.g. df[row, col] ?Extract # for additional information on indexing using column vectors. So: video[ ,length(video)] #should return the last column vector although it will be no longer be named. The rest of the dataframe with intact column names could be obtained with: video[ ,-length(video)] -- David Winsemius On Sep 6, 2008, at 3:00 PM, drflxms wrote: Dear R-colleagues, another question from a newbie: I am creating a lot of simple pivot-charts from my raw data using the reshape-package. In these charts we have medical doctors judging videos in the columns and the videos they judge in the rows. Simple example of chart/data.frame input with two categories 1/0: video 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 1010 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I recently learned, that I can easily create a confusion matrix out of this data using the following commands: pairs-data.frame(pred=factor(unlist(input[2:21])),ref=factor(input[, 22])) pred-pairs$pred ref - pairs$ref library (caret) confusionMatrix(pred, ref, positive=1) - where column 21 is the reference/goldstandard. My problem is now, that I analyse data.frames with an unknown count of columns. So to get rid of the first and last column for the pred variable and to select the last column for the ref variable, I have to look at the data.frame before doing the above commands to set the proper column numbers. It would be very comfortable, if I could address the last column not by number (where I have to count beforehand) but by a variable last column. Probably there is a more easy solution for this problem using the names of the columns as well: the reference is always number 21 the first column is always called video. So I tried: attach(input) pairs-data.frame(pred=factor(unlist(input[[,-c(video, 21)]])),ref=factor(input[[21]])) which does not work unfortunately :-(. I'd be very happy in case someone could help me out, cause I am really tired of counting - there are a lot of tables to analyse... Cheers and greetings from Munich, Felix __ 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] LME prediction - object not subsettable?
I fit a random effects linear model to data, and then tried to use it to predict, but I got this error: predict(lmeObject, newdata, level=0) Error in eval(mCall$fixed)[-2] : object is not subsettable This is a new error for me. It still occurs if I change the level to 1 or if I change the data for prediction back to the original dataset to which the lme model was fitted. Can anyone help? Thank you in advance! Rebecca __ 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 address last and all but last column in dataframe
Hello Mr. Burns, Hello Mr. Dwinseminus thank you very much for your incredible quick and efficient reply! I was completely successful with the following command: pairs-data.frame(pred=factor(unlist(input[,-c(1,ncol(input))])),ref=factor(input[,ncol(input)])) In case of the input example data.frame I sent with my question the above code is equivalent to: pairs-data.frame(pred=factor(unlist(input[2:17])),ref=factor(input[,18])) Great! That is exactly what I was looking for! This simple code will save me hours! Patrick, your book looks in fact very interesting and will be my perfect reading material for the following nights :-) (probably not only the first chapter ;-). Thanks for the hint - and the free book of course. David, the input data.frame is the result of the reshape-command I performed. I just copied it from the R-console into the e-mail. In fact the first column video is not part of the data, but needed for analysis with kappam.fleiss function of the irr-package. Sorry, you are absolutely correct, I should have mentioned this in my question. I will improve when I ask my next question :-). Again I like to thank you for your help and wish you a pleasant Sunday. Greetings from Munich, Felix Patrick Burns wrote: If I understand properly, you want input[, -c(1, ncol(input))] rather than input[[, -c(video, 21)]] Chapter 1 of S Poetry might be of interest to you. Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) drflxms wrote: Dear R-colleagues, another question from a newbie: I am creating a lot of simple pivot-charts from my raw data using the reshape-package. In these charts we have medical doctors judging videos in the columns and the videos they judge in the rows. Simple example of chart/data.frame input with two categories 1/0: video 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 1010 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I recently learned, that I can easily create a confusion matrix out of this data using the following commands: pairs-data.frame(pred=factor(unlist(input[2:21])),ref=factor(input[,22])) pred-pairs$pred ref - pairs$ref library (caret) confusionMatrix(pred, ref, positive=1) - where column 21 is the reference/goldstandard. My problem is now, that I analyse data.frames with an unknown count of columns. So to get rid of the first and last column for the pred variable and to select the last column for the ref variable, I have to look at the data.frame before doing the above commands to set the proper column numbers. It would be very comfortable, if I could address the last column not by number (where I have to count beforehand) but by a variable last column. Probably there is a more easy solution for this problem using the names of the columns as well: the reference is always number 21 the first column is always called video. So I tried: attach(input) pairs-data.frame(pred=factor(unlist(input[[,-c(video,21)]])),ref=factor(input[[21]])) which does not work unfortunately :-(. I'd be very happy in case someone could help me out, cause I am really tired of counting - there are a lot of tables to analyse... Cheers and greetings from Munich, Felix __ 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] I don't know how to run a r-code written in emacs
did you install ess for emacs? On Sat, Sep 6, 2008 at 11:52 AM, Luisa [EMAIL PROTECTED] wrote: Hi, I just installed R, I'm work in UBUNTU and I don't have idea about how to run a r-code written in emacs into the shell. Well I am in a shell, and obviously I can run simple commands over there, Must I compile the program? if yes, How must I do that? what is the extension? I really appreciate your help -- View this message in context: http://www.nabble.com/I-don%27t-know-how-to-run-a-r-code-written-in-emacs-tp19348030p19348030.html Sent from the R help mailing list archive at Nabble.com. [[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. -- === WenSui Liu Acquisition Risk, Chase Email : [EMAIL PROTECTED] Blog : statcompute.spaces.live.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] subsetting a data frame
Hi Joseph, Try this: # Data set DF=read.table(textConnection(V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 be2:2:6 fc5:5:0),header=TRUE) closeAllConnections() target=10 DF[sapply(strsplit(as.character(DF$V3), :), function(x) sum(as.numeric(x))== target), ] HTH, Jorge On Sat, Sep 6, 2008 at 2:25 PM, joseph [EMAIL PROTECTED] wrote: Hi Jorge I got the rows where V3 looks like this 10:10:10; Ithe sum here is 30 and not 10. I want the rows where the sum is 10 for exaple 5:5:0 and 2:2:6 thanks Joseph - Original Message From: Jorge Ivan Velez [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Sent: Saturday, September 6, 2008 10:43:09 AM Subject: Re: [R] subsetting a data frame Dear Joseph, Try DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) == 10)), ] HTH, Jorge On Sat, Sep 6, 2008 at 1:24 PM, joseph [EMAIL PROTECTED] wrote: Hello How can I change the function to get the rows with the sum (x+y+z) = 10? Thank you very much Joseph - Original Message From: Marc Schwartz [EMAIL PROTECTED] To: joseph [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Wednesday, September 3, 2008 3:24:58 PM Subject: Re: [R] subsetting a data frame on 09/03/2008 05:06 PM joseph wrote: I have a data frame that looks like this: V1 V2 V3 ab0:1:12 df1:2:1 cd1:0:9 where V3 is in the form x:y:z Can someone show me how to subset the rows where the values of x, y and z = 10: V1 V2 V3 df1:2:1 cd1:0:9 Thanks Joseph How about this: DF[sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)), ] V1 V2V3 2 d f 1:2:1 3 c d 1:0:9 Basically, use strsplit() to break apart 'V3': strsplit(as.character(DF$V3), :) [[1]] [1] 0 1 12 [[2]] [1] 1 2 1 [[3]] [1] 1 0 9 The use sapply() to crawl the list, converting the elements to numerics and do the value comparison: sapply(strsplit(as.character(DF$V3), :), function(i) all(as.numeric(i) = 10)) [1] FALSE TRUE TRUE The above then returns the logical vector to subset the rows of 'DF'. HTH, Marc Schwartz [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot a list
i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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] I don't know how to run a r-code written in emacs
On Saturday, 06 September 2008, 08:52 (UTC-0700), Luisa wrote: Hi, I just installed R, I'm work in UBUNTU and I don't have idea about how to run a r-code written in emacs into the shell. Well I am in a shell, and obviously I can run simple commands over there, Must I compile the program? Which program? Emacs? If you've installed R, you can read a txt file of R code into your R session with source(/path/if/any/file.R) what is the extension? not sure what you mean. It makes sense to save files with R commands with the .R extension, but AFAIK it's not obligatory. If you don't have ess, I think you can get that, too, with aptitude install ess. Then, in Emacs, you can open R with M-x R. You can now copypaste (killyank) R code from your text file to the R buffer; alternatively there are special commands available, look at the menu in Emacs. If you have Emacs anyway, ESS is much more convenient than running from the console. m. -- Marianne Promberger Graduate student in Psychology http://www.psych.upenn.edu/~mpromber __ 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] Axis tick label format and rotation
Please tell me how to format data in a data frame so when currency amount is displayed in a chart the axis tick labels contain leading $ signs. Please also tell me if it is possible to rotate x axis labels using ggplot2. Thank you, Kurt __ 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] Test for equality of complicatedly related average correlations
Hi Ralph, I had the same problem you do a few months ago, and realized that the question I had (does time show a different effect for X than Y) was not best modeled as differences between correlations across individuals, but as whether time interacts with condition. I answered this question with library(nlme) lme(obs ~ cond*time, random=~cond*time|subj) ...where obs is the responses on the X or Y variable, cond is a factor of either X or Y, and subj is your subject variable. This fits a heirarchical linear model to the data. The relationship between X and time is sig. diff. from the relationship between Y and time if the cond:time fixed effect is true. This approach makes better use of your data, because when you correlate the observations, you're effectively losing variability (because correlations are doubly standardized) as well as degrees of freedom (you have 9 df within each individual, but each correlation is only one number). --Adam On Sat, 6 Sep 2008, Ralph79 wrote: Dear R-Users, I am currently looking for a way to test the equality of two correlations that are related in a very special way. Let me describe the situation with an example. - There are 100 respondents, and there are 2 points in time, t=1 and t=2. - For each of the respondents and at each of the time points, I have information on 10 X-variables and on 10 Y-variables. - Based on this information, I calculate two correlations for each respondent: cor(X[t=1],X[t=2]) and cor(Y[t=1],Y[t=2]), with X and Y being the vectors of the corresponding 10 variables. - Now I get the average correlations over the whole sample using Fishers Z-transformation, i.e. I have mean(cor(X[t=1],X[t=2])) and mean(cor(X[t=1],X[t=2])) and want to know if the mean correlations are significantly different! I haven't found any test that deals with exactly my situation. Therefore, I simply apply a paired t-test based on the individual z-correlations. From my point of view this should be ok, because of the z's normality. However, I am unsure if there is a better way to test the hypothesis that I am interested in? I'd be grateful for any comment or hint. Thank you very much, Ralph - Ralph Wirth University Erlangen-Nuremberg, Chair of Statistics GfK Group, Department of Methods and Product Development -- View this message in context: http://www.nabble.com/Test-for-equality-of-complicatedly-related-average-correlations-tp19346312p19346312.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] I don't know how to run a r-code written in emacs
2008/9/6 Luisa [EMAIL PROTECTED]: Hi, I just installed R, I'm work in UBUNTU and I don't have idea about how to run a r-code written in emacs into the shell. Well I am in a shell, and obviously I can run simple commands over there, Must I compile the program? if yes, How must I do that? what is the extension? You don't need to compile anything and you can use .R as an extension for your R scripts (I believe someone else has suggested this already though). What I usually do to run my scripts, is something like the following: R foo.R, where foo.R is the name of the script. The man page for R suggests this usage (admittedly, though, it's perhaps not obvious) and there are options listed in it that you may want to use. For what it's worth, it makes no difference which editor you use to write your code in. Regards, Nicky Chorley __ 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] Area of density
BIG THANKS!!! )) -- View this message in context: http://www.nabble.com/Area-of-density-tp19338958p19351299.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] re ferring to a group of vectors without explicit enumeration
I am trying to define 25 vectors of varying lengths, say y1 to y25 in a loop, and then store the results of some computations in them. My problem is about using some sort of concatenation for names. For example, instead of initializing each of y1 through y25, I would like to do it in a loop. Similar to cat and paste for texts, is there anyway of using yi for the vector name where i ranges from 1 to 25, so ultimately it refers to the vector y1,..,y25? Varying lengths is not a problem. To start with each has only length 1 and then I will be adding to each vector based on some results. -- View this message in context: http://www.nabble.com/referring-to-a-group-of-vectors-without-explicit-enumeration-tp19351518p19351518.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] re ferring to a group of vectors without explicit enumeration
shalu shahlar at hotmail.com writes: I am trying to define 25 vectors of varying lengths, say y1 to y25 in a loop, and then store the results of some computations in them. My problem is about using some sort of concatenation for names. For example, instead of initializing each of y1 through y25, I would like to do it in a loop. Similar to cat and paste for texts, is there anyway of using yi for the vector name where i ranges from 1 to 25, so ultimately it refers to the vector y1,..,y25? Varying lengths is not a problem. To start with each has only length 1 and then I will be adding to each vector based on some results. I think this is essentially http://cran.r-project.org/doc/FAQ/R-FAQ.html #How-can-I-turn-a-string-into-a-variable? [URL broken in order to make Gmane happy, reassemble it in your browser] the short answer: assign(), but it would work better to use a list instead. Ben Bolker __ 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] Sweave and/or beamer issue
Dear Friends,
I not sure whether this is an Sweave or a beamer problem.
The Rnw file:
\documentclass[compress,smaller]{beamer}
%\documentclass{article}
%\usepackage{beamerarticle}
\usepackage{Sweave}
\title{Psychophysics II}
\date{September 9, 2008}
\begin{document}
\frame{
\begin{Schunk}
\begin{Sinput}
ro - 0.2
c - seq(from = -3, to = 4, by = 0.1)
fn - 1 - pnorm(c)
fo - 1 - pnorm(c, mean = 1)
h - fo + ro - ro * fo
f - fn
plot(h ~ f, type = l, asp = 1, xlim = c(0.01, 1.01), ylim =
c(0.01, 1.01), las = 1, xlab = p(yes | old),
+ ylab = p(yes | new), main = Dual-process model,
p(recollection) = 0.2)
\end{Sinput}
\end{Schunk}
\includegraphics{20080909test-model}
}
\end{document}
The resulting LaTeX
***FAILS: *
Runaway argument?
ro - 0.2 c - seq(from = -3, to = 4, by = 0.1) fn - 1 - pnorm
\ETC.
! Paragraph ended before [EMAIL PROTECTED] was complete.
to be read again
\par
l.27 }
?
**
But when the Rnw file starts:
**
%\documentclass[compress,smaller]{beamer}
\documentclass{article}
\usepackage{beamerarticle}
***IT DOES NOT FAIL: *
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
[[alternative HTML version deleted]]
__
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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] loop
I would suggest that you use a 'list' since it seems that the result
is more than a single value:
bbayes-list()
for(i in 1:100) {
xx-t(X1_10)%*%X1_10
xxmeno1-solve(xx)
V-xxmeno1*i
Vmeno1-solve(V)
tx-t(X1_10)
prpar-solve(Vmeno1+xx)
snpar-tx%*%y
bbayes[[i]]-prpar%*%snpar
}
On Sat, Sep 6, 2008 at 9:29 AM, Davide Crapis [EMAIL PROTECTED] wrote:
I have to calculate a formula that gives me a ten components vector. I want
to see how the components behave at varying the variable i.
But when i run the following function:
bbayes-c()
for(i in 1:100) {
+ xx-t(X1_10)%*%X1_10
+ xxmeno1-solve(xx)
+ V-xxmeno1*i
+ Vmeno1-solve(V)
+ tx-t(X1_10)
+ prpar-solve(Vmeno1+xx)
+ snpar-tx%*%y
+ bbayes-prpar%*%snpar
+ }
it oly gives me the vector calculated with the last value of the sequence,
in this case 100.
Could you please suggest how to print all the 100 vectors calculated for i
in 1:100.
Thank you in advance.
Davide Crapis
--
View this message in context:
http://www.nabble.com/loop-tp19346683p19346683.html
Sent from the R help mailing list archive at Nabble.com.
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+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] request: most repeated sequnce
This may come closer since it removes the zeros before comparison:
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
+ 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
x=array(x,dim=c(3,6,5))
apply(x,3,function(.mat){
+rows - table(apply(.mat,1,function(z){
+# remove the zeros
+z - z[z != 0]
+if (length(z) == 0) return(NULL)
+paste(z,collapse=' ')
+}))
+names(rows[which.max(rows)])
+ })
[1] 1 1 2 3 1 2 3 4 1 2 3 4 2 2 3 4
On Sat, Sep 6, 2008 at 12:48 PM, Muhammad Azam [EMAIL PROTECTED] wrote:
Dear R community
Initially i thought my problem has been solved but one thing which i found
e.g. if
1. All the elements of a sector are zero e.g
, , 7
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]000000000 0
[2,]000000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
2. Majority of the rows consist of zeros e.g.
, , 5
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]440000000 0
[2,]440000000 0
[3,]000000000 0
[4,]000000000 0
[5,]000000000 0
Actually
zeros are not my values. I get values and fill the remaining parts with
zeros like x=array(0,dim=c(3,6,5)). Now according to first strategy
000000000 0 are most repeated
sequence of rows in both of above cases. But i don't want to consider
cases where all elements are zeros and interested to get 44
0000000 0 or just 4 4 in case 2.
Thanks and best regards
Muhammad Azam
- Original Message
From: jim holtman [EMAIL PROTECTED]
To: Muhammad Azam [EMAIL PROTECTED]
Cc: R Help r-help@r-project.org; R-help request [EMAIL PROTECTED]
Sent: Saturday, September 6, 2008 2:39:19 PM
Subject: Re: [R] request: most repeated sequnce
Here is a start. You can delete the zeros:
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
+ 0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
x=array(x,dim=c(3,6,5))
apply(x,3,function(.mat){
+ rows - table(apply(.mat,1,function(z){
+ paste(z,collapse=' ')
+ }))
+ names(rows[which.max(rows)])
+ })
[1] 1 0 0 0 0 0 1 2 3 0 0 0 1 2 3 4 0 0 1 2 3 4 0 0 2 2 3 4 0 0
On Sat, Sep 6, 2008 at 4:54 AM, Muhammad Azam [EMAIL PROTECTED] wrote:
Dear R community
Hope every one be in best of his/her health. I have a situation in which
there are s-sectors. Each sector is further divided into r-rows and
c-columns. All it makes an array having dimension (r,c,s). e.g.
x=c(1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,0,0,0,0,0,0,0,0,0,0,1,1,1,2,2,3,3,3,4,4,4,0,0,0,0,0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,
0,0,0,0,0,0,1,2,2,2,2,2,0,3,3,0,4,4,0,0,0,0,0,0)
x=array(x,dim=c(3,6,5))
x
, , 1
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]100000
[2,]100000
[3,]100000
, , 2
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123000
[2,]123000
[3,]120000
, , 3
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123400
[2,]123400
[3,]134000
, , 4
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]123400
[2,]123400
[3,]123400
, , 5
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]120000
[2,]223400
[3,]223400
I want to get the most repeated sequence (row-wise) of values in each
sector. e.g. in sector 1 i.e. , , 1
the most repeated sequence is 1 (ignoring zeros). In , , 2 the most repeated
sequence is 1 2 3. Similarly in last sector i.e.
, , 5 such sequence is 2 2 3 4. Any body can help to solve this
problem. Thanks
best regards
Muhammad Azam
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Re: [R] re ferring to a group of vectors without explicit enumeration
I would suggest that you use a list to store the values since it is easier to create and reference: output - list() for (i in 1:10) output[[i]] - seq(i) output [[1]] [1] 1 [[2]] [1] 1 2 [[3]] [1] 1 2 3 [[4]] [1] 1 2 3 4 [[5]] [1] 1 2 3 4 5 [[6]] [1] 1 2 3 4 5 6 [[7]] [1] 1 2 3 4 5 6 7 [[8]] [1] 1 2 3 4 5 6 7 8 [[9]] [1] 1 2 3 4 5 6 7 8 9 [[10]] [1] 1 2 3 4 5 6 7 8 9 10 On Sat, Sep 6, 2008 at 5:33 PM, shalu [EMAIL PROTECTED] wrote: I am trying to define 25 vectors of varying lengths, say y1 to y25 in a loop, and then store the results of some computations in them. My problem is about using some sort of concatenation for names. For example, instead of initializing each of y1 through y25, I would like to do it in a loop. Similar to cat and paste for texts, is there anyway of using yi for the vector name where i ranges from 1 to 25, so ultimately it refers to the vector y1,..,y25? Varying lengths is not a problem. To start with each has only length 1 and then I will be adding to each vector based on some results. -- View this message in context: http://www.nabble.com/referring-to-a-group-of-vectors-without-explicit-enumeration-tp19351518p19351518.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave and/or beamer issue
Hi Michael,
I think that beamer needs that frames that contain any verbatim text
or R output to be declared as fragile.
Try
\begin{frame}[fragile]
...
\end{frame}
On Sat, Sep 06, 2008 at 06:22:41PM -0400, Michael Kubovy wrote:
Dear Friends,
I not sure whether this is an Sweave or a beamer problem.
The Rnw file:
\documentclass[compress,smaller]{beamer}
%\documentclass{article}
%\usepackage{beamerarticle}
\usepackage{Sweave}
\title{Psychophysics II}
\date{September 9, 2008}
\begin{document}
\frame{
\begin{Schunk}
\begin{Sinput}
ro - 0.2
c - seq(from = -3, to = 4, by = 0.1)
fn - 1 - pnorm(c)
fo - 1 - pnorm(c, mean = 1)
h - fo + ro - ro * fo
f - fn
plot(h ~ f, type = l, asp = 1, xlim = c(0.01, 1.01), ylim =
c(0.01, 1.01), las = 1, xlab = p(yes | old),
+ ylab = p(yes | new), main = Dual-process model,
p(recollection) = 0.2)
\end{Sinput}
\end{Schunk}
\includegraphics{20080909test-model}
}
\end{document}
The resulting LaTeX
***FAILS: *
Runaway argument?
ro - 0.2 c - seq(from = -3, to = 4, by = 0.1) fn - 1 - pnorm
\ETC.
! Paragraph ended before [EMAIL PROTECTED] was complete.
to be read again
\par
l.27 }
?
**
But when the Rnw file starts:
**
%\documentclass[compress,smaller]{beamer}
\documentclass{article}
\usepackage{beamerarticle}
***IT DOES NOT FAIL: *
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
[[alternative HTML version deleted]]
__
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.
--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-6410
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/
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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] re ferring to a group of vectors without explicit enumeration
On Sep 6, 2008, at 6:07 PM, Ben Bolker wrote: shalu shahlar at hotmail.com writes: I am trying to define 25 vectors of varying lengths, say y1 to y25 in a loop, and then store the results of some computations in them. My problem is about using some sort of concatenation for names. For example, instead of initializing each of y1 through y25, I would like to do it in a loop. Similar to cat and paste for texts, is there anyway of using yi for the vector name where i ranges from 1 to 25, so ultimately it refers to the vector y1,..,y25? Varying lengths is not a problem. To start with each has only length 1 and then I will be adding to each vector based on some results. I think this is essentially http://cran.r-project.org/doc/FAQ/R-FAQ.html #How-can-I-turn-a-string-into-a-variable? [URL broken in order to make Gmane happy, reassemble it in your browser] the short answer: assign(), but it would work better to use a list instead. It may help to consider these options: varnames - paste(y, 1:25, sep=) varnames [1] y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 y12 y13 y14 y15 y16 y17 y18 y19 [20] y20 y21 y22 y23 y24 y25 varlist - list(paste(varbl, 1:25, sep=)) varlist [[1]] [1] varbl1 varbl2 varbl3 varbl4 varbl5 varbl6 varbl7 varbl8 varbl9 varbl10 varbl11 [12] varbl12 varbl13 varbl14 varbl15 varbl16 varbl17 varbl18 varbl19 varbl20 varbl21 varbl22 [23] varbl23 varbl24 varbl25 -- David Winsemius __ 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 address last and all but last column in dataframe
On Sep 6, 2008, at 4:24 PM, drflxms wrote: Hello Mr. Burns, Hello Mr. Dwinseminus snip David, the input data.frame is the result of the reshape-command I performed. I just copied it from the R-console into the e-mail. In fact the first column video is not part of the data, but needed for analysis with kappam.fleiss function of the irr-package. Sorry, you are absolutely correct, I should have mentioned this in my question. I will improve when I ask my next question :-). I have not yet picked up facility with reshape. Is there a method using reshape to take its screen output and return a dataframe? -- David Winsemius, MD Again I like to thank you for your help and wish you a pleasant Sunday. Greetings from Munich, Felix Patrick Burns wrote: If I understand properly, you want input[, -c(1, ncol(input))] rather than input[[, -c(video, 21)]] Chapter 1 of S Poetry might be of interest to you. Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) drflxms wrote: Dear R-colleagues, another question from a newbie: I am creating a lot of simple pivot-charts from my raw data using the reshape-package. In these charts we have medical doctors judging videos in the columns and the videos they judge in the rows. Simple example of chart/data.frame input with two categories 1/0: video 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 5 5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 8 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 9 9 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 1 0 1010 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 I recently learned, that I can easily create a confusion matrix out of this data using the following commands: pairs- data.frame(pred=factor(unlist(input[2:21])),ref=factor(input[,22])) pred-pairs$pred ref - pairs$ref library (caret) confusionMatrix(pred, ref, positive=1) - where column 21 is the reference/goldstandard. My problem is now, that I analyse data.frames with an unknown count of columns. So to get rid of the first and last column for the pred variable and to select the last column for the ref variable, I have to look at the data.frame before doing the above commands to set the proper column numbers. It would be very comfortable, if I could address the last column not by number (where I have to count beforehand) but by a variable last column. Probably there is a more easy solution for this problem using the names of the columns as well: the reference is always number 21 the first column is always called video. So I tried: attach(input) pairs-data.frame(pred=factor(unlist(input[[,-c(video, 21)]])),ref=factor(input[[21]])) which does not work unfortunately :-(. I'd be very happy in case someone could help me out, cause I am really tired of counting - there are a lot of tables to analyse... Cheers and greetings from Munich, Felix __ 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] plot a list
Please read the last line to every message to r-help. In particular this question needs to include a cut down version of the data. I'll take a guess at what it looks like: library(zoo) L - list(a = zoo(1:3), b = zoo(4:5)) plot(do.call(merge, L)) On Sat, Sep 6, 2008 at 4:47 PM, stephen sefick [EMAIL PROTECTED] wrote: i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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. __ 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] Sweave and/or beamer issue
On 06/09/2008 6:38 PM, Andrew Robinson wrote:
Hi Michael,
I think that beamer needs that frames that contain any verbatim text
or R output to be declared as fragile.
Try
\begin{frame}[fragile]
...
\end{frame}
The declaration I usually use is containsverbatim, but it probably
does the same thing as fragile.
Duncan Murdoch
On Sat, Sep 06, 2008 at 06:22:41PM -0400, Michael Kubovy wrote:
Dear Friends,
I not sure whether this is an Sweave or a beamer problem.
The Rnw file:
\documentclass[compress,smaller]{beamer}
%\documentclass{article}
%\usepackage{beamerarticle}
\usepackage{Sweave}
\title{Psychophysics II}
\date{September 9, 2008}
\begin{document}
\frame{
\begin{Schunk}
\begin{Sinput}
ro - 0.2
c - seq(from = -3, to = 4, by = 0.1)
fn - 1 - pnorm(c)
fo - 1 - pnorm(c, mean = 1)
h - fo + ro - ro * fo
f - fn
plot(h ~ f, type = l, asp = 1, xlim = c(0.01, 1.01), ylim =
c(0.01, 1.01), las = 1, xlab = p(yes | old),
+ ylab = p(yes | new), main = Dual-process model,
p(recollection) = 0.2)
\end{Sinput}
\end{Schunk}
\includegraphics{20080909test-model}
}
\end{document}
The resulting LaTeX
***FAILS: *
Runaway argument?
ro - 0.2 c - seq(from = -3, to = 4, by = 0.1) fn - 1 - pnorm
\ETC.
! Paragraph ended before [EMAIL PROTECTED] was complete.
to be read again
\par
l.27 }
?
**
But when the Rnw file starts:
**
%\documentclass[compress,smaller]{beamer}
\documentclass{article}
\usepackage{beamerarticle}
***IT DOES NOT FAIL: *
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
[[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] plot a list
the plot(do.call(merge, z.l)) works on the following data well. Is there a way to get control of xlim so that it plots each individual graph shows only the one day (figures the x axis on the range of the data for each plot individually) and control labeling? Thanks in advance, and sorry for not putting an example in the first email. #here is a made up example library(chron) library(zoo) t1 - chron(1/1/2006, 00:01:00) t2 - chron(1/1/2006, 23:46:00) deltat - times(00:15:00) tt - seq(t1, t2, by = times(00:15:00)) DO - rnorm(96) Temp - rnorm(96) a - cbind(Temp, DO) t3 - chron(12/1/2006, 00:01:00) t4 - chron(12/1/2006, 23:46:00) deltat - times(00:15:00) tt.2 - seq(t3, t4, by = times(00:15:00)) DO.2 - rnorm(96) Temp.2- rnorm(96) b - cbind(Temp.2, DO.2) z1 - zoo(a, tt) z2 - zoo(b, tt.2) z.l - list(z1,z2) plot(do.call(merge, z.l)) On Sat, Sep 6, 2008 at 7:22 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Please read the last line to every message to r-help. In particular this question needs to include a cut down version of the data. I'll take a guess at what it looks like: library(zoo) L - list(a = zoo(1:3), b = zoo(4:5)) plot(do.call(merge, L)) On Sat, Sep 6, 2008 at 4:47 PM, stephen sefick [EMAIL PROTECTED] wrote: i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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] plot a list
Try xyplot.zoo with scale = list(relation = free) specifying xlim as shown below: library(zoo) library(lattice) zm - do.call(merge, z.l) xlim - lapply(zm, function(x) range(time(na.omit(x xyplot(zm, xlim = xlim, scale = list(relation = free)) On Sat, Sep 6, 2008 at 7:49 PM, stephen sefick [EMAIL PROTECTED] wrote: the plot(do.call(merge, z.l)) works on the following data well. Is there a way to get control of xlim so that it plots each individual graph shows only the one day (figures the x axis on the range of the data for each plot individually) and control labeling? Thanks in advance, and sorry for not putting an example in the first email. #here is a made up example library(chron) library(zoo) t1 - chron(1/1/2006, 00:01:00) t2 - chron(1/1/2006, 23:46:00) deltat - times(00:15:00) tt - seq(t1, t2, by = times(00:15:00)) DO - rnorm(96) Temp - rnorm(96) a - cbind(Temp, DO) t3 - chron(12/1/2006, 00:01:00) t4 - chron(12/1/2006, 23:46:00) deltat - times(00:15:00) tt.2 - seq(t3, t4, by = times(00:15:00)) DO.2 - rnorm(96) Temp.2- rnorm(96) b - cbind(Temp.2, DO.2) z1 - zoo(a, tt) z2 - zoo(b, tt.2) z.l - list(z1,z2) plot(do.call(merge, z.l)) On Sat, Sep 6, 2008 at 7:22 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Please read the last line to every message to r-help. In particular this question needs to include a cut down version of the data. I'll take a guess at what it looks like: library(zoo) L - list(a = zoo(1:3), b = zoo(4:5)) plot(do.call(merge, L)) On Sat, Sep 6, 2008 at 4:47 PM, stephen sefick [EMAIL PROTECTED] wrote: i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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] Hopefully an easy error bar question
Ben Bolker wrote: Martin Maechler maechler at stat.math.ethz.ch writes: FEH == Frank E Harrell f.harrell at vanderbilt.edu on Sat, 06 Sep 2008 07:19:33 -0500 writes: FEH See http://biostat.mc.vanderbilt.edu/DynamitePlots for many reasons not FEH to use dynamite plots. Ah! Very good! Let's hope this eventually leads to progress i.e., to much less use of these.. Thank you, Frank! Martin Maechler, ETH Zurich FEH -- Yes, but: http://emdbolker.wikidot.com/blog:dynamite (repeats Frank's criticisms, but also raises the point that boxplots show the range of the data, but not much about the statistical inferences on the data ...) comments welcome. Ben Bolker Hi Ben, I enjoyed the blog. Depending on the sample size I prefer extended box plots (e.g., ?panel.bpplot in the Hmisc package) with raw data superimposed. The 3-number summary from the box plot does not contain enough information when the sample size is moderate or large IMHO. But for small samples, raw data points supplemented by the 3 quartiles and the mean work well. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Label 2 groups in PCA different colours
Hi, I'm wanting to do a PCA on some data which is comprised of two different groups (to see how well the groups are discriminated). Is there a way to change the colour of the datapoints in a biplot so that I can easily see which group is which (eg objects 1-100, red, 101-200, black). Might be simple, but I'm new to R and can't seem to find how to do this. Thanks. Paul -- View this message in context: http://www.nabble.com/Label-2-groups-in-PCA-different-colours-tp19354077p19354077.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] Corrupted PDF files
Its a FAQ:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-do-lattice_002ftrellis-graphics-not-work_003f
On Sat, Sep 6, 2008 at 9:47 PM, Nathan Teuscher [EMAIL PROTECTED] wrote:
I have the following code that when executed from the command line
works properly and produces a proper PDF. When the script is executed,
the PDF produced is considered corrupt. I am using R 2.7.2 on Mac OSX
10.5.4. Thank you in advance for the help!
library(lattice)
pdf(file=CLDiag2.pdf)
xyplot(
CL ~ HT + WT + AGE + CREA + SEX,
data=data2,
outer=TRUE,
scales=list(x=list(relation=free)),
panel=function(...){
panel.loess(..., col=red)
panel.xyplot(..., pch=.)
}
)
dev.off()
Nathan Teuscher
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
__
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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.
