[R] Incompleteness (or bug ?) in mAr.est for VAR estimation
Hi all, I feel there is a incompleteness in mAr.est function in mAr package for VAR estimation. I does not cheak whether there is multicolinearity in data set. Here I used mAr.est function for following dataset : head(log(data1) + ) V1 V2 V3 V4 V5 V6 1 2.859340 3.909620 3.913622 3.913622 3.954699 3.699572 2 2.865624 3.910021 3.901569 3.901569 3.931433 3.708437 3 2.876949 3.912023 3.908617 3.908617 3.943134 3.708437 4 2.873000 3.909620 3.897518 3.897518 3.921973 3.716981 5 2.856470 3.912623 3.875981 3.875981 3.901569 3.716981 6 2.857619 3.915617 3.856087 3.856087 3.884241 3.716981 clearly V3 and V4 are same. However it gives VAR estimate without any problem whereas EViews notified about this problem. - [[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] Inverse t-distribution(TINV function in Excel)
Felipe Carrillo wrote: I can't actually tell which function in R is the equivalent to TINV function in Excel. Anyone familiar with that function? Thanks help.search(t distribution) I think the quantile function is what you are looking for: qt Try qt(pt(5,10),10) Berend -- View this message in context: http://www.nabble.com/Inverse-t-distribution-tp16243933p16246200.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] What is the correct model formula for the results of piecewise linear function?
Dear friends, I used the B-spline (degree=1) method to fit the piecewise linear function and the results are listed below. m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202 )),family=binomial(logit),data=point) summary(m.glm) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 - 1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 - 1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2 -10.348 2.025 -5.110 3.22e-07 *** bs(distance, degree = 1, knots = c(16.13, 24))3 -3.530 3.752 - 0.941 0.346850 bs(y, degree = 1, knots = c(-0.4357, -0.3202))1 -6.828 1.836 - 3.719 0.000200 *** bs(y, degree = 1, knots = c(-0.4357, -0.3202))2 -4.426 1.614 - 2.742 0.006105 ** bs(y, degree = 1, knots = c(-0.4357, -0.3202))3-11.216 2.861 - 3.920 8.86e-05 *** I think bs(degree=1) has fitted the piecewise linear functions. Now i hope to write the model formua for the above results, but not very sure about how to write the correct results for *distance* and *y,* because they have three parts . Logitp=12.104 + 5.815*x + 6.654*elevation - 6.755*elevation^2 -1.291*distance1 - 10.348*distance2 - 3.53*distance3 -6.828*y1 - 4.426*y2 - 11.216*y3 In the above formula, distance1-3 denotes the three linear functions for*distance *, my problem is how to replace distace1-3 using original variable *distance *? The same question is for *y*. Could anybody please show me the correct model formula according to the above result? Thanks a lot. -- With Kind Regards, Zhi Jie [[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] Question about as.numeric with tclvalue
It's not clear what is going on here, but one possibility is as.numeric(pi) [1] NA Warning message: NAs introduced by coercion OTOH, Inf works for me. The difference is that 'Inf' is a standard C99 value (like 1.23), whereas 'pi' is a variable in R. So if you want to allow expressions you need to parse and eval them, something like inp - pi assign(a, eval(parse(text=inp)), .GlobalEnv) As ever, please see the message footer and remember to supply a reproducible example. On Sun, 23 Mar 2008, Erin Hodgess wrote: Dear R People: I have an interactive menu via an Rcmdr extension package which asks for lower and upper limit to evaluate. Typically, I use: assign(a,as.numeric(tclvalue(lowlim)),envir=.GlobalEnv) and that's fine. However, if I try to use pi or Inf or -Inf, I get either coerced NAs or NaN. Does anyone have any suggestions, please? Thanks, Erin PS Happy Easter if you celebrate Easter. -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@r-project.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.
[R] update win FAQ q1?
Hi, on http://cran.at.r-project.org/bin/windows/base/rw-FAQ.html#Where-can-I-find-the-latest-version_003f it says: ... The current release is distributed as an installer `R-2.6.1-win32.exe' of about 30Mb. ... however I find this file as well: http://cran.at.r-project.org/bin/windows/base/R-2.6.2-win32.exe I guess that in the FAQ it should read as 2.6.2 Have a nice holiday, Thomas __ R-help@r-project.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] intraday OHLC plot
Dear all, thank you very much for continuing to look into this issue! I checked the versions and in Gabor's code the spaces and tabs, but still the problem remains (see code and error below). To install quantmod, I used this time Rforge and it did not work. I had to install the Defaults package from cran first by hand (ie not like zoo and xts where it recognizes the dependence). And to get the most recent version of the package I had to upgrade R as well (my mistake of course). The link mentioned by Jeff points to 0.3.4, but I got by the command install.packages(quantmod,repos=http://R-Forge.R-project.org;) the version 0.3.3 as you see below. Lines - '17-03-2008 00:00:001,57641,57661,57471,5750 + 17-03-2008 00:05:001,57491,57501,57411,5744 + 17-03-2008 00:10:001,57451,57621,57411,5749' DF - read.delim2(textConnection(Lines)) library(quantmod) Lade nötiges Paket: Defaults quantmod: Quantitative Financial Modelling Framework Version 0.3-3 http://www.quantmod.com Warning message: ungenutzte Verbindung 3 (Lines) geschlossen z - read.zoo(textConnection(Lines), # replace with myfile.dat +header = F, sep = \t, dec = ,, +format = %d-%m-%Y %H:%M:%S, tz = ) Fehler in read.zoo(textConnection(Lines), header = F, sep = \t, dec = ,, : index contains NAs q - as.quantmod.OHLC(z, col.names = c(Open, High, Low, Close)) Fehler in ncol(x) : objekt z nicht gefunden chartSeries(q) Fehler in UseMethod(as.xts) : keine anwendbare Methode für as.xts sessionInfo() R version 2.6.2 (2008-02-08) i386-pc-mingw32 locale: LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC_NUMERIC=C;LC_TIME=German_Austria.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] quantmod_0.3-3 Defaults_1.1-1 xts_0.0-11 zoo_1.4-2 loaded via a namespace (and not attached): [1] grid_2.6.2 lattice_0.17-4 tools_2.6.2 Anyway, after the upgarde and Jeff's hint to use the chartSeries function it works. Now I still have to play around to make the labels look nice... Thanks a lot for all your help, Thomas __ R-help@r-project.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 assign multiple return values
Hi,
I am moving from MATLAB, where one can easily assign a number of
output values from a function like this:
[x,y] = myfun(a,b)
Then variables x and y can be directly used in the caller workspace.
I understand that R functions return a single argument, which could be
a list. This in a way makes it possible to return multiple values with
a single function call, but accessing the list variables is a little
bit awkward.
mylist -myfun(a,b) {
return(list(x = a, y = b))
}
x = mylist$x; y = mylist$y
I found some discussion going back to 2004. Is there anything new to
add to what was said at the time?
http://thread.gmane.org/gmane.comp.lang.r.general/21223/focus=21226
http://thread.gmane.org/gmane.comp.lang.r.general/42875/focus=42876
While at it, somewhere in the Introduction to R, I think, I read
that use of return is considered a bad practice. Why is that? I
thought making clear what is returned by the function is a positive
thing...
Regards,
TL
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Re: [R] inheritence in S4
Hi Martin
I am re reading all the mail we exchange with new eyes because of all
the thing I learn in the past few weeks. That very interesting and some
new question occurs...
***
Once, you speak about callGeneric :
setClass(A, representation(x=numeric))
setClass(C, contains=c(A))
setMethod(show, A, function(object) cat(A\n))
setMethod(show, C, function(object) {
callGeneric(as(object, A))
cat(C\n)
})
new(C)
Considere the following definition (that you more or less teach me with
your yesterday remarques...) :
setMethod(show, C, function(object) {
callNextMethod()
cat(C\n)
})
In this case, is there any difference between the former and the latter ?
Which one would you use ?
(I get that in more complicate case, for example if
setClass(C, contains=c(A,B)), it might be more complicate to use
the latter, right ?)
*
This works :
setMethod(initialize,B,
function(.Object,..., yValue){
callNextMethod(.Object, ..., y=yValue)
return(.Object)
})
new(B,yValue=3)
but this does not :
setMethod(initialize,B,
function(.Object, yValue){
callNextMethod(.Object, y=yValue)
return(.Object)
})
new(B,yValue=3)
Why ?
Is there any help page about ... ?
**
showMethods gives the list of all the method. Is there a way to see all
the method for a specific signature IN THE ORDER they will be call by
callNextMethod ?
If ANY - D - E, a method that will gives :
Function initialize:
.Object = E
.Object = D
.Object = ANY
Thanks for your help
And happy easter eggs !
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
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Re: [R] update win FAQ q1?
On Mon, 24 Mar 2008, Thomas Steiner wrote: Hi, on http://cran.at.r-project.org/bin/windows/base/rw-FAQ.html#Where-can-I-find-the-latest-version_003f it says: ... The current release is distributed as an installer `R-2.6.1-win32.exe' of about 30Mb. ... however I find this file as well: http://cran.at.r-project.org/bin/windows/base/R-2.6.2-win32.exe I guess that in the FAQ it should read as 2.6.2 It does in the copy which comes with R 2.6.2! That page says, right at the top R for Windows FAQ Version for R-2.6.1 However, http://cran.at.r-project.org/bin/windows/ told you where to report this (it was not R-help, and never is for the content of CRAN). -- 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] Great difference for piecewise linear function between R and SAS
On 24/03/2008 5:23 AM, zhijie zhang wrote: Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. You're using different bases. Compare the predictions, not the coefficients. To see the bases that R uses, do this: matplot(distance, bs(distance,degree=1,knots=c(16.13,24))) matplot(y, bs(y,degree=1,knots=c(-0.4357,-0.3202))) Duncan Murdoch #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202 )),family=binomial(logit),data=point) summary(m.glm) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 - 1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 - 1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2-10.348 2.025 - 5.110 3.22e-07 *** bs(distance, degree = 1, knots = c(16.13, 24))3-3.530 3.752 - 0.941 0.346850 bs(y, degree = 1, knots = c(-0.4357, -0.3202))1-6.828 1.836 - 3.719 0.000200 *** bs(y, degree = 1, knots = c(-0.4357, -0.3202))2-4.426 1.614 - 2.742 0.006105 ** bs(y, degree = 1, knots = c(-0.4357, -0.3202))3-11.216 2.861 - 3.920 8.86e-05 *** #SAS codes data b; set a; if distance 16.13 then d1=1; else d1=0; distance2=d1*(distance - 16.13); if distance 24 then d2=1; else d2=0; distance3=d2*(distance - 24); if y-0.4357 then dd1=1; else dd1=0; y2=dd1*(y+0.4357); if y-0.3202 then dd2=1; else dd2=0; y3=dd2*(y+0.3202); run; proc logistic descending data=b; model mark =x elevation elevation*elevation distance distance2 distance3 y y2 y3; run; The LOGISTIC Procedure Analysis of Maximum Likelihood Estimates Standard Wald Parameter DFEstimate Error Chi-SquarePr ChiSq Intercept1 -2.6148 2.1445 1.4867 0.2227 x1 5.8146 1.9872 8.5615 0.0034 elevation1 0.4457 0.1506 8.7545 0.0031 elevation*elevation 1 -0.0279 0.0142 3.8533 0.0496 distance 1 -0.1091 0.0963 1.2836 0.2572 distance21 -1.0418 0.2668 15.2424 .0001 distance31 2.8633 0.7555 14.3625 0.0002 y1-16.2032 4.3568 13.8314 0.0002 y2 1 36.9974 10.3219 12.8476 0.0003 y3 1-58.4938 14.0279 17.3875 .0001 Q: What is the problem? which one is correct for the piecewise linear function? Thanks very much. __ R-help@r-project.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 assign multiple return values
On 24/03/2008 5:09 AM, Tribo Laboy wrote:
Hi,
I am moving from MATLAB, where one can easily assign a number of
output values from a function like this:
[x,y] = myfun(a,b)
Then variables x and y can be directly used in the caller workspace.
I understand that R functions return a single argument, which could be
a list. This in a way makes it possible to return multiple values with
a single function call, but accessing the list variables is a little
bit awkward.
mylist -myfun(a,b) {
return(list(x = a, y = b))
}
x = mylist$x; y = mylist$y
I found some discussion going back to 2004. Is there anything new to
add to what was said at the time?
http://thread.gmane.org/gmane.comp.lang.r.general/21223/focus=21226
http://thread.gmane.org/gmane.comp.lang.r.general/42875/focus=42876
I don't think so.
While at it, somewhere in the Introduction to R, I think, I read
that use of return is considered a bad practice. Why is that? I
thought making clear what is returned by the function is a positive
thing...
Where did you see that? It would be good advice if it said not to hide
many calls to return() deep within a long function. I'd be neutral
about it if it said not to bother with return() in a very short
function. In intermediate cases, I think using return() is good practice.
Duncan Murdoch
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Re: [R] (no subject)
Donna Tucker wrote: Is there any way to use more than one color or shape in the same plot. I would like to make the points different colors for various levels of a variable. I have tried a simple 'if' statement in the plot command, but I get an error message. Here is what I have tried and the error message I get: plot(test, if(test[,1]8) col=steelblue2 else col=wheat2)Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differIn addition: Warning message:In if (test[, 1] 8) col = steelblue2 else col = wheat2 : the condition has length 1 and only the first element will be usedI know 'x' and 'y' lenghths do NOT differ. If I just do plot(test), it works perfectly. Is there any way to this?Thanks,Donna Hi Donna, What is happening is that you have passed two arguments to plot without naming them. The first one (test) is interpreted as the x coordinates, and the second one: if(test[,1]8) col=steelblue2 else col=wheat2 finishes up as steelblue2 if test[1,1] is less than 8 or wheat2 if not. This is interpreted as the y coordinates. Unless you have only one element in test (and of course we can infer that you have more), those lengths will differ. What you can do is to specify that your second argument is meant to be a color vector like this: plot(test,col=ifelse(test[,1]8,steelblue,wheat2)) Now of course I am assuming that test is a matrix or data frame with at least two columns. If it isn't, the above example will fall over and whinge. How well do you know your celebrity gossip? I'm sorry, but I can't help you at all with that. 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] little subplot in corner
Henrique, Although your solution is perfect, I had later a problem with setting the background. Perhaps you know how to get this as well: set.seed(24032008) plot(rnorm(10),type=l,col=red) grid() front=c(0.5, .97, 0.5, .97) par(fig=front, new=T)#, bg=skyblue does not work rec=c(par(usr)[1]+(par(usr)[2]-par(usr)[1])*front[1], par(usr)[3]+(par(usr)[4]-par(usr)[3])*front[3], par(usr)[1]+(par(usr)[2]-par(usr)[1])*front[3], par(usr)[1]+(par(usr)[2]-par(usr)[1])*front[4]) #rect(rec[1],rec[2],rec[3],rec[4],col=yellow) plot(rnorm(4),type=l,col=blue,xlab=,ylab=) as you can see it would be good to be able to set the background of the subfigure. as the par(bg=) does not work, I tried to plot a rectangle. Just uncomment the send-last line and you will see that the original plot disappears after some strange rectangle is drawn. If you have any idea again, it would be evry much appreaciated to let me know. Thomas PS: to see my first attempts, look at http://commons.wikimedia.org/wiki/Image:Eurusd-id.png __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Great difference for piecewise linear function between R and SAS
Dear Murdoch, Compare the predictions, not the coefficients., this is the key point. I have checked their predicted probability and their results are the same. What do you mean by You're using different bases? Could you please give me a little more info on it, so that i can go to find some materials? Thanks a lot. On 3/24/08, Duncan Murdoch [EMAIL PROTECTED] wrote: On 24/03/2008 5:23 AM, zhijie zhang wrote: Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. You're using different bases. Compare the predictions, not the coefficients. To see the bases that R uses, do this: matplot(distance, bs(distance,degree=1,knots=c(16.13,24))) matplot(y, bs(y,degree=1,knots=c(-0.4357,-0.3202))) Duncan Murdoch #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13 ,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202 )),family=binomial(logit),data=point) summary(m.glm) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 - 1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 - 1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2-10.348 2.025 - 5.110 3.22e-07 *** bs(distance, degree = 1, knots = c(16.13, 24))3-3.530 3.752 - 0.941 0.346850 bs(y, degree = 1, knots = c(-0.4357, -0.3202))1-6.828 1.836 - 3.719 0.000200 *** bs(y, degree = 1, knots = c(-0.4357, -0.3202))2-4.426 1.614 - 2.742 0.006105 ** bs(y, degree = 1, knots = c(-0.4357, -0.3202))3-11.216 2.861 - 3.920 8.86e-05 *** #SAS codes data b; set a; if distance 16.13 then d1=1; else d1=0; distance2=d1*(distance - 16.13); if distance 24 then d2=1; else d2=0; distance3=d2*(distance - 24); if y-0.4357 then dd1=1; else dd1=0; y2=dd1*(y+0.4357); if y-0.3202 then dd2=1; else dd2=0; y3=dd2*(y+0.3202); run; proc logistic descending data=b; model mark =x elevation elevation*elevation distance distance2 distance3 y y2 y3; run; The LOGISTIC Procedure Analysis of Maximum Likelihood Estimates Standard Wald Parameter DFEstimate Error Chi-SquarePr ChiSq Intercept1 -2.6148 2.1445 1.4867 0.2227 x1 5.8146 1.9872 8.5615 0.0034 elevation1 0.4457 0.1506 8.7545 0.0031 elevation*elevation 1 -0.0279 0.0142 3.8533 0.0496 distance 1 -0.1091 0.0963 1.2836 0.2572 distance21 -1.0418 0.2668 15.2424 .0001 distance31 2.8633 0.7555 14.3625 0.0002 y1-16.2032 4.3568 13.8314 0.0002 y2 1 36.9974 10.3219 12.8476 0.0003 y3 1-58.4938 14.0279 17.3875 .0001 Q: What is the problem? which one is correct for the piecewise linear function? Thanks very much. -- With Kind Regards, oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [***] Zhi Jie,Zhang ,PHD Tel:+86-21-54237149 Dept. of Epidemiology,School of Public Health,Fudan University Address:No. 138 Yi Xue Yuan Road,Shanghai,China Postcode:200032 Email:[EMAIL PROTECTED] Website: www.statABC.com [***] oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [[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] Inverse t-distribution
Bert Gunter wrote: Yes. And, amazingly, one can find out about it by typing help.search(t distribution) . Why don't you try it for yourself? Maybe because the obvious search is help.search(t) :-) Alberto Monteiro __ R-help@r-project.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] intraday OHLC plot
You need zoo 1.5-0 and you are using 1.4-2. Also note that the DF - line is not needed (though it doesn't hurt either) and was just there to check your similar statement. On Mon, Mar 24, 2008 at 4:48 AM, Thomas Steiner [EMAIL PROTECTED] wrote: Dear all, thank you very much for continuing to look into this issue! I checked the versions and in Gabor's code the spaces and tabs, but still the problem remains (see code and error below). To install quantmod, I used this time Rforge and it did not work. I had to install the Defaults package from cran first by hand (ie not like zoo and xts where it recognizes the dependence). And to get the most recent version of the package I had to upgrade R as well (my mistake of course). The link mentioned by Jeff points to 0.3.4, but I got by the command install.packages(quantmod,repos=http://R-Forge.R-project.org;) the version 0.3.3 as you see below. Lines - '17-03-2008 00:00:001,57641,57661,57471,5750 + 17-03-2008 00:05:001,57491,57501,57411,5744 + 17-03-2008 00:10:001,57451,57621,57411,5749' DF - read.delim2(textConnection(Lines)) library(quantmod) Lade nötiges Paket: Defaults quantmod: Quantitative Financial Modelling Framework Version 0.3-3 http://www.quantmod.com Warning message: ungenutzte Verbindung 3 (Lines) geschlossen z - read.zoo(textConnection(Lines), # replace with myfile.dat +header = F, sep = \t, dec = ,, +format = %d-%m-%Y %H:%M:%S, tz = ) Fehler in read.zoo(textConnection(Lines), header = F, sep = \t, dec = ,, : index contains NAs q - as.quantmod.OHLC(z, col.names = c(Open, High, Low, Close)) Fehler in ncol(x) : objekt z nicht gefunden chartSeries(q) Fehler in UseMethod(as.xts) : keine anwendbare Methode für as.xts sessionInfo() R version 2.6.2 (2008-02-08) i386-pc-mingw32 locale: LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC_NUMERIC=C;LC_TIME=German_Austria.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] quantmod_0.3-3 Defaults_1.1-1 xts_0.0-11 zoo_1.4-2 loaded via a namespace (and not attached): [1] grid_2.6.2 lattice_0.17-4 tools_2.6.2 Anyway, after the upgarde and Jeff's hint to use the chartSeries function it works. Now I still have to play around to make the labels look nice... Thanks a lot for all your help, Thomas __ R-help@r-project.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] Great difference for piecewise linear function between R and SAS
On 24/03/2008 7:06 AM, zhijie zhang wrote: Dear Murdoch, Compare the predictions, not the coefficients., this is the key point. I have checked their predicted probability and their results are the same. What do you mean by You're using different bases? Could you please give me a little more info on it, so that i can go to find some materials? There are many ways to represent a piecewise linear function. R and your SAS code have both chosen linear combinations of basis functions, but you have used different basis functions. R uses the B-spline basis. You have what is sometimes called the truncated power basis, but maybe not commonly in the linear context. I don't know if it has a name here. Duncan Murdoch Thanks a lot. On 3/24/08, Duncan Murdoch [EMAIL PROTECTED] wrote: On 24/03/2008 5:23 AM, zhijie zhang wrote: Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. You're using different bases. Compare the predictions, not the coefficients. To see the bases that R uses, do this: matplot(distance, bs(distance,degree=1,knots=c(16.13,24))) matplot(y, bs(y,degree=1,knots=c(-0.4357,-0.3202))) Duncan Murdoch #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13 ,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202 )),family=binomial(logit),data=point) summary(m.glm) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 - 1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 - 1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2-10.348 2.025 - 5.110 3.22e-07 *** bs(distance, degree = 1, knots = c(16.13, 24))3-3.530 3.752 - 0.941 0.346850 bs(y, degree = 1, knots = c(-0.4357, -0.3202))1-6.828 1.836 - 3.719 0.000200 *** bs(y, degree = 1, knots = c(-0.4357, -0.3202))2-4.426 1.614 - 2.742 0.006105 ** bs(y, degree = 1, knots = c(-0.4357, -0.3202))3-11.216 2.861 - 3.920 8.86e-05 *** #SAS codes data b; set a; if distance 16.13 then d1=1; else d1=0; distance2=d1*(distance - 16.13); if distance 24 then d2=1; else d2=0; distance3=d2*(distance - 24); if y-0.4357 then dd1=1; else dd1=0; y2=dd1*(y+0.4357); if y-0.3202 then dd2=1; else dd2=0; y3=dd2*(y+0.3202); run; proc logistic descending data=b; model mark =x elevation elevation*elevation distance distance2 distance3 y y2 y3; run; The LOGISTIC Procedure Analysis of Maximum Likelihood Estimates Standard Wald Parameter DFEstimate Error Chi-SquarePr ChiSq Intercept1 -2.6148 2.1445 1.4867 0.2227 x1 5.8146 1.9872 8.5615 0.0034 elevation1 0.4457 0.1506 8.7545 0.0031 elevation*elevation 1 -0.0279 0.0142 3.8533 0.0496 distance 1 -0.1091 0.0963 1.2836 0.2572 distance21 -1.0418 0.2668 15.2424 .0001 distance31 2.8633 0.7555 14.3625 0.0002 y1-16.2032 4.3568 13.8314 0.0002 y2 1 36.9974 10.3219 12.8476 0.0003 y3 1-58.4938 14.0279 17.3875 .0001 Q: What is the problem? which one is correct for the piecewise linear function? Thanks very much. __ R-help@r-project.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] Great difference for piecewise linear function between R and SAS
On 3/24/2008 9:03 AM, zhijie zhang wrote: Dear Murdoch, Thanks very much for your rapid response. It helps me greatly. If i want to write the model formula according to the estimets from R, Is it correct for the below formula? I'm not very sure about it, and i also hope that you can recommend a good book on this area. I want to learn it systematically. Thanks a million. Logit/p/=12.10+5.815*x+6.654*elevation-6.755*elevation^2 -1.291*distance_1 -10.348*distance2-3.53*distance3 _ -6.828*_y1 _ -4.426*_y2 _ -11.216*_y3 _ I would use predict() instead. What you have there doesn't look as though it uses the B-spline basis. The reference given in the ?bs help page is a reasonable starting point, but just about any book that covers splines should handle the B-spline basis and the linear case. Duncan Murdoch #R code and the estimate results--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202)),family=binomial(logit),data=point) summary(m.glm) Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 -1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 -1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2-10.348 2.025 -5.110 3.22e-07 *** bs(distance, degree = 1, knots = c(16.13, 24))3-3.530 3.752 -0.941 0.346850 bs(y, degree = 1, knots = c(-0.4357, -0.3202))1-6.828 1.836 -3.719 0.000200 *** bs(y, degree = 1, knots = c(-0.4357, -0.3202))2-4.426 1.614 -2.742 0.006105 ** bs(y, degree = 1, knots = c(-0.4357, -0.3202))3-11.216 2.861 -3.920 8.86e-05 *** Thanks again. On Mon, Mar 24, 2008 at 8:08 PM, Duncan Murdoch [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: On 24/03/2008 7:06 AM, zhijie zhang wrote: Dear Murdoch, Compare the predictions, not the coefficients., this is the key point. I have checked their predicted probability and their results are the same. What do you mean by You're using different bases? Could you please give me a little more info on it, so that i can go to find some materials? There are many ways to represent a piecewise linear function. R and your SAS code have both chosen linear combinations of basis functions, but you have used different basis functions. R uses the B-spline basis. You have what is sometimes called the truncated power basis, but maybe not commonly in the linear context. I don't know if it has a name here. Duncan Murdoch Thanks a lot. On 3/24/08, Duncan Murdoch [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: On 24/03/2008 5:23 AM, zhijie zhang wrote: Dear Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. You're using different bases. Compare the predictions, not the coefficients. To see the bases that R uses, do this: matplot(distance, bs(distance,degree=1,knots=c(16.13,24))) matplot(y, bs(y,degree=1,knots=c(-0.4357,-0.3202))) Duncan Murdoch #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13 ,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202 )),family=binomial(logit),data=point) summary(m.glm) Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 12.104 3.138 3.857 0.000115 *** x 5.815 1.987 2.926 0.003433 ** poly(elevation, 2)1 6.654 4.457 1.493 0.135444 poly(elevation, 2)2 -6.755 3.441 - 1.963 0.049645 * bs(distance, degree = 1, knots = c(16.13, 24))1 -1.291 1.139 - 1.133 0.257220 bs(distance, degree = 1, knots = c(16.13, 24))2-10.348 2.025 -
[R] ARCH(1,0) with t-residuals
Dear R users, I need to estimate an ARCH(1,0) model with t-residuals. To do this I use garchFit function from fGarch library. However, I get the following error message: Error in.garchInitParameters (formula.mean = formula.mean, formula.var = formula.var, ): object alpha not found I tried to estimate this model with different series, but I always get this error message. For example, data(EuStockMarkets) dax - diff(log(EuStockMarkets))[,DAX] g=garchFit(formula=~garch(0,1),data=dax,cond.dist=dstd,include.mean=FALSE) Series Initialization: ARMA model:arma Formula mean: ~ arma(0, 0) GARCH model: garch Formula var: ~ garch(0, 1) ARMA Order:0 0 Max ARMA Order:0 GARCH Order: 0 1 Max GARCH Order: 1 Maximum Order: 1 h.start: 2 llh.start: 1 Length of Series: 1859 Recursion Init:mci Series Scale: 0.01030084 Error in .garchInitParameters(formula.mean = formula.mean, formula.var = formula.var, : object alpha not found If I use GARCH(0,1) model I get the following error message: g=garchFit(formula=~garch(0,1),data=dax,cond.dist=dstd,include.mean=FALSE) Series Initialization: ARMA model:arma Formula mean: ~ arma(0, 0) GARCH model: garch Formula var: ~ garch(1, 0) ARMA Order:0 0 Max ARMA Order:0 GARCH Order: 1 0 Max GARCH Order: 1 Maximum Order: 1 h.start: 2 llh.start: 1 Length of Series: 1859 Recursion Init:mci Series Scale: 0.01030084 Error in sum(beta) : invalid 'type' (closure) of argument Do fGarch function allow to estimate GARCH(p,q) with p and q 0 only? Are there any other functions in order to estimate ARCH(1,0) with t-residuals? Thank you. Regards, Elena Advertisement: Lietuviams reikia saugios pasto dezutes. Dideles ir saugios. INBOX.LT - 10 GB nemokamas pastas! [[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] vlookup in R
Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[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] vlookup in R
I think that merge is what you want: set.seed(24032008) x - data.frame(ID=sample(10), Value=rnorm(10)) idx - sample(5) merge(idx, x, by.x=1, by.y=1) On 24/03/2008, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vlookup in R
?findInterval On 3/24/08, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem 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] [PS] vlookup in R
Another way: If x is a two column matrix, as suggested by Henrique D., IDValue 1 7 0.000656733 2 6 0.201764789 3 1 0.671113391 4 10 -0.739727826 5 9 -1.111310154 6 5 -0.859455833 7 2 -1.408229877 8 8 0.993126295 9 3 -0.171906808 10 4 -0.140107677 And you are looking up the value corresponding to ID ID - 4 x[(1:dim(x)[1])[x[,1]==ID],2] will also do it, and you can vary the value of the 2 in order to query the column of interest, much as you can do with vlookup in the E program. Ben -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Sachin J Sent: Monday, March 24, 2008 9:25 AM To: r-help@r-project.org Subject: [PS] [R] vlookup in R Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin __ R-help@r-project.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] vlookup in R
Sachin J [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Is there are function similar to excel vlookup in R. Please let me
know.
Caveat: definition of VLOOKUP done from memory and by checking OO.o
Calc function of same name. (Don't have Excel on this machine.)
VLOOKUP looks up a single value in the first column of an Excel range
and returns a column value (offset by a given integer) from the first
matching row in that range. The indexing functions (extract or
[ ) can be used:
df4
V1 V2 V3
1 4.56 1 0.1
2 8.42 1 0.2
3 0.79 3 0.3
4 5.39 3 0.4
5 0.95 4 0.5
6 7.73 5 0.6
7 7.17 6 0.7
8 3.89 7 0.8
9 0.54 10 1.0
10 9.53 9 0.9
df4[df4$V1==0.79,2]
[1] 3
vlookup - function(val, df, row){
df[df[1] == val, row][1] }
vlookup(0.79, df4, 2)
[1] 3
I thought there was an optional 4th argument to VLOOKUP that specifies
the action to be taken if there is no exact match. You may need to
change the equality in that function to an inequality and identify the
first column value that is less than or equal to val. If I remember
correctly, Excel assumes that the first column is ordered ascending.
--
David Winsemius
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] [PS] Re: vlookup in R
Quite right, there is an optional 4th argument, and the table must be
sorted ascending on the first column in Excel. Thus these functions
only approximately duplicate the Excel functions (improve on them IMHO).
BTW, I pasted the wrong formula in my reply; though it works, simpler is
ID - 4 #for example, find value corresponding to 4
x[x[,1]==ID,2]
Ben
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of David Winsemius
Sent: Monday, March 24, 2008 10:20 AM
To: [EMAIL PROTECTED]
Subject: [PS] Re: [R] vlookup in R
Sachin J [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Is there are function similar to excel vlookup in R. Please let me
know.
Caveat: definition of VLOOKUP done from memory and by checking OO.o
Calc function of same name. (Don't have Excel on this machine.)
VLOOKUP looks up a single value in the first column of an Excel range
and returns a column value (offset by a given integer) from the first
matching row in that range. The indexing functions (extract or
[ ) can be used:
df4
V1 V2 V3
1 4.56 1 0.1
2 8.42 1 0.2
3 0.79 3 0.3
4 5.39 3 0.4
5 0.95 4 0.5
6 7.73 5 0.6
7 7.17 6 0.7
8 3.89 7 0.8
9 0.54 10 1.0
10 9.53 9 0.9
df4[df4$V1==0.79,2]
[1] 3
vlookup - function(val, df, row){
df[df[1] == val, row][1] }
vlookup(0.79, df4, 2)
[1] 3
I thought there was an optional 4th argument to VLOOKUP that specifies
the action to be taken if there is no exact match. You may need to
change the equality in that function to an inequality and identify the
first column value that is less than or equal to val. If I remember
correctly, Excel assumes that the first column is ordered ascending.
--
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.
__
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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] inheritence in S4
[EMAIL PROTECTED] wrote:
Hi Martin
I am re reading all the mail we exchange with new eyes because of all
the thing I learn in the past few weeks. That very interesting and some
new question occurs...
***
Once, you speak about callGeneric :
setClass(A, representation(x=numeric))
setClass(C, contains=c(A))
setMethod(show, A, function(object) cat(A\n))
setMethod(show, C, function(object) {
callGeneric(as(object, A))
cat(C\n)
})
new(C)
Considere the following definition (that you more or less teach me with
your yesterday remarques...) :
setMethod(show, C, function(object) {
callNextMethod()
cat(C\n)
})
In this case, is there any difference between the former and the latter ?
Which one would you use ?
callNextMethod is the right thing to do for this case. callGeneric is
useful in a very specific case -- when dispatching from within a
so-called 'group generic' function. But this is an advanced topic.
(I get that in more complicate case, for example if
setClass(C, contains=c(A,B)), it might be more complicate to use
the latter, right ?)
The right thing to do in this case is to sit down with the rules of
method dispatch, and figure out what the 'next' method will be. A common
alternative paradigm is to have a plain function (not visible to the
user, e.g., not exported in a package name space) that several different
methods all invoke, after mapping their arguments appropriately. The
methods provide a kind of structured interface to the function, making
sure arguments are of the appropriate type, etc. The function does the
work, confident that the arguments are appropriate.
*
This works :
setMethod(initialize,B,
function(.Object,..., yValue){
callNextMethod(.Object, ..., y=yValue)
return(.Object)
})
new(B,yValue=3)
but this does not :
setMethod(initialize,B,
function(.Object, yValue){
callNextMethod(.Object, y=yValue)
return(.Object)
})
new(B,yValue=3)
Why ?
Is there any help page about ... ?
Both 'work' in the sense that an object is returned (by the way, no need
to use 'return' explicitly). And actually the examples on some of the
man pages do not include '...', so this is really my opinion rather than
the 'right' way to do things.
In an object-oriented sense, initialize,B-method should really just deal
with it's own slots; it shouldn't have to 'know' about either classes
that it extends (A) or classes that extend it. And it shouldn't do work
that inherited methods (i.e., initialize,ANY-method) do. In the second
form above, without the ..., there is no way for the initialize,A-method
to see arguments that might be relevant to it (e.g., values to be used
to initialize its slots). So initialize,B-method would have to do all
the work of initializing A. This is not good design.
**
showMethods gives the list of all the method. Is there a way to see all
the method for a specific signature IN THE ORDER they will be call by
callNextMethod ?
If ANY - D - E, a method that will gives :
Function initialize:
.Object = E
.Object = D
.Object = ANY
There is, but I have never been able to figure it out in detail or to
feel confident that I was using functions that were meant to be used for
this purpose by the user (as opposed to by the methods package).
John Chambers posted recently to the R-devel mailing list about changes
to the internal representation of methods and classes.
https://stat.ethz.ch/pipermail/r-devel/2008-March/048729.html
I have not explored the new functions Dr. Chambers mentions; to use them
requires the 'devel' version of R, not 2.6.2. Any questions they
generate should definitely be addressed to the R-devel mailing list.
Best,
Martin
Thanks for your help
And happy easter eggs !
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
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Re: [R] vlookup in R
Hi Henrique, This is what I am trying to accomplish: I want to read values in V1 and V2 and populate the column V4 with R or B based on the values in V3 i.e. if its A,B,C then its R else if D,F then B. You can assume V1, V2 are dataframe1 and V3, V4 are in dataframe2 (note V4 is empty initially i.e. its a new column). df1 V1V2 A R BR CR DB FB df2 V3V4 AR CR DB FB AR CR BR BR BR BR AR DB Thanks in advance for your help. - Original Message From: Henrique Dallazuanna [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Monday, March 24, 2008 10:41:52 AM Subject: Re: [R] vlookup in R I think that merge is what you want: set.seed(24032008) x - data.frame(ID=sample(10), Value=rnorm(10)) idx - sample(5) merge(idx, x, by.x=1, by.y=1) On 24/03/2008, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] function for the average or expected range?; CORECTION
Well it looks like you found your answer. Further the fact that my
suggestions of possibilities did not help and the fact that noone else
has chimed in would suggest that there is not any easier way to get your
answer.
I was thinking that taking into account the correlation between the min
and the max may give different answers than your formula, but so far my
tests have not shown enough of a difference to matter. If you use this
with a distribution rather than the normal, you may want to do a couple
of simulations just to check, but otherwise your formula seems to be
working fine.
Hope this helped,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: Spencer Graves [mailto:[EMAIL PROTECTED]
Sent: Saturday, March 22, 2008 9:40 AM
To: Greg Snow
Cc: r-help@r-project.org
Subject: Re: [R] function for the average or expected range?;
CORECTION
Hi, Greg:
1. I did the integration in Excel for four reasons:
First, it's easier (even for me) to see what's happening and
debug for something that simple. Second, my audience were
people who were probably not R literate, and they could
likely understand and use the Excel file easier than than an
R script. Third, my experience with the R 'integrate' has
been less than satisfactory, especially when integrating from
(-Inf) to Inf. Finally, to check my work, I often program
things like that first in Excel then in R. If I get the same
answer in both, I feel more confident in my R results. I
haven't programmed this result in R yet, but if I do, the
fact that I already did it in Excel will make it easier for
me to be confident of the answers. The function
getParamerFun{qAnalyst} gets the correct answer from n =
2:25 but returns wrong answers outside that range.
2. I think the CORRECTION TO CORRECTION included a correct
formula:
E(R) = n*integral{-Inf to Inf of x*[(F(x))**(n-1)
- (1-F(x))**(n-1)]*dF(x).
The CORRECTION omitted the x*. The first version
had many more problems.
Am I communicating?
Best Wishes,
Spencer
Greg Snow wrote:
Why do the integration in Excel instead of using the integrate
function in R? The R function will allow integration from
-Inf to Inf.
What was the correction to the formula? The last one you showed
looked like the difference between the average min and average max,
but did not take into account the correlation between the
max and min
(going from memory, don't have my theory books handy). For
large n the
correlation is probably small enough that it makes a good
approximation.
--
--
*From:* Spencer Graves [mailto:[EMAIL PROTECTED]
*Sent:* Fri 3/21/2008 3:39 PM
*To:* Greg Snow
*Cc:* r-help@r-project.org
*Subject:* Re: [R] function for the average or expected range?;
CORECTION
Hi, Greg:
Thanks very much for the reply.
1. The 'ptukey' and 'qtukey' function are the
distribution of
the studentized range, not the range. I tried sum(ptukey(x, 2,
df=Inf, lower=FALSE))*.1 and got 1.179 vs. 1.128 in the standard
table of d2 for n = 2 observations per subgroup.
2. I tried simulation and found that I needed 1e7 or
1e8 random
normal deviates to get the accuracy of the published table.
3. Then I programmed in Excel the integral over
seq(-5, 5, .1)
using a correction to the formula I got from Kendall and Stuart and
got the exact numbers in the published table except in one
case where
it was off by 1 in the last significant digit.
Thanks again,
Spencer
Greg Snow wrote:
The ptukey and qtukey functions may be what you want (or at
least in the right direction).
You could also easily estimate this by simulation.
Hope this helps,
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Re: [R] vlookup in R
# Setup the translation map transMap - c(A=R, B=R, C=R, D=B, F=B); # The input data v3 - c(A, C, D, F, A, C, B, B, B, B, A, D); # The translated data v4 - transMap[v3]; df2 - data.frame(V3=v3, V4=v4); I recommend you to read 'An Introduction to R' that comes with any R installation. There is a lot of useful stuff in there; it takes many reads to grasp parts of it but it is worth it if you're going to use R for more than, say, two weeks. You can find it via help.start(). /Henrik On Mon, Mar 24, 2008 at 8:25 AM, Sachin J [EMAIL PROTECTED] wrote: Hi Henrique, This is what I am trying to accomplish: I want to read values in V1 and V2 and populate the column V4 with R or B based on the values in V3 i.e. if its A,B,C then its R else if D,F then B. You can assume V1, V2 are dataframe1 and V3, V4 are in dataframe2 (note V4 is empty initially i.e. its a new column). df1 V1V2 A R BR CR DB FB df2 V3V4 AR CR DB FB AR CR BR BR BR BR AR DB Thanks in advance for your help. - Original Message From: Henrique Dallazuanna [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Monday, March 24, 2008 10:41:52 AM Subject: Re: [R] vlookup in R I think that merge is what you want: set.seed(24032008) x - data.frame(ID=sample(10), Value=rnorm(10)) idx - sample(5) merge(idx, x, by.x=1, by.y=1) On 24/03/2008, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] vlookup in R
Is this what you want? control - data.frame(V1=c(A,B,C,D,F), V2=c(R,R,R,B,B)) test - data.frame(V3=sample(c(A,B,C,D,F), 10, TRUE), v4='') test V3 v4 1 A 2 F 3 B 4 F 5 B 6 B 7 C 8 F 9 F 10 B test$V4 - control$V2[match(test$V3, control$V1)] test V3 v4 V4 1 A R 2 F B 3 B R 4 F B 5 B R 6 B R 7 C R 8 F B 9 F B 10 B R On 3/24/08, Sachin J [EMAIL PROTECTED] wrote: Hi Henrique, This is what I am trying to accomplish: I want to read values in V1 and V2 and populate the column V4 with R or B based on the values in V3 i.e. if its A,B,C then its R else if D,F then B. You can assume V1, V2 are dataframe1 and V3, V4 are in dataframe2 (note V4 is empty initially i.e. its a new column). df1 V1V2 A R BR CR DB FB df2 V3V4 AR CR DB FB AR CR BR BR BR BR AR DB Thanks in advance for your help. - Original Message From: Henrique Dallazuanna [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Monday, March 24, 2008 10:41:52 AM Subject: Re: [R] vlookup in R I think that merge is what you want: set.seed(24032008) x - data.frame(ID=sample(10), Value=rnorm(10)) idx - sample(5) merge(idx, x, by.x=1, by.y=1) On 24/03/2008, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem 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] spreading out a numeric vector
Levi,
Here is one possible function:
spread - function(x, mindiff) {
df - x[-1] - x[-length(x)]
i - 1
while (any(df mindiff)) {
x[c(df mindiff, FALSE)] - x[c(df mindiff, FALSE)] - mindiff/10
x[c(FALSE, df mindiff)] - x[c(FALSE, df mindiff)] + mindiff/10
df - x[-1] - x[-length(x)]
i - i + 1
if (i 100) {
break
}
}
x
}
I have tried experimenting with using optim to minimize a function of
the distances between new points and old points penealized for being to
close, but it sometimes gave me some weird results, the above has worked
fairly well for me (the above is based on some of the code inside the
triplot function in the TeachingDemos package).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Levi Waldron
Sent: Saturday, March 22, 2008 7:03 PM
To: R-help mailing list
Subject: [R] spreading out a numeric vector
I am creating a timeline plot, but running into a problem in
positions where the values to plot are too close together to
print the text without overlap. The simplest way I can think
of to solve this (although there may be other ways?) is to
create a new numeric vector whose values are as close as
possible to the original vector, but spread out to a given
minimum difference. For example, take the following vector:
x - c(1,2,seq(2.1,2.3,0.1),3,4)
x
[1] 1.0 2.0 2.1 2.2 2.3 3.0 4.0
However, suppose I don't want to plot any of these points
with less than 0.5 between them. The problem could be solved
by a function that behaved something like this:
x2 - spread(x,mindiff=0.5)
x2
[1] 0.5 1.0 1.5 2.0 2.5 3.0 4.0
Or for a minimum difference of 0.2,
x2 - spread(x,mindiff=0.2)
x2
[1] 1.0 1.8 2.0 2.2 2.4 3.0 4.0
Thus, if there is a cluster of close values, spreading the
values may require changing values which previously weren't
too close to their nearest neighbors.
Then I could use segments() to draw lines from the timeline
to where the shifted text is printed, labeling tics on the
timeline accurately but keeping the text from overlapping.
Any ideas on how to program this or how to do it using
built-in functions would be greatly appreciated.
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Re: [R] vlookup in R
Thank you all for your help. I will definately go through the suggested document. Thanks again. Sachin - Original Message From: Henrik Bengtsson [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED] Cc: Henrique Dallazuanna [EMAIL PROTECTED]; r-help@r-project.org Sent: Monday, March 24, 2008 12:43:42 PM Subject: Re: [R] vlookup in R # Setup the translation map transMap - c(A=R, B=R, C=R, D=B, F=B); # The input data v3 - c(A, C, D, F, A, C, B, B, B, B, A, D); # The translated data v4 - transMap[v3]; df2 - data.frame(V3=v3, V4=v4); I recommend you to read 'An Introduction to R' that comes with any R installation. There is a lot of useful stuff in there; it takes many reads to grasp parts of it but it is worth it if you're going to use R for more than, say, two weeks. You can find it via help.start(). /Henrik On Mon, Mar 24, 2008 at 8:25 AM, Sachin J [EMAIL PROTECTED] wrote: Hi Henrique, This is what I am trying to accomplish: I want to read values in V1 and V2 and populate the column V4 with R or B based on the values in V3 i.e. if its A,B,C then its R else if D,F then B. You can assume V1, V2 are dataframe1 and V3, V4 are in dataframe2 (note V4 is empty initially i.e. its a new column). df1 V1V2 A R BR CR DB FB df2 V3V4 AR CR DB FB AR CR BR BR BR BR AR DB Thanks in advance for your help. - Original Message From: Henrique Dallazuanna [EMAIL PROTECTED] To: Sachin J [EMAIL PROTECTED] Cc: r-help@r-project.org Sent: Monday, March 24, 2008 10:41:52 AM Subject: Re: [R] vlookup in R I think that merge is what you want: set.seed(24032008) x - data.frame(ID=sample(10), Value=rnorm(10)) idx - sample(5) merge(idx, x, by.x=1, by.y=1) On 24/03/2008, Sachin J [EMAIL PROTECTED] wrote: Hi, Is there are function similar to excel vlookup in R. Please let me know. Thanks, Sachin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] inheritence in S4
callGeneric is an advanced topic.
Ok, when I will be older :-)
*
This works :
setMethod(initialize,B,
function(.Object,..., yValue){
callNextMethod(.Object, ..., y=yValue)
return(.Object)
})
new(B,yValue=3)
but this does not :
setMethod(initialize,B,
function(.Object, yValue){
callNextMethod(.Object, y=yValue)
return(.Object)
})
new(B,yValue=3)
Why ?
Is there any help page about ... ?
Both 'work' in the sense that an object is returned
Well yes, but the second one does return an object without assigning
the value 3, that is not realy working...
In an object-oriented sense, initialize,B-method should really just
deal with it's own slots; it shouldn't have to 'know' about either
classes that it extends (A) or classes that extend it. And it
shouldn't do work that inherited methods (i.e.,
initialize,ANY-method) do.
I get your point and I agree : I am developing B, you are developing A,
I do not want to know what is in A so B should not initialize its 'A
part'
On the other hand, I do not like the ... . ... can be anything,
there is no controle at all, no type checking.
I would prefers to initialize B giving its value for its own slot AND
an object class A. So I send you the value for A, you send me an objets
'aaa' of class A then I initialize B with some value and aaa. This way,
B keep its role but does not transmit anythink to A without controling
it.
Best,
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
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[R] as.numeric with tclvalue redux
Hi again R People: This works fine: library(tcltk) a - tclVar(4.5) as.numeric(tclvalue(a)) [1] 4.5 #But if you have: b - tclVar(pi) as.numeric(tclvalue(b)) [1] NA Warning message: NAs introduced by coercion Is anyone aware of a way around this, please? thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@r-project.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] make error: ../../../bin/R: bad substitution
Hi, I am getting this error when I run 'make' under src/library/base: ../../../library/base/R/base is unchanged ../../../bin/R: bad substitution make: *** [all] Error 1 I traced it down to the following line in src/library/base/Makefile: @cat $(srcdir)/makebasedb.R | \ R_DEFAULT_PACKAGES=NULL LC_ALL=C $(R_EXE) --slave /dev/null I am trying to build R-2.6.2 on a Solaris 9 box. Can anyone advise on what may go wrong? Thank you. [[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] inheritence in S4
[EMAIL PROTECTED] wrote:
callGeneric is an advanced topic.
Ok, when I will be older :-)
*
This works :
setMethod(initialize,B,
function(.Object,..., yValue){
callNextMethod(.Object, ..., y=yValue)
return(.Object)
})
new(B,yValue=3)
but this does not :
setMethod(initialize,B,
function(.Object, yValue){
callNextMethod(.Object, y=yValue)
return(.Object)
})
new(B,yValue=3)
Why ?
Is there any help page about ... ?
Both 'work' in the sense that an object is returned
Well yes, but the second one does return an object without assigning the
value 3, that is not realy working...
I wasn't paying enough attention to your code. callNextMethod is like
any R function -- it will not change .Object 'in place', but makes a
copy, modifies the copy, and returns the copy. So either
setMethod(initialize,B,
function(.Object,..., yValue){
.Object - callNextMethod(.Object, ..., y=yValue)
return(.Object)
})
or more compactly
setMethod(initialize,B,
function(.Object,..., yValue){
callNextMethod(.Object, ..., y=yValue)
})
The code example is incomplete, so I don't really know why one version
assigned y=3 for you and the other did not; for me, neither version did
the assignment.
Martin
In an object-oriented sense, initialize,B-method should really just
deal with it's own slots; it shouldn't have to 'know' about either
classes that it extends (A) or classes that extend it. And it
shouldn't do work that inherited methods (i.e., initialize,ANY-method)
do.
I get your point and I agree : I am developing B, you are developing A,
I do not want to know what is in A so B should not initialize its 'A part'
On the other hand, I do not like the ... . ... can be anything,
there is no controle at all, no type checking.
I would prefers to initialize B giving its value for its own slot AND an
object class A. So I send you the value for A, you send me an objets
'aaa' of class A then I initialize B with some value and aaa. This way,
B keep its role but does not transmit anythink to A without controling it.
Best,
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109
Location: Arnold Building M2 B169
Phone: (206) 667-2793
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Re: [R] as.numeric with tclvalue redux
Erin Hodgess said the following on 3/24/2008 10:39 AM:
Hi again R People:
This works fine:
library(tcltk)
a - tclVar(4.5)
as.numeric(tclvalue(a))
[1] 4.5
#But if you have:
b - tclVar(pi)
as.numeric(tclvalue(b))
[1] NA
Warning message:
NAs introduced by coercion
Is anyone aware of a way around this, please?
thanks,
Erin
Does this help?
eval.tclvalue - function(x, ...) {
x - type.convert(tclvalue(x), as.is = TRUE)
if(is.character(x) exists(x, ...)) {
get(x)
} else {
x
}
}
a - tclVar(4.5)
b - tclVar(pi)
c - tclVar(abcd)
eval.tclvalue(a)
eval.tclvalue(b)
eval.tclvalue(c)
HTH,
--sundar
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Re: [R] [Rd] resampling from string when it runs across multiple lines
try this: y - as.matrix(read.table(textConnection( A C G T T G C A G C A C G F F F F F F G A C G S S S S S G A A C G T T G C A G G A B B B B B B A G T ), stringsAsFactors = FALSE)) ind - sample(length(y), 20, TRUE) y[ind] I hope it helps. Best, Dimitris ps, it would be best that you send that kind of e-mails in R-help not R-devel; check http://www.r-project.org/mail.html for more info regarding the different R-mailing-lists. Dimitris Rizopoulos Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/(0)16/336899 Fax: +32/(0)16/337015 Web: http://med.kuleuven.be/biostat/ http://www.student.kuleuven.be/~m0390867/dimitris.htm Quoting Suraaga Kulkarni [EMAIL PROTECTED]: Hi, I need to resample from a long string, which is written in many lines with carriage-return marks at the end of each line. Perhaps because the data looks like a matrix, using the code: sample(data, 25, replace=T) gives me 25 columns of characters from the data because it is resampling whole columns. What I would like it to do is to treat the data as a vector that has just been spread across many lines, and pick single characters from random positions in randomly chosen lines. I am reproducing a sample dataset, the command and the output here: y X..1. X..2. X..3. X..4. X..5. X..6. X..7. X..8. X..9. X..10. [1,] A C G T T G C A G C [2,] A C G F F F F F F G [3,] A C GS S S S S G A [4,] A C G T T G C A G G [5,] A B B B B B B A G T sample(y, 20, replace=T) X..9. X..4. X..2. X..7. X..9..1 X..3. X..3..1 X..9..2 X..9..3 X..4..1 X..3..2 X..8. X..9..4 X..3..3 X..6. X..7..1 [1,] G T C C G G G G G T G A G G G C [2,] F FC FF G G F F F G F F G F F [3,] G SC S G G G G G S G S G G S S [4,] G T C C G G G G G T G A G G G C [5,] G B B B G B B G G B B A G B B B X..6..1 X..3..4 X..7..2 X..10. [1,] G G C C [2,] F GF G [3,] S G S A [4,] G G C G [5,] B B B T I wanted to try the bootstrap approach (since that's what I am doing - resampling with replacement) but that requires a statistic and I don't know what sense that makes for character data. Any help will be greatly appreciated. Thanks, S. [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-devel Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm __ R-help@r-project.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] as.numeric with tclvalue redux
Erin Hodgess wrote: Hi again R People: This works fine: library(tcltk) a - tclVar(4.5) as.numeric(tclvalue(a)) [1] 4.5 #But if you have: b - tclVar(pi) as.numeric(tclvalue(b)) [1] NA Warning message: NAs introduced by coercion Is anyone aware of a way around this, please? Brian already told you: Parse and eval. eval(parse(text=as.character(tclvalue(tclVar(pi) [1] 3.141593 (Beware: Depending on the environment in which you eval(), strange things can happen if you use the name of an internal variable in your function) What happened with the Inf issue? I can't reproduce that: as.numeric(as.tclObj(Inf)) [1] Inf -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.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] as.numeric with tclvalue redux
Thank you! On 3/24/08, Peter Dalgaard [EMAIL PROTECTED] wrote: Erin Hodgess wrote: Hi again R People: This works fine: library(tcltk) a - tclVar(4.5) as.numeric(tclvalue(a)) [1] 4.5 #But if you have: b - tclVar(pi) as.numeric(tclvalue(b)) [1] NA Warning message: NAs introduced by coercion Is anyone aware of a way around this, please? Brian already told you: Parse and eval. eval(parse(text=as.character(tclvalue(tclVar(pi) [1] 3.141593 (Beware: Depending on the environment in which you eval(), strange things can happen if you use the name of an internal variable in your function) What happened with the Inf issue? I can't reproduce that: as.numeric(as.tclObj(Inf)) [1] Inf -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@r-project.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] make error: ../../../bin/R: bad substitution
Since you have posted about this before with errors earlier in the build process and not told us those were resolved, very likely something prior to this is the problem. You will find people here more responsive if you don't disregard the posting guide, e.g. by sending HTML mail. On Mon, 24 Mar 2008, [EMAIL PROTECTED] wrote: Hi, I am getting this error when I run 'make' under src/library/base: ../../../library/base/R/base is unchanged ../../../bin/R: bad substitution make: *** [all] Error 1 I traced it down to the following line in src/library/base/Makefile: @cat $(srcdir)/makebasedb.R | \ R_DEFAULT_PACKAGES=NULL LC_ALL=C $(R_EXE) --slave /dev/null I am trying to build R-2.6.2 on a Solaris 9 box. Can anyone advise on what may go wrong? Thank you. That doesn't help at all: we already know the problem is in bin/R, which will affect any attempt to run R. You will have to track down the 'bad substitution' (and no one else is going to be able to do it for you). BTW, R 2.6.2 runs perfectly well on Solaris 8 and 10, and we know R 2.6.0 works on Solaris 9 because sunfreeware.com distribute it. This points pretty firmly to something non-standard about your system. [[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 https://stat.ethz.ch/mailman/listinfo/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] inheritence in S4
The code example is incomplete, so I don't really know why one
version assigned y=3 for you and the other did not; for me, neither
version did the assignment.
I probably add the return in the mail without imagining il will change things.
My question was more on the use of ... versus the absence of ...
You anwer me by correcting my bug. So I can use callNextMethod with or
without ... :
setClass(B,representation(y=numeric))
setMethod(initialize,B,
function(.Object,..., yValue){
return(callNextMethod(.Object, ..., y=yValue))
})
new(B,yValue=3) #1
try(new(B,yValueee=3))#2 try(new(B,yValue=3,yValueee=3)) #3
setMethod(initialize,B,
function(.Object, yValue){
return(callNextMethod(.Object, y=yValue))
})
new(B,yValue=3)#4
try(new(B,yValueee=3)) #5
try(new(B,yValue=3,yValueee=3))#6
I undersand that 1 and 4 work. I understand that 2 and 5 do not work
since yValue is missing
I understand that 6 does not work since yValueee is not a valid argument
But I would expect that 3 will work since it get a value for yValue and
yValueee can be one of the ...
It does not...
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
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Re: [R] Great difference for piecewise linear function between R and SAS
On Mon, Mar 24, 2008 at 01:09:35PM -0500, Paul Johnson wrote: On Mon, Mar 24, 2008 at 8:11 AM, Duncan Murdoch [EMAIL PROTECTED] wrote: On 3/24/2008 9:03 AM, zhijie zhang wrote: Dear Murdoch, Thanks very I would use predict() instead. What you have there doesn't look as though it uses the B-spline basis. The reference given in the ?bs help page is a reasonable starting point, but just about any book that covers splines should handle the B-spline basis and the linear case. Duncan Murdoch Dear Duncan and others: Can you please refer us to an understandable book that explains about b-splines? I've tried a few and the math pretty quickly becomes unintelligible to a non-math major. I think you should try Simon Wood's Generalized Additive Models: An introduction in R. 2006 Chapman and Hall. I reviewed it recently and I think it's really very good. Cheers Andrew -- 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/ __ R-help@r-project.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] inheritence in S4
[EMAIL PROTECTED] wrote:
The code example is incomplete, so I don't really know why one version
assigned y=3 for you and the other did not; for me, neither version
did the assignment.
I probably add the return in the mail without imagining il will change
things.
My question was more on the use of ... versus the absence of ...
You anwer me by correcting my bug. So I can use callNextMethod with or
without ... :
setClass(B,representation(y=numeric))
setMethod(initialize,B,
function(.Object,..., yValue){
return(callNextMethod(.Object, ..., y=yValue))
})
new(B,yValue=3) #1
try(new(B,yValueee=3))#2
try(new(B,yValue=3,yValueee=3)) #3
setMethod(initialize,B,
function(.Object, yValue){
return(callNextMethod(.Object, y=yValue))
})
new(B,yValue=3)#4
try(new(B,yValueee=3)) #5
try(new(B,yValue=3,yValueee=3))#6
I undersand that 1 and 4 work. I understand that 2 and 5 do not work
since yValue is missing
I understand that 6 does not work since yValueee is not a valid argument
But I would expect that 3 will work since it get a value for yValue and
yValueee can be one of the ...
The first challenge is to understand how ... works (maybe from 'An
Introduction to R', section 10.4?)
f - function(...) names(list(...))
f(x=1)
[1] x
f(x=1, y=2)
[1] x y
Probably ok so far. Now
g - function(..., x) f(..., z=x) # 'g' transmits 'x' as 'z'
g(x=1)
[1] z
g(y=1,x=2)
[1] y z
or
h - function(..., x) f(...) # 'h' consumes 'x'
h()
NULL
h(x=1)
NULL
h(y=1)
[1] y
h(y=1,x=1)
[1] y
For instance, g passes an argument to f that consisting of (what g
received as) ... and z. f is expects ..., and z ends up as part of that
list.
What happens with initialize? The default method is described on
?initialize, where the signature is
initialize(.Object, ...)
with
...: Data to include in the new object. Named arguments
correspond to slots in the class definition.
[snip]
in #3, your callNextMethod results in the default method seeing ...
containing an argument yValueee=3. Since yValueee does not match a slot,
R responds with an error
Error in .nextMethod(.Object, ..., y = yValue) :
invalid names for slots of class B: yValueee
With the ..., a class extending B can rely on the default method to fill
in slots with provided arguments.
setClass(C, representation=representation(c=numeric), contains=B)
new(C, yValue=1, c=2) # works with ... in B's initialize method
Without ..., a class extending B would have to write an initialize
method that fills in slots (and checks validity) itself, repeating the
work already implemented in initialize,ANY-method.
It does not...
Christophe
Ce message a ete envoye par IMP, grace a l'Universite Paris 10 Nanterre
--
Martin Morgan
Computational Biology / Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N.
PO Box 19024 Seattle, WA 98109
Location: Arnold Building M2 B169
Phone: (206) 667-2793
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Re: [R] Simple problem in R
On Mon, 24 Mar 2008, Chaser wrote:
I found a package on www.bioconductor.com that allows me to install using
this line:
source(http://bioconductor.org/biocLite.R;)
biocLite(MassSpecWavelet)
The prompt showed me the following message:
Running biocinstall version 2.1.10 with R version 2.6.2
Your version of R requires version 2.1 of Bioconductor.
trying URL
'http://bioconductor.org/packages/2.1/bioc/bin/windows/contrib/2.6/MassSpecWavelet_1.4.0.zip'
Content type 'application/zip' length 2129961 bytes (2.0 Mb)
opened URL
downloaded 2.0 Mb
package 'MassSpecWavelet' successfully unpacked and MD5 sums checked
The downloaded packages are in
C:\Documents and Settings\Yuan\Local
Settings\Temp\RtmpIjgEwP\downloaded_packages
updating HTML package descriptions
In their help package, they have listed commands that will allow a sample
runthrough of their algorithm using a sample dataset.
I entered these codes in but was returned the following error messages:
data(exampleMS)
Warning message:
In data(exampleMS) : data set 'exampleMS' not found
SNR.Th - 3
peakInfo - peakDetectionCWT(exampleMS, SNR.Th=SNR.Th)
Error: could not find function peakDetectionCWT
majorPeakInfo = peakInfo$majorPeakInfo
Error: object peakInfo not found
peakIndex - majorPeakInfo$peakIndex
Error: object majorPeakInfo not found
plotPeak(exampleMS, peakIndex, main=paste('Identified peaks with SNR ',
SNR.Th))
Did I make a mistake somewhere or is the code they wrote flawed?
Did you remember
library( MassSpecWavelet )
??
HTH,
Chuck
--
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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Re: [R] spreading out a numeric vector
*Thank you* Greg, this function works perfectly! I had imagined that
the ideal solution would iteratively modify the vector to fix new
violations of mindiff created by each subsequent spreading of tight
clusters, but couldn't figure how to do it. A small note, the vector
x must be sorted before using this function, which is a reasonable
requirement.
Thanks also to Jim Lemon for pointing out the very useful plotrix
package - the spread.labels function didn't work so well for this
application because the timelines looked strange with the labels
spread evenly, but I like a number of the functions provided by the
package.
-levi
On Mon, Mar 24, 2008 at 12:43 PM, Greg Snow [EMAIL PROTECTED] wrote:
Levi,
Here is one possible function:
spread - function(x, mindiff) {
df - x[-1] - x[-length(x)]
i - 1
while (any(df mindiff)) {
x[c(df mindiff, FALSE)] - x[c(df mindiff, FALSE)] - mindiff/10
x[c(FALSE, df mindiff)] - x[c(FALSE, df mindiff)] + mindiff/10
df - x[-1] - x[-length(x)]
i - i + 1
if (i 100) {
break
}
}
x
}
I have tried experimenting with using optim to minimize a function of
the distances between new points and old points penealized for being to
close, but it sometimes gave me some weird results, the above has worked
fairly well for me (the above is based on some of the code inside the
triplot function in the TeachingDemos package).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
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[R] Newbie help with Sweave
I think I've gotten my Emacs/Sweave/R system set up correctly, thanks to
Vincent and Jim, but I haven't been successful getting my first document
produced. I'm trying to use one of Friedrich Leisch's examples,
http://www.ci.tuwien.ac.at/~leisch/Sweave/example-1.Snw. I cut and
pasted the text into a document sweaveexample.Rnw in Emacs. It seemed to
be processed successfully with R:
Sweave(sweaveexample.Rnw)
Writing to file sweaveexample.tex
Processing code chunks ...
You can now run LaTeX on 'sweaveexample.tex'
However, when I try to open the file sweaveexample.tex and process it
with Latex in Emacs, I get this error:
ERROR: Missing \endcsname inserted.
--- TeX said ---
to be read again
\protect
l.7 \begin
{document}
--- HELP ---
From the .log file...
The control sequence marked to be read again should
not appear between \csname and \endcsname.
I've tried a variety of examples, but the error messages are the same.
Can anyone point out my errors or mistakes? I've pasted in the full
files below. Thanks so much for your help and advice.
-Kevin
Kevin Zembower
Internet Services Group manager
Center for Communication Programs
Bloomberg School of Public Health
Johns Hopkins University
111 Market Place, Suite 310
Baltimore, Maryland 21202
410-659-6139
==
sweaveexample.tex:
==
\documentclass[a4paper]{article}
\title{Sweave Example 1}
\author{Friedrich Leisch}
\usepackage{C:/PROGRA~1/R/R-26~1.2/share/texmf/Sweave}
\begin{document}
\maketitle
In this example we embed parts of the examples from the
\texttt{kruskal.test} help page into a \LaTeX{} document:
\begin{Schunk}
\begin{Sinput}
data(airquality)
library(ctest)
kruskal.test(Ozone ~ Month, data = airquality)
\end{Sinput}
\begin{Soutput}
Kruskal-Wallis rank sum test
data: Ozone by Month
Kruskal-Wallis chi-squared = 29.2666, df = 4, p-value = 6.901e-06
\end{Soutput}
\end{Schunk}
which shows that the location parameter of the Ozone
distribution varies significantly from month to month. Finally we
include a boxplot of the data:
\begin{center}
\includegraphics{sweaveexample-002}
\end{center}
\end{document}
sweaveexample.Rnw:
==
\documentclass[a4paper]{article}
\title{Sweave Example 1}
\author{Friedrich Leisch}
\begin{document}
\maketitle
In this example we embed parts of the examples from the
\texttt{kruskal.test} help page into a \LaTeX{} document:
=
data(airquality)
library(ctest)
kruskal.test(Ozone ~ Month, data = airquality)
@
which shows that the location parameter of the Ozone
distribution varies significantly from month to month. Finally we
include a boxplot of the data:
\begin{center}
fig=TRUE,echo=FALSE=
boxplot(Ozone ~ Month, data = airquality)
@
\end{center}
\end{document}
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Re: [R] spreading out a numeric vector
Would this function be worth submitting to some existing cran library? I'd be willing to document it and add error-checking if there is somewhere that it belongs (with credit to Greg of course, unless Greg wants to submit it himself). I am using it in conjunction with a function I wrote which draws tics on the timeline and draws a line from the tic to the label when they aren't aligned. I have tried experimenting with using optim to minimize a function of the distances between new points and old points penealized for being to close, but it sometimes gave me some weird results, the above has worked fairly well for me (the above is based on some of the code inside the triplot function in the TeachingDemos package). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 __ R-help@r-project.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] Plotting matrix rows with lattice graphics?
Sorry if this is an FAQ, but I haven't found the answer (yet)... I'm trying to plot each row of a simple numeric matrix in a separate panel using the layout features of lattice, but can't make it work - help would be appreciated! Example: m - matrix(seq(1:20), nrow=4) m [,1] [,2] [,3] [,4] [,5] [1,]159 13 17 [2,]26 10 14 18 [3,]37 11 15 19 [4,]48 12 16 20 The following will plot one row in one box: xyplot(m[1,] ~ seq(1:5)) but this gives all rows in the same box: xyplot(m ~ seq(1:5)) Since there are no factors, how can I set conditioning variable to get what I want, or is there a better way to do this? Thanks in advance Ron Ron Bonner [EMAIL PROTECTED] skype: rbonners [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R GUI question
Thanks so much for all the input; I appreciate it. I think I'll stick with Tcl/Tk for now primarily on the basis of installation issues. I'm not sure that many of these users would have Java, so the other alternatives all seemed to add another level of complexity on to the installation task (again, this is aimed at a very non-computer savvy audience). Thanks again for the advice! Jeff -- View this message in context: http://www.nabble.com/R-GUI-question-tp16149624p16262820.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] Running rpy produces the error message undefined symbol: KillAllDevices
Hello All, I've been using R for a while and decided to test R from the svn trunk. I know that trunk is not stable, but I wanted to see if I can install and run the latest rpy on my Ubuntu 7.10 box. It installed ok, but I got the runtime error message: $ python Python 2.5.1 (r251:54863, Mar 7 2008, 04:10:12) [GCC 4.1.3 20070929 (prerelease) (Ubuntu 4.1.2-16ubuntu2)] on linux2 Type help, copyright, credits or license for more information. import rpy Traceback (most recent call last): File stdin, line 1, in module File /usr/lib/python2.5/site-packages/rpy.py, line 134, in module % RVERSION) RuntimeError: /usr/lib/python2.5/site-packages/_rpy2070.so: undefined symbol: KillAllDevices RPy module can not be imported. Please check if your rpy installation supports R 2.7.0. If you have multiple R versions installed, you may need to set RHOME before importing rpy. For example: from rpy_options import set_options set_options(RHOME='c:/progra~1/r/rw2011/') from rpy import * Btw, RHOME on my computer is located at /usr/local/lib/R and I have compiled R shared library directory in my ld.so.conf. Bob [[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] Simple problem in R
I never knew to enter that command in. Thanks guys! -- View this message in context: http://www.nabble.com/Simple-problem-in-R-tp16259116p16260713.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] derivatives in R
Hi, I posted this message earlier in Rmetrics and I don't know whether I posted in the wrong place, so I'm posting it again in Rhelp. I have a function in x and y and let's call it f(x,y). I need to get the Hessian matrix. i.e I need (d^2f/dx^2), (d^2f/dxdy), (d^2f/dydx), (d^2f/dy^2).I can get these using the D function. now I need to evaluste the hessian matrix for -5x5 and -2y2 and I also need to print the output. I tried using a for loop,but it is giving me an error. Thanks -- View this message in context: http://www.nabble.com/derivatives-in-R-tp16265419p16265419.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] derivatives in R
Lavan [EMAIL PROTECTED] wrote in news:[EMAIL PROTECTED]: Hi, I posted this message earlier in Rmetrics and I don't know whether I posted in the wrong place, so I'm posting it again in Rhelp. I have a function in x and y and let's call it f(x,y). I need to get the Hessian matrix. i.e I need (d^2f/dx^2), (d^2f/dxdy), (d^2f/dydx), (d^2f/dy^2).I can get these using the D function. now I need to evaluste the hessian matrix for -5x5 and -2y2 and I also need to print the output. I tried using a for loop,but it is giving me an error. I was wondering if the fact that you provide neither an example nor even the text of the error that could be the explanation for no response. -- 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.
[R] Passing (Optional) Arguments
Dear List:
In short, I am writing a number of functions as building blocks for
other functions and have some questions about scoping and passing arguments.
Suppose I have functions foo1, foo2, and foo3 such that:
foo1-function(a=1,b=TRUE,c=FALSE){#do stuff};
foo2-function(x=1,y=FALSE,z=c(1,2,3,4)){#do stuff};
foo3-function(lambda,...){lambda*foo1()*foo2()};
I want to be able to pass a,b,c,x,y,z to the functions foo1 and foo2
though foo3 (whether I define default values or not). How do I do this?
I read a bit in the wiki about problems with partial argument matching
and argument matching (lesson: make argument names not match truncated
versions of other argument names in the target function).
To get a better feel for things I've been playing with examples such as:
b-c(0.25,0.25);
fun-function(a=1,...){a*b};
fun() returns 0.25 0.25 as expected.
fun(a=2) returns 0.5 0.5 as expected.
However, fun(b=1) returns 0.25 0.25 when I want to overwrite b with the
value 1 and have it return 1.
Likewise with
fun-function(a=1,...){a*return(b)};
any argument I supply for b seems to be ignored.
I understand as b is not defined within the function when I enter
fun()
lexical scoping means R looks for b up one level and, finding b, uses it.
Thanks for any/all help.
Sincerely,
Jason Q. McClintic
--
Jason Q McClintic
UST MB 1945
2115 Summit Avenue
St. Paul, MN 55105
[EMAIL PROTECTED]
[EMAIL PROTECTED]
It is insufficient to protect ourselves with laws, we must protect
ourselves with mathematics.--Bruce Schneier
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and provide commented, minimal, self-contained, reproducible code.
[R] peak finding
Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x=c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y=somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] peak finding
It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.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 with rowsum/aggregate type functions
Hi-- This is a question with a trivial and obvious answer, I'm sure, but I can't seem to find it in the help files and books that I have handy. I have a dataframe consisting of two columns, Gene_Name, a list of gene symbols, and Number, a numeric measure of how frequently a tag representing that gene showed up in a SAGE library. Several of the genes are represented by multiple tags, and therefore are present more than once in the list, e.g.: 1167 Zcchc8 6 1168 Zcwpw1 5 1169 Zdhhc18 6 1170 Zdhhc20 5 1171 Zdhhc3 6 1172 Zdhhc3 5 1173 Zeb29 1174 Zeb26 What I want is to collapse the list by gene name, such that duplicates are summed up and appear only once in the final version: Zcchc8 6 Zcwpw1 5 Zdhhc18 6 Zdhhc20 5 Zdhhc3 11 Zeb2 15 The only way I can figure out to do this is via rowsum: rowsum (Number,Gene_Name) gives me exactly what I want, *except* that in the end, I am left with a matrix containing the Number values and with the Gene_Names used as row names (the output therefore looks exactly as printed above) -- what I want is a dataframe equivalent to the starting table, with numbered rows and separate, accessible columns containing the Gene_Name and Number values. I was able to put such a dataframe together manually, by cobbling together the row names of the above list with the values: genes.unique - data.frame (rownames (rowsum(Number,Gene_Name)), rowsum(Number,Gene_Name)) but then I have to manually replace the row names of the dataframe with numbers, to get back to what I wanted in the first place. I hope this makes some sort of sense. Is there an easier way to do this? Thanks in advance! Charlie Murtaugh = L. Charles Murtaugh Assistant Professor University of Utah Dept. of Human Genetics 15 N. 2030 E. Rm. 2100 Salt Lake City, UT 84112 tel 801-581-5958 fax 801-581-6463 email [EMAIL PROTECTED] [[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] peak finding
Hi Thanks for replying. I meant x[4] is the start of a peak shape and x[14] is the end of that peak and x[9] is the maxima of the peak. Thanks, John On Mon, Mar 24, 2008 at 11:09 PM, [EMAIL PROTECTED] wrote: It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] peak finding
Then the answer is pretty simple: 'no'. The idea probably needs a lot more refining to make it workable, too. Why do you discard he peak at 2, with the shape starting at 1 and finishing at 4, for example? In thinking about this it might be useful for you to look at signs of successive differences: sign(diff(c(-Inf, x))) [1] 1 1 -1 -1 1 1 1 1 1 -1 -1 -1 -1 -1 1 -1 That's perhaps a starting point. You seem to want to know (roughly) where do the runs of '1's start and the following run of '-1's end? The function rle(), for run length encoding, might be useful in this regard, too. Bill Venables. -Original Message- From: Research Scholar [mailto:[EMAIL PROTECTED] Sent: Tuesday, 25 March 2008 1:24 PM To: Venables, Bill (CMIS, Cleveland) Cc: r-help@r-project.org Subject: Re: [R] peak finding Hi Thanks for replying. I meant x[4] is the start of a peak shape and x[14] is the end of that peak and x[9] is the maxima of the peak. Thanks, John On Mon, Mar 24, 2008 at 11:09 PM, [EMAIL PROTECTED] wrote: It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.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] peak finding
John, There is a peak finding algorithm attributed to Prof. Ripley in the R-help archive. It has a span parameter which might give you something close to what you seem to be looking for. http://finzi.psych.upenn.edu/R/Rhelp02a/archive/33097.html -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Monday, March 24, 2008 11:24 PM To: [EMAIL PROTECTED] Cc: r-help@r-project.org Subject: Re: [R] peak finding Hi Thanks for replying. I meant x[4] is the start of a peak shape and x[14] is the end of that peak and x[9] is the maxima of the peak. Thanks, John On Mon, Mar 24, 2008 at 11:09 PM, [EMAIL PROTECTED] wrote: It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/p osting-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-help@r-project.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] Partition data into clusters
Hi,
Thanks for your reply. I have tried yr suggestions with success. TQVM.
I have another query, say I want to write each cluster into a csv file such
as follows:
clus1 - my.clusters[[1]]
write.csv(clus1, file = clus1.csv)
.
.
clus10 - my.clusters[[10]]
write.csv(clus10,file = clus10.csv)
I can write the functions to do that for all 10 clusters by repeatedly
calling write.csv as above. Is there a more elegant way of doing it by
using a loop (for example).
for(i in 1:xx) {
clusn - my.clusters[[n]]
write.csv(clusn, file = clusn.csv)
}
I am trying something using list as well, as folows:
names( my.clusters ) - paste('Cluster_',1:10, sep='')
Now I can use
my.clusters$Cluster_1
The above will return all members of Cluster_1.
I have an idea to use lapply or sapply to do write.csv on each
components(Cluster_1.Cluster_n) from my.clusters, that will do the same
as the loop example. Maybe this is the better way..
Any comments are appreciated.
Thanks!
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[R] Output of order() incorrectly ordered?
Hello, I have a data frame consisting of four columns and would like to sort based on the first column and then write the sorted data frame to a file. df - read.table(file.txt, sep=\t) where file.txt is simply a tab-delimited file containing 4 columns of data (first 2 numeric, second 2 character). I then do, df_ordered - df[order(df$V1), ] OR, I assume equivalently, df_ordered - df[ do.call(order, df), ] and then, write.table(df_ordered, file=newfile.txt, ...) The input data file looks like this: 0.083044375.276 680220 majority 5.50816e-09 2.48914e-05 26377 conformation 0.000169618 0.7665051546938 interaction 3.90425e-05 0.1764331655338 vitamin 0.0378182 170.9 1510941 array 3.00359e-07 0.00135732 69421 oligo(dT)-cellulose 1.01517e-13 4.58754e-10 699918 elastase ... I'd like the output file to look the same except sorted by the first column. The output of the commands above give me something that is sorted in some places but not sorted in others: [sorted section] ... 1.87276e-07 0.000846299 1142090 vitamin K 1.89026e-07 0.000854207 917889 leader peptide 1.90884e-07 0.000862605 31206 s 0.00536062 24.2246 1706420 prevent 5.42648e-05 0.2452231513041 measured 5.42648e-05 0.2452231513040 measured 0.01993990.1044 12578 fly 0.00135512 6.12377 61688 GPI 0.00124421 5.62257 681915 content 0.0128271 57.9655 681916 estimated ... [sorted section] ... [unsorted section] ... [etc] I'm not sure if this is a problem with the input data or with order() or what. I am only doing this in R because many of my numeric values are expressed in exponential notation and UNIX sort does not handle this to my knowledge, but this behavior baffles me. I am pretty new to R so it's possible I'm missing something. Any insight would be greatly appreciated! Thanks, -Shirley graduate student Stanford 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] Subset of matrix
Dear R users I have a big matrix like 6021118879029011741015199066136288 100714 602110.6580.6880.4740.2620.1630.1370.32 0.2520.206 11880.65810.9170.2450.3310.1220.1480.194 0.1680.171 7900.6880.91710.2430.310.1220.150.19 0.1710.174 2900.4740.2450.24310.390.3190.1870.4 0.3110.235 11740.2620.3310.310.3910.2950.3520.345 0.3060.308 10150.1630.1220.1220.3190.29510.4210.343 0.4420.435 19900.1370.1480.150.1870.3520.42110.313 0.380.395 66130.320.1940.190.40.3450.3430.31310.58 0.429 62880.2520.1680.1710.3110.3060.4420.380.58 10.723 1007140.2060.1710.1740.2350.3080.4350.395 0.4290.7231 59510.220.1680.1710.2670.3180.4780.4090.478 0.8440.85 59600.2320.1830.1860.2720.3240.420.3830.522 0.6860.711 2360.2590.2130.2180.2830.3570.3820.40.549 0.6550.608 649560.1990.1720.1750.250.3550.4440.4750.523 0.6330.614 2390.1880.1690.1730.2370.3620.4630.50.469 0.5960.692 330320.250.1670.1690.3080.3010.4070.3460.486 0.6550.64 62620.2540.1780.1820.3140.3290.3960.360.459 0.6790.633 59610.2620.1890.1940.3110.3420.3620.3770.468 0.6210.574 59620.2430.1770.180.2980.3240.3890.3530.453 0.6670.62 63060.2260.1680.1710.2740.3060.3890.380.514 0.6980.62 61060.2260.1680.1710.2740.3060.3890.380.514 0.6980.62 62870.2080.1750.1780.250.3280.4290.4220.529 0.7710.689 59500.1930.1750.1790.230.3440.4650.50.448 0.7330.816 8330.3120.2440.2420.4080.4440.3280.3170.526 0.5620.517 7500.1680.1550.1570.2080.3390.5140.5150.385 0.5780.722 61370.2010.1680.1710.2390.3060.3890.380.453 0.6980.62 1457420.3170.2290.2360.3490.3250.3390.305 0.5190.5810.534 58100.3640.2160.2210.4110.2940.3380.2690.603 0.6310.47 8250.3640.2160.2210.4110.2940.3380.2690.603 0.6310.47 74050.3130.2460.2440.3550.3670.290.3170.537 0.5070.484 10450.1330.1240.1350.1980.3330.4410.3440.277 0.3620.341 11230.110.1060.1040.1570.2620.4150.3680.227 0.3040.333 5880.2810.3120.340.3030.4190.2780.40.287 0.3120.316 with more than thousand number of rows and column I want to extract or subset a child matrix from above with specific row names or col names. e.g. For above matrix I need only rows which has following row names. 5951 236 6306 5950 145742 1123 I would appreciate if you can use this matrix as input. Thanks in advance. Dinesh -- Dinesh Kumar Barupal Research Associate Metabolomics Fiehn Lab UCD Genome Center 451 East Health Science Drive GBSF Builidng University of California DAVIS 95616 http://fiehnlab.ucdavis.edu/staff/kumar [[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] peak finding
Another approach that involves more infrastructure is to fit a smooth line to your points, compute the first derivative, and look for change in sign in the first derivative. eg x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) smoothed.dx - predict(smooth.spline(x), deriv=1)$y which(c(smoothed.dx,NA) 0 c(NA, smoothed.dx) 0) My experience is that this approach sometimes requires some fine-tuning, eg in fitting the smooth. I hope that this helps Andrew On Mon, Mar 24, 2008 at 11:23:57PM -0400, Research Scholar wrote: Hi Thanks for replying. I meant x[4] is the start of a peak shape and x[14] is the end of that peak and x[9] is the maxima of the peak. Thanks, John On Mon, Mar 24, 2008 at 11:09 PM, [EMAIL PROTECTED] wrote: It's hard to see how positions 4 and 14 correspond to 'peaks', they look like troughs to me. So perhaps this is what you mean: x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - which(x == min(x)) y [1] 4 14 as a function: somefunction - function(x) which(x == min(x)) Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Research Scholar Sent: Tuesday, 25 March 2008 12:54 PM To: r-help@r-project.org Subject: [R] peak finding Hi all Is there a function that can find the start and end position of peaks in a set of numbers. eg. x - c(14,15,12,11,12,13,14,15,16,15,14,13,12,11,14,12) y - somefunction(x) y 4 14 Thanks John [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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/ __ R-help@r-project.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] derivatives in R
hi thanks,
pls see the code below. the elements of matrix Il should be the derivatives
in matrix h (in Boldface) plus some additional terms, here I did calculate
derivatoves in the matrix, but I want R to calculate it for me.
pid- function (tau1,tau2,sigma=10,n=100,x_bar=20) {
library(akima)
library(lattice)
library(rgl)
a1 - 1.5
a2 - 4
t1 -seq(-1,1);
t2 -seq(3,4);
l -length(t1);
m -length(t2);
N1 - matrix(NA,l,m);
N2 - matrix(NA,l,m);
wp - matrix(NA,l,m);
rr - matrix(NA,l,m);
N - matrix(NA,l,m);
mu - matrix(NA,l,m);
Il - matrix(NA,l,m);
iIl - matrix(NA,l,m);
theta -expression(x+y);
dx -D(theta, x);
dy -D(theta, y);
d2x -D(dx, x);
dxy -D(dx, y);
dyx -D(dy, x);
d2y -D(dy, y);
h - matrix(c(d2x,dxy,dyx,d2y),nrow=2,ncol=2);
for (i in 1:l)
{
for (j in 1:m)
{
wp[i,j]-((t1[i]-a1)/tau1)^2+((t2[j]-a2)/tau2)^2
mu[i,j] -(t1[i] + t2[j]);
Il[i,j]
-matrix(c(n/(sigma^2)+1/(tau1^2),n/(sigma^2),n/(sigma^2),n/(sigma^2)+1/
tau2^2)),2,2);
iIl -solve(Il);
ul[i,j] -
(c(n*(x_bar-mu[i,j])/(sigma^2)-(t1[i]-a1)/(tau1^2),n*(x_bar-mu[i,j])/
(sigma^2)-(t2[j]-a2)/(tau2^2)));
ul -matrix(c(n*(x_bar-mu)/(sigma^2)-(theta1-a1)/(tau1^2),n*(x_bar-mu)/
(sigma^2)-(theta2-a2)/(tau2^2)),2,1);
ult -t(ul);
wl[i,j] -t(ul)%*%solve(Il,ul);
rr[i,j] -(wp[i,j]/wl[i,j]);
N[i,j] -(wl[i,j]-wp[i,j])/wl[i,j]
}
}
list(Test Statistics Wp= wp,Stat Ratio=rr, N=N,N1=N1,
N2=N2, mean=mu, Il=Il,
inverse of Il=iIl)
}
--
View this message in context:
http://www.nabble.com/derivatives-in-R-tp16265419p16267397.html
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Re: [R] Partition data into clusters
Hi,
Just to inform you that I found the solution for the write.csv task I
mentioned earlier. Searching thru the r-help leads me to this solution..
###Your soln starts
xx - max(cl2$cl.kmr10.cluster)
# find out how many clusters there are.
my.clusters - list(NULL) # set up an empty list for storage
# Loop through the dataset and subset by cluster.
for(i in 1:xx) {
my.clusters[[i]] - subset(cl2,
cl2$cl.kmr10.cluster==i)
}
# Eh voilà!
my.clusters
### Your solution ends
# Soln for writing the clustered results into multiple files using loop.
for(i in 1:xx) {
write.csv(my.clusters[[i]], paste(mycluster,i,sep=.))
}
Regards
suhaila
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