Re: [R] conditional probability
Hello Jumlong, For Normal distribution see the help page for pnorm. For dealing with unknown (empirical) distributions, look at ecdf. Hope this helps Michael On 8 November 2010 16:29, Jumlong Vongprasert jumlong.u...@gmail.com wrote: Dear all I have problem with calculate probability, I have data x1,...,x10, I want to calculate probability x11 given x1,...,x10 with two conditions. 1. x is normal 2. unknow distribution How I can do this. Many Thanks. Jumlong -- Jumlong Vongprasert Assist, Prof. Institute of Research and Development Ubon Ratchathani Rajabhat University Ubon Ratchathani THAILAND 34000 [[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] How to eliminate this for loop ?
Hi, I would like to create a list recursively and eliminate my for loop :
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
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Re: [R] How to eliminate this for loop ?
Whenever you use a recursion (that cannot be expressed otherwise), you
always need a (for) loop.
Apply and the like do not allow to use the intermediary results (i.e. a[i-1]
to calculate a[i]).
So: no, it cannot be avoided in your case, I guess.
Nick Sabbe
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of PLucas
Sent: maandag 8 november 2010 10:26
To: r-help@r-project.org
Subject: [R] How to eliminate this for loop ?
Hi, I would like to create a list recursively and eliminate my for loop :
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-eliminate-this-for-loop-tp3031667p30316
67.html
Sent from the R help mailing list archive at Nabble.com.
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] About 5.1 Arrays
Hi Joshua, .You can think of a 3d array kind of like a journal (I don't know if this is helpful, but I kind of like the analogy so...). Each page holds a two dimensional table, so I could tell you to look at row 4, column 3 on page 16. Nevertheless, at any given point, it is just a flat page. A good example. It can be regarded as a book with multiple pages. Each page holds a table of 2 dimensions. No, it is correct. You cannot assume that you may use the same indices to access an array when you have created it with different dimensions. Consider: array(1:24, dim = c(3, 4, 2))[1, 2, 1] [1] 4 array(1:24, dim = c(1, 2, 1))[1, 2, 1] [1] 2 I understand now, e.g. array(1:24, dim = c(3, 4, 2))[2,3,2] [1] 20 Not array(1:24, dim = c(2, 3, 2))[2,3,2] [1] 12 Lot of thanks for your advice and effort. B.R. Stephen L - Original Message From: Joshua Wiley jwiley.ps...@gmail.com To: Stephen Liu sati...@yahoo.com Cc: r-help@r-project.org Sent: Mon, November 8, 2010 8:33:50 AM Subject: Re: [R] About 5.1 Arrays On Sat, Nov 6, 2010 at 7:38 AM, Stephen Liu sati...@yahoo.com wrote: Hi Joshua, Thanks for your advice. 1) Re your advice:-[quote] a3d , , 1 --- this is the first position of the third dimension ***THIS IS THE THIRD DIMENSION*** [,1] [,2] [,3] [,4] --- positions 1, 2, 3, 4 of the second dimension [1,]147 10 [2,]258 11 [3,]369 12 ^ the first dimension , , 2 --- the second position of the third dimension ***THIS IS THE THIRD DIMENSION*** ... [/quote] Where is the third dimension? I pointed to the third dimension above. You can think of a 3d array kind of like a journal (I don't know if this is helpful, but I kind of like the analogy so...). Each page holds a two dimensional table, so I could tell you to look at row 4, column 3 on page 16. Nevertheless, at any given point, it is just a flat page. 2) Re your advice:-[quote] so you can think that in the original vector a: 1 maps to a[1, 1, 1] in the 3d array 2 maps to a[2, 1, 1]. 3 maps to a[3, 1, 1] 4 maps to a[1, 2, 1] 12 maps to a[3, 4, 1] 20 maps to a[2, 3, 2] 24 maps to a[3, 4, 2] [/quote] My finding; # 1 maps to a[1, 1, 1] in the 3d array a3d - array(a, dim = c(1, 1, 1)) a3d , , 1 [,1] [1,]1 Correct # 2 maps to a[2, 1, 1]. a3d - array(a, dim = c(2, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 Correct # 3 maps to a[3, 1, 1] a3d - array(a, dim = c(3, 1, 1)) a3d , , 1 [,1] [1,]1 [2,]2 [3,]3 Correct # 4 maps to a[1, 2, 1] a3d - array(a, dim = c(1, 2, 1)) a3d , , 1 [,1] [,2] [1,]12 Incorrect. It is 2 No, it is correct. You cannot assume that you may use the same indices to access an array when you have created it with different dimensions. Consider: array(1:24, dim = c(3, 4, 2))[1, 2, 1] [1] 4 array(1:24, dim = c(1, 2, 1))[1, 2, 1] [1] 2 # 12 maps to a[3, 4, 1] a3d - array(a, dim = c(3, 4, 1)) a3d , , 1 [,1] [,2] [,3] [,4] [1,]147 10 [2,]258 11 [3,]369 12 Correct # 20 maps to a[2, 3, 2] a3d - array(a, dim = c(2, 3, 2)) a3d , , 1 [,1] [,2] [,3] [1,]135 [2,]246 , , 2 [,1] [,2] [,3] [1,]79 11 [2,]8 10 12 Incorrect. It is 12 See my above comment about not expecting things in the same location when you change the space they live in. Sorry this was so slow in coming, I missed the email somehow. Cheers, Josh [snip] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] unknown dimensions for loglm
Jason Hwa wrote: Dear R-help community, I am working with multidimensional contingency tables and I am having trouble getting loglm to run on all dimensions without typing out each dimension. I have generated random data and provided output for the results I want below: d1.c1 - rnorm(20, .10, .02) d1.c2 - rnorm(20, .10, .02) d2.c1 - rnorm(20, .09, .02) d2.c2 - rnorm(20, .09, .02) d3.c1 - rnorm(20, .11, .02) d3.c2 - rnorm(20, .11, .02) group1 - cbind(1, d1.c1, d2.c1, d3.c1) group2 - cbind(2, d1.c2, d2.c2, d3.c2) colnames(group1) - colnames(group2) - c(group, dim1, dim2, dim3) combined - rbind(group1, group2) combined[,2:4] - combined[,2:4] .1 ctables - xtabs(~., data = combined) loglm(~group+dim1+dim2+dim3, data=ctables) Call: loglm(formula = ~group + dim1 + dim2 + dim3, data = ctables) Statistics: X^2 df P( X^2) Likelihood Ratio 12.29856 11 0.3416253 Pearson 10.28058 11 0.5053391 However, the number and the names of the dimensions change for each dataset. What I want is to be able to run the following line at the end of the code: loglm(~., data=ctables), but it always prints the following error: Error in terms.formula(formula - denumerate(formula)) : '.' in formula and no 'data' argument Can anyone help me out? Thank you, Jason Presumably you're using loglm from package MASS (which you should indicate). To use the dot notation, you need to put your data in dataframe form. This works: d - data.frame(ftable(ctables)) loglm(Freq ~ ., data = d) Or you could use the integer notation for the formula: loglm( ~ 1 + 2 + 3 + 4, data = ctables) You could generate the formula in your code: form - as.formula(paste(~, paste(1:4, collapse=+))) and then use loglm(form, data = ctables) -Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 remove border while saving R plots ?
I am saving R graph as a jpeg image and the border is plotted defaultly when
you open the image.How can I avoid the border?
small R code is as follows :-
volume- seq(100,10,-10)
jpeg(filename = mygraph.jpg,width=366,height=284,units=px)
{
pie(volume,main=title,radius=radius ,bty=n,clockwise =
TRUE,col=color,
font.main = 2,cex.main = 1.8,cex.lab=1.5,border = FALSE)
}
dev.off()
when I open this image automaticaly border is plotted around the graph.
How to avoid this?
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[R] [R-pkgs] RGtk2 2.20.x
This is to announce the release of the RGtk2 2.20.x series. RGtk2 is an interface between R and the GTK+ GUI toolkit and friends. The new release updates the bindings to support up to GTK+ 2.20 (and remains backwards compatible to 2.8). Previously, the interface supported only up to GTK+ 2.12, which is several years old. New features in GTK 2.20 (relative to 2.12) include: * GtkInfoBar widget: Displays transient popup notifications, generally at the bottom of the window (near the status bar). * Off-screen rendering of widgets, e.g., automatically save a widget or visualization to disk. * GtkSpinner: Animated widget for indicating activity * GtkToolPalette: Essentially a multi-row toolbar with drag and drop of toolbar items between the palette and other tool item containers. The bindings to Pango and Cairo were also updated to the latest stable versions. RGtk2 also adds a new library to its interface: GIO (2.22), a low-level library supporting asynchronous streaming I/O, mounting of volumes, networking, etc. Might be useful. There is also a new mode of error handling available, that is disabled by default. By setting the option RGtk2::newErrorHandling to TRUE, errors in the libraries will result in errors in R (instead of warnings), and the error object is no longer returned, which often simplifies return values by no longer requiring a list to hold both the error and the original return value. This is probably a more sensible strategy. Windows users: The new binaries for Windows are being built against 2.20, so you'll need to get GTK+ 2.20 (or higher, 2.22 is released but does not really add any new features) for Windows at: http://gtk-win.sourceforge.net/home/index.php/en/Downloads. Thanks, Michael [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Create Matrix of 2 Dim from two vectors
Hello, I have two data. x-c(1, 2, 3) y-c(4,5,6) How do i create matrix of 3 by 3 from this two, such that (1,4) (1,5) (1,6) (2,4) (2,5) (2,6) (3,4) (3,5) (3,6) I tried some thing like this: xy - as.data.frame(c(0,0,0), dim=c(3,3)) for(i in 1:3) for(j in 1:3) xy[i][j]-c(x[i],y[j]) but i got errors.. any help would appreciate -- View this message in context: http://r.789695.n4.nabble.com/Create-Matrix-of-2-Dim-from-two-vectors-tp3031718p3031718.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] Create Matrix of 2 Dim from two vectors
Try x1-matrix(1,3,1)%x%x y1-y%x%matrix(1,3,1) Z-cbind(x1,y1) And later you need to move towards list and matrix On Mon, Nov 8, 2010 at 11:15 AM, abotaha yaseen0...@gmail.com wrote: Hello, I have two data. x-c(1, 2, 3) y-c(4,5,6) How do i create matrix of 3 by 3 from this two, such that (1,4) (1,5) (1,6) (2,4) (2,5) (2,6) (3,4) (3,5) (3,6) I tried some thing like this: xy - as.data.frame(c(0,0,0), dim=c(3,3)) for(i in 1:3) for(j in 1:3) xy[i][j]-c(x[i],y[j]) but i got errors.. any help would appreciate -- View this message in context: http://r.789695.n4.nabble.com/Create-Matrix-of-2-Dim-from-two-vectors-tp3031718p3031718.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to eliminate this for loop ?
Hi:
?Recall
HTH,
Dennis
On Mon, Nov 8, 2010 at 1:25 AM, PLucas plucasplucasplu...@yopmail.comwrote:
Hi, I would like to create a list recursively and eliminate my for loop :
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-eliminate-this-for-loop-tp3031667p3031667.html
Sent from the R help mailing list archive at Nabble.com.
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Create Matrix of 2 Dim from two vectors
Hi, First, create a MATRIX of the correct size: xy - matrix(nrow=3, ncol=3) Then, in your for loop, you need to index each cell correctly, like this: xy[i,j] Finally, if you want to assign 1,4 to each cell, you need to paste x[i] and y[j] together, like this: xy[i,j]-paste(x[i],y[j], sep=,) With the brackets if you want: xy[i,j]-paste((, x[i], ,, y[j], ), sep=) Or easier like this: xy - matrix(rep(x,3),nrow=3, ncol=3) apply(xy, 1, FUN=paste, y, sep=,) HTH, Ivan Le 11/8/2010 11:15, abotaha a écrit : Hello, I have two data. x-c(1, 2, 3) y-c(4,5,6) How do i create matrix of 3 by 3 from this two, such that (1,4) (1,5) (1,6) (2,4) (2,5) (2,6) (3,4) (3,5) (3,6) I tried some thing like this: xy- as.data.frame(c(0,0,0), dim=c(3,3)) for(i in 1:3) for(j in 1:3) xy[i][j]-c(x[i],y[j]) but i got errors.. any help would appreciate -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create Matrix of 2 Dim from two vectors
Hi:
If you want the literal character strings, this works:
x-c(1, 2, 3)
y-c(4,5,6)
outer(x, y, function(x, y) paste('(', x, ',', y, ')', sep = '') )
[,1][,2][,3]
[1,] (1,4) (1,5) (1,6)
[2,] (2,4) (2,5) (2,6)
[3,] (3,4) (3,5) (3,6)
However, I have the sense you want to use the values of x and y as
coordinates in a function. This is also a job for outer(); e.g.,
f - function(x, y) x + 0.5 * y
outer(x, y, f)
[,1] [,2] [,3]
[1,]3 3.54
[2,]4 4.55
[3,]5 5.56
If you want to apply a two-dimensional function to a pair of vectors,
outer() may be what you're after.
HTH,
Dennis
On Mon, Nov 8, 2010 at 2:15 AM, abotaha yaseen0...@gmail.com wrote:
Hello,
I have two data.
x-c(1, 2, 3)
y-c(4,5,6)
How do i create matrix of 3 by 3 from this two, such that
(1,4) (1,5) (1,6)
(2,4) (2,5) (2,6)
(3,4) (3,5) (3,6)
I tried some thing like this:
xy - as.data.frame(c(0,0,0), dim=c(3,3))
for(i in 1:3)
for(j in 1:3)
xy[i][j]-c(x[i],y[j])
but i got errors..
any help would appreciate
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[R] nonparametric multiple comparisons
Hello everyone, I apologize if this question is out of the scope of this forum I have a small dataset with a dependent continuous variable which I want to compare amongst 3 levels of a factor, which I analyzed using a Kruskall-Wallis test. I am wondering which posthoc multiple comparison I should use? Pairwise wilcoxon test or the Steel test within the package npmc? Thanks! Ellen [[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] Computing ergodic mean with CODA
Hi all, I would like to compute ergodic mean using MCMC output from WinBUGS. I tried using CODA package, but it seems that it is not implemented yet. Could anyone help me to compute this? Kind regards, Raquel -- Raquel Rangel de Meireles Guimarães Doutoranda em Demografia raq...@cedeplar.ufmg.br http://ufmg.academia.edu/RaquelGuimaraes Cedeplar - Centro de Desenvolvimento e Planejamento Regional Avenida Antônio Carlos, 6627, Sala 2090, Pampulha, BH-MG Telefones: (31) 3409-7144 - (31) 9732-2132 Faculdade de Ciências Econômicas, UFMG (http://www.cedeplar.ufmg.br) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Rserve alternative?
The Rserve documentation at http://rosuda.org/Rserve/doc.shtml#start states that even when making multiple connections to the Rserve, Windows won't separate workspaces physically and share environments, which will obviously cause problems and should therefore not be used. Are there any alternatives for the windows platform? Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Create Matrix of 2 Dim from two vectors
Thanks a lot guys...all of your solution are useful, and good. once again thanks. -- View this message in context: http://r.789695.n4.nabble.com/Create-Matrix-of-2-Dim-from-two-vectors-tp3031718p3031797.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] finding the last day of the month
Dear R Help, I am trying to get fields showing the last day of each month for a monthly closing project. In order to find the last day of the previous month, I subtract the number of days from the current month. For all months my code works; however, for October, my code doesn't work...it returns 2010-09-*29* instead of 2010-09-*30*. format(strptime(2010-10-31, %Y-%m-%d)-(as.numeric(format(as.Date(2010-10-31),%d))*24*3600),%Y-%m-%d) I appreciate any suggestions or tips. Thank you. Ben [[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] finding the last day of the month
Benjamin Williams ben at bwmcct.com writes: I am trying to get fields showing the last day of each month for a monthly closing project. In order to find the last day of the previous month, I subtract the number of days from the current month. For all months my code works; however, for October, my code doesn't work...it returns 2010-09-*29* instead of 2010-09-*30*. format(strptime(2010-10-31, %Y-%m-%d)-(as.numeric(format(as.Date(2010-10-31),%d))*24*3600), %Y-%m-%d) Very timely question. This is a daylight savings time issue; subtracting 24 hours gives a different answer when there are only 23 hours in a particular day (spring forward, fall back). Search the R-help archives for the last week or so for a similar issue. (Hint: try subtracting days from dates rather than converting to seconds first.) Ben Bolker __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using changing names in loop in R
Josh, thanks so much for your detailed reply. This is exactly what I was looking for. -- View this message in context: http://r.789695.n4.nabble.com/Using-changing-names-in-loop-in-R-tp3030132p3031899.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] Visualization of histograms
Dear r-user,
I've wrote a small function to visualize the creation of a histogram.
Therefore I create a vector with random values and let them plot in
histograms. I better give a small example in R-Code:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),min(temp$breaks)),
ylim=c(0,max(temp$counts)+0.1*max(temp$counts)),col=white)
for(i in 1:1000)
{
hist(funVec[1:i],add=TRUE,col=white)
Sys.sleep(1/25)
}
I don't find any mistake in the example above and nevertheless there
something strange about the result. R should plot 1000 histograms with time
breaks of 1/25 seconds between the single plots. But the program visualize
not even 100 steps. I don't understand that circumstance. If I don't use the
add parameter, R visualize all 1000 steps.
I hope somebody could help me or explain me what ist happening there. I m
looking forward to any hints or clues.
Greetings and thanks in advance
Etienne
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Re: [R] finding the last day of the month
On Mon, Nov 8, 2010 at 7:06 AM, Benjamin Williams b...@bwmcct.com wrote: Dear R Help, I am trying to get fields showing the last day of each month for a monthly closing project. In order to find the last day of the previous month, I subtract the number of days from the current month. For all months my code works; however, for October, my code doesn't work...it returns 2010-09-*29* instead of 2010-09-*30*. format(strptime(2010-10-31, %Y-%m-%d)-(as.numeric(format(as.Date(2010-10-31),%d))*24*3600),%Y-%m-%d) There are many ways to do this but here is one using as.yearmon in zoo. This converts the Date variable, today, to yearmon class which is a year and a fraction representing a month. Then it converts it back to Date class using frac = 1 which means all the way to the end of the month (frac = 0, the default, means 1st of month, etc.): library(zoo) today - Sys.Date() as.Date(as.yearmon(today), frac = 1) [1] 2010-11-30 If you are trying to show monthly data another option might be just to not show the day at all: format(as.yearmon(today)) [1] Nov 2010 # or without yearmon format(today, %b %Y) [1] Nov 2010 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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] help to sum up data frame
Nice thing about R is there is more than one way of doing something:
x
name ip Bsent Breceived
1 a 1 0.00 0.00
2 a 2 1.43 19.83
3 a 1 0.00 0.00
4 a 2 1.00 1.00
5 b 1 0.00 2.00
6 b 3 0.00 2.00
7 b 2 2.00 0.00
8 b 2 2.00 0.00
9 b 1 24.40 22.72
10c 1 1.00 1.00
11c 1 2.00 1.00
12c 1 2.00 1.00
13c 1 90.97 15.70
14d 0 0.00 0.00
15d 1 30.00 17.14
require(sqldf)
sqldf('select name, sum(ip) as ip, sum(Bsent) as Bsent,
+ sum(Breceived) as Breceived
+ from x
+ group by name')
name ip Bsent Breceived
1a 6 2.43 20.83
2b 9 28.40 26.72
3c 4 95.97 18.70
4d 1 30.00 17.14
On Sun, Nov 7, 2010 at 8:59 AM, Mohan L l.mohanphys...@gmail.com wrote:
Dear All,
I have a data frame like this:
name ip Bsent Breceived
a 1 0.00 0.00
a 2 1.43 19.83
a 1 0.00 0.00
a 2 1.00 1.00
b 1 0.00 2.00
b 3 0.00 2.00
b 2 2.00 0.00
b 2 2.00 0.00
b 1 24.40 22.72
c 1 1.00 1.00
c 1 2.00 1.00
c 1 2.00 1.00
c 1 90.97 15.70
d 0 0.00 0.00
d 1 30.00 17.14
I want to sum up the similar name into one row, like :
name ip Bsent Breceived
a 6 2.43 20.83
b 9 28.40 26.72
c
d
I need help to sum up. Thanks for your time.
Thanks Rg
Mohan L
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Jim Holtman
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+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] Rserve causes Perl error
Could be both. Do you have perl installed and is it on a path that R can find. On Mon, Nov 8, 2010 at 1:32 AM, Ralf B ralf.bie...@gmail.com wrote: Hi all, I tried to run Rserve: I installed it from CRAN using install.packages(Rserve) and tried to run it from the command line using: R CMD Rserve I am getting an error telling me that the command perl cannot be found. What is wrong and what can I do to fix this? Do I need to install any other packages or is it just a path problem? Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rserve alternative?
You might want to look at statconnWS (available from rcom.univie.ac.at). Warning: This project is not open source. On 11/8/2010 12:40 PM, Ralf B wrote: The Rserve documentation at http://rosuda.org/Rserve/doc.shtml#start states that even when making multiple connections to the Rserve, Windows won't separate workspaces physically and share environments, which will obviously cause problems and should therefore not be used. Are there any alternatives for the windows platform? Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] irregular grid
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On 10-11-08 08:05 AM, wolfgang.pola...@gmail.com wrote: Thanks, what if a I want to add to the plot a second column of observations, like a type h line from the x,y plane? Wolfgang see segments3d. rgl is somewhat more do-it-yourself/construct-your-own-geometrical objects than the various 2D graphic systems (a bit more than base graphics, certainly more than lattice or ggplot) [forwarding back to r-help] -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.10 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/ iEYEARECAAYFAkzX+lsACgkQc5UpGjwzenNy2QCfbA6YKcbr8uqXZI0EdYtnxMd0 DaAAni7Lo1ml+xu1BlThktyajvcmP7TE =z0h8 -END PGP SIGNATURE- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Rserve alternative?
Alternatives to Windows platform are Linux. Try Ubuntu download from http://www.ubuntu.com/getubuntu/downloadmirrors if you are a Linux newbie to use Rserve Also see Rapache at http://rapache.net/ for using Apache Web Server and R together. The third R interface on web is Rweb see http://www.jstatsoft.org/v04/i01/paper or http://www.math.montana.edu/Rweb/ Regards Ajay Websites- http://decisionstats.com http://dudeofdata.com http://kushohri.com Linkedin- www.linkedin.com/in/ajayohri On Mon, Nov 8, 2010 at 5:10 PM, Ralf B ralf.bie...@gmail.com wrote: The Rserve documentation at http://rosuda.org/Rserve/doc.shtml#start states that even when making multiple connections to the Rserve, Windows won't separate workspaces physically and share environments, which will obviously cause problems and should therefore not be used. Are there any alternatives for the windows platform? Ralf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Visualization of histograms
Dear r-user,
I've wrote a small function to visualize the creation of a histogram.
Therefore I create a vector with random values and let them plot in
histograms. I better give a small example in R-Code:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),min(temp$breaks)),
ylim=c(0,max(temp$counts)+0.1*max(temp$counts)),col=white)
for(i in 1:1000)
{
hist(funVec[1:i],add=TRUE,col=white)
Sys.sleep(1/25)
}
I don't find any mistake in the example above and nevertheless there
something strange about the result. R should plot 1000 histograms with time
breaks of 1/25 seconds between the single plots. But the program visualize
not even 100 steps. I don't understand that circumstance. If I don't use the
add parameter, R visualize all 1000 steps.
I hope somebody could help me or explain me what ist happening there. I m
looking forward to any hints or clues.
Greetings and thanks in advance
Etienne
[[alternative HTML version deleted]]
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Re: [R] Visualization of histograms
On 08-Nov-10 12:56:26, Etienne Stockhausen wrote:
Dear r-user,
I've wrote a small function to visualize the creation of a histogram.
Therefore I create a vector with random values and let them plot in
histograms. I better give a small example in R-Code:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),min(temp$breaks)),
ylim=c(0,max(temp$counts)+0.1*max(temp$counts)),col=white)
for(i in 1:1000)
{
hist(funVec[1:i],add=TRUE,col=white,)
Sys.sleep(1/25)
}
I don't find any mistake in the example above and nevertheless there
something strange about the result. R should plot 1000 histograms
with time breaks of 1/25 seconds between the single plots. But the
program visualize not even 100 steps. I don't understand that
circumstance. If I don't use the add parameter, R visualize all 1000
steps.
I hope somebody could help me or explain me what ist happening there.
I m looking forward to any hints or clues.
Greetings and thanks in advance
Etienne
I tried your code, and observed what you expected to see (not what
you say you saw). It may be a problem with your system. As a check,
I added a couple of lines to your code, as follows:
i=0
funVec=rnorm(1000)
temp=hist(funVec,plot=FALSE)
plot(1,xlim=c(max(temp$breaks),min(temp$breaks)),
ylim=c(0,max(temp$counts)+0.1*max(temp$counts)),col=white)
maxx-max(temp$breaks) ; maxy-max(temp$counts) ## upper x y lims
for(i in 1:1000)
{
hist(funVec[1:i],add=TRUE,col=white,)
points(i*maxx/1000,i*maxy/1000,pch=+) ## plot a tracker point
Sys.sleep(1/25)
}
and saw the tracker move steadily from start to finish, with
clearly far more than a mere 100 steps, and at each new tracker
point there is also a change in the histogram. So in my case the
results were plotted as you wanted.
My system:
sessionInfo()
R version 2.11.0 (2010-04-22)
i486-pc-linux-gnu
locale:
[1] LC_CTYPE=en_GB.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_GB.UTF-8LC_COLLATE=en_GB.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_GB.UTF-8
[7] LC_PAPER=en_GB.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_GB.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] tools_2.11.0
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Fax-to-email: +44 (0)870 094 0861
Date: 08-Nov-10 Time: 13:34:10
-- XFMail --
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Re: [R] how to remove border while saving R plots ?
On Nov 8, 2010, at 5:13 AM, vikrant wrote:
I am saving R graph as a jpeg image and the border is plotted
defaultly when
you open the image.How can I avoid the border?
small R code is as follows :-
volume- seq(100,10,-10)
jpeg(filename =
mygraph.jpg,width=366,height=284,units=px)
{
pie(volume,main=title,radius=radius ,bty=n,clockwise =
TRUE,col=color,
font.main = 2,cex.main = 1.8,cex.lab=1.5,border = FALSE)
}
dev.off()
when I open this image automaticaly border is plotted around the
graph.
How to avoid this?
Read the help pages? The pie help page Argument definition for border
refers you to the polygon help page... see that handy little link?
border
the color to draw the border. The default, NULL, means to use
par(fg). Use border = NA to omit borders.
I get an error complaining about missing a color vector when I try
to test border=NA with your code, but presumably a more complete
example would have demonstrated success. ( I did try putting in
something for col= but then radius was also missing so back to you,
booby.)
--
David Winsemius, MD
West Hartford, CT
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[R] Working with semantic data triplets
Dear R Help List, I have the following data set: id date factor eg. 1, date11, f1 1, date12, f2 1, date13, f3 [...] 2, date21, fi 2, date22, fj […] f1 – fn are various levels of a factor variable. Each ID may contain 1 to many entries. These represent basically semantic data (triplets). I want to construct the corresponding contingency table, detailing all direct transitions: factor_i = factor_j; in other words: __ f1 f2 f3 f4 … fn f1 f2 f3 … fn Beyond the simple count, I would also want to compute various statistics on the time intervals between the (fi, fj) transition, including mean, sd, median and various quantiles. (fi, fi) transitions are only possible if only one entry is available for that ID, and consequently this represents the number of unique IDs with only one triplet (and that triplet having associated fi) – but the solution is not required to compute this (I am not very interested in this data; it is more for pedantic reasons). However, the solution should also compute: all (fi, fj) direct transitions all (fi, fj) direct transitions, where fi is the baseline value (baseline = smallest date); it may be easy to create such a data subset, once the first problem is solved, and the true issue is only solving for the general problem What would be the best way to do this? I am invariably ending up with solutions based on loops, but I feel this is entirely wrong. Also, the dataset contains over 600,000 triplets and is growing fast, so loops would be pretty unusable on my home computer (my senses are telling me that loops are very slow). I can sort the data externally (by ID, DATE) to ease computations. Thank you very much for your help. Sincerely, Leonard -- GRATIS! Movie-FLAT mit über 300 Videos. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Computing ergodic mean with CODA
Hola Raquel, As Ben already told you, ergodic means are pretty easy to compute from scratch. If you are lazy, or you want something more, like estimated standard errors, you can check out the functions ergMean and mcmcMean in package dlm. Best, Giovanni Petris On Sun, 2010-11-07 at 14:34 -0200, Raquel Rangel de Meireles Guimarães wrote: Hi all, I would like to compute ergodic mean using MCMC output from WinBUGS. I tried using CODA package, but it seems that it is not implemented yet. Could anyone help me to compute this? Attached to this email are my output and index files. Kind regards, Raquel -- Raquel Rangel de Meireles Guimarães Doutoranda em Demografia raq...@cedeplar.ufmg.br http://ufmg.academia.edu/RaquelGuimaraes Cedeplar - Centro de Desenvolvimento e Planejamento Regional Avenida Antônio Carlos, 6627, Sala 2090, Pampulha, BH-MG Telefones: (31) 3409-7144 - (31) 9732-2132 Faculdade de Ciências Econômicas, UFMG (http://www.cedeplar.ufmg.br) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] conditional probability
If you want a conditional distribution of x11 given x1,..x10, you need to make some assumptions about the joint distribution. It seems from the original post that items 1. and 2. refer to the marginal distribution of each observation. In general, it would help to specify the question a little more carefully. Giovanni Petris On Mon, 2010-11-08 at 19:31 +1100, Michael Bedward wrote: Hello Jumlong, For Normal distribution see the help page for pnorm. For dealing with unknown (empirical) distributions, look at ecdf. Hope this helps Michael On 8 November 2010 16:29, Jumlong Vongprasert jumlong.u...@gmail.com wrote: Dear all I have problem with calculate probability, I have data x1,...,x10, I want to calculate probability x11 given x1,...,x10 with two conditions. 1. x is normal 2. unknow distribution How I can do this. Many Thanks. Jumlong -- Jumlong Vongprasert Assist, Prof. Institute of Research and Development Ubon Ratchathani Rajabhat University Ubon Ratchathani THAILAND 34000 [[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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 eliminate this for loop ?
On Nov 8, 2010, at 4:30 AM, Nick Sabbe wrote:
Whenever you use a recursion (that cannot be expressed otherwise), you
always need a (for) loop.
Not necessarily true ... assuming a is of length n:
a[2:n] - a[1:(n-1))]*b + cc[1:(n-1)]
# might work if b and n were numeric vectors of length 1 and cc had
length = n. (Never use c as a vector name.)
# it won't work if there are no values for the nth element at the
beginning and you are building up a element by element.
And you always need to use operations that appropriate to the object
type. So if a really is a list, this will always fail since
arithmetic does not work on list elements. If on the other hand, the
OP were incorrect in calling this a list and a were a numeric
vector, there might be a chance of success if the rules of indexing
were adhered to. The devil is in the details and the OP has not
supplied enough code to tell what might happen.
--
David.
Apply and the like do not allow to use the intermediary results
(i.e. a[i-1]
to calculate a[i]).
So: no, it cannot be avoided in your case, I guess.
Nick Sabbe
--
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link: http://biomath.ugent.be
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-- Do Not Disapprove
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
] On
Behalf Of PLucas
Sent: maandag 8 november 2010 10:26
To: r-help@r-project.org
Subject: [R] How to eliminate this for loop ?
Hi, I would like to create a list recursively and eliminate my for
loop :
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context:
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67.html
Sent from the R help mailing list archive at Nabble.com.
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David Winsemius, MD
West Hartford, CT
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Re: [R] RMark error: only 0's may be mixed with negative subscripts
Umesh- You should contact the package maintainer or a SIG with a question about a package. RMark isn't posted on CRAN so it is unlikely most folks will even know about it. There is a support forum for MARK and RMark on the phidot site where you got them. I'll look at your data and problem and will get back to you. I suggest that you subscribe to the phidot forum. Jeff Laake (RMark maintainer) On 11/7/2010 11:03 PM, Umesh Srinivasan wrote: Hi all, I have just started using RMark to analyse capture-recapture data. I am trying to analyse a simulated data set using the Robust Design (two primary periods with three secondary sessions in each) to estimate apparent survival. On specifying the time intervals (that tell R about the primary and secondary sampling sessions), this is the error I get: time.int- c(0,0,1,0,0) model- mark(dat, model = Robust, time.intervals = time.int) Error in time.intervals[1:(i - 1)] : only 0's may be mixed with negative subscripts This is what I tried to do overall: ### library(RMark) dat- import.chdata(/home/umesh/Desktop/capture history.txt) head(dat) time.int- c(0,0,1,0,0) model- mark(dat, model = Robust, time.intervals = time.int) ### I am attaching the simulated capture histories in the file 'capture history.txt' as well if that will help. Thanks in advance, Umesh Srinivasan National Centre for Biological Sciences Bangalore, India __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] try (nls stops unexpectedly because of chol2inv error
Hi,
I am running simulations that does multiple comparisons to control.
For each simulation, I need to model 7 nls functions. I loop over 7 to do
the nls using try
if try fails, I break out of that loop, and go to next simulation.
I get warnings on nls failures, but the simulation continues to run, except
when the internal call (internal to nls) of the chol2inv fails.
Error in chol2inv(object$m$Rmat()) :
element (2, 2) is zero, so the inverse cannot be computed
In addition: Warning messages:
1: In nls(myModel.nlm, fData, start = initialValues, control =
nls.control(warnOnly = TRUE), :
number of iterations exceeded maximum of 50
2: In nls(myModel.nlm, fData, start = initialValues, control =
nls.control(warnOnly = TRUE), :
singular gradient
===
Any suggestions on how to prevent chol2inv from breaking my simulation...
The point of the simulation is to address power. As our data goes down to
N, of the 100 simulations, only 53 are good simulations because we don't
have enough data for nls or chol2inv to work correctly.
monte
{x:
###
## case I ## EQUAL SAMPLE SIZES and design points
nsim = 100;
N_i = M_i = 10; ## also try (10, 30, 50, 100, 200)
r = M_i / N_i;
X.start = 170; # 6 design points, at 170,180,190, etc. where each point has
N_i elements
X.increment = 10;
X.points = 6;
X.end = 260;
Xval = seq(X.start,length.out=X.points,by=X.increment );
Xval = seq(X.start,X.end,length.out=X.points);
L = 7; ## 6 + control
k = 3;
varY = 0.15;
### for each simulation, we need to record all of this information, write to
a table or file.
### Under the null of simulation, we assign all locations to have same model
## we assume these are the true parameters
b = 2.87; d = 0.0345; t = 173;
B = seq(2.5,4.5,length.out=21);
#B = seq(2.75,3.25,length.out=21);
#B = seq(2.85,2.95,length.out=21);
#B = seq(2.8,3.0,length.out=21);
B = seq(2.5,3.2,length.out=21);
D = seq(0.02,0.04,length.out=21);
T = seq(165,185,length.out=21);
alpha = .05;
nu = k; ## number of parameters
tr = L-1; ## number of treatments (including control)
rho = 1/(1+r); ## dependency parameter
myCritical = qmvchi(alpha,nu,tr,rho);
## we change one parameter at a time until the results fail most of the
time.
## do independent for now, but let's store the parameters and quantiles???
INFO for one location
# beta delta tau nsim %Reject(V.pooled) %Reject(V.total) [Simulation level]
resultS
# beta delta tau i of nsim max(V.pooled) max(V.total) [Individual level]
resultI
resultS = resultI = NULL;
for(p1 in 1:length(D))
{
print(paste(p1, [D] of ,length(D))); flush.console();
print(D[p1]);
myReject.pooled = myReject.pooled.1 = MAX.pooled = rep(-1,nsim);
gsim = 0; ## good simulations
for(i in 1:nsim)
{
doubleBreak = F;
print(paste(i, of ,nsim)); flush.console();
tData = NULL;
pooledNum = matrix(0,nrow=k,ncol=k); ##numerator as weighted sum AS
(n_k-1)cov.scaled
pooledDen = 0; ##denominator as correction AS N-k
#Sigma_pooled = ((omit.1-1)*summary.nls.1$cov.scaled +
(omit.2-1)*summary.nls.2$cov.scaled +
(omit.L-1)*summary.nls.L$cov.scaled)/(sum(omit.1,omit.2,omit.L)-L);
for(j in 1:L)
{
Y = numeric(N_i);
X = createDomain(Xval,N_i); noise = rnorm(N_i, mean=0,sd=sqrt(varY) );
if(j==1)
{
## location #1 is different
Y = noise + evaluateModel(X,b,D[p1],t);
beta = b;
delta = D[p1];
tau = t;
} else {
Y = noise + evaluateModel(X,b,d,t);
}
print(paste(j, location NLS of ,L)); flush.console();
fData = as.data.frame(cbind(Y,X)); colnames(fData)=c(Y,X); unique =
doUnique(fData);
initialValues = list(b=3,d=0.04,t=180);
#plot(X,Y,main=j);
# http://stackoverflow.com/questions/2963729/r-catching-errors-in-nls
try.fit = try(
nls(
myModel.nlm ,
fData,
start = initialValues,
control = nls.control(warnOnly = TRUE),
trace=T
)
);
if(class(try.fit) == try-error)
{
doubleBreak = T;
print(doubleBreak);
break; ## skip to next simulation?
} else {
fit.nls = try.fit;
summary.nls = summary(fit.nls);
summary.nls$cov.scaled = scaledCOV(summary.nls);
pooledDen = pooledDen + dim(fData)[1];
pooledNum = pooledNum + (dim(fData)[1]-1)*summary.nls$cov.scaled;
results = list(data=fData,fit.nls=fit.nls,summary.nls=summary.nls);
tData = rbind(tData,fData); ## total data
}
if(j==L)
{
myStr = nls.L;
} else {
myStr = paste(nls.,j,sep=);
}
assign(myStr,results);
}
if(doubleBreak==T)
{
# break from outer loop if fit didn't work [SKIP simulation]
print(doubleBreak);
doubleBreak = F;
next;
}
gsim = gsim + 1;
# http://www.maths.bris.ac.uk/~mazjcr/SGP.R
COV.pooled = pooledNum/pooledDen;
## loop back through, use COV.t and COV.pooled to do tests and record reject
or not
CONTROL = nls.L$summary.nls$parameters[,1];
Vp = numeric(L-1);
for(j in 1:(L-1))
{
myStr = paste(nls.,j,sep=);
myData = get(myStr);
Diff = myData$summary.nls$parameters[,1]-nls.L$summary.nls$parameters[,1];
Re: [R] help! kennard-stone algorithm in soil.spec packages does not work for my dataset!!!
http://r.789695.n4.nabble.com/file/n3032045/rsv1.txt rsv1.txt I am very grateful to David's suggestion, here , I upload my dataset rsv1.txt, also the question, ks-ken.sto(rsv1,per=TRUE,per.n=0.3,va=FALSE,sav=FALSE) it does not work, all results are NULL, i do not known why it is ? hope, friends can give me a hand! thanks kevin -- View this message in context: http://r.789695.n4.nabble.com/help-kennard-stone-algorithm-in-soil-spec-packages-does-not-work-for-my-dataset-tp3031344p3032045.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] Maria Teresa Torres Muñoz/Gestion de Intangi bles/LineaDirecta está ausente de la oficina.
Estaré ausente de la oficina desde el 08/11/2010 y no volveré hasta el 16/11/2010. Responderé a su mensaje cuando regrese, si es urgente enviar mail a INNOVACION Visítenos en: www.lineadirecta.com Este mensaje y los documentos que, en su caso, lleve anexos, pueden contener información confidencial. Por ello, se informa a quien lo reciba por error que la información contenida en el mismo es reservada y su uso no autorizado está prohibido legalmente, por lo que en tal caso le rogamos se abstenga de realizar copias del mensaje, leerlo, remitirlo o difundirlo y proceda a borrarlo inmediatamente. This message is intended only for the use of the individual to whom it is addressed and may contain information that is confidential. If you have received this communication, by error, you are hereby notified that any distribution or copying of this communication is prohibited. [[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] Add text to a stacked barplot
Hi All,
I need some help in putting text in a stacked barplot. The barplot is filled
with 5 levels and now I would like to put text to each level in the stacked
barplot. However, it seems that the code that I am using is not placing the
text at the correct hight (centered at each fill) in the barplot. Any
suggestions to improve the code to make it work?
barchart(FREQ ~ VISIT |which*as.factor(TRTN), data = stuffa,par.strip.text =
list(cex = 0.35),
groups = RES, xlab=Visit,ylab=Frequency,
layout = c(4,2),
par.settings=list(superpose.polygon=list(col=colors()[c(636,96,256,92,27,376)])),
stack = T,
auto.key = list(points = FALSE, rectangles = TRUE,
space = top),
scales = list(x = list(alternating=c(1,1),tck=c(1,0),abbreviate = TRUE,
minlength
= 5, rot = 0),y=list(alternating=c(1,1),tck=c(1,0))),
panel = function(y,x,...){
panel.grid(h = -1, v = 0, col = gray, lty = 3,lwd=1)
panel.barchart(x,y,...)
panel.text(x,y,label = round(y,1),cex=0.48)
}
)
Kind regards,
Ashraf Yassen
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Add text to a stacked barplot
Please post some data (fake data is fine) for stuffa. The example doesn't
execute as is.
Rich
On Mon, Nov 8, 2010 at 10:27 AM, Ashraf Yassen ashraf.yas...@gmail.comwrote:
Hi All,
I need some help in putting text in a stacked barplot. The barplot is
filled
with 5 levels and now I would like to put text to each level in the stacked
barplot. However, it seems that the code that I am using is not placing the
text at the correct hight (centered at each fill) in the barplot. Any
suggestions to improve the code to make it work?
barchart(FREQ ~ VISIT |which*as.factor(TRTN), data = stuffa,par.strip.text
=
list(cex = 0.35),
groups = RES, xlab=Visit,ylab=Frequency,
layout = c(4,2),
par.settings=list(superpose.polygon=list(col=colors()[c(636,96,256,92,27,376)])),
stack = T,
auto.key = list(points = FALSE, rectangles = TRUE,
space = top),
scales = list(x = list(alternating=c(1,1),tck=c(1,0),abbreviate = TRUE,
minlength
= 5, rot = 0),y=list(alternating=c(1,1),tck=c(1,0))),
panel = function(y,x,...){
panel.grid(h = -1, v = 0, col = gray, lty = 3,lwd=1)
panel.barchart(x,y,...)
panel.text(x,y,label = round(y,1),cex=0.48)
}
)
Kind regards,
Ashraf Yassen
[[alternative HTML version deleted]]
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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]]
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and provide commented, minimal, self-contained, reproducible code.
[R] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
Hi, I'm try to compute the minimum sample size needed to have at least an 80% of power, with alpha=0.05. The problem is that empirical proportions are really small: 0.00154 in one case and 0.00234. These are the estimated failure proportion of two medical treatments. Thomas and Conlon (1992) suggested Fisher's exact test and proposed a computational method, which according to their table gives a sample size of roughly 2. Unfortunately I cannot find any software applying their method. -Does anyone know how to estimate sample size on Fisher's exact test by using R? -Even better, does anybody know other, maybe optimal, methods for such a situation (small p1 and p2) and the corresponding R software? Thanks in advance, Giulio [[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] Add values of rlm coefficients to xyplot
Hello,
I have a simple xyplot with rlm lines.
I would like to add the a and b coefficients (y=ax+b) of the rlm calculation
in each panel.
I know I can do it 'outside' the xyplot command but I would like to do all
at the same time.
I found some posts with the same question, but no answer.
Is it impossible ?
Thanks in advance for your help.
Ptit Bleu.
x11(15,12)
xyplot(df1$col2 ~ df1$col3 | df1$Name,
panel = function(x, y,...) {
panel.abline(h=seq(20,40,5), col=gray)
panel.abline(v=seq(-10,60,5), col=gray)
panel.xyplot(x, y, type=p, col=red, pch=20,...)
panel.abline(rlm(y ~ x), col=blue)
}, scales=list(cex=1.2),
xlab=list(X, cex=1.4), ylab=list(Y, cex=1.4),
xlim=c(-10,60), ylim=c(20,40))
--
View this message in context:
http://r.789695.n4.nabble.com/Add-values-of-rlm-coefficients-to-xyplot-tp3032166p3032166.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Error: could not find function extract in package raster
Hi, I would like to use the function extract from package raster and i get the following error: m01e - extract(marsh01, p) Error: could not find function extract marsh01 is a raster object and p is an intersectExtent object that is not null. The package is installed and loaded, i use a Windows machine 64 bit and R x64 2.11.0. Do i really need to update my R to use this function? Thanks, Monica [[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] Error: could not find function extract in package raster
Hi Monica, Evidently, there is not an extract function in your search path (base R, loaded packages, etc.). Given that you are talking about a function from a package (rather than base R), it would probably help us more if you mentioned what version of raster you have installed. You can get this easily by loading it and reporting the results of sessionInfo() (per the posting guide). On my current system (see specs below), there is an extract function and it seems to work., Cheers, Josh sessionInfo() R version 2.12.0 (2010-10-15) Platform: i486-pc-linux-gnu (32-bit) attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.6-10 sp_0.9-72 loaded via a namespace (and not attached): [1] grid_2.12.0 lattice_0.19-13 tools_2.12.0 On Mon, Nov 8, 2010 at 8:08 AM, Monica Pisica pisican...@hotmail.com wrote: Hi, I would like to use the function extract from package raster and i get the following error: m01e - extract(marsh01, p) Error: could not find function extract marsh01 is a raster object and p is an intersectExtent object that is not null. The package is installed and loaded, i use a Windows machine 64 bit and R x64 2.11.0. Do i really need to update my R to use this function? Thanks, Monica [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
Not with R, but look for G*Power3, a free tool for power calc, includes FIsher's test. http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3 On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni perimessagg...@hotmail.com wrote: Hi, I'm try to compute the minimum sample size needed to have at least an 80% of power, with alpha=0.05. The problem is that empirical proportions are really small: 0.00154 in one case and 0.00234. These are the estimated failure proportion of two medical treatments. Thomas and Conlon (1992) suggested Fisher's exact test and proposed a computational method, which according to their table gives a sample size of roughly 2. Unfortunately I cannot find any software applying their method. -Does anyone know how to estimate sample size on Fisher's exact test by using R? -Even better, does anybody know other, maybe optimal, methods for such a situation (small p1 and p2) and the corresponding R software? Thanks in advance, Giulio [[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] How to plot a normal distribution curve and a shaded tail with alpha?
I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha-tp3032194p3032194.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] varclus in Hmisc vs SAS PROC VARCLUS
My impression is that varclus from the Hmisc library is very convenient to use when you want to get an idea of the correlation among predictor variables in a regression setting, but if you want to perform cluster analysis in general, you may be better off using a different function in R, such as hclust or kmeans. Darin On Sun, Nov 07, 2010 at 11:00:00AM -0500, Lars Bishop wrote: Hi, I'll apreciate your guidance on how can I re-create the output from SAS PROC VARCLUS in R. I've found the varclus function in Hmisc. However, is it possible to use that function to compute for each variable the 1-R**2 ratio (this is the ratio of 1 minus the R-squared with Own Cluster to one minus the R-squared in the Next Closest cluster)? Thanks in advance for any help, Lars. [[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] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
On Nov 8, 2010, at 11:16 AM, Mitchell Maltenfort wrote: Not with R, Really? require(sos) findFn(power exact test) found 54 matches; retrieving 3 pages 2 3 These look on point: http://finzi.psych.upenn.edu/R/library/statmod/html/power.html http://finzi.psych.upenn.edu/R/library/binom/html/cloglog.sample.size.html Would also think that methods based on a poisson model of rare events could be informative: http://finzi.psych.upenn.edu/R/library/asypow/html/asypow.n.html -- David. but look for G*Power3, a free tool for power calc, includes FIsher's test. http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3 On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni perimessagg...@hotmail.com wrote: Hi, I'm try to compute the minimum sample size needed to have at least an 80% of power, with alpha=0.05. The problem is that empirical proportions are really small: 0.00154 in one case and 0.00234. These are the estimated failure proportion of two medical treatments. Thomas and Conlon (1992) suggested Fisher's exact test and proposed a computational method, which according to their table gives a sample size of roughly 2. Unfortunately I cannot find any software applying their method. -Does anyone know how to estimate sample size on Fisher's exact test by using R? -Even better, does anybody know other, maybe optimal, methods for such a situation (small p1 and p2) and the corresponding R software? Thanks in advance, Giulio David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot a normal distribution curve and a shaded tail with alpha?
On Nov 8, 2010, at 11:18 AM, Wu Gong wrote: I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg I think you could have titled this request: Yet Another Request to Do My Searching For Me: http://search.r-project.org/cgi-bin/namazu.cgi?query=plot+normal+curve+with+shadingmax=100result=normalsort=scoreidxname=functionsidxname=Rhelp08idxname=Rhelp10idxname=Rhelp02 The R search space is littered with examples and that search string is not likely to uncover them all, but is rather just one example. -- David Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha-tp3032194p3032194.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. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Centre of gravity of a mountain
Hi all, I have a matrix of a mountain of form 21x21 and values in them are height (Z). Using the persp function I can view this mountain in 3D. Now, I am trying to find a measure to find the centre of gravity (maybe centroid?) of this mountain. Any idea what would be the best way to go? -- View this message in context: http://r.789695.n4.nabble.com/Centre-of-gravity-of-a-mountain-tp3032319p3032319.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] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
Hi,
I don't have access to the article, but must presume that they are doing
something radically different if you are only getting a total sample size
of 20,000. Or is that 20,000 per arm?
Using the G*Power app that Mitchell references below (which I have used
previously, since they have a Mac app):
Exact - Proportions: Inequality, two independent groups (Fisher's exact test)
Options:Exact distribution
Analysis: A priori: Compute required sample size
Input: Tail(s) = Two
Proportion p1 = 0.00154
Proportion p2 = 0.00234
α err prob = 0.05
Power (1-β err prob)= 0.8
Allocation ratio N2/N1 = 1
Output: Sample size group 1 = 49851
Sample size group 2 = 49851
Total sample size = 99702
Actual power= 0.8168040
Actual α= 0.0462658
Using the base R power.prop.test() function:
power.prop.test(p1 = 0.00154, p2 = 0.00234, power = 0.8)
Two-sample comparison of proportions power calculation
n = 47490.34
p1 = 0.00154
p2 = 0.00234
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
Using Frank's bsamsize() function in Hmisc:
bsamsize(p1 = 0.00154, p2 = 0.00234, fraction = .5, alpha = .05, power = .8)
n1 n2
47490.34 47490.34
Finally, throwing together a quick Monte Carlo simulation using the FET, I get:
TwoSampleFET - function(n, p1, p2, power = 0.85,
R = 5000, correct = FALSE)
{
MCSim - function(n, p1, p2)
{
Control - rbinom(n, 1, p1)
Treat - rbinom(n, 1, p2)
fisher.test(cbind(table(Control), table(Treat)))$p.value
}
# Run MC Replicates
MC.res - replicate(R, MCSim(n, p1, p2))
# Get p value at power quantile
quantile(MC.res, power)
}
# 50,000 per arm
TwoSampleFET(5, p1 = 0.00154, p2 = 0.00234, power = 0.8, R = 500)
80%
0.04628263
So all four of these are coming back with numbers in the 48,000 to 50,000
***per arm***.
HTH,
Marc Schwartz
On Nov 8, 2010, at 10:16 AM, Mitchell Maltenfort wrote:
Not with R, but look for G*Power3, a free tool for power calc,
includes FIsher's test.
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni
perimessagg...@hotmail.com wrote:
Hi,
I'm try to compute the minimum sample size needed to have at least an 80% of
power, with alpha=0.05. The problem is that empirical proportions are really
small: 0.00154 in one case and 0.00234. These are the estimated failure
proportion of two medical treatments.
Thomas and Conlon (1992) suggested Fisher's exact test and proposed a
computational method, which according to their table gives a sample size of
roughly 2. Unfortunately I cannot find any software applying their
method.
-Does anyone know how to estimate sample size on Fisher's exact test by
using R?
-Even better, does anybody know other, maybe optimal, methods for such a
situation (small p1 and p2) and the corresponding R software?
Thanks in advance,
Giulio
__
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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] Centre of gravity of a mountain
Weighted mean of x and y coordinates (sorry for the pun :)), that is something like n = 21 y = matrix( c(1:n), n, n) x = matrix( c(1:n), n, n, byrow = TRUE) # These are the Center of mass coordinates: xCenter = sum(x * Z)/sum(Z); yCenter = sum(y * Z)/sum(Z); If you also need the z coordinate, it simply the mean of the matrix Z. zCenter = mean(Z) Peter On Mon, Nov 8, 2010 at 9:07 AM, Ab Hu master.rs...@yahoo.com wrote: Hi all, I have a matrix of a mountain of form 21x21 and values in them are height (Z). Using the persp function I can view this mountain in 3D. Now, I am trying to find a measure to find the centre of gravity (maybe centroid?) of this mountain. Any idea what would be the best way to go? -- View this message in context: http://r.789695.n4.nabble.com/Centre-of-gravity-of-a-mountain-tp3032319p3032319.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: could not find function extract in package raster
Hi Joshua, OK, my session info is this: R version 2.11.0 (2010-04-22) x86_64-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.5-16 sp_0.9-64 loaded via a namespace (and not attached): [1] grid_2.11.0lattice_0.18-8 I also have installed the 32 bit version of R under windows 64 bit and there the raster package has the function extract and session info is: R version 2.11.0 (2010-04-22) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.6-10 sp_0.9-62 loaded via a namespace (and not attached): [1] grid_2.11.0lattice_0.18-5 Obviously i would like to work with the R64 bit because i have more memory available. It is not clear to me why the function extract exists in the package raster for 32 bit but not for 64 bit. Is there any reason? Thanks, Monica Date: Mon, 8 Nov 2010 08:17:48 -0800 Subject: Re: [R] Error: could not find function extract in package raster From: jwiley.ps...@gmail.com To: pisican...@hotmail.com CC: r-help@r-project.org Hi Monica, Evidently, there is not an extract function in your search path (base R, loaded packages, etc.). Given that you are talking about a function from a package (rather than base R), it would probably help us more if you mentioned what version of raster you have installed. You can get this easily by loading it and reporting the results of sessionInfo() (per the posting guide). On my current system (see specs below), there is an extract function and it seems to work., Cheers, Josh sessionInfo() R version 2.12.0 (2010-10-15) Platform: i486-pc-linux-gnu (32-bit) attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.6-10 sp_0.9-72 loaded via a namespace (and not attached): [1] grid_2.12.0 lattice_0.19-13 tools_2.12.0 On Mon, Nov 8, 2010 at 8:08 AM, Monica Pisica pisican...@hotmail.com wrote: Hi, I would like to use the function extract from package raster and i get the following error: m01e - extract(marsh01, p) Error: could not find function extract marsh01 is a raster object and p is an intersectExtent object that is not null. The package is installed and loaded, i use a Windows machine 64 bit and R x64 2.11.0. Do i really need to update my R to use this function? Thanks, Monica [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to eliminate this for loop ?
1. The for loop is probably as or more efficient than recursion,
especially for large n (comments/corrections on this claim from
cogniscenti are welcome);
2. One can **Always** express a for loop recursively -- that is the
nature of computer languages (comments/corrections again welcome).
3. Here is code that does it: Note that I have changed argument name
c to const to avoid confusion with the c() function
f - function(n,b,const)
{
if(length(const) n-1)stop(const is too short)
g - function(i){
if(i==1) 1
else b*g(i-1)+ const[i-1]
}
g(n)
}
As David intimated, this produces a vector, not a list. Cast with
as.list if that's what you want.
Cheers,
Bert
On Mon, Nov 8, 2010 at 6:03 AM, David Winsemius dwinsem...@comcast.net wrote:
On Nov 8, 2010, at 4:30 AM, Nick Sabbe wrote:
Whenever you use a recursion (that cannot be expressed otherwise), you
always need a (for) loop.
Not necessarily true ... assuming a is of length n:
a[2:n] - a[1:(n-1))]*b + cc[1:(n-1)]
# might work if b and n were numeric vectors of length 1 and cc had length
= n. (Never use c as a vector name.)
# it won't work if there are no values for the nth element at the beginning
and you are building up a element by element.
And you always need to use operations that appropriate to the object type.
So if a really is a list, this will always fail since arithmetic does not
work on list elements. If on the other hand, the OP were incorrect in
calling this a list and a were a numeric vector, there might be a chance
of success if the rules of indexing were adhered to. The devil is in the
details and the OP has not supplied enough code to tell what might happen.
--
David.
Apply and the like do not allow to use the intermediary results (i.e.
a[i-1]
to calculate a[i]).
So: no, it cannot be avoided in your case, I guess.
Nick Sabbe
--
ping: nick.sa...@ugent.be
link: http://biomath.ugent.be
wink: A1.056, Coupure Links 653, 9000 Gent
ring: 09/264.59.36
-- Do Not Disapprove
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of PLucas
Sent: maandag 8 november 2010 10:26
To: r-help@r-project.org
Subject: [R] How to eliminate this for loop ?
Hi, I would like to create a list recursively and eliminate my for loop :
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-eliminate-this-for-loop-tp3031667p30316
67.html
Sent from the R help mailing list archive at Nabble.com.
__
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide
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David Winsemius, MD
West Hartford, CT
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--
Bert Gunter
Genentech Nonclinical Biostatistics
467-7374
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Re: [R] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
Yep, it is 20.000 per arm, sorry. The reference it's about an application of
the method, and I cannot download the paper with the main algorithm, so I don't
know exactly how they did.
Thanks everybody for the rich and interesting suggestions. Through free web
software (PS, others) I found also an N around 47.000 per arm. I guess these
are the values (also seen Marc's Monte Carlo).
Maybe the Poisson models approach suggested by David can be an alternative,
even if I guess at this point I won't get big differences in numbers. Would I?
Thanks a lot everybody again for your suggestions,
if anybody has other comments, they are always welcome.
Best,
Giulio
Subject: Re: [R] Sample size calculation for differences between two very
small proportions (Fisher's exact test or others)?
From: marc_schwa...@me.com
Date: Mon, 8 Nov 2010 11:13:12 -0600
CC: perimessagg...@hotmail.com; r-h...@stat.math.ethz.ch
To: mmal...@gmail.com
Hi,
I don't have access to the article, but must presume that they are doing
something radically different if you are only getting a total sample size
of 20,000. Or is that 20,000 per arm?
Using the G*Power app that Mitchell references below (which I have used
previously, since they have a Mac app):
Exact - Proportions: Inequality, two independent groups (Fisher's exact test)
Options: Exact distribution
Analysis: A priori: Compute required sample size
Input:Tail(s) = Two
Proportion p1 = 0.00154
Proportion p2 = 0.00234
á err prob = 0.05
Power (1-â err prob)= 0.8
Allocation ratio N2/N1 = 1
Output: Sample size group 1 = 49851
Sample size group 2 = 49851
Total sample size = 99702
Actual power= 0.8168040
Actual á= 0.0462658
Using the base R power.prop.test() function:
power.prop.test(p1 = 0.00154, p2 = 0.00234, power = 0.8)
Two-sample comparison of proportions power calculation
n = 47490.34
p1 = 0.00154
p2 = 0.00234
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
Using Frank's bsamsize() function in Hmisc:
bsamsize(p1 = 0.00154, p2 = 0.00234, fraction = .5, alpha = .05, power = .8)
n1 n2
47490.34 47490.34
Finally, throwing together a quick Monte Carlo simulation using the FET, I
get:
TwoSampleFET - function(n, p1, p2, power = 0.85,
R = 5000, correct = FALSE)
{
MCSim - function(n, p1, p2)
{
Control - rbinom(n, 1, p1)
Treat - rbinom(n, 1, p2)
fisher.test(cbind(table(Control), table(Treat)))$p.value
}
# Run MC Replicates
MC.res - replicate(R, MCSim(n, p1, p2))
# Get p value at power quantile
quantile(MC.res, power)
}
# 50,000 per arm
TwoSampleFET(5, p1 = 0.00154, p2 = 0.00234, power = 0.8, R = 500)
80%
0.04628263
So all four of these are coming back with numbers in the 48,000 to 50,000
***per arm***.
HTH,
Marc Schwartz
On Nov 8, 2010, at 10:16 AM, Mitchell Maltenfort wrote:
Not with R, but look for G*Power3, a free tool for power calc,
includes FIsher's test.
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni
perimessagg...@hotmail.com wrote:
Hi,
I'm try to compute the minimum sample size needed to have at least an 80%
of power, with alpha=0.05. The problem is that empirical proportions are
really small: 0.00154 in one case and 0.00234. These are the estimated
failure proportion of two medical treatments.
Thomas and Conlon (1992) suggested Fisher's exact test and proposed a
computational method, which according to their table gives a sample size
of roughly 2. Unfortunately I cannot find any software applying their
method.
-Does anyone know how to estimate sample size on Fisher's exact test by
using R?
-Even better, does anybody know other, maybe optimal, methods for such a
situation (small p1 and p2) and the corresponding R software?
Thanks in advance,
Giulio
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[R] Several lattice plots on one page
Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,4), var4 = c(10,2,3,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1))) pdf(test1.pdf, width=20, heigh=27, paper=a4) print(plot(groupedData(var1 ~ day | all, data = df), main = var1, xlab=, ylab=), split=c(1,1,2,4), more=TRUE) print(plot(groupedData(var2 ~ day | all, data = df), main = var2, xlab=, ylab=), split=c(1,2,2,4), more=TRUE) print(plot(groupedData(var3 ~ day | all, data = df), main = var3, xlab=, ylab=), split=c(1,3,2,4), more=TRUE) print(plot(groupedData(var4 ~ day | all, data = df), main = var4, xlab=, ylab=), split=c(1,4,2,4), more=TRUE) print(plot(groupedData(var5 ~ day | all, data = df), main = var5, xlab=, ylab=), split=c(2,1,2,4), more=TRUE) print(plot(groupedData(var6 ~ day | all, data = df), main = var6, xlab=, ylab=), split=c(2,2,2,4), more=TRUE) print(plot(groupedData(var7 ~ day | all, data = df), main = var7, xlab=, ylab=), split=c(2,3,2,4), more=TRUE) print(plot(groupedData(var8 ~ day | all, data = df), main = var8, xlab=, ylab=), split=c(2,4,2,4)) dev.off() My problem is that the separate plots all have different sizes. (Some are tall, but very small, or the other way around. The target is to have equally tall and wide graphs. (The variables have different scales. Grouping does not work.) Optimally, the plots would use the complete pdf page. Any ideas how to adjust height and width? Best Marcus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: could not find function extract in package raster
Users of R 2.11.1 (let alone 2.11.0) x64 Windows were asked to update in August, and we switched off the last remnants of support when 2.12.0 was released. Please do as the posting guide asked you to do before posting: update to R 2.12.0 (or 2.12.0 patched), and update all your packages. lattics is 0.19-13 raster is 1.6-15 sp is 0.9-72 On Mon, 8 Nov 2010, Monica Pisica wrote: Hi Joshua, OK, my session info is this: R version 2.11.0 (2010-04-22) x86_64-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.5-16 sp_0.9-64 These are not the same versions as below. loaded via a namespace (and not attached): [1] grid_2.11.0lattice_0.18-8 I also have installed the 32 bit version of R under windows 64 bit and there the raster package has the function extract and session info is: R version 2.11.0 (2010-04-22) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.6-10 sp_0.9-62 loaded via a namespace (and not attached): [1] grid_2.11.0lattice_0.18-5 Obviously i would like to work with the R64 bit because i have more memory available. It is not clear to me why the function extract exists in the package raster for 32 bit but not for 64 bit. Is there any reason? Thanks, Monica Date: Mon, 8 Nov 2010 08:17:48 -0800 Subject: Re: [R] Error: could not find function extract in package raster From: jwiley.ps...@gmail.com To: pisican...@hotmail.com CC: r-help@r-project.org Hi Monica, Evidently, there is not an extract function in your search path (base R, loaded packages, etc.). Given that you are talking about a function from a package (rather than base R), it would probably help us more if you mentioned what version of raster you have installed. You can get this easily by loading it and reporting the results of sessionInfo() (per the posting guide). On my current system (see specs below), there is an extract function and it seems to work., Cheers, Josh sessionInfo() R version 2.12.0 (2010-10-15) Platform: i486-pc-linux-gnu (32-bit) attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] raster_1.6-10 sp_0.9-72 loaded via a namespace (and not attached): [1] grid_2.12.0 lattice_0.19-13 tools_2.12.0 On Mon, Nov 8, 2010 at 8:08 AM, Monica Pisica pisican...@hotmail.com wrote: Hi, I would like to use the function extract from package raster and i get the following error: m01e - extract(marsh01, p) Error: could not find function extract marsh01 is a raster object and p is an intersectExtent object that is not null. The package is installed and loaded, i use a Windows machine 64 bit and R x64 2.11.0. Do i really need to update my R to use this function? Thanks, Monica [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ [[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, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot a normal distribution curve and a shaded tail with alpha?
Look at power.examp in the TeachingDemos package. If that plot is not good enough, you can steal the code and modify to your specifications. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Wu Gong Sent: Monday, November 08, 2010 9:18 AM To: r-help@r-project.org Subject: [R] How to plot a normal distribution curve and a shaded tail with alpha? I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to- plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha- tp3032194p3032194.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Difference scores
Hi all, I have a couple of questions that are general statistics questions rather than being R-specific. I'm interested in figuring out how to compute something like standard error for difference scores (in particular, differences scores of reaction times). Does anyone know if there is a standard way to do this? Any references would be very much appreciate. Secondly, I'm interested in figure out something like standard error for data points contributing to a correlation. I am looking at the correlation between the means of reaction times in two tasks by individual participants, and for some participants, the scores are going to have more or less error associated with them (and it is also possible that task A might have more error associated with it than task B, etc.). It seems logical to weight each set of points by this error when computing the correlation somehow, but I'm not sure if there might be a standardized way of doing this. Does anyone have any ideas or know of a method that is typically used? Any pointers are greatly appreciate, thanks very much! M -- View this message in context: http://r.789695.n4.nabble.com/Difference-scores-tp3032484p3032484.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to rbind list of vectors with unequal vector lengths?
Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a121 b123 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 rbind list of vectors with unequal vector lengths?
Try this: t(sapply(l, '[', 1:max(sapply(l, length On Mon, Nov 8, 2010 at 5:05 PM, johannes rara johannesr...@gmail.comwrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a121 b123 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] How to rbind list of vectors with unequal vector lengths?
What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a121 b123 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to rbind list of vectors with unequal vector lengths?
Thanks, data.frame or matrix. -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a 1 2 1 b 1 2 3 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to rbind list of vectors with unequal vector lengths?
So what do you want the matrix to look like, since the number of columns will be different between the two rows? johannes rara wrote: Thanks, data.frame or matrix. -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a121 b123 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Several lattice plots on one page
Hi: Is this what you had in mind? library(reshape2) df - df[, -10] # last variable is superfluous dm - melt(df, id = 'day') head(dm) day variable value 1 1 var1 1 2 2 var1 2 3 3 var1 3 4 4 var1 4 5 1 var2 100 6 2 var2 200 xyplot(value ~ day | variable, data = dm, scales = list(y = list(relation = 'free'))) melt() is useful in this problem because it stacks the data in 'long' form and creates a factor 'variable' whose levels are the variable names. The variable 'value' contains the responses. Conditioning on 'variable' in dm then gets you individual plots of value ~ day by variable. You can use layout = to specify the arrangement of plots per page, and perhaps as.table = TRUE if you want the plots rendered from top left rather than bottom left. HTH, Dennis On Mon, Nov 8, 2010 at 10:19 AM, Marcus Drescher dresc...@tum.de wrote: Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,4), var4 = c(10,2,3,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1))) pdf(test1.pdf, width=20, heigh=27, paper=a4) print(plot(groupedData(var1 ~ day | all, data = df), main = var1, xlab=, ylab=), split=c(1,1,2,4), more=TRUE) print(plot(groupedData(var2 ~ day | all, data = df), main = var2, xlab=, ylab=), split=c(1,2,2,4), more=TRUE) print(plot(groupedData(var3 ~ day | all, data = df), main = var3, xlab=, ylab=), split=c(1,3,2,4), more=TRUE) print(plot(groupedData(var4 ~ day | all, data = df), main = var4, xlab=, ylab=), split=c(1,4,2,4), more=TRUE) print(plot(groupedData(var5 ~ day | all, data = df), main = var5, xlab=, ylab=), split=c(2,1,2,4), more=TRUE) print(plot(groupedData(var6 ~ day | all, data = df), main = var6, xlab=, ylab=), split=c(2,2,2,4), more=TRUE) print(plot(groupedData(var7 ~ day | all, data = df), main = var7, xlab=, ylab=), split=c(2,3,2,4), more=TRUE) print(plot(groupedData(var8 ~ day | all, data = df), main = var8, xlab=, ylab=), split=c(2,4,2,4)) dev.off() My problem is that the separate plots all have different sizes. (Some are tall, but very small, or the other way around. The target is to have equally tall and wide graphs. (The variables have different scales. Grouping does not work.) Optimally, the plots would use the complete pdf page. Any ideas how to adjust height and width? Best Marcus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to rbind list of vectors with unequal vector lengths?
This is the ideal result (data.frame): result names X1 X2 X3 1 a 1 2 NA 2 b 1 2 3 2010/11/8 Erik Iverson er...@ccbr.umn.edu: So what do you want the matrix to look like, since the number of columns will be different between the two rows? johannes rara wrote: Thanks, data.frame or matrix. -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a 1 2 1 b 1 2 3 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sample size calculation for differences between two very small proportions (Fisher's exact test or others)?
On Nov 8, 2010, at 1:00 PM, Giulio Di Giovanni wrote:
Yep, it is 20.000 per arm, sorry. The reference it's about an
application of the method, and I cannot download the paper with the
main algorithm, so I don't know exactly how they did.
Thanks everybody for the rich and interesting suggestions. Through
free web software (PS, others) I found also an N around 47.000 per
arm. I guess these are the values (also seen Marc's Monte Carlo).
Maybe the Poisson models approach suggested by David can be an
alternative, even if I guess at this point I won't get big
differences in numbers. Would I?
I certainly would not expect remarkable differences. With 50,000/arm
you would be expecting:
c(p1 = 0.00154, p2 = 0.00234)*5
p1 p2
77 117
# with a rate ratio of:
0.00234/0.00154
[1] 1.519481
A difference of 30 in expected counts would seem to give fairly
significant power. It seems that a Poisson structured test might give
you smaller numbers but probably not as small as 20,000
c(p1 = 0.00154, p2 = 0.00234)*2
p1 p2
30.8 46.8
(The sd() of a Poisson variable is sqrt(mean()) so that 31 is well
within any sensibly constructed CI around 47.)
If you look up Table 7.5 in Breslow and Day (vol2, page 283) with a
relative risk of 1.5, the necessary expected value in the control
group using and equal sized control group ( for 80% power at 5%
significance) is 64.9. So that a bit lower than the 77 above but
implies that 42,207 would be needed.
--
David.
Thanks a lot everybody again for your suggestions,
if anybody has other comments, they are always welcome.
Best,
Giulio
Subject: Re: [R] Sample size calculation for differences between
two very small proportions (Fisher's exact test or others)?
From: marc_schwa...@me.com
Date: Mon, 8 Nov 2010 11:13:12 -0600
CC: perimessagg...@hotmail.com; r-h...@stat.math.ethz.ch
To: mmal...@gmail.com
Hi,
I don't have access to the article, but must presume that they are
doing something radically different if you are only getting a
total sample size of 20,000. Or is that 20,000 per arm?
Using the G*Power app that Mitchell references below (which I have
used previously, since they have a Mac app):
Exact - Proportions: Inequality, two independent groups (Fisher's
exact test)
Options:Exact distribution
Analysis: A priori: Compute required sample size
Input: Tail(s) = Two
Proportion p1 = 0.00154
Proportion p2 = 0.00234
α err prob = 0.05
Power (1-β err prob) = 0.8
Allocation ratio N2/N1 =1
Output: Sample size group 1 = 49851
Sample size group 2 = 49851
Total sample size = 99702
Actual power = 0.8168040
Actual α = 0.0462658
Using the base R power.prop.test() function:
power.prop.test(p1 = 0.00154, p2 = 0.00234, power = 0.8)
Two-sample comparison of proportions power calculation
n = 47490.34
p1 = 0.00154
p2 = 0.00234
sig.level = 0.05
power = 0.8
alternative = two.sided
NOTE: n is number in *each* group
Using Frank's bsamsize() function in Hmisc:
bsamsize(p1 = 0.00154, p2 = 0.00234, fraction = .5, alpha = .05,
power = .8)
n1 n2
47490.34 47490.34
Finally, throwing together a quick Monte Carlo simulation using
the FET, I get:
TwoSampleFET - function(n, p1, p2, power = 0.85,
R = 5000, correct = FALSE)
{
MCSim - function(n, p1, p2)
{
Control - rbinom(n, 1, p1)
Treat - rbinom(n, 1, p2)
fisher.test(cbind(table(Control), table(Treat)))$p.value
}
# Run MC Replicates
MC.res - replicate(R, MCSim(n, p1, p2))
# Get p value at power quantile
quantile(MC.res, power)
}
# 50,000 per arm
TwoSampleFET(5, p1 = 0.00154, p2 = 0.00234, power = 0.8, R =
500)
80%
0.04628263
So all four of these are coming back with numbers in the 48,000 to
50,000 ***per arm***.
HTH,
Marc Schwartz
On Nov 8, 2010, at 10:16 AM, Mitchell Maltenfort wrote:
Not with R, but look for G*Power3, a free tool for power calc,
includes FIsher's test.
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3
On Mon, Nov 8, 2010 at 10:52 AM, Giulio Di Giovanni
perimessagg...@hotmail.com wrote:
Hi,
I'm try to compute the minimum sample size needed to have at
least an 80% of power, with alpha=0.05. The problem is that
empirical proportions are really small: 0.00154 in one case and
0.00234. These are the estimated failure proportion of two medical
treatments.
Thomas and Conlon (1992) suggested Fisher's exact test and
proposed a computational method, which according to their table
gives a sample size of roughly 2. Unfortunately I cannot find
any software applying their method.
-Does anyone know how to estimate sample size on Fisher's exact
test by using R?
-Even better, does anybody know other, maybe optimal, methods
for such a situation
Re: [R] How to rbind list of vectors with unequal vector lengths?
Then one solution is to use rbind.fill from the plyr package. johannes rara wrote: This is the ideal result (data.frame): result names X1 X2 X3 1 a 1 2 NA 2 b 1 2 3 2010/11/8 Erik Iverson er...@ccbr.umn.edu: So what do you want the matrix to look like, since the number of columns will be different between the two rows? johannes rara wrote: Thanks, data.frame or matrix. -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a121 b123 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to rbind list of vectors with unequal vector lengths?
I have tried it, but it does not seem to work with vectors, only data.frames do.call(rbind.fill, l) NULL -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: Then one solution is to use rbind.fill from the plyr package. johannes rara wrote: This is the ideal result (data.frame): result names X1 X2 X3 1 a 1 2 NA 2 b 1 2 3 2010/11/8 Erik Iverson er...@ccbr.umn.edu: So what do you want the matrix to look like, since the number of columns will be different between the two rows? johannes rara wrote: Thanks, data.frame or matrix. -J 2010/11/8 Erik Iverson er...@ccbr.umn.edu: What class of object / structure do you exactly want in the end? A matrix, a data.frame, a vector? johannes rara wrote: Hi, How to rbind these vectors from a list?: l - list(a = c(1, 2), b = c(1, 2, 3)) l $a [1] 1 2 $b [1] 1 2 3 do.call(rbind, l) [,1] [,2] [,3] a 1 2 1 b 1 2 3 Warning message: In function (..., deparse.level = 1) : number of columns of result is not a multiple of vector length (arg 1) -J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lookup in R - possible to avoid loops?
Hello!
Hope there is a nifty way to speed up my code by avoiding loops.
My task is simple - analogous to the vlookup formula in Excel. Here is
how I programmed it:
# My example data frame:
set.seed(1245)
my.df-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0))
my.df-my.df[order(my.df$names),]
my.df$names-as.character(my.df$names)
(my.df)
# My example lookup table:
my.lookup-data.frame(names=letters[1:3],category=c(AAA,BBB,CCC))
my.lookup$names-as.character(my.lookup$names)
my.lookup$category-as.character(my.lookup$category)
(my.lookup)
# Just adding an extra column to my.df that contains the categories of
the names in the column names:
my.df2-my.df
my.df2$category-NA
for(i in unique(my.df$names)){
my.df2$category[my.df2$names %in%
i]-my.lookup$category[my.lookup$names %in% i]
}
(my.df2)
It does what I need, but it's way too slow - I need to run it for
hundreds and hundreds of names in 100 of huge files (tens of
thousands of rows in each).
Any way to speed it up?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
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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] lookup in R - possible to avoid loops?
Try this:
merge(my.df, my.lookup)
On Mon, Nov 8, 2010 at 5:43 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
Hope there is a nifty way to speed up my code by avoiding loops.
My task is simple - analogous to the vlookup formula in Excel. Here is
how I programmed it:
# My example data frame:
set.seed(1245)
my.df-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0))
my.df-my.df[order(my.df$names),]
my.df$names-as.character(my.df$names)
(my.df)
# My example lookup table:
my.lookup-data.frame(names=letters[1:3],category=c(AAA,BBB,CCC))
my.lookup$names-as.character(my.lookup$names)
my.lookup$category-as.character(my.lookup$category)
(my.lookup)
# Just adding an extra column to my.df that contains the categories of
the names in the column names:
my.df2-my.df
my.df2$category-NA
for(i in unique(my.df$names)){
my.df2$category[my.df2$names %in%
i]-my.lookup$category[my.lookup$names %in% i]
}
(my.df2)
It does what I need, but it's way too slow - I need to run it for
hundreds and hundreds of names in 100 of huge files (tens of
thousands of rows in each).
Any way to speed it up?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] lookup in R - possible to avoid loops?
Dimitri -
While merge is most likely the fastest way to solve
your problem, I just want to point out that you can use
a named vector as a lookup table. For your example:
categories = my.lookup$category
names(categories) = my.lookup$names
creates the lookup table, and
my.df$category = categories[my.df$names]
creates the category column.
- Phil
On Mon, 8 Nov 2010, Dimitri Liakhovitski wrote:
Hello!
Hope there is a nifty way to speed up my code by avoiding loops.
My task is simple - analogous to the vlookup formula in Excel. Here is
how I programmed it:
# My example data frame:
set.seed(1245)
my.df-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0))
my.df-my.df[order(my.df$names),]
my.df$names-as.character(my.df$names)
(my.df)
# My example lookup table:
my.lookup-data.frame(names=letters[1:3],category=c(AAA,BBB,CCC))
my.lookup$names-as.character(my.lookup$names)
my.lookup$category-as.character(my.lookup$category)
(my.lookup)
# Just adding an extra column to my.df that contains the categories of
the names in the column names:
my.df2-my.df
my.df2$category-NA
for(i in unique(my.df$names)){
my.df2$category[my.df2$names %in%
i]-my.lookup$category[my.lookup$names %in% i]
}
(my.df2)
It does what I need, but it's way too slow - I need to run it for
hundreds and hundreds of names in 100 of huge files (tens of
thousands of rows in each).
Any way to speed it up?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] A more efficient way to roll values in an irregular time series dataset?
Does anyone recommend a more efficient way to roll values in a time series
dataset?
I merged a bunch of different time series datasets (10's of thousands of them)
whose observation dates and sampling interval differ. Some time series
observations are reported at the beginning of the month, some at the end, some
on Mondays, some on Wednesday, some annually, etc.
In the process of merging all of the irregular time series (by date observed),
a significant number of NA's appear in the dataset where I really want the last
reported value 'rolled' forward.
To use a concrete example, a time series that has reported values at the
beginning of every month shows NA's for every day except the date it was
reported (in this case, the first of the month). I want the value to roll
forward so that NA's after the first of the month are replaced with a last
reported value.
I wrote the following for loop to accomplish the task on the object 'dataset',
however it is far to slow too process 10's of thousands of different time
series with 15,000 observations each. At this rate it is going, it would take
weeks to complete.
for(j in 1:length(names(dataset)))
{
last-NA;
for(i in 1:length(row.names(dataset)))
ifelse(is.na(dataset[i,j]), test[i,j] - last,
last-dataset[i,j]);
}
One would think a rather simple operation as this could perform much faster.
My sense is using the apply function is the way to go, however I just can't
get my head around a function that would reference the last reported value.
Any guidance is appreciated.
-Richard
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to eliminate this for loop ?
If you are willing to shift the c vector by 1 and have 1 (the initial value) as
the start of c, then you can just do:
cumsum( cc * b^( (n-1):0 ) ) / b^( (n-1):0 )
to compare:
cc - c(1, rnorm(999) )
b - 0.5
n - length(cc)
a1 - numeric(100)
a1[1] - 1
system.time(for(i in 2:n ) {
a1[i] - b*a1[i-1] + cc[i]
})
system.time(a2 - cumsum( cc * b^( (n-1):0 ) ) / b^( (n-1):0 ))
all.equal(a1,a2)
Though you could have problems with the b^ part if the length gets too long.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of PLucas
Sent: Monday, November 08, 2010 2:26 AM
To: r-help@r-project.org
Subject: [R] How to eliminate this for loop ?
Hi, I would like to create a list recursively and eliminate my for loop
:
a-c()
a[1] - 1; # initial value
for(i in 2:N) {
a[i]-a[i-1]*b - c[i-1] # b is a value, c is another vector
}
Is it possible ?
Thanks
--
View this message in context: http://r.789695.n4.nabble.com/How-to-
eliminate-this-for-loop-tp3031667p3031667.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A more efficient way to roll values in an irregular time series dataset?
Hi:
Look into the zoo package and its rollapply() function. The package is
designed to handle irregular and multiple series.
HTH,
Dennis
On Mon, Nov 8, 2010 at 12:16 PM, Richard Vlasimsky
richard.vlasim...@imidex.com wrote:
Does anyone recommend a more efficient way to roll values in a time
series dataset?
I merged a bunch of different time series datasets (10's of thousands of
them) whose observation dates and sampling interval differ. Some time
series observations are reported at the beginning of the month, some at the
end, some on Mondays, some on Wednesday, some annually, etc.
In the process of merging all of the irregular time series (by date
observed), a significant number of NA's appear in the dataset where I really
want the last reported value 'rolled' forward.
To use a concrete example, a time series that has reported values at the
beginning of every month shows NA's for every day except the date it was
reported (in this case, the first of the month). I want the value to roll
forward so that NA's after the first of the month are replaced with a last
reported value.
I wrote the following for loop to accomplish the task on the object
'dataset', however it is far to slow too process 10's of thousands of
different time series with 15,000 observations each. At this rate it is
going, it would take weeks to complete.
for(j in 1:length(names(dataset)))
{
last-NA;
for(i in 1:length(row.names(dataset)))
ifelse(is.na(dataset[i,j]), test[i,j] - last,
last-dataset[i,j]);
}
One would think a rather simple operation as this could perform much
faster. My sense is using the apply function is the way to go, however I
just can't get my head around a function that would reference the last
reported value.
Any guidance is appreciated.
-Richard
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Re: [R] A more efficient way to roll values in an irregular time series dataset?
On Mon, Nov 8, 2010 at 3:16 PM, Richard Vlasimsky
richard.vlasim...@imidex.com wrote:
Does anyone recommend a more efficient way to roll values in a time series
dataset?
I merged a bunch of different time series datasets (10's of thousands of
them) whose observation dates and sampling interval differ. Some time series
observations are reported at the beginning of the month, some at the end,
some on Mondays, some on Wednesday, some annually, etc.
In the process of merging all of the irregular time series (by date
observed), a significant number of NA's appear in the dataset where I really
want the last reported value 'rolled' forward.
To use a concrete example, a time series that has reported values at the
beginning of every month shows NA's for every day except the date it was
reported (in this case, the first of the month). I want the value to roll
forward so that NA's after the first of the month are replaced with a last
reported value.
I wrote the following for loop to accomplish the task on the object
'dataset', however it is far to slow too process 10's of thousands of
different time series with 15,000 observations each. At this rate it is
going, it would take weeks to complete.
for(j in 1:length(names(dataset)))
{
last-NA;
for(i in 1:length(row.names(dataset)))
ifelse(is.na(dataset[i,j]), test[i,j] - last,
last-dataset[i,j]);
}
One would think a rather simple operation as this could perform much faster.
My sense is using the apply function is the way to go, however I just can't
get my head around a function that would reference the last reported value.
Any guidance is appreciated.
Don't know if its fast enough for you but in zoo you can merge and
carry the last occurrence forward like this:
# suppose z1, z2, z3 are zoo series
na.locf(merge(z1, z2, z3)) # as many as you like
or
L - list(z1, z2, z3)
na.locf(do.call(merge, L))
which produces a multivariate series, one per column with NAs filled in.
--
Statistics Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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Re: [R] How to plot a normal distribution curve and a shaded tail with alpha?
Here's an example of a PDF with the tails shaded: http://stackoverflow.com/q/3494593/37751 -JD On Mon, Nov 8, 2010 at 10:18 AM, Wu Gong w...@mtmail.mtsu.edu wrote: I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha-tp3032194p3032194.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot a normal distribution curve and a shaded tail with alpha?
I wrote this a bit ago, its far from perfect, but might give you
ideas/serve your purpose.
HTH,
Josh
plot.dist - function(alpha, from = -5, to = 5, n = 1000, filename = NULL,
alternative = c(two.tailed, greater, lesser),
distribution = c(normal, t, F, chisq, binomial),
colour = black, fill = skyblue2, ...) {
alternative - match.arg(alternative)
## Calculate alpha level given hypothesis
alt.alpha - switch(alternative,
two.tailed = alpha/2,
greater = alpha,
lesser = alpha)
## use a ’switch’ to pick the right functions based on distribution
my.den - switch(distribution,
normal = dnorm,
t = dt,
F = df,
chisq = dchisq,
binomial = dbinom)
my.dist - switch(distribution,
normal = qnorm,
t = qt,
F = qf,
chisq = qchisq,
binomial = qbinom)
## Additional arguments passed via ’...’ e.g., degrees of freedom
crit.lower - my.dist(p = alt.alpha, lower.tail = TRUE, ...)
crit.upper - my.dist(p = alt.alpha, lower.tail = FALSE, ...)
## Calculate alpha (lower) region coordinates
cord.x1 - c(from, seq(from = from, to = crit.lower,
length.out = 100), crit.lower)
cord.y1 - c(0, my.den(x = seq(from = from, to = crit.lower,
length.out = 100), ...), 0)
## Calculate alpha (upper) region coordinates
cord.x2 - c(crit.upper, seq(from = crit.upper, to = to,
length.out = 100), to)
cord.y2 - c(0, my.den(x = seq(from = crit.upper, to = to,
length.out = 100), ...), 0)
## Logic test to choose which graphic device to open
if(is.null(filename)) {
dev.new()
} else {
pdf(file = filename)
}
## plot distribution
curve(my.den(x, ...), from = from, to = to,
n = n, col = colour, lty = 1, lwd = 2,
ylab = Density, xlab = Values)
## Add alpha region(s) based on given hypothesis
if(!identical(alternative, greater)) {
polygon(x = cord.x1, y = cord.y1, col = fill)
}
if(!identical(alternative, lesser)) {
polygon(x = cord.x2, y = cord.y2, col = fill)
}
## If the PDF device was started, shut it down
if(!is.null(filename)) {dev.off()}
}
On Mon, Nov 8, 2010 at 8:18 AM, Wu Gong w...@mtmail.mtsu.edu wrote:
I want to create a graph to express the idea of the area under a pdf curve,
like
http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg
Thank you for any help.
-
A R learner.
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha-tp3032194p3032194.html
Sent from the R help mailing list archive at Nabble.com.
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
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[R] Random Sample
Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then runif(10) is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[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] lookup in R - possible to avoid loops?
Thanks a lot - extremely heplful!
While I'll definitely try to use merge in the future, in my situation
I run into problems with memory (files are too large).
However, Phil's suggestion is perfect for me - sped me up considerably!
Thank you, again!
Dimitri
On Mon, Nov 8, 2010 at 2:51 PM, Phil Spector spec...@stat.berkeley.edu wrote:
Dimitri -
While merge is most likely the fastest way to solve
your problem, I just want to point out that you can use
a named vector as a lookup table. For your example:
categories = my.lookup$category
names(categories) = my.lookup$names
creates the lookup table, and
my.df$category = categories[my.df$names]
creates the category column.
- Phil
On Mon, 8 Nov 2010, Dimitri Liakhovitski wrote:
Hello!
Hope there is a nifty way to speed up my code by avoiding loops.
My task is simple - analogous to the vlookup formula in Excel. Here is
how I programmed it:
# My example data frame:
set.seed(1245)
my.df-data.frame(names=rep(letters[1:3],3),value=round(rnorm(9,mean=20,sd=5),0))
my.df-my.df[order(my.df$names),]
my.df$names-as.character(my.df$names)
(my.df)
# My example lookup table:
my.lookup-data.frame(names=letters[1:3],category=c(AAA,BBB,CCC))
my.lookup$names-as.character(my.lookup$names)
my.lookup$category-as.character(my.lookup$category)
(my.lookup)
# Just adding an extra column to my.df that contains the categories of
the names in the column names:
my.df2-my.df
my.df2$category-NA
for(i in unique(my.df$names)){
my.df2$category[my.df2$names %in%
i]-my.lookup$category[my.lookup$names %in% i]
}
(my.df2)
It does what I need, but it's way too slow - I need to run it for
hundreds and hundreds of names in 100 of huge files (tens of
thousands of rows in each).
Any way to speed it up?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
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Ninah Consulting
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Re: [R] Random Sample
Use set.seed() -- Jonathan P. Daily Technician - USGS Leetown Science Center 11649 Leetown Road Kearneysville WV, 25430 (304) 724-4480 Is the room still a room when its empty? Does the room, the thing itself have purpose? Or do we, what's the word... imbue it. - Jubal Early, Firefly From: Xiaoxi Gao rhel...@hotmail.com To: R Help r-help@r-project.org Date: 11/08/2010 03:59 PM Subject: [R] Random Sample Sent by: r-help-boun...@r-project.org Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then runif(10) is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[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] Random Sample
?set.seed is what you're looking for Xiaoxi Gao wrote: Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then runif(10) is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[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] Random Sample
Hi Xiaoxi, Take a look at the following: set.seed(123) rnorm(10) [1] -0.56047565 -0.23017749 1.55870831 0.07050839 0.12928774 1.71506499 [7] 0.46091621 -1.26506123 -0.68685285 -0.44566197 rnorm(10) [1] 1.2240818 0.3598138 0.4007715 0.1106827 -0.5558411 1.7869131 0.4978505 [8] -1.9666172 0.7013559 -0.4727914 set.seed(123) rnorm(10) [1] -0.56047565 -0.23017749 1.55870831 0.07050839 0.12928774 1.71506499 [7] 0.46091621 -1.26506123 -0.68685285 -0.44566197 and check ?set.seed(). HTH, Jorge On Mon, Nov 8, 2010 at 3:57 PM, Xiaoxi Gao wrote: Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then runif(10) is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with getting ?match to not sort
Hello all, I think I am missing something about the sorting parameter in the match command/ Here is an example: a1 - data.frame(name = c(D, B, C, A, A, C)) a2 - data.frame(name = c(A, B, C, D), num = 1:4) a1 a2 merge(a1, a2, sort = F, by.x = T) The result is: name num 1D 4 2B 2 3C 3 4C 3 5A 1 6A 1 While I wish my rows to be in the same order as in a1, they are having some other order. What am I missing here? Thanks. Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to plot a normal distribution curve and a shaded tail with alpha?
You can use the normal.and.t.dist() function in the HH package, available from CRAN with install.packages(HH) This works on all platforms. There is menu access to the function through Rcmdr if you install install.packages(RcmdrPlugin.HH) If you are on Windows, you can additionally control that function interactively from an Excel spreadsheet with the RExcel package. Install with the RthroughExcelWorkbooksInstaller package from CRAN. Rich On Mon, Nov 8, 2010 at 11:18 AM, Wu Gong w...@mtmail.mtsu.edu wrote: I want to create a graph to express the idea of the area under a pdf curve, like http://r.789695.n4.nabble.com/file/n3032194/w7295e04.jpg Thank you for any help. - A R learner. -- View this message in context: http://r.789695.n4.nabble.com/How-to-plot-a-normal-distribution-curve-and-a-shaded-tail-with-alpha-tp3032194p3032194.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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to represent factor levels ordiellipses using different colors?
Dear Jari Oksanen, I am Trinadh Kumar, a student of Biotechnology from Texas Tech University. I have a question for you regarding constrained ordination graphs. I had previously plotted constrained ordination graphs using distance based redundancy analysis for species data on patients. The species matrix consists of 70 patients and 274 species of bacteria.The metadata matrix consists of 10 variables(environmental matrix) My Professor asked me to do the following.To plot a constrained ordination plot constrained by tobacco habit(which is one of the factors). However he asked me to represent the graph as follows.. Represent the significant species (species that are farthest from the origin in the direction of the environmental variable) with biplot arrows pointing from origin to the species and hide the insignificant species(closer to the origin) on the graph. Also he asked me to use colored solid circles for plotting patients(instead of patient names) with the three levels of the tobacco factor( for example: red circle to represent tobacco0, green circles to represent tobacco1, black circles to represent tobacco2).He asked me to shade the 95% confidence ellipses with the same color as the levels of tobacco factor. Please advice. Thanks, Trinadh [[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] Help with getting ?match to not sort
Missing: a closer reading of the help page -- Value A data frame. The rows are by default lexicographically sorted on the common columns, but for sort = FALSE are in an unspecified order. So sort = FALSE says unspecified. If you want the original order, then add a column to the dataframe with the order and then sort the result. On Mon, Nov 8, 2010 at 4:09 PM, Tal Galili tal.gal...@gmail.com wrote: Hello all, I think I am missing something about the sorting parameter in the match command/ Here is an example: a1 - data.frame(name = c(D, B, C, A, A, C)) a2 - data.frame(name = c(A, B, C, D), num = 1:4) a1 a2 merge(a1, a2, sort = F, by.x = T) The result is: name num 1 D 4 2 B 2 3 C 3 4 C 3 5 A 1 6 A 1 While I wish my rows to be in the same order as in a1, they are having some other order. What am I missing here? Thanks. Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[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 that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sapply
y=sapply(1:nrow(x), function(i) sapply(1:ncol(x), function(j)
getrc(i, j)))
Arvin Zolfaghari, Ph.D.
Aon Benfield | Aon Benfield Analytics
55 Bishopsgate, London, EC2N 3BD, UK
t: +44 20 7522 8241 | f: +44 20 7522 3846| M: +44 78 8940 6637
e: arvin.zolfagh...@aonbenfield.com | w:
www.AonBenfield.comhttp://www.aonbenfield.com/
From: Arvin Zolfaghari
Sent: 08 November 2010 14:35
To: 'markus.gesm...@lloyds.com'
Subject: question
Hi List,
I am trying to apply a
function to all elements of a matrix. I can use two consequent ' for'
loops, but it is really slow process. I can also use the sapply as follows,
but it is 10% faster, is there any other faster way (like fft)?
I got a matrix of x=matrix(0,1000,2000)
I want to load it with myfunction
first option:
For ( i in nrow(x)){
For(j in nrow(x)){
x=myfunction(i,j)
}}
second option:
t(sapply(1:nrow(x), function(i) sapply(1:ncol(x), function(j)
getrc(i, j
both option 1 and 2 work, but so slow. I appreciate your help.
thanks
Alireza
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Re: [R] Help with getting ?match to not sort
Ted - If you want to retain the exact order of the rows of one of the merged matrices, I think you have to merge them by hand: cbind(a1,num=a2[match(a1$name,a2$name),'num']) name num 1D 4 2B 2 3C 3 4A 1 5A 1 6C 3 - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Mon, 8 Nov 2010, Tal Galili wrote: Hello all, I think I am missing something about the sorting parameter in the match command/ Here is an example: a1 - data.frame(name = c(D, B, C, A, A, C)) a2 - data.frame(name = c(A, B, C, D), num = 1:4) a1 a2 merge(a1, a2, sort = F, by.x = T) The result is: name num 1D 4 2B 2 3C 3 4C 3 5A 1 6A 1 While I wish my rows to be in the same order as in a1, they are having some other order. What am I missing here? Thanks. Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] try (nls stops unexpectedly because of chol2inv error
Date: Mon, 8 Nov 2010 06:18:49 -0800
From: monte.shaf...@gmail.com
To: r-help@r-project.org
Subject: [R] try (nls stops unexpectedly because of chol2inv error
Hi,
I am running simulations that does multiple comparisons to control.
For each simulation, I need to model 7 nls functions. I loop over 7 to do
the nls using try
if try fails, I break out of that loop, and go to next simulation.
I get warnings on nls failures, but the simulation continues to run, except
when the internal call (internal to nls) of the chol2inv fails.
Error in chol2inv(object$m$Rmat()) :
element (2, 2) is zero, so the inverse cannot be computed
In addition: Warning messages:
1: In nls(myModel.nlm, fData, start = initialValues, control =
nls.control(warnOnly = TRUE), :
number of iterations exceeded maximum of 50
2: In nls(myModel.nlm, fData, start = initialValues, control =
nls.control(warnOnly = TRUE), :
singular gradient
===
Any suggestions on how to prevent chol2inv from breaking my simulation...
Since no one else has answered, let me supply some thoughts and google hits.
I'm not sure what your question is- the error message suggests the
matrix has no inverse as in A*A-1 =I can't be found- usually these things happen
because the data is not a good fit to the model. Is the message not
literally true as in you know that A has an inverse? It does seem
you posted a good complete example but it may take a bit of effort
for someone to debug.
The reason it is non-invertible probably has to do with the gradient issue,
in any case some good hits on google like this may help,
https://stat.ethz.ch/pipermail/r-help/2008-March/158329.html
( http://www.google.com/?#hl=enq=r+nls+singular+gradientfp=1 )
Personally I
tend to use SVD in my c++ code since it is the only method I know
that provides a good diagnostic on how close I came to having an
ill posed model. In your case, presumably either your model or data or code is
creating an exactly singular matrix, this may be easier to find
than the almost singular situations that often create odd results :)
I would just ask however if anyone
has more thoughts on inverting mtricies for model fits as someone
previously mentioned that R uses QR decomposition for one task to qualify my
generic response to a question.
The point of the simulation is to address power. As our data goes down to
N, of the 100 simulations, only 53 are good simulations because we don't
have enough data for nls or chol2inv to work correctly.
monte
{x:
###
## case I ## EQUAL SAMPLE SIZES and design points
nsim = 100;
N_i = M_i = 10; ## also try (10, 30, 50, 100, 200)
r = M_i / N_i;
X.start = 170; # 6 design points, at 170,180,190, etc. where each point has
N_i elements
X.increment = 10;
X.points = 6;
X.end = 260;
Xval = seq(X.start,length.out=X.points,by=X.increment );
Xval = seq(X.start,X.end,length.out=X.points);
L = 7; ## 6 + control
k = 3;
varY = 0.15;
### for each simulation, we need to record all of this information, write to
a table or file.
### Under the null of simulation, we assign all locations to have same model
## we assume these are the true parameters
b = 2.87; d = 0.0345; t = 173;
B = seq(2.5,4.5,length.out=21);
#B = seq(2.75,3.25,length.out=21);
#B = seq(2.85,2.95,length.out=21);
#B = seq(2.8,3.0,length.out=21);
B = seq(2.5,3.2,length.out=21);
D = seq(0.02,0.04,length.out=21);
T = seq(165,185,length.out=21);
alpha = .05;
nu = k; ## number of parameters
tr = L-1; ## number of treatments (including control)
rho = 1/(1+r); ## dependency parameter
myCritical = qmvchi(alpha,nu,tr,rho);
## we change one parameter at a time until the results fail most of the
time.
## do independent for now, but let's store the parameters and quantiles???
INFO for one location
# beta delta tau nsim %Reject(V.pooled) %Reject(V.total) [Simulation level]
resultS
# beta delta tau i of nsim max(V.pooled) max(V.total) [Individual level]
resultI
resultS = resultI = NULL;
for(p1 in 1:length(D))
{
print(paste(p1, [D] of ,length(D))); flush.console();
print(D[p1]);
myReject.pooled = myReject.pooled.1 = MAX.pooled = rep(-1,nsim);
gsim = 0; ## good simulations
for(i in 1:nsim)
{
doubleBreak = F;
print(paste(i, of ,nsim)); flush.console();
tData = NULL;
pooledNum = matrix(0,nrow=k,ncol=k); ##numerator as weighted sum AS
(n_k-1)cov.scaled
pooledDen = 0; ##denominator as correction AS N-k
#Sigma_pooled = ((omit.1-1)*summary.nls.1$cov.scaled +
(omit.2-1)*summary.nls.2$cov.scaled +
(omit.L-1)*summary.nls.L$cov.scaled)/(sum(omit.1,omit.2,omit.L)-L);
for(j in 1:L)
{
Y = numeric(N_i);
X = createDomain(Xval,N_i); noise = rnorm(N_i, mean=0,sd=sqrt(varY) );
if(j==1)
{
## location #1 is different
Y =
[R] : unusual combinations of categorical data
Regarding unusual combinations of factors in categorical data. Are there any R packages that can be used to identify the outliers i.e. unusual combinations in categorical datasets ? Thanks. Notice of Confidentiality This transmission contains information that may be confidential and that may also be privileged. Unless you are the intended recipient of the message (or authorised to receive it for the intended recipient) you may not copy, forward, or otherwise use it, or disclose it or its contents to anyone else. If you have received this transmission in error please notify us immediately and delete it from your system. RSA Insurance Group plc. Registered in England No. 2339826. The Registered Office is 9th Floor, One Plantation Place, 30 Fenchurch Street, London EC3M 3BD [[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] : unusual combinations of categorical data
Hi, On Mon, Nov 8, 2010 at 2:25 PM, Alan Chalk alan.ch...@gcc.rsagroup.com wrote: Regarding unusual combinations of factors in categorical data. where all variables are categorical? Are there any R packages that can be used to identify the outliers i.e. unusual combinations in categorical datasets ? outlier or unusual tends to be rather variable, that is something unusual in one data set may not be in another. If you are dealing with strictly categorical variables, I am not certain how you would define an outlier. The categories only have the meaning attached to them, so it seems like they would only indicate outliers if you decided that an entire category was an outlier (e.g., males, females, half-man-half-ox). If you have one continuous variable in mind by different levels of a factor, then you could just use some simple plots (e.g., ggplot() + geom_point() + facet_grid(factor ~ .) or something similar). You could also z-score the values by each factor level and then extract zscores more extreme than +/- 3 or whatever value you like. It might be easier to give you feedback if you have a more specific example. Cheers, Josh Thanks. Notice of Confidentiality This transmission contains information that may be confidential and that may also be privileged. Unless you are the intended recipient of the message (or authorised to receive it for the intended recipient) you may not copy, forward, or otherwise use it, or disclose it or its contents to anyone else. If you have received this transmission in error please notify us immediately and delete it from your system. RSA Insurance Group plc. Registered in England No. 2339826. The Registered Office is 9th Floor, One Plantation Place, 30 Fenchurch Street, London EC3M 3BD [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how do i plot this hist?
Hi all, I have the following data in abc.dat === 50 0 1 0 0 55 114 0 1 60 786 0 3 6522 324 2 3 7058 1035 1 7 7530 2568 034 80 9 293615 162 8527 216946 365 9080 1439 212 432 95 236 1670 521 281 100 332 827 709 172 105 156 311 556 103 1106949 14444 11526103617 120 2 9 3 3 125 1 6 1 1 130 014 0 0 135 0 5 0 0 140 0 0 0 0 145 0 0 0 0 150 0 0 0 0 155 0 0 0 0 160 0 0 0 0 165 0 0 0 0 170 0 0 0 0 175 0 0 0 0 180 0 0 0 0 185 0 0 0 0 190 0 0 0 1 195 0 0 0 0 200 0 0 0 0 205 0 0 0 0 210 0 0 0 0 === which i have used abc=read.table(abc.dat) to read the table into R. There are two problems: 1- I want the first column of the data to be the 'column names', how should i read the data? 2- I want to plot the histogram, using the first column as 'x' values, and the 2nd,3rd,4th and 5th columns as the frequencies. How do I plot it? I have tried to add a 'row' of variable names to it, and then read with 'header=T', then the first column become 'col.names' as I was expecting it to be. However, when I plot it using 'hist', R uses the 2nd column as the 'x value', where it should be used as 'frequency'. (the 50,55,60,65,70... should be on the x-axis) Thanks! Casper -- View this message in context: http://r.789695.n4.nabble.com/how-do-i-plot-this-hist-tp3032796p3032796.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] Random Sample
Here is an example of what I think the original poster wanted to achieve. rnorm(10) [1] -1.2165869 -0.4698460 -0.4209811 -1.4856052 0.3765774 -1.3822470 [7] 0.2818458 0.5500957 -1.1474455 -1.2221257 x - .Random.seed runif(10) [1] 0.5610780 0.5911841 0.5868183 0.3833801 0.7397059 0.4973270 0.6544219 [8] 0.1257485 0.2756357 0.5672207 .Random.seed - x runif(10) [1] 0.5610780 0.5911841 0.5868183 0.3833801 0.7397059 0.4973270 0.6544219 [8] 0.1257485 0.2756357 0.5672207 Best, Giovanni Petris On Mon, 2010-11-08 at 16:05 -0500, Jorge Ivan Velez wrote: Hi Xiaoxi, Take a look at the following: set.seed(123) rnorm(10) [1] -0.56047565 -0.23017749 1.55870831 0.07050839 0.12928774 1.71506499 [7] 0.46091621 -1.26506123 -0.68685285 -0.44566197 rnorm(10) [1] 1.2240818 0.3598138 0.4007715 0.1106827 -0.5558411 1.7869131 0.4978505 [8] -1.9666172 0.7013559 -0.4727914 set.seed(123) rnorm(10) [1] -0.56047565 -0.23017749 1.55870831 0.07050839 0.12928774 1.71506499 [7] 0.46091621 -1.26506123 -0.68685285 -0.44566197 and check ?set.seed(). HTH, Jorge On Mon, Nov 8, 2010 at 3:57 PM, Xiaoxi Gao wrote: Hello R users, Here is my question about generating random sample. How to set the random seed to recreate the same random numbers? For example, 10 random numbers is generated from N(0,1), then runif(10) is used.What if I want to get the same 10 random numbers when I run runif(10) again? Is it possible?I think .Random.seed should be used here. Thanks. Xiaoxi [[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. -- Giovanni Petris gpet...@uark.edu Associate Professor Department of Mathematical Sciences University of Arkansas - Fayetteville, AR 72701 Ph: (479) 575-6324, 575-8630 (fax) http://definetti.uark.edu/~gpetris/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with getting ?match to not sort
Tena koe Tal
sort: logical. Should the results be sorted on the 'by' columns?
Thus it is clear what sort=TRUE does, sort=FALSE does not do this. That
doesn't mean it leaves the result in the same order as x (your a1), although
many people (including me at first) assume it does. It is easy enough to write
a wrapper function to retain the order of x in cases like yours. For example:
mergeRO - function (x, y, ...)
{
myMerge - merge(cbind(x, myOrder = 1:nrow(x)), y, ...)
attr(myMerge, creation time) - Sys.time()
myMerge[order(myMerge$myOrder), colnames(myMerge) != myOrder]
}
but I'm not sure this will work in all circumstances in which people might use
merge().
HTH ...
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Tal Galili
Sent: Tuesday, 9 November 2010 10:10 a.m.
To: r-help@r-project.org
Subject: [R] Help with getting ?match to not sort
Hello all,
I think I am missing something about the sorting parameter in the
match
command/
Here is an example:
a1 - data.frame(name = c(D, B, C, A, A, C))
a2 - data.frame(name = c(A, B, C, D), num = 1:4)
a1
a2
merge(a1, a2, sort = F, by.x = T)
The result is:
name num
1D 4
2B 2
3C 3
4C 3
5A 1
6A 1
While I wish my rows to be in the same order as in a1, they are having
some
other order.
What am I missing here?
Thanks.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew)
|
www.r-statistics.com (English)
---
---
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Re: [R] how do i plot this hist?
try this:
x - read.table('clipboard')
x
V1 V2 V3 V4 V5
1 50 01 0 0
2 55 1 14 0 1
3 60 7 86 0 3
4 65 22 324 2 3
5 70 58 1035 1 7
6 75 30 2568 0 34
7 80 9 2936 15 162
8 85 27 2169 46 365
9 90 80 1439 212 432
10 95 236 1670 521 281
11 100 332 827 709 172
12 105 156 311 556 103
13 110 69 49 144 44
14 115 26 10 36 17
15 120 29 3 3
16 125 16 1 1
17 130 0 14 0 0
18 135 05 0 0
19 140 00 0 0
20 145 00 0 0
21 150 00 0 0
22 155 00 0 0
23 160 00 0 0
24 165 00 0 0
25 170 00 0 0
26 175 00 0 0
27 180 00 0 0
28 185 00 0 0
29 190 00 0 1
30 195 00 0 0
31 200 00 0 0
32 205 00 0 0
33 210 00 0 0
x.m - as.matrix(x) # dataframe - matrix
barplot(t(x.m), names.arg = x.m[,1], las=2)
On Mon, Nov 8, 2010 at 5:42 PM, casperyc caspe...@hotmail.co.uk wrote:
Hi all,
I have the following data in abc.dat
===
50 0 1 0 0
55 1 14 0 1
60 7 86 0 3
65 22 324 2 3
70 58 1035 1 7
75 30 2568 0 34
80 9 2936 15 162
85 27 2169 46 365
90 80 1439 212 432
95 236 1670 521 281
100 332 827 709 172
105 156 311 556 103
110 69 49 144 44
115 26 10 36 17
120 2 9 3 3
125 1 6 1 1
130 0 14 0 0
135 0 5 0 0
140 0 0 0 0
145 0 0 0 0
150 0 0 0 0
155 0 0 0 0
160 0 0 0 0
165 0 0 0 0
170 0 0 0 0
175 0 0 0 0
180 0 0 0 0
185 0 0 0 0
190 0 0 0 1
195 0 0 0 0
200 0 0 0 0
205 0 0 0 0
210 0 0 0 0
===
which i have used
abc=read.table(abc.dat)
to read the table into R.
There are two problems:
1- I want the first column of the data to be
the 'column names', how should i read the data?
2- I want to plot the histogram, using the first column as
'x' values, and the 2nd,3rd,4th and 5th columns as the frequencies.
How do I plot it?
I have tried to add a 'row' of variable names to it,
and then read with 'header=T', then the first column
become 'col.names' as I was expecting it to be.
However, when I plot it using 'hist',
R uses the 2nd column as the 'x value', where it should be used as
'frequency'.
(the 50,55,60,65,70... should be on the x-axis)
Thanks!
Casper
--
View this message in context:
http://r.789695.n4.nabble.com/how-do-i-plot-this-hist-tp3032796p3032796.html
Sent from the R help mailing list archive at Nabble.com.
__
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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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] how do i plot this hist?
Typing too fast; last line should be: barplot(t(x.m[, 2:5]), names.arg = x.m[,1], las=2) On Mon, Nov 8, 2010 at 5:42 PM, casperyc caspe...@hotmail.co.uk wrote: Hi all, I have the following data in abc.dat === 50 0 1 0 0 55 1 14 0 1 60 7 86 0 3 65 22 324 2 3 70 58 1035 1 7 75 30 2568 0 34 80 9 2936 15 162 85 27 2169 46 365 90 80 1439 212 432 95 236 1670 521 281 100 332 827 709 172 105 156 311 556 103 110 69 49 144 44 115 26 10 36 17 120 2 9 3 3 125 1 6 1 1 130 0 14 0 0 135 0 5 0 0 140 0 0 0 0 145 0 0 0 0 150 0 0 0 0 155 0 0 0 0 160 0 0 0 0 165 0 0 0 0 170 0 0 0 0 175 0 0 0 0 180 0 0 0 0 185 0 0 0 0 190 0 0 0 1 195 0 0 0 0 200 0 0 0 0 205 0 0 0 0 210 0 0 0 0 === which i have used abc=read.table(abc.dat) to read the table into R. There are two problems: 1- I want the first column of the data to be the 'column names', how should i read the data? 2- I want to plot the histogram, using the first column as 'x' values, and the 2nd,3rd,4th and 5th columns as the frequencies. How do I plot it? I have tried to add a 'row' of variable names to it, and then read with 'header=T', then the first column become 'col.names' as I was expecting it to be. However, when I plot it using 'hist', R uses the 2nd column as the 'x value', where it should be used as 'frequency'. (the 50,55,60,65,70... should be on the x-axis) Thanks! Casper -- View this message in context: http://r.789695.n4.nabble.com/how-do-i-plot-this-hist-tp3032796p3032796.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to eliminate this for loop ?
Reduce(function(x1,x2)b*x1-x2,c,init=1,accum=TRUE) might be what you are looking for. This is not fully tested, so you should test it before you want to use it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 detect if a vector is FP constant?
Hi all, What's the equivalent to length(unique(x)) == 1 if want to ignore small floating point differences? Should I look at diff(range(x)) or sd(x) or something else? What cut off should I use? If it helps to be explicit, I'm interested in detecting when a vector is constant for the purpose of visual display. In other words, if I rescale x to [0, 1] do I have enough precision to get at least 100 unique values. Thanks! Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to detect if a vector is FP constant?
how about all.equal(x,rep(mean(x),length(x))) or all.equal(x,rep(mean(x),length(x), tolerance=...) albyn On Mon, Nov 08, 2010 at 06:45:00PM -0600, Hadley Wickham wrote: Hi all, What's the equivalent to length(unique(x)) == 1 if want to ignore small floating point differences? Should I look at diff(range(x)) or sd(x) or something else? What cut off should I use? If it helps to be explicit, I'm interested in detecting when a vector is constant for the purpose of visual display. In other words, if I rescale x to [0, 1] do I have enough precision to get at least 100 unique values. Thanks! Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Albyn Jones Reed College jo...@reed.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] cannot find system Renviron Fatal error: unable to open the base package
hi people I need to make an application with Java and R. I installed the library rJava using the command R - install.packages (rJava), and i configured my $ R_HOME = / Library / Frameworks / R.framework / Resources. My SO is OSX. When I run an exemple in eclipse, it gives me the following error: cannot find system Renviron Fatal error: unable to open the base package I need help. Thanks -- View this message in context: http://r.789695.n4.nabble.com/cannot-find-system-Renviron-Fatal-error-unable-to-open-the-base-package-tp3032754p3032754.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.
