Re: [R] Modifying the embed-results
Again my example was't very clear: there were not enough same numbers in the
VECTOR.
What I need is something like this:
VECTOR-c(0,3,6,3,11,2,11,4,3,4,7,7,6,4,8)
MATRIX-c()
for(i in 1:length(VECTOR)){
v-(unique(VECTOR[i:length(VECTOR)])[1:5])
MATRIX-rbind(MATRIX,v)
}
MATRIX-na.omit(MATRIX)
MATRIX
data.frame(MATRIX, row.names=NULL)
X1 X2 X3 X4 X5
1 0 3 6 11 2
2 3 6 11 2 4
3 6 3 11 2 4
4 3 11 2 4 7
5 11 2 4 3 7
6 2 11 4 3 7
7 11 4 3 7 6
8 4 3 7 6 8
9 3 4 7 6 8
So, there are no duplicates in rows.
VECTOR is always scanned forward as long as the number of items (here 5)
becomes full.
Atte
Try:
embed(VECTOR, 5)[,5:1]
On 8/25/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Hi,
Here is a vector and the result from the embed-command:
VECTOR=c(0,3,6,3,11,2,4,3,7,6,4,5,10,2,3,5,8)
embed(VECTOR, dimension=5)
[,1] [,2] [,3] [,4] [,5]
[1,] 113630
[2,]2 11363
[3,]42 1136
[4,]342 113
[5,]7342 11
[6,]67342
[7,]46734
[8,]54673
[9,] 105467
[10,]2 10546
[11,]32 1054
[12,]532 105
[13,]8532 10
Is there a way to little modify the algorithm so that the result
looks like this:
[1] 0 3 6 11 2 - beginning from the first number of the VECTOR
[1] 3 6 11 2 4 - beginning from the second number of the
VECTOR etc
[1] 6 3 11 2 4
[1] 3 11 2 4 7
[1] 11 2 4 3 7
[1] 2 4 3 7 6
[1] 4 3 7 6 5
[1] 3 7 6 4 5
[1] 7 6 4 5 10
[1] 6 4 5 10 2
[1] 4 5 10 2 3
[1] 5 10 2 3 8
[1] 10 2 3 5 8
Every row consists of next five unique(!) member of the VECTOR. I
made this example result with a time consuming algorithm which
uses for-loops
and whiles.
How to do this better??
Thanks in advance!
Atte Tenkanen
University of Turku
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[R] Problem on Histogtam
Dear all, May be question seems trivial for most of the R users, but really at least for me, this comes out to be very problematic. Suppose I have the following data: r [1] -0.0008179960 -0.0277968529 -0.0105731583 -0.0254050262 0.0321847131 0.0328170674 [7] 0.0431894392 -0.0217614918 -0.0218366946 0.0048939739 -0.0012212499 0.0032533579 [13] -0.0081533269 -0.0098725606 -0.0099710008 -0.0130281313 0.0121927094 0.0029209284 [19] -0.0206279678 -0.0210624593 -0.0100415600 -0.0266801063 -0.0241744954 0.0178453603 [25] 0.0452204703 0.0133478839 0.0274220937 0.0091135676 0.0380304730 -0.0087719861 [31] -0.0412893310 -0.0092243841 -0.0016863410 -0.0080491856 -0.0293504020 0.0411898638 [37] -0.0037902762 -0.0076239251 0.0193690266 -0.0012518257 0.0259647803 0.0241169245 [43] 0.0446589449 0.0536073638 -0.0028839241 0.0168251687 0.0095390230 -0.0213227695 [49] -0.0519741069 -0.0287872498 -0.0035108287 0.0197336642 -0.0065347148 -0.0011576308 [55] -0.0085304899 0.0162228355 0.0087736592 0.0007593015 0.0113208756 0.0229980462 [61] 0.0370746078 0.0679558204 0.0401218485 -0.005759 0.0188922496 -0.0230330422 [67] -0.0261938985 -0.0503969590 -0.0069156569 -0.0031277176 -0.0235967983 -0.0100287373 [73] -0.0014409224 -0.0307491283 -0.0240794177 -0.0091813958 0.0076570053 0.0621042105 [79] 0.0156475454 0.0353589695 0.0078032627 0.0070719271 0.0026809668 -0.0111056164 [85] -0.0088375830 0.0128902735 0.0279180644 0.0268407382 -0.0038363218 -0.0194055249 [91] -0.0284881803 0.0016786977 -0.0145297369 0.0081356381 -0.0187430943 0.0123078477 [97] 0.0148400700 -0.0070552955 0.0103975446 0.0109508312 0.0147422817 0.0218728566 [103] -0.0041447533 -0.0096308931 0.0289322358 0.0216995495 0.0499431039 0.0717506367 [109] 0.0032529169 -0.0040678022 -0.0331522073 -0.0164263110 -0.0322093707 0.0542379373 [115] 0.0259181864 0.0387076215 0.0044415488 -0.0078513881 0.0411792612 0.0062837960 [121] -0.0243481467 0.0328312315 -0.0520127424 -0.0137024542 -0.0367509610 -0.0149753215 [127] 0.0256530410 0.0376798081 0.0106952891 0.0525638932 0.0281572000 0.0324963843 [133] -0.0170584812 -0.0063837550 0.0294482602 -0.0118152654 -0.0169216729 0.0061436866 [139] -0.0245630568 0.0093716894 -0.0122730454 0.0234516610 -0.0090283341 0.0059488575 [145] -0.0104937762 -0.0123026519 0.0263488297 -0.0087855249 0.0021441342 0.0204930031 [151] -0.0079588435 -0.0021173987 0.0135674596 0.0053296384 -0.0004623209 0.0199159610 [157] -0.0034056112 -0.0052445679 -0.0142760886 -0.0158956129 -0.0137605967 -0.0007169316 [163] 0.0236250062 0.0002334540 0.0007000350 -0.0023353584 -0.0210285362 -0.0178275018 [169] -0.0663226429 0.0098217383 -0.0111212836 -0.0597472435 0.0371266051 -0.0247790755 [175] 0.0337833736 0.0265364166 -0.0160458929 0.0085725944 0.0161645598 0.0700273748 [181] -0.0308458190 0.0362889752 -0.0195430433 0.0284716741 0.0097766142 -0.0037131628 [187] 0.0275183610 0.0276618128 0.0109410282 -0.0302685208 0.0191074465 -0.0086197904 [193] 0.0013309674 0.0169253408 -0.0010903937 0.0073913380 0.0330203607 0.0116667990 [199] -0.0099917567 0.0035501760 -0.0231980036 0.0023441673 -0.0150120769 -0.0165581125 [205] -0.0101568548 -0.0031118051 -0.0051333669 0.0312800833 -0.0050005540 -0.0109578129 [211] -0.0222825618 -0.0388242167 -0.0037549915 -0.0185102580 -0.0120483385 0.0336106003 [217] 0.0146599484 0.0146757965 0.0117675040 0.0196041030 -0.0030932415 -0.0024371343 [223] -0.0158751238 -0.0145292000 0.0050182587 0.0061245513 0.0139234228 -0.0116645558 [229] -0.0475950303 -0.0346688548 0.0041559771 -0.0246987117 0.0195623423 0.0167785091 [235] -0.0133513332 -0.0113051093 0.0215167225 -0.0009680543 -0.0141431277 -0.0150936925 [241] -0.0183676066 -0.0117498159 0.0165629981 -0.0248189789 -0.0133004960 -0.0132137229 [247] -0.0429413799 -0.0055694937 0.0005589715 0.0168908903 0.0321518982 0.0007973422 [253] 0.0573060001 -0.0078076206 -0.0253502245 0.0210426035 0.0025361414 -0.0091603694 [259] -0.0064111003 -0.0126862543 -0.0067974118 0.0023581827 -0.0359703092 0.0185464260 [265] -0.0007992541 -0.0193762790 0.0225694695 0.0052994295 -0.0039719369 0.0144872809 [271] 0.0002614721 -0.0076105865 -0.0153971717 0.0056022555 -0.0104292963 -0.0189965007 [277] 0.0079138085 0.0254944708 -0.0165646320 -0.0162958522 -0.0107365106 0.0049696403 [283] 0.0041225838 0.00 -0.0035719230 -0.0200173489 -0.0022490873 -0.0039481157 [289] -0.0182497223 0.0157033919 -0.0154156635 -0.0023041485 -0.0028876717 0.00 [295] -0.0195655241 0.0210104023 -0.0020234145 -0.0026075634 0.0149729259 0.0045623119 [301] 0.0431521368 0.0228966231 -0.0077530130 0.0285733724 -0.0091707704 -0.0137823544 [307] 0.0092974504 -0.0076974501 0.0106017426 0.0481245566 0.0164469194 0.0356877294 [313] 0.0002384643 0.0151446340 -0.0184883199 0.00 Now I want to plot a histogram: histo = hist(r,freq=F,breaks=50)
Re: [R] generating an expression for a formula automatically
On Fri, 25 Aug 2006, Maria Montez wrote:
Thank you for your answers yesterday. I now have another question!
Suppose that instead of creating a formula for a regression model I
simply wanted to add the variables. I believe I cannot use the
as.formula anymore. Again I tried to use expression to no avail. I get
an expression but I can't use it.
fit.sum - function(x) {
fo - expression(paste(x, collapse = +))
eval( fo)
}
fit.sum(c(x1,x2))
Basically what I need is to learn how to use variables when what is
given to me are their names (character list).
I think we do need to tell you that parse(text=) is how to turn a
character vector into an expression, although it is rarely a good way (see
the fortune in the 'fortunes' package about it) and you need to be careful
where you evaluate:
fit.sum - function(x) eval.parent(parse(text=paste(x, collapse = +)))
As an alternative, the answer to
Basically what I need is to learn how to use variables when what is
given to me are their names (character list).
is most often 'get'. So for two objects
fit.sum2 - function(x) {
env - parent.frame()
do.call(+, lapply(x, get, envir=env))
}
but again you need to be careful where you get() from.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Memory usage decreases drastically after save workspace, quit, restart, load workspace
On Sat, 26 Aug 2006, Klaus Thul wrote: Dear all, I have the following problem: - I have written a program in R which runs out of physical memory on my MacBook Pro with 1.5 GB RAM How does R know about physical memory on a virtual-memory OS? I presume the symptom is swapping by your OS, but how do you attribute that to R? - The memory usage is much higher then I would expect from the actual data in my global environment - When I save the workspace, quit R, restart R and load the workspace again, memory usage is much less then before (~50 MB instead of ~1.5 GB), although nothing seems to be missing. - It doesn't seem to be possible to reduce the amount of memory used by calling gc() - the behavior is independent of if I use R in command line or from GUI, so I can exclude a problem of the Mac GUI Any suggestions what the problem might be or how I could debug this? How are you measuring memory usage? If this is by gc(), note that lazyloading is one-way, and memory increases after you use a lazyloaded object in a session. This can be dramatic where datasets are involved. If you are asking the OS about memory usage, return of memory to the OS is a rather haphazard and OS-specific issue. On some OSes virtual memory is not actually reclaimed from applications until it is needed (or ever). It is certainly possible that something you are using has a memory leak, but valgrind has been used to plug those on the most-used parts of R. I at least need more precise indications of what you observed to be able to comment more. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
sub.m - lapply(m, function(x)x[x2])
sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
sub4.m - lapply(m, function(x)x[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
xpRt.wannabe wrote:
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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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 you are trying to solve?
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Re: [R] issues with Sweave and inclusion of graphics in a document
On Fri, 25 Aug 2006 11:05:48 -0700 (PDT),
Thomas Harte (TH) wrote:
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
savePlot is just an internal version of dev.copy, part of the support for
the menus on the windows() graphics device.
It is described in `An Introduction to R' (the most basic R manual).
the most basic R manual doesn't quite answer my question.
by itself, dev.copy doesn't copy the width and height of the device whereas
savePlot
copies whatever is displayed on the screen giving me
what-you-see-is-what-you-save
capabilities (but only under the Windows OS).
i can get pretty close to this in linux by writing a function to save the
plot to a pdf device:
label=first.ar.1, results=hide=
# no savePlot in Linux ... so write my own function
savePlotAsPdf- function(pdfname, from=dev.cur()) {
from- from
pdf(pdfname, width=width, height=height)
to- dev.cur()
dev.set(from)
dev.copy(which=to)
dev.off()
}
# a long AR process is best viewed in a wide window ...
# width height are now variables
width- 20; height- 5
x11(width=width, height=height)
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
# width height via dynamic scoping in savePlotAsPdf
savePlotAsPdf(ar.pdf)
@
Umm, maybe I don't get your point, but in what way does the complicated
code above do anythingdifferent from
**
results=hide=
sp- make.ar.1(alpha=.5, n=800)
@
ar,fig=true,width=20,height=5,include=false=
plot(sp, type=l, col=blue)
@
\begin{figure}
\includegraphics[width=14.5cm]{myprefix-ar}
\end{figure}
**
???
Best,
--
---
Prof. Dr. Friedrich Leisch
Institut für Statistik Tel: (+49 89) 2180 3165
Ludwig-Maximilians-Universität Fax: (+49 89) 2180 5308
Ludwigstraße 33
D-80539 München http://www.stat.uni-muenchen.de/~leisch
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem on Histogtam
try thie: r - rnorm(100,4,1) hist(r, freq=FALSE) r.f - function(x)dnorm(x, mean(r), sd(r)) curve(r.f, from=min(r), to=max(r), add=TRUE, col='red') On 8/26/06, stat stat [EMAIL PROTECTED] wrote: Dear all, May be question seems trivial for most of the R users, but really at least for me, this comes out to be very problematic. Suppose I have the following data: r [1] -0.0008179960 -0.0277968529 -0.0105731583 -0.0254050262 0.0321847131 0.0328170674 [7] 0.0431894392 -0.0217614918 -0.0218366946 0.0048939739 -0.0012212499 0.0032533579 [13] -0.0081533269 -0.0098725606 -0.0099710008 -0.0130281313 0.0121927094 0.0029209284 [19] -0.0206279678 -0.0210624593 -0.0100415600 -0.0266801063 -0.0241744954 0.0178453603 [25] 0.0452204703 0.0133478839 0.0274220937 0.0091135676 0.0380304730 -0.0087719861 [31] -0.0412893310 -0.0092243841 -0.0016863410 -0.0080491856 -0.0293504020 0.0411898638 [37] -0.0037902762 -0.0076239251 0.0193690266 -0.0012518257 0.0259647803 0.0241169245 [43] 0.0446589449 0.0536073638 -0.0028839241 0.0168251687 0.0095390230 -0.0213227695 [49] -0.0519741069 -0.0287872498 -0.0035108287 0.0197336642 -0.0065347148 -0.0011576308 [55] -0.0085304899 0.0162228355 0.0087736592 0.0007593015 0.0113208756 0.0229980462 [61] 0.0370746078 0.0679558204 0.0401218485 -0.005759 0.0188922496 -0.0230330422 [67] -0.0261938985 -0.0503969590 -0.0069156569 -0.0031277176 -0.0235967983 -0.0100287373 [73] -0.0014409224 -0.0307491283 -0.0240794177 -0.0091813958 0.0076570053 0.0621042105 [79] 0.0156475454 0.0353589695 0.0078032627 0.0070719271 0.0026809668 -0.0111056164 [85] -0.0088375830 0.0128902735 0.0279180644 0.0268407382 -0.0038363218 -0.0194055249 [91] -0.0284881803 0.0016786977 -0.0145297369 0.0081356381 -0.0187430943 0.0123078477 [97] 0.0148400700 -0.0070552955 0.0103975446 0.0109508312 0.0147422817 0.0218728566 [103] -0.0041447533 -0.0096308931 0.0289322358 0.0216995495 0.0499431039 0.0717506367 [109] 0.0032529169 -0.0040678022 -0.0331522073 -0.0164263110 -0.0322093707 0.0542379373 [115] 0.0259181864 0.0387076215 0.0044415488 -0.0078513881 0.0411792612 0.0062837960 [121] -0.0243481467 0.0328312315 -0.0520127424 -0.0137024542 -0.0367509610 -0.0149753215 [127] 0.0256530410 0.0376798081 0.0106952891 0.0525638932 0.0281572000 0.0324963843 [133] -0.0170584812 -0.0063837550 0.0294482602 -0.0118152654 -0.0169216729 0.0061436866 [139] -0.0245630568 0.0093716894 -0.0122730454 0.0234516610 -0.0090283341 0.0059488575 [145] -0.0104937762 -0.0123026519 0.0263488297 -0.0087855249 0.0021441342 0.0204930031 [151] -0.0079588435 -0.0021173987 0.0135674596 0.0053296384 -0.0004623209 0.0199159610 [157] -0.0034056112 -0.0052445679 -0.0142760886 -0.0158956129 -0.0137605967 -0.0007169316 [163] 0.0236250062 0.0002334540 0.0007000350 -0.0023353584 -0.0210285362 -0.0178275018 [169] -0.0663226429 0.0098217383 -0.0111212836 -0.0597472435 0.0371266051 -0.0247790755 [175] 0.0337833736 0.0265364166 -0.0160458929 0.0085725944 0.0161645598 0.0700273748 [181] -0.0308458190 0.0362889752 -0.0195430433 0.0284716741 0.0097766142 -0.0037131628 [187] 0.0275183610 0.0276618128 0.0109410282 -0.0302685208 0.0191074465 -0.0086197904 [193] 0.0013309674 0.0169253408 -0.0010903937 0.0073913380 0.0330203607 0.0116667990 [199] -0.0099917567 0.0035501760 -0.0231980036 0.0023441673 -0.0150120769 -0.0165581125 [205] -0.0101568548 -0.0031118051 -0.0051333669 0.0312800833 -0.0050005540 -0.0109578129 [211] -0.0222825618 -0.0388242167 -0.0037549915 -0.0185102580 -0.0120483385 0.0336106003 [217] 0.0146599484 0.0146757965 0.0117675040 0.0196041030 -0.0030932415 -0.0024371343 [223] -0.0158751238 -0.0145292000 0.0050182587 0.0061245513 0.0139234228 -0.0116645558 [229] -0.0475950303 -0.0346688548 0.0041559771 -0.0246987117 0.0195623423 0.0167785091 [235] -0.0133513332 -0.0113051093 0.0215167225 -0.0009680543 -0.0141431277 -0.0150936925 [241] -0.0183676066 -0.0117498159 0.0165629981 -0.0248189789 -0.0133004960 -0.0132137229 [247] -0.0429413799 -0.0055694937 0.0005589715 0.0168908903 0.0321518982 0.0007973422 [253] 0.0573060001 -0.0078076206 -0.0253502245 0.0210426035 0.0025361414 -0.0091603694 [259] -0.0064111003 -0.0126862543 -0.0067974118 0.0023581827 -0.0359703092 0.0185464260 [265] -0.0007992541 -0.0193762790 0.0225694695 0.0052994295 -0.0039719369 0.0144872809 [271] 0.0002614721 -0.0076105865 -0.0153971717 0.0056022555 -0.0104292963 -0.0189965007 [277] 0.0079138085 0.0254944708 -0.0165646320 -0.0162958522 -0.0107365106 0.0049696403 [283] 0.0041225838 0.00 -0.0035719230 -0.0200173489 -0.0022490873 -0.0039481157 [289] -0.0182497223 0.0157033919 -0.0154156635 -0.0023041485 -0.0028876717 0.00 [295] -0.0195655241 0.0210104023 -0.0020234145 -0.0026075634 0.0149729259 0.0045623119 [301] 0.0431521368 0.0228966231 -0.0077530130
Re: [R] Modifying the embed-results
You can replace the for with lapply like this:
VECTOR - c(0, 3, 6, 3, 11, 2, 11, 4, 3, 4, 7, 7, 6, 4, 8)
f - function(i) unique(tail(VECTOR, length(VECTOR)-i+1))[1:5]
out - do.call(rbind, lapply(seq(along = VECTOR), f))
na.omit(rbind(rep(NA, 5), out))
Note that a matrix with zero rows is returned in the case
that VECTOR has zero length and in the case that VECTOR
has fewer than 5 unique elements.
On 8/26/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Again my example was't very clear: there were not enough same numbers in the
VECTOR.
What I need is something like this:
VECTOR-c(0,3,6,3,11,2,11,4,3,4,7,7,6,4,8)
MATRIX-c()
for(i in 1:length(VECTOR)){
v-(unique(VECTOR[i:length(VECTOR)])[1:5])
MATRIX-rbind(MATRIX,v)
}
MATRIX-na.omit(MATRIX)
MATRIX
data.frame(MATRIX, row.names=NULL)
X1 X2 X3 X4 X5
1 0 3 6 11 2
2 3 6 11 2 4
3 6 3 11 2 4
4 3 11 2 4 7
5 11 2 4 3 7
6 2 11 4 3 7
7 11 4 3 7 6
8 4 3 7 6 8
9 3 4 7 6 8
So, there are no duplicates in rows.
VECTOR is always scanned forward as long as the number of items (here 5)
becomes full.
Atte
Try:
embed(VECTOR, 5)[,5:1]
On 8/25/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Hi,
Here is a vector and the result from the embed-command:
VECTOR=c(0,3,6,3,11,2,4,3,7,6,4,5,10,2,3,5,8)
embed(VECTOR, dimension=5)
[,1] [,2] [,3] [,4] [,5]
[1,] 113630
[2,]2 11363
[3,]42 1136
[4,]342 113
[5,]7342 11
[6,]67342
[7,]46734
[8,]54673
[9,] 105467
[10,]2 10546
[11,]32 1054
[12,]532 105
[13,]8532 10
Is there a way to little modify the algorithm so that the result
looks like this:
[1] 0 3 6 11 2 - beginning from the first number of the VECTOR
[1] 3 6 11 2 4 - beginning from the second number of the
VECTOR etc
[1] 6 3 11 2 4
[1] 3 11 2 4 7
[1] 11 2 4 3 7
[1] 2 4 3 7 6
[1] 4 3 7 6 5
[1] 3 7 6 4 5
[1] 7 6 4 5 10
[1] 6 4 5 10 2
[1] 4 5 10 2 3
[1] 5 10 2 3 8
[1] 10 2 3 5 8
Every row consists of next five unique(!) member of the VECTOR. I
made this example result with a time consuming algorithm which
uses for-loops
and whiles.
How to do this better??
Thanks in advance!
Atte Tenkanen
University of Turku
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Re: [R] issues with Sweave and inclusion of graphics in a document
--- Friedrich Leisch [EMAIL PROTECTED] wrote:
On Fri, 25 Aug 2006 11:05:48 -0700 (PDT),
Thomas Harte (TH) wrote:
i can get pretty close to this in linux by writing a function to save the
plot to a pdf device:
label=first.ar.1, results=hide=
# no savePlot in Linux ... so write my own function
savePlotAsPdf- function(pdfname, from=dev.cur()) {
from- from
pdf(pdfname, width=width, height=height)
to- dev.cur()
dev.set(from)
dev.copy(which=to)
dev.off()
}
# a long AR process is best viewed in a wide window ...
# width height are now variables
width- 20; height- 5
x11(width=width, height=height)
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
# width height via dynamic scoping in savePlotAsPdf
savePlotAsPdf(ar.pdf)
@
Umm, maybe I don't get your point, but in what way does the complicated
code above do anythingdifferent from
**
results=hide=
sp- make.ar.1(alpha=.5, n=800)
@
ar,fig=true,width=20,height=5,include=false=
plot(sp, type=l, col=blue)
@
\begin{figure}
\includegraphics[width=14.5cm]{myprefix-ar}
\end{figure}
**
hallo, friedrich, and thanks for your reply.
if i Stangle your code i get:
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
whereas if i Stangle my code, i get:
width- 20; height- 5
x11(width=width, height=height)
sp- make.ar.1(alpha=.5, n=800)
plot(sp, type=l, col=blue)
savePlotAsPdf(ar.pdf)
this, i think, is the problem in a nutshell.
i want my R code to look like this because i (often) want to open a separate
device (for example, to have specific dimensions as above) and i may wish to
save its contents. on the other hand, i may not wish to save a device's
contents:
i may wish to open several devices for comparison purposes and i may wish to
save only
certain devices, in keeping with the Sweave options echo=false and results=hide.
if i use your, admittedly uncomplicated, code chunk, then i am limited to
seeing the
output in the final document. i can't cut and paste the code chunk into an R
session and
see the results as i wish to see them before saving.
currently for each project i work on, i usually end up with a very lengthy .R
file with
all the code (and embedded mathematical annotation in LaTeX), and then i have
to write a
set of notes or a report in LaTeX duplicating or expanding on the mathematics.
i'm hoping
to supplant this, rather inefficient, method with a single .Rnw file and the
use of
Sweave. this, i believe, is what Sweave was designed to do. but i do wish to
write my R
code (warts and all) the way that i want to in the .Rnw file and only include
the output
(graphics or text) of certain parts of that code in the document. i hope that
seems
reasonable.
perhaps i was not clear, or succinct enough, in my original post
(i tried to get the point across by providing example .Rnw files,
but i'm aware that it's asking a lot to wade through a lengthy
example to retrieve a point).
thanks again for your reply.
cheers,
thomas.
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[R] problems with loop
Dear all,
I am trying to evaluate the optimisation behaviour of a function. Originally
I have optimised a model with real data and got a set of parameters. Now I
am creating simulated data sets based on these estimates. With these
simulations I am estimating the parameters again to see how variable the
estimation is. To this end I have written a loop which should generate a new
simulated data set each time. However, the optimisation algorithm, which
works fine if only one data set is used, does not recognise the simulated
data in the loop. Can anyone tell me where the error is? The code is below.
Thanks for your help
Simon
# The loop in which nothing works anymore. The NLL in the optim function
appears not to recognise the new data set. The functions used by this loop
are given below.
sim.estim=function(r)
{
res=matrix(nrow=r,ncol=5)
for (s in 1:r)
{
new=new.set()
Min=optim(c(0.5,0.5,0.01,0.1),NLL,method=L-BFGS-B,lower=c(0,0,0,0))
res[s,1]=Min$par[1]
res[s,2]=Min$par[2]
res[s,3]=Min$par[3]
res[s,4]=Min$par[4]
res[s,5]=Min$value[1]
}
return(res)
}
# function to create a new data set. Works fine on its own
new.set=function()
{
tmp=matrix(nrow=510,ncol=2)
v=numeric()
sim=Model_1(1.2761707, 0.1953354, 2.7351930, 0.1032929)
tmp[,1]=sample(sim[,1],510,replace=T)
v=runif(510,0,1)
for (k in 1 : 510)
{
for (n in 1 : length(sim[,1]))
{
if (tmp[k,1]==sim[n,1] v[k]=sim[n,2]){tmp[k,2]=1}
if(tmp[k,1]==sim[n,1] v[k]sim[n,2]) {tmp[k,2]=0}
}
}
return(tmp)
}
# The model to be fitted
Model_1=function(beta0_mod,mu_mod,k_mod,I_mod)
{
parms = c(beta0=beta0_mod, mu=mu_mod, k=k_mod)
my.atol = 1e-6
times1=c(0,0.002884615,0.003846154,0.005769231,0.009615385,0.01,0.019230769)
times2=c(0.028846154,0.038461538,0.057692308,
0.076923077,0.08333,0.096153846)
times3=c(0.115384615,0.125,0.134615385,0.16667,0.25,0.41667,0.5)
times4=c(0.5833,0.7,0.75,0.8333,0.91667,1)
times5=c( 1.16667,1.25,1.5,1.6,1.75,2,2.5)
times6=seq(3,20)
times = c(times1,times2,times3,times4,times5,times6)
ODE_sys = function(t, x, w) #w=c(beta0,mu,k)
{
I=x[1]
dI=-w[mu]*I+w[beta0]*exp(-w[k]*t)*(1-I)
list(c(dI),c(S=1-I))
}
output = lsoda(c(I_mod),times,ODE_sys, parms, rtol=1e-6, atol=
my.atol)
return(output)
}
# The negative log likelihood of this model given a data set new. This
works fine if it is used in the optim() routine with only one data set.
NLL=function(coef)
{
beta0=coef[1]
mu=coef[2]
k=coef[3]
I=coef[4]
data_sim=Model_1(beta0,mu,k,I)
LLsum=0 #log likelihood sum
for (j in 1:length(data_sim[,1]))
{
if (data_sim[j,2]0 || data_sim[j,3]0)
{return(1500)}
for (i in 1:length(new[,1]))
{
if (new[i,1]==data_sim[j,1] is.na(new[i,1])==F
is.na(data_sim[j,1])==F)
# if age of real data = age of simulated model and not
equal to NA
{
if (is.na(new[i,2])==F is.na(data_sim[j,2])==F
is.na(data_sim[j,3])==F)
# if state of real data and state of simulated model
are not equal to NA
{
tmp1=new[i,2]*log(data_sim[j,2])
tmp2=(1-new[i,2])*log(data_sim[j,3])
if (tmp1==-Inf || tmp1==NaN){tmp1=-700}
if (tmp2==-Inf || tmp2==NaN){tmp2=-700}
LLsum=LLsum+tmp1+tmp2
}
}
}
}
return(-LLsum)
}
Simon Ruegg
Dr.med.vet., PhD student
Institute for Parasitology
Winterthurstr. 266a
8057 Zurich
Switzerland
phone: +41 44 635 85 93
fax: +41 44 635 89 07
e-mail: [EMAIL PROTECTED]
[[alternative HTML version deleted]]
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[R] Adding a footnote in plot-window in R
Dear all R users, Suppose, x = rnorm(1000) y = rt(1000,3) plot(range(1:1000),range(x,y),type=n,xlab=NA,ylab=NA) lines(x,col=red) lines(y,col=blue) Now I want to put a footnote in the plot window to tell that RED lines represents the random numbers from normal-dist and blue line represents the random numbers from t-dist. Can anyone please tell me how to do that? Sincerely yours, Arun [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding a footnote in plot-window in R
?legend __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] issues with Sweave and inclusion of graphics in a document
On Sat, 26 Aug 2006 05:38:58 -0700 (PDT), Thomas Harte (TH) wrote: hallo, friedrich, and thanks for your reply. if i Stangle your code i get: sp- make.ar.1(alpha=.5, n=800) plot(sp, type=l, col=blue) whereas if i Stangle my code, i get: width- 20; height- 5 x11(width=width, height=height) sp- make.ar.1(alpha=.5, n=800) plot(sp, type=l, col=blue) savePlotAsPdf(ar.pdf) this, i think, is the problem in a nutshell. i want my R code to look like this because i (often) want to open a separate device (for example, to have specific dimensions as above) and i may wish to save its contents. on the other hand, i may not wish to save a device's contents: i may wish to open several devices for comparison purposes and i may wish to save only certain devices, in keeping with the Sweave options echo=false and results=hide. if i use your, admittedly uncomplicated, code chunk, then i am limited to seeing the output in the final document. i can't cut and paste the code chunk into an R session and see the results as i wish to see them before saving. currently for each project i work on, i usually end up with a very lengthy .R file with all the code (and embedded mathematical annotation in LaTeX), and then i have to write a set of notes or a report in LaTeX duplicating or expanding on the mathematics. i'm hoping to supplant this, rather inefficient, method with a single .Rnw file and the use of Sweave. this, i believe, is what Sweave was designed to do. but i do wish to write my R code (warts and all) the way that i want to in the .Rnw file and only include the output (graphics or text) of certain parts of that code in the document. i hope that seems reasonable. perhaps i was not clear, or succinct enough, in my original post (i tried to get the point across by providing example .Rnw files, but i'm aware that it's asking a lot to wade through a lengthy example to retrieve a point). OK, no I think I understand what you want: running the *tangled* code should open windows with the same height/width ratio as in the Latex document, right? Your first posting suggested to me more the other way round ... No, what you want is currently not directly possible with Sweave, you will have to use workarounds like the one you posted. You can even easily write your own tangle function doing it automatically (Rtangle is really trivial can be easily adjusted to ones personal needs). Best, Fritz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
lapply(m,function(x)x[x2]) [[1]] [1] 3 4 [[2]] [1] 4 5 [[3]] [1] 4 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Implementing EM Algorithm in R!
Hi All, I need some help in how one can implement maximumlikelihood estimation for models with discrete hidden variables in EM in R. Regards [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
Thank you!
On 8/26/06, Patrick Burns [EMAIL PROTECTED] wrote:
sub.m - lapply(m, function(x)x[x2])
sub.m
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
[[1]]
[1] 3 4
[[2]]
[1] 4 5
sub4.m - lapply(m, function(x)x[x4])
sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
[[1]]
[1] 5
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
xpRt.wannabe wrote:
Yet another question:
Let's say I do the following:
set.seed(123)
tmpf - function(){
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n, tmpf())
m
sub.m - lapply(m, function(x)x[x2])
'sub.m' gives me:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
The question is: What do I need to do such that I can extract
componets of length 2 in 'sub.m' into another sublist, which would
look like this:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
I think that's all the questions I can think of -- for now.
Many, many thanks!!!
On 8/25/06, xpRt. wannabe [EMAIL PROTECTED] wrote:
Jim and Patrick,
Both of you made the same suggestion, which works great!
A follow-up question: Suppose I change the condition 'x2' in 'lapply'
to 'x4', as follows:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
sub.m - lapply(m, function(x)x[x4]) # was x2
As a result, I'd get:
sub.m
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
However, what would I need to do such that 'sub.m' contains only the
non-zero length component; namely, the 'sub.m[[2]]'? In essence, I'd
like to drop all the components of zero length such that 'sub.m'
results in:
[[1]]
[1] 5
My best effort was to use 'lapply' again:
lapply(sub.m, function(x)x[length(x)0])
which still gives me:
[[1]]
numeric(0)
[[2]]
[1] 5
[[3]]
numeric(0)
Again, any help would be greately appreciated.
p.s. Sorry to bug you again. I should have thought through a little
more prior to composing an example that would represent all possible
scenarios.
On 8/25/06, jim holtman [EMAIL PROTECTED] wrote:
try this:
set.seed(123)
tmpf - function() {
+ x - rpois(rpois(1,4),2)
+ }
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
lapply(m, function(x)x[x2])
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
On 8/25/06, xpRt.wannabe [EMAIL PROTECTED] wrote:
Dear List,
The following code produces a list, which is what I what:
set.seed(123)
tmpf - function() {
x - rpois(rpois(1,4),2)
}
n - 3
m - replicate(n,tmpf())
m
[[1]]
[1] 3 2 4
[[2]]
[1] 0 2 4 2 2 5 2
[[3]]
[1] 2 0 4 1 0
Now I need something that would to extract iteratively (or as many
times as
the size of 'n') the values that are greater than 2 in each component
of
'm' into another list such that the sub-list would be:
[[1]]
[1] 3 4
[[2]]
[1] 4 5
[[3]]
[1] 4
Below is what I tried:
for(i in 1:3)
sub.list - lapply(m,subset,m[[i]]2)
sub.list
which gives me something different from what I want:
[[1]]
[1] 4
[[2]]
[1] 4
[[3]]
[1] 4
Any help would be appreciated.
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major2
minor2.1
year 2005
month12
day 20
svn rev 36812
language R
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] problems with loop
I think it's a scoping problem. Your function NLL() looks for new in
the environment in which NLL() was defined, but you generate your
simulated datasets in a different environment (local to sim.estim()).
There are a number of ways to deal with this:
- pass the dataset as a parameter to NLL()
- manipulate environments with assign() or -
- put new in its own environment and use assign() and get()
- define NLL() within sim.estim() after generating new, like NLL -
negative.log.likelihood(new) where negative.log.likelihood() is a
function like
negative.log.likelihood - function(new) {
NLL - function(coef) {
... (your definition here)...
}
NLL
}
This defines a new NLL() each time, using the dataset new just generated.
Reid Huntsinger
Simon Ruegg wrote:
Dear all,
I am trying to evaluate the optimisation behaviour of a function. Originally
I have optimised a model with real data and got a set of parameters. Now I
am creating simulated data sets based on these estimates. With these
simulations I am estimating the parameters again to see how variable the
estimation is. To this end I have written a loop which should generate a new
simulated data set each time. However, the optimisation algorithm, which
works fine if only one data set is used, does not recognise the simulated
data in the loop. Can anyone tell me where the error is? The code is below.
Thanks for your help
Simon
# The loop in which nothing works anymore. The NLL in the optim function
appears not to recognise the new data set. The functions used by this loop
are given below.
sim.estim=function(r)
{
res=matrix(nrow=r,ncol=5)
for (s in 1:r)
{
new=new.set()
Min=optim(c(0.5,0.5,0.01,0.1),NLL,method=L-BFGS-B,lower=c(0,0,0,0))
res[s,1]=Min$par[1]
res[s,2]=Min$par[2]
res[s,3]=Min$par[3]
res[s,4]=Min$par[4]
res[s,5]=Min$value[1]
}
return(res)
}
# function to create a new data set. Works fine on its own
new.set=function()
{
tmp=matrix(nrow=510,ncol=2)
v=numeric()
sim=Model_1(1.2761707, 0.1953354, 2.7351930, 0.1032929)
tmp[,1]=sample(sim[,1],510,replace=T)
v=runif(510,0,1)
for (k in 1 : 510)
{
for (n in 1 : length(sim[,1]))
{
if (tmp[k,1]==sim[n,1] v[k]=sim[n,2]){tmp[k,2]=1}
if(tmp[k,1]==sim[n,1] v[k]sim[n,2]) {tmp[k,2]=0}
}
}
return(tmp)
}
# The model to be fitted
Model_1=function(beta0_mod,mu_mod,k_mod,I_mod)
{
parms = c(beta0=beta0_mod, mu=mu_mod, k=k_mod)
my.atol = 1e-6
times1=c(0,0.002884615,0.003846154,0.005769231,0.009615385,0.01,0.019230769)
times2=c(0.028846154,0.038461538,0.057692308,
0.076923077,0.08333,0.096153846)
times3=c(0.115384615,0.125,0.134615385,0.16667,0.25,0.41667,0.5)
times4=c(0.5833,0.7,0.75,0.8333,0.91667,1)
times5=c( 1.16667,1.25,1.5,1.6,1.75,2,2.5)
times6=seq(3,20)
times = c(times1,times2,times3,times4,times5,times6)
ODE_sys = function(t, x, w) #w=c(beta0,mu,k)
{
I=x[1]
dI=-w[mu]*I+w[beta0]*exp(-w[k]*t)*(1-I)
list(c(dI),c(S=1-I))
}
output = lsoda(c(I_mod),times,ODE_sys, parms, rtol=1e-6, atol=
my.atol)
return(output)
}
# The negative log likelihood of this model given a data set new. This
works fine if it is used in the optim() routine with only one data set.
NLL=function(coef)
{
beta0=coef[1]
mu=coef[2]
k=coef[3]
I=coef[4]
data_sim=Model_1(beta0,mu,k,I)
LLsum=0 #log likelihood sum
for (j in 1:length(data_sim[,1]))
{
if (data_sim[j,2]0 || data_sim[j,3]0)
{return(1500)}
for (i in 1:length(new[,1]))
{
if (new[i,1]==data_sim[j,1] is.na(new[i,1])==F
is.na(data_sim[j,1])==F)
# if age of real data = age of simulated model and not
equal to NA
{
if (is.na(new[i,2])==F is.na(data_sim[j,2])==F
is.na(data_sim[j,3])==F)
# if state of real data and state of simulated model
are not equal to NA
{
tmp1=new[i,2]*log(data_sim[j,2])
tmp2=(1-new[i,2])*log(data_sim[j,3])
if (tmp1==-Inf || tmp1==NaN){tmp1=-700}
if (tmp2==-Inf || tmp2==NaN){tmp2=-700}
LLsum=LLsum+tmp1+tmp2
}
}
}
}
return(-LLsum)
}
[R] for() loop question
Dear Lister, If I have a list of number, say x-c(0.1, 0.5, 0.6...), how to use a for() to loop through each number in x one by one? Thank you so much! wensui [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] issues with Sweave and inclusion of graphics in a document
--- Friedrich Leisch [EMAIL PROTECTED] wrote: On Sat, 26 Aug 2006 05:38:58 -0700 (PDT), Thomas Harte (TH) wrote: hallo, friedrich, and thanks for your reply. if i Stangle your code i get: sp- make.ar.1(alpha=.5, n=800) plot(sp, type=l, col=blue) whereas if i Stangle my code, i get: width- 20; height- 5 x11(width=width, height=height) sp- make.ar.1(alpha=.5, n=800) plot(sp, type=l, col=blue) savePlotAsPdf(ar.pdf) this, i think, is the problem in a nutshell. i want my R code to look like this because i (often) want to open a separate device (for example, to have specific dimensions as above) and i may wish to save its contents. on the other hand, i may not wish to save a device's contents: i may wish to open several devices for comparison purposes and i may wish to save only certain devices, in keeping with the Sweave options echo=false and results=hide. if i use your, admittedly uncomplicated, code chunk, then i am limited to seeing the output in the final document. i can't cut and paste the code chunk into an R session and see the results as i wish to see them before saving. currently for each project i work on, i usually end up with a very lengthy .R file with all the code (and embedded mathematical annotation in LaTeX), and then i have to write a set of notes or a report in LaTeX duplicating or expanding on the mathematics. i'm hoping to supplant this, rather inefficient, method with a single .Rnw file and the use of Sweave. this, i believe, is what Sweave was designed to do. but i do wish to write my R code (warts and all) the way that i want to in the .Rnw file and only include the output (graphics or text) of certain parts of that code in the document. i hope that seems reasonable. perhaps i was not clear, or succinct enough, in my original post (i tried to get the point across by providing example .Rnw files, but i'm aware that it's asking a lot to wade through a lengthy example to retrieve a point). OK, no I think I understand what you want: running the *tangled* code should open windows with the same height/width ratio as in the Latex document, right? yes, that is correct. the goal is to develop R code in exactly the same way as i always do (a monolithic script with lots of inroads into a problem, solutions i tried and rejected, helpful pointers to the real solution, and so on), but to have a document (.pdf, .ps, whatever) that goes along with the monolithic script noting all the key points. hence, six months after writing the code i want to be able to go back, refresh my memory with the document and then delve into the R code if i have to. maintaining both the R code and the LaTeX documentation in one .Rnw file is the genius of your Sweave package and this is what i want to adopt as my standard development methodology. Your first posting suggested to me more the other way round ... mea maxima culpa. No, what you want is currently not directly possible with Sweave, you will have to use workarounds like the one you posted. You can even easily write your own tangle function doing it automatically (Rtangle is really trivial can be easily adjusted to ones personal needs). thank you for the clarification. i feel a lot less awkward in using that complicated workaround code ;) cheers / gruss, thomas. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] for() loop question
On Sat, 2006-08-26 at 13:06 -0400, Wensui Liu wrote:
Dear Lister,
If I have a list of number, say x-c(0.1, 0.5, 0.6...), how to use a for()
to loop through each number in x one by one?
Thank you so much!
wensui
Two options:
x - c(0.1, 0.5, 0.6)
for (i in x) {print (i)}
[1] 0.1
[1] 0.5
[1] 0.6
for (i in seq(along = x)) {print (x[i])}
[1] 0.1
[1] 0.5
[1] 0.6
Which approach you take tends to depends upon what else you are doing
within the loop.
I would also take a look at ?sapply, depending up what is it you are
doing.
HTH,
Marc Schwartz
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Re: [R] for() loop question
let us know what you want to do because the beauty of R is that, in many cases, you may not have to loop. - Original Message - From: Wensui Liu [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Saturday, August 26, 2006 1:06 PM Subject: [R] for() loop question Dear Lister, If I have a list of number, say x-c(0.1, 0.5, 0.6...), how to use a for() to loop through each number in x one by one? Thank you so much! wensui [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Can R compute the expected value of a random variable?
Dear All Can R compute the expected value of a random variable? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Can R compute the expected value of a random variable?
Yes. On 8/26/06, Paul Smith [EMAIL PROTECTED] wrote: Dear All Can R compute the expected value of a random variable? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Regards, Mike Nielsen __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] fitting a gaussian to some x,y data
Success! The line I needed was: gcoeffs -nls(y~(a/b)*exp(-(x-c)^2/(2*b^2)),start=list (a=0.4,b=2,c=-10), trace=TRUE) I also needed to provide good guesses for a, b and c. The attached PNG should explain what I was going after, which is the line in the center of that curve. I am sorry for my confusing and perhaps incorrect usage of terminology. Cheers, Michael __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Implementing EM Algorithm in R!
Implementing the EM algorithm will be easy if you know what the algorithm is for your particular problem. This will be very specific to your problem. The trick is to augment your data to get something for which there is an easy ML estimate. I do not believe there is a unique recipe to perform the EM algorithm for any problem. On 8/26/06, Pushkar Kumar [EMAIL PROTECTED] wrote: Hi All, I need some help in how one can implement maximumlikelihood estimation for models with discrete hidden variables in EM in R. Regards [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ritwik Sinha Graduate Student Epidemiology and Biostatistics Case Western Reserve University http://darwin.cwru.edu/~rsinha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] fitting a gaussian to some x,y data
Whoops, I forgot to add, many thanks to all who replied publicly and privately. I am very appreciative of all comments and suggestions. Mark Leeds was especially kind in terms of clarifying why my description of the situation was so confusing. Cheers, Michael On Aug 26, 2006, at 2:51 PM, Michael Koppelman wrote: Success! The line I needed was: gcoeffs -nls(y~(a/b)*exp(-(x-c)^2/(2*b^2)),start=list (a=0.4,b=2,c=-10), trace=TRUE) I also needed to provide good guesses for a, b and c. The attached PNG should explain what I was going after, which is the line in the center of that curve. I am sorry for my confusing and perhaps incorrect usage of terminology. Cheers, Michael __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] boxplot( ..args.., axes=FALSE, frame.plot=TRUE) doesn't frame the plot
I'm running R 2.3.1 on Windows. When calling boxplot(), shouldn't the axes and frame.plot arguments get passed down to bxp()? A specific example where the plot is not framed (but seems like it should be): X - data.frame(x=as.factor(rep(c(1,2,3), 10)), y=rnorm(30)) boxplot(y~x, data=X, frame.plot=TRUE, axes=FALSE) And this doesn't seem to affect the default at all: boxplot(y~x, data=X, frame.plot=FALSE) However, this does the trick: y - boxplot(y~x, data=X, plot=FALSE) bxp(y, frame.plot=TRUE, axes=FALSE) Any ideas? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Type II and III sum of square in Anova (R, car package)
Hello everybody, I have some questions on ANOVA in general and on ANOVA in R particularly. I am not Statistician, therefore I would be very appreciated if you answer it in a simple way. 1. First of all, more general question. Standard anova() function for lm() or aov() models in R implements Type I sum of squares (sequential), which is not well suited for unbalanced ANOVA. Therefore it is better to use Anova() function from car package, which was programmed by John Fox to use Type II and Type III sum of squares. Did I get the point? 2. Now more specific question. Type II sum of squares is not well suited for unbalanced ANOVA designs too (as stated in STATISTICA help), therefore the general rule of thumb is to use Anova() function using Type II SS only for balanced ANOVA and Anova() function using Type III SS for unbalanced ANOVA? Is this correct interpretation? 3. I have found a post from John Fox in which he wrote that Type III SS could be misleading in case someone use some contrasts. What is this about? Could you please advice, when it is appropriate to use Type II and when Type III SS? I do not use contrasts for comparisons, just general ANOVA with subsequent Tukey post-hoc comparisons. Thank you in advance, Amasco [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Capture of iterative integration output
Hello,
I am a novice R user and am having difficulty retrieving the values
from 21 iterations of the R function integrate.
The only way I have found is to do a write.table and then a read.table
as shown in the code below. I would rather capture the 21 values inside
the braces ( sapply might work, but I can't set it up without getting an
error in function) so I could compute the negative log-likelihood inside
the braces. My goal is to then use optim to find the best fitting mu
and sigma. In the code that follows mu and sigma are given, but
normally they would not be known.
Any help on y$value capture would be appreciated. I would be ecstatic
for help on implementation of the optim function using finite
differences for the following code.
By the way in case anyone is interested function(x) is the
Poisson-lognormal pdf.
q - read.table(file=c:/Bayes/FigA3-1.prn,header=TRUE)
attach(q)
V - c(volume)
c - c(count)
n - length(count)
mu - -8.202900475
sigma - 1.609437912
for(i in c(1:n))
{
integrand - function(x)
{(x*V[[i]])^c[[i]]*exp(-x*V[[i]])/factorial(c[[i]])*0.3989422804014327/
(x*sigma)*exp(-.5*((log(x)-mu)/sigma)^2)}
y - integrate(integrand, lower = 0, upper = Inf)
write.table(y[[1]],file=c:/Bayes/PLNA3-1.out,append=TRUE,
quote=FALSE,row.names=FALSE,col.names=FALSE)
}
z -read.table(file=c:/Bayes/PLNA3-1.out)
z - c(z)
LL - log(c(z$V1))
#negative Log-Likelihood
print(-sum(LL),digits=6)
file.remove(c:/Bayes/PLNA3-1.out)
--
Terry W. Schulz
Applied Statistician
1218 Pomegranate Lane
Golden, CO 80401
USA
[EMAIL PROTECTED]
(303) 526-1461
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[R] Permanently changing gui preferences
Greetings, I made changes to my gui preferences and saved them. When I close and then open R, it reverts back to default preferences. How do I permanently change gui preferences? Thanks in advance. David -- David Kaplan, Ph.D. Professor Department of Educational Psychology University of Wisconsin - Madison Educational Sciences, Room 1061 1025 W. Johnson Street Madison, WI 53706 email: [EMAIL PROTECTED] Web: http://www.education.wisc.edu/edpsych/facstaff/kaplan/kaplan.htm Phone: 608-262-0836 Fax: 608-262-0843 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Importing data from clipboard on Mac OSX
Dear R users, I am trying to get data from the clipboard into R on MacOSX. I tried the following, but got an error message: read.delim(clipboard) Error in file(file, r) : unable to open connection In addition: Warning message: unable to contact X11 display Obviously, I'm not running R using X11. I'm wondering, can I import data from the clipboard on MacosX? Thanks in advance, Rense Nieuwenhuis version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.0 year 2006 month 04 day24 svn rev37909 language R version.string Version 2.3.0 (2006-04-24) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 iteratively extract elements out of a list
On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote:
| sub.m - lapply(m, function(x)x[x2])
| sub.m
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| [[3]]
| [1] 4
|
| sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))]
| [[1]]
| [1] 3 4
|
| [[2]]
| [1] 4 5
|
| sub4.m - lapply(m, function(x)x[x4])
| sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))]
| [[1]]
| [1] 5
Or slightly shorter in this case:
sub.m[sapply(sub.m, function(x) length(x) == 2)]
etc.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to iteratively extract elements out of a list
On 8/26/06, Patrick Connolly [EMAIL PROTECTED] wrote: On Sat, 26-Aug-2006 at 09:57AM +0100, Patrick Burns wrote: | sub.m - lapply(m, function(x)x[x2]) | sub.m | [[1]] | [1] 3 4 | | [[2]] | [1] 4 5 | | [[3]] | [1] 4 | | sub.m[unlist(lapply(sub.m, function(x) length(x) == 2))] | [[1]] | [1] 3 4 | | [[2]] | [1] 4 5 | | sub4.m - lapply(m, function(x)x[x4]) | sub4.m[unlist(lapply(sub4.m, function(x) length(x) 0))] | [[1]] | [1] 5 Or slightly shorter in this case: sub.m[sapply(sub.m, function(x) length(x) == 2)] or even shorter: sub.m[sapply(sub.m, length) == 2] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Capture of iterative integration output
Thanks for the reply Sego, Landon H. The vector is created as you
say. However, only the first iteration of integrate is captured.
The other twenty vector values remain at 0. Integrate output includes
several attributes and I only want the first one which is under $names
and called value
Terry W. Schulz
Applied Statistician
1218 Pomegranate Lane
Golden, CO 80401
USA
[EMAIL PROTECTED]
(303) 526-1461
Sego, Landon H wrote:
Here's one way to do this:
Before you begin the loop, define a vector to capture the output. Then
each iteration of the loop will be assigned to the corresponding element
in that vector:
capture - double(n) # creates a vector of length n, all with values of
0
for (i in 1:n) {
_ some code _
capture[i] - integrate( _your arguments_ )$value
}
capture should be the same as your z.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Terry W. Schulz
Sent: Saturday, August 26, 2006 3:42 PM
To: r-help@stat.math.ethz.ch
Subject: [R] Capture of iterative integration output
Hello,
I am a novice R user and am having difficulty retrieving the
values from 21 iterations of the R function integrate.
The only way I have found is to do a write.table and then a
read.table as shown in the code below. I would rather
capture the 21 values inside the braces ( sapply might work,
but I can't set it up without getting an error in function)
so I could compute the negative log-likelihood inside the
braces. My goal is to then use optim to find the best
fitting mu and sigma. In the code that follows mu and sigma
are given, but normally they would not be known.
Any help on y$value capture would be appreciated. I would be
ecstatic for help on implementation of the optim function
using finite differences for the following code.
By the way in case anyone is interested function(x) is the
Poisson-lognormal pdf.
q - read.table(file=c:/Bayes/FigA3-1.prn,header=TRUE)
attach(q)
V - c(volume)
c - c(count)
n - length(count)
mu - -8.202900475
sigma - 1.609437912
for(i in c(1:n))
{
integrand - function(x)
{(x*V[[i]])^c[[i]]*exp(-x*V[[i]])/factorial(c[[i]])*0.39894228
04014327/
(x*sigma)*exp(-.5*((log(x)-mu)/sigma)^2)}
y - integrate(integrand, lower = 0, upper = Inf)
write.table(y[[1]],file=c:/Bayes/PLNA3-1.out,append=TRUE,
quote=FALSE,row.names=FALSE,col.names=FALSE)
}
z -read.table(file=c:/Bayes/PLNA3-1.out)
z - c(z)
LL - log(c(z$V1))
#negative Log-Likelihood
print(-sum(LL),digits=6)
file.remove(c:/Bayes/PLNA3-1.out)
--
Terry W. Schulz
Applied Statistician
1218 Pomegranate Lane
Golden, CO 80401
USA
[EMAIL PROTECTED]
(303) 526-1461
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[R] how to create many objects with sequencial names?
Dear Lister, Is there a way to create many objects with sequencial names, say lm1, lm2...lm100? Thanks. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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 create many objects with sequencial names?
Yes there is with statements like:
assign(paste('m', i, sep='', value)
But I would suggest that you put the values in a list to make it
easier to access since all the data is in a single object. You could
do it in a loop:
result - list()
for(i in 1:100){
..computation.
result[[i]] - yourResult
}
On 8/26/06, Wensui Liu [EMAIL PROTECTED] wrote:
Dear Lister,
Is there a way to create many objects with sequencial names, say lm1,
lm2...lm100?
Thanks.
__
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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 you are trying to solve?
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Re: [R] how to create many objects with sequencial names?
Hi, Jim,
It is you again. I couldn't remember how many times you answered my
silly questions. ^_^
I am not sure assign() is what I want. Say, if I want to create 1000
linear model objects with names lm1, lm2lm1000, it seems assign
can't solve it.
But your second solution is close to what I am looking for.
Any idea?
Thanks.
wesui
On 8/26/06, jim holtman [EMAIL PROTECTED] wrote:
Yes there is with statements like:
assign(paste('m', i, sep='', value)
But I would suggest that you put the values in a list to make it
easier to access since all the data is in a single object. You could
do it in a loop:
result - list()
for(i in 1:100){
..computation.
result[[i]] - yourResult
}
On 8/26/06, Wensui Liu [EMAIL PROTECTED] wrote:
Dear Lister,
Is there a way to create many objects with sequencial names, say lm1,
lm2...lm100?
Thanks.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
--
WenSui Liu
(http://spaces.msn.com/statcompute/blog)
Senior Decision Support Analyst
Health Policy and Clinical Effectiveness
Cincinnati Children Hospital Medical Center
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[R] Simulations in R during power failure
Hi everyone, I recently ran a simulation on a computer using R that was hooked up to a UPS. There was one time when the power was out for length and the computer shut down. I was worried that I had lost the simulation, but upon booting the machine up, I heard the processor kick in. It sounded like the simulation had resumed. Does anyone have any experience with this? Because I live in Florida and there is a possibility of another hurricane. And I am running a huge simulation which may or may not finish before this hurricane finishes. What options are there in saving a simulation and it picking up where it left off? Thanks, Brandon __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Students With Writing Difficulties?
If you have undergraduates who have trouble writing a decent term paper, or postgraduates who are having problems with their theses or publications, we may be able to help you. We offer a personalised, online, tuition service designed to help students and academics produce solid, well argued texts that are fit for their intended purpose and appropriate for their intellectual discourse community, whether it be mechanical engineering or medieval Danish. For more information about who we are and what we do, click here http://www.discourse.com.ar/ Best Regards Eamonn MacDonagh -- If you never want to hear from us again just click on the link below. If you´d rather we did it for you, just ask. http://www.dscr.com.ar/dada/mail.cgi/u/dis_cons_01/ If the above URL is inoperable, make sure that you have copied the entire address. Some mail readers will wrap a long URL and thus break this automatic unsubscribe mechanism. If you're still having trouble, please contact us at: mailto:[EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Modifying the embed-results
Thanks!
I see, that do.call-function is used often in R-algorithms... Looking over
some extra do.call-examples seems useful. This tail-function is also new for
me.
Is there some reason to use seq(along = VECTOR) instead of 1:length(VECTOR)?
Atte
You can replace the for with lapply like this:
VECTOR - c(0, 3, 6, 3, 11, 2, 11, 4, 3, 4, 7, 7, 6, 4, 8)
f - function(i) unique(tail(VECTOR, length(VECTOR)-i+1))[1:5]
out - do.call(rbind, lapply(seq(along = VECTOR), f))
na.omit(rbind(rep(NA, 5), out))
Note that a matrix with zero rows is returned in the case
that VECTOR has zero length and in the case that VECTOR
has fewer than 5 unique elements.
On 8/26/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Again my example was't very clear: there were not enough same
numbers in the VECTOR.
What I need is something like this:
VECTOR-c(0,3,6,3,11,2,11,4,3,4,7,7,6,4,8)
MATRIX-c()
for(i in 1:length(VECTOR)){
v-(unique(VECTOR[i:length(VECTOR)])[1:5])
MATRIX-rbind(MATRIX,v)
}
MATRIX-na.omit(MATRIX)
MATRIX
data.frame(MATRIX, row.names=NULL)
X1 X2 X3 X4 X5
1 0 3 6 11 2
2 3 6 11 2 4
3 6 3 11 2 4
4 3 11 2 4 7
5 11 2 4 3 7
6 2 11 4 3 7
7 11 4 3 7 6
8 4 3 7 6 8
9 3 4 7 6 8
So, there are no duplicates in rows.
VECTOR is always scanned forward as long as the number of items
(here 5) becomes full.
Atte
Try:
embed(VECTOR, 5)[,5:1]
On 8/25/06, Atte Tenkanen [EMAIL PROTECTED] wrote:
Hi,
Here is a vector and the result from the embed-command:
VECTOR=c(0,3,6,3,11,2,4,3,7,6,4,5,10,2,3,5,8)
embed(VECTOR, dimension=5)
[,1] [,2] [,3] [,4] [,5]
[1,] 113630
[2,]2 11363
[3,]42 1136
[4,]342 113
[5,]7342 11
[6,]67342
[7,]46734
[8,]54673
[9,] 105467
[10,]2 10546
[11,]32 1054
[12,]532 105
[13,]8532 10
Is there a way to little modify the algorithm so that the result
looks like this:
[1] 0 3 6 11 2 - beginning from the first number of the
VECTOR [1] 3 6 11 2 4 - beginning from the second number
of the
VECTOR etc
[1] 6 3 11 2 4
[1] 3 11 2 4 7
[1] 11 2 4 3 7
[1] 2 4 3 7 6
[1] 4 3 7 6 5
[1] 3 7 6 4 5
[1] 7 6 4 5 10
[1] 6 4 5 10 2
[1] 4 5 10 2 3
[1] 5 10 2 3 8
[1] 10 2 3 5 8
Every row consists of next five unique(!) member of the
VECTOR. I
made this example result with a time consuming algorithm which
uses for-loops
and whiles.
How to do this better??
Thanks in advance!
Atte Tenkanen
University of Turku
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Re: [R] Permanently changing gui preferences
David Kaplan a écrit : Greetings, I made changes to my gui preferences and saved them. When I close and then open R, it reverts back to default preferences. How do I permanently change gui preferences? one way is using options() and also using your Rprofile.site file (in ~/etc). edit it. See also .First(), source(), etc you may also directly edit your Rconsole file (also in ~/etc) or use the config editor in R and save hih __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Simulations in R during power failure
[EMAIL PROTECTED] a écrit : Hi everyone, I recently ran a simulation on a computer using R that was hooked up to a UPS. There was one time when the power was out for length and the computer shut down. I was worried that I had lost the simulation, but upon booting the machine up, I heard the processor kick in. It sounded like the simulation had resumed. Does anyone have any experience with this? Because I live in Florida and there is a possibility of another hurricane. And I am running a huge simulation which may or may not finish before this hurricane finishes. What options are there in saving a simulation and it picking up where it left off? Thanks, Brandon write your results, let's say every hour or 30' on your disk. something like if (substr(date(),...)=='30') write.table(...) where it could take place __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/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] Type II and III sum of square in Anova (R, car package)
1. First of all, more general question. Standard anova() function for lm() or aov() models in R implements Type I sum of squares (sequential), which is not well suited for unbalanced ANOVA. Therefore it is better to use Anova() function from car package, which was programmed by John Fox to use Type II and Type III sum of squares. Did I get the point? 2. Now more specific question. Type II sum of squares is not well suited for unbalanced ANOVA designs too (as stated in STATISTICA help), therefore the general rule of thumb is to use Anova() function using Type II SS only for balanced ANOVA and Anova() function using Type III SS for unbalanced ANOVA? Is this correct interpretation? 3. I have found a post from John Fox in which he wrote that Type III SS could be misleading in case someone use some contrasts. What is this about? Could you please advice, when it is appropriate to use Type II and when Type III SS? I do not use contrasts for comparisons, just general ANOVA with subsequent Tukey post-hoc comparisons. There are many threads on this list that discuss this issue. Not being a great statistician myself, I would suggest you read through some of these as a start. As I understand, the best philosophy with regards to types of sums of squares is to use the type that tests the hypothesis you want. They were developed as a convenience to test many of the hypotheses a person might want automatically, and put it into a nice, neat little table. However, with an interactive system like R it is usually even easier to test a full model vs. a reduced model. That is if I want to test the significance of an interaction, I would use anova(lm.fit2, lm.fit1) where lm.fit2 contains the interaction and lm.fit2 does not. The anova function will return the appropriate F-test. The danger with worrying about what type of sums of squares to use is that often we do not think about what hypotheses we are testing and if those hypotheses make sense in our situation. Mark Lyman __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
