Re: [R] order function
I agree with Duncan. may be you need this format(output[order(output[,3]),],scientific=T) Pratap --- On Sat, 2/2/13, Duncan Murdoch murdoch.dun...@gmail.com wrote: From: Duncan Murdoch murdoch.dun...@gmail.com Subject: Re: [R] order function To: Kripa R kripa...@hotmail.com Cc: r-help@r-project.org Date: Saturday, 2 February, 2013, 4:19 AM On 13-02-01 12:15 PM, Kripa R wrote: Hi I'm having a simple issue with the order function. When I use the following code my data is not ordered correctly output[order(output[,3]),] Name beta pval 881 9.09303277751237 0.000100253350483199 7402 6.40553461638365 0.00010228641631914 4879 -8.88509881106217 0.000103251645995887 That's correctly ordered in the pval column. However when I export the data and sort it in excel I see the following: Name beta pval pval 25037 -5.70737 2.48E-07 34294 -19.6931 1.04E-05 36002 -12.2478 1.63E-05 That is a different dataset, also correctly ordered in the third column. Excel seems to have added a 4th heading for some reason. Any suggestions on how I can get this sort to work properly on data in scientifict format? You need to explain what is wrong before we can fix it. Duncan Murdoch .kripa [[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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question on plotCI function
On 02/02/2013 02:21 AM, li li wrote: Thanks so much for the reply, Ista. I used plotrix library. Here is my example: xx- seq(0.05, 0.95, by=0.05) lower- c(-2.896865, -2.728416, -2.642574, -2.587724, -2.548672, -2.518994, -2.495409, -2.476031, -2.459662, -2.445513, -2.433014, -2.421739, -2.411344, -2.401536, -2.392040, -2.382571, -2.372786, -2.362198, -2.349891) upper- c(2.311539, 2.372006, 2.423280, 2.469220, 2.511851, 2.552421, 2.591797, 2.630657, 2.669579, 2.709135, 2.749928, 2.792670, 2.838268, 2.887976, 2.943683, 3.008502, 3.088240, 3.195954, 3.373528) library(plotrix) plotCI(xx,ui=upper,li=lower,err=y,pch=NA, xlab=, ylab=, ylim=c(-5, 5), slty=solid,scol=blue, lwd=2) My question is how can I change the xllim to be c(0,1) which corresponds to the xx values. Hanna 2013/2/1 Ista Zahnistaz...@gmail.com There are many plotCI functions in many different packages... which one are you referring to? Also please construct a reproducible example illustrating your problem. Best, Ista On Thu, Jan 31, 2013 at 11:58 PM, li lihannah@gmail.com wrote: Hi all, In my plotCI function, the argument x is chosen to be seq(0.05, 0.95, by=0.05). However, when I make the plot, the plot has the x coordinate goes 1:19. Does anyone know how to make the x coordinate to be (0,0.5, 0.1, ..., 0.95). Hi Hanna, You are getting what you request in the example. xx is interpreted as the y values and the x values default to the number of y values (1:19). Perhaps what you want is something like this: x-seq(0.05,0.95,by=0.05) xx-seq(0.05,0.95,by=0.05) lower-c(-2.896865,-2.728416,-2.642574,-2.587724,-2.548672,-2.518994, -2.495409,-2.476031,-2.459662,-2.445513,-2.433014,-2.421739,-2.411344, -2.401536, -2.392040, -2.382571, -2.372786, -2.362198, -2.349891) upper-c(2.311539,2.372006,2.423280,2.469220,2.511851,2.552421, 2.591797, 2.630657, 2.669579, 2.709135, 2.749928, 2.792670, 2.838268, 2.887976, 2.943683, 3.008502, 3.088240, 3.195954, 3.373528) library(plotrix) plotCI(x,xx,ui=upper,li=lower,err=y,pch=NA,xlab=,ylab=, slty=solid,scol=blue,lwd=2) Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Mixed Models: Contribution of random variable to final estimate
Dear all, We want to test if the invasiveStatus is predicted by the amount (quant) of animals arriving to a country of a certain species (taxonid). We are using lmer to perform the model. The model is: lmer(invasiveStatus~I(log(quant+1))+I(log(inDegree+1))+(1|taxonid)+(1|country), family=binomial,data=td), where invasiveStatus is a binary variable, quant and inDegree are integer variables, and taxonid and country are factor variables. The fixef output is (Intercept)I(log(quant + 1)) I(log(inDegree + 1)) -15.63382880.31980742.1566502 and the ranef output is, sorted from higher to lower, andshowing only the first 10 lines, $taxonid T16 9.51 T2588.36 T3888.24 T9617.98 T76 7.48 T4707.46 T1087.17 T84 7.15 T2926.91 T1896.65 ... $country US 3.23 JP 2.45 ES 2.35 IT 2.14 BM 1.63 IL 1.41 SI 1.39 LB 1.06 FR 1.05 VE 0.996 ... Our problem is that the coefficients to the final estimate of invasiveStatus are higher for the random variables than the fixed ones. We think this is a result of the confound effect between quant, and country and taxonid. In other words, the higher the number of animals of a given species(taxonid) arriving to given country, the higher the probability of other species to arrive to the same country. Are we formulating the model correctly? Is there a way to avoid that the contribution of the random variables is the most contributing part to the final estimate? Thanks, Luis Reino [[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] Armadillo error in R extension
Dear Michael,
during the last days of programming I came across RcppArmadillo (Rcpp) and for
my kind of work this is indeed pretty interesting. So, I will take a closer
look today what it is about and how it works.
As a C++ programmer though, I am still interested why compilation of my package
fails at this point of code. I hope I get a little more insight on the
Rcpp-devel list suggested by Dirk!
Thanks for your comment!
Simon
On Feb 2, 2013, at 12:14 AM, Michael Weylandt michael.weyla...@gmail.com
wrote:
Look up the terribly wonderful RcppArmadillo package.
MW
On Feb 1, 2013, at 8:38 PM, Simon Zehnder szehn...@uni-bonn.de wrote:
Is there anyway with some experience in using armadillo in R C++ extensions?
My problem is the following:
I programmed a function in a header looking like
#include armadillo
inline arma::vec foo(input) {
... do something
return an arma::vec object
}
compiling this via R CMD INSTALL packagename (PKG_CXXFLAGS =
-I/folder/of/armadillo and armadillo_bits in my package)
I get the following error
error: expected initializer before 'foo'
The exact line and word is where inline arma::vec ends.
Does anyone know what this error could be?
Best
Simon
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Re: [R] Mixed Models: Contribution of random variable to final estimate
Luis Reino luisreino at isa.utl.pt writes:
Dear all,
We want to test if the invasiveStatus is predicted by the amount
(quant) of animals arriving to a country of a certain species
(taxonid). We are using lmer to perform the model.
In general lmer questions belong on r-sig-mixed-mod...@r-project.org,
but I think this
The model is:
lmer(invasiveStatus~I(log(quant+1))+I(log(inDegree+1))+
(1|taxonid)+(1|country),
family=binomial,data=td)
You don't need I() around those terms -- you only need it to
protect expressions such as x^2 that would be interpreted differently
in the formula context.
where invasiveStatus is a binary variable, quant and inDegree are
integer variables, and taxonid and country are factor variables.
The fixef output is
(Intercept)I(log(quant + 1)) I(log(inDegree + 1))
-15.63382880.31980742.1566502
and the ranef output is, sorted from higher to lower, andshowing
only the first 10 lines,
$taxonid
T16 9.51
T258 8.36
[snip]
$country
US3.23
JP2.45
ES2.35
[snip]
Our problem is that the coefficients to the final estimate of
invasiveStatus are higher for the random variables than the fixed
ones. We think this is a result of the confound effect between
quant, and country and taxonid. In other words, the higher the
number of animals of a given species(taxonid) arriving to given
country, the higher the probability of other species to arrive to
the same country. Are we formulating the model correctly? Is there
a way to avoid that the contribution of the random variables is the
most contributing part to the final estimate? Thanks, Luis Reino
This might be an issue of parameter scaling.
The idea is that your coefficients measure the effect of
the parameters *per unit*. Thus the random effects are
measured in log-odds units, while the effects of quant and inDegree
are measured in units of log-odds change **per log-unit change in
the variable**, i.e. multiplying by e is expected to make 1 log-odds
change in the outcome. You might try scaling your variables
(see e.g. Schielzeth 2010 Methods in Ecology Evolution).
(Of course, you can make the fixed effects look as big as you
want by scaling the predictor appropriately ...)
It worries me a little that your intercept is so small --
suggests that the average fraction invasive when quant=0
and inDegree=0 is 3 x 10^{-7} ...
Follow-ups to r-sig-mixed-models
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Re: [R] use name (not values!) of a dataframe inside a funktion
Another R-ish way of modifying an object with a function is to use
'replacement functions' (there must be other names, I'm not sure
what the standard is) that let you use syntax like
someProperty(myData, ...) - newProperty
To do this define a function called `someProperty-` whose last argument
is named 'value' and which returns a modified version of its first argument.
When R sees the above syntax it does the equivalent of
myData - `someProperty-`(myData, ..., value=newProperty)
E.g.,
`useLog2Scale-` - function(dataframe, value) {
if (value) dataframe[] - lapply(dataframe, log2)
else dataframe[] - lapply(dataframe, function(x)2^x)
dataframe
}
d - data.frame(x=1:5, y=1/(1:5))
useLog2Scale(d) - TRUE
d
x y
1 0.00 0.00
2 1.00 -1.00
3 1.584963 -1.584963
4 2.00 -2.00
5 2.321928 -2.321928
useLog2Scale(d) - FALSE
d
x y
1 1 1.000
2 2 0.500
3 3 0.333
4 4 0.250
5 5 0.200
(Usually such a function is used to set a property in the data, such as a
logScale
flag, that is used or queried later.)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Greg Snow
Sent: Friday, February 01, 2013 8:54 AM
To: Winfried Moser
Cc: r-help
Subject: Re: [R] use name (not values!) of a dataframe inside a funktion
It is strongly discouraged in R to have functions that change data values
in the global workspace (or any location other than their local
environment).
The usual procedure in R is to have your function return a modified version
of the object and the user then decides what to do with it. They can
assign it back to the same original object so that there is still only one
copy and it has changed (but the user made that decision, not the
programmer), or they can save it to a different name and not lose the
original.
If you really want to change the original copy (and there are sometimes
when the exception to the rule makes sense) then you can either use
environments (which don't copy on modify) or use macros instead of
functions. Given your examples I would look at the macro approach first.
There is a 'defmacro' function in the 'gtools' package and the reference
on the help page for 'defmacro' leads to the original R news (now R
Journal) article describing the use of macros in R (definitely read this if
you are considering this approach).
On Thu, Jan 31, 2013 at 7:34 AM, Winfried Moser
winfried.mo...@gmail.comwrote:
Dear Listers,
can anyone help me, please.
Since several days i try to figure out, how to assign values, vectors,
functions etc to variables with dynamically generated names inside of
functions.
Sometimes I succeed, but the success is rather arbitrary, it seems. up to
now i don't fully understand, why things like get, assign, - etc do
sometimes work, and sometimes not.
here's one of my daily examples, i am stuck with: Example 1 does work, but
example 2 doesn't?
How kann i tell R, that i want it to expand the string dfb to dfb[,2]
inside the function.
In the end i want the function to change the second variable of the
dataframe dfb permanently to factor (not just inside the function).
Thanks in advance!
Winfried
Example 1:
dfa - data.frame(a=c(1:4),b=c(1:4))
dfa[,2] - factor(dfa[,2])
is.factor(dfa[,2])
TRUE
Example 2:
dfb - data.frame(a=c(1:4),b=c(1:4))
f.fact - function(x) {x[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
PS: I tried a whole lot of other things like, ...
I really don't know where to keep on searching.
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(get(x)[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x[,2]) - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {assign(x[,2], factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Ungültiges erstes Argument
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {quote(x)[,2], factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Unerwartetes ','
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(quote(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(get(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Falsche Anzahl von Dimensionen
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact -
Re: [R] Loading a list into the environment
Thanks all! list2env was exactly what I was looking for. As an FYI (and
please correct me if I'm wrong), if you want to load a list into the
current environment, use:
myvariables - list(a=1:10,b=20)
loadenv - list2env(myvariables ,envir=environment())
a
b
--j
On Fri, Feb 1, 2013 at 5:49 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Something like this?
myfun - function(x, envir = .GlobalEnv){
nm - names(x)
for(i in seq_along(nm))
assign(nm[i], x[[i]], envir)
}
myvariables - list(a=1:10,b=20)
myfun(myvariables)
a
b
Hope this helps,
Rui Barradas
Em 01-02-2013 22:24, Jonathan Greenberg escreveu:
R-helpers:
Say I have a list:
myvariables - list(a=1:10,b=20)
Is there a way to load the list components into the environment as
variables based on the component names? i.e. by applying this theoretical
function to myvariables I would have the variables a and b loaded into the
environment without having to explicitly define them.
--j
--
Jonathan A. Greenberg, PhD
Assistant Professor
Global Environmental Analysis and Remote Sensing (GEARS) Laboratory
Department of Geography and Geographic Information Science
University of Illinois at Urbana-Champaign
607 South Mathews Avenue, MC 150
Urbana, IL 61801
Phone: 217-300-1924
http://www.geog.illinois.edu/~jgrn/
AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307, Skype: jgrn3007
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[R] vectorisation
Hi
I'm trying to set up a simulation problem without resorting to (m)any loops. I
want to set entries in a data frame of zeros ('starts' in the code below) to 1
at certain points and the points have been randomly generated and stored in a
separate data.frame ('sl'), which has the same number of columns.
An example of the procedure is as follows:
ml - data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm - apply(ml, 2, cumsum)
starts- data.frame(matrix(0,600,4))
I can achieve the result I want with a loop:
for (i in 1:4){
lstarts[,i][mm[,i]] -1
}
But as I want to use a large number of columns I would like to do away with the
loop
Can anyone suggest how this might be done?
Thanks in advance
Brett Robinson
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[R] Reading .rdata files
Hello, I am a developer working on extracting information from .rdata files using java. I have been looking through the documentation a lot trying to figure out how to read these files, but am a bit stuck at the moment. I have generated two .RDATA files from StatTransfer and have noticed that they both begin in similar ways. *File 1* RDA2 A 2 131585 66560 1026 1 4105 10 .Traceback 254 1026 1 4105 *8 - these are the only difference* SPSSTest -FileName 787 * * *File 2* RDA2 A 2 131585 66560 1026 1 4105 10 .Traceback 254 1026 1 4105 *9 **- these are the only difference* statatest -FileName 787 Because of the remarkable similarity between these two beginning segments I assume that this is some sort of header. After examining the R Internals Manual (Section 1.8) I was able to discover that the first line describes the file RDA2 means its an ascii data file opposed to XDR or word order binary and the 2 indicates that it is version 2 of R. After this first line there are supposed to be 3 integers that also describe the version. However this is pretty much where section 1.8 ends and even though I searched through the rest of the document (and a lot of Google) I couldn't find anything about what the rest of the bytes/numbers indicate. I may just be missing something (I have seen a lot of people reference the Internals Manual) but is there any documentation on the different data files (RDA, RDX, RDB) and how they are organized. Any info would be greatly appreciated. Thanks a bunch! -Andrew [[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] Choice of text for intermediate level R programming course
I recommend The Art of R Programming: A Tour of Statistical Software Design, by Norman Matloff, No Starch Press. It is only 40 bucks so your students will appreciate that. As far as content goes, it covers just about everything, you can view a sample chapter, detailed ToC, and index at the website: http://nostarch.com/artofr.htm. The only drawback to using it as a textbook is that it doesn't contain any excercises, so you would have to create your own. I think that the no exercises drawback is an acceptable price to pay considering the quality of the book. Here's some anecdotal evidence for you: I used R for my M.S. thesis and at first I spent my time searching on the internet for how to do things in R. When I found this book, I took some time off from my thesis work and chugged through the book full time- I think it would take a full semester in a 3 hour course. After getting through the book I found that I wasn't worrying about R anymore and I could concentrate on the intellectual part of my work. I won't say that content of my thesis was that great but the code worked and I managed to finish on time. -- Cheers, Harris Butler On Fri, Feb 1, 2013 at 5:43 PM, William Grove grove...@umn.edu wrote: The Subject line mostly says it. I'm designing it as a semester-long, 3 hours per week, course that takes in students who got the basics of R in stats classes, but don't really know how to program in it. Translation: if their own examples don't look enough like examples from previous work, they're stumped. Does anybody have a text for an intermediate R course (but not too highbrow, no total guide to object oriented programming, and class s3 vs. s4 objects), they would recommend. I read 24 books some ielementary some advance not many intermediate, and I have two I like but for quite difference reasons. I won't name them so as not to prejudice anybody. It needs to be English language and I greatly prefer a good index like \LaTeX\ produces by the makeindex() prograrm. Regards, William Grove -- William M. Grove, Ph.D. Psychology Department University of Minnesota My public encryption key.is: -BEGIN PGP PUBLIC KEY BLOCK- Version: GnuPG v1.2.2 (MingW32) mQGiBE+tS9ERBACCTYW1uclqHbUvBvP92rUN8zOlMWe2zr3NfRp6ELNyYa60H5PK +Xx/vbHGWF0MXhGLLVFMiRNBRomJ4S30Ify4exTg6vWpPrVsmDvyKLmYilWSOuw1 0yoimDdr4DphGQSKqVD/O/tLUhO1wyEKJI89qtJmFG4usnn7f/Bvt4cRZwCgq76W LRhGATk0E2GYl0n+2UeFU+8D/ipLpFYNJEkz3y492/9orimBsxsAdnXmK5bpOBG5 pQaoFuvvPMtkUYhVNYUqMyjJniM3F6mnv7OX52sHHcYn89ScO9S+WtWAYsekwTTE xb1wvwB9u7PoGDC46tihXhTl533oUcMFGMwlVKvFZ4zpWYo5W3i0DmOAtjcMPzl2 4wcfA/0fgbUe2JIkPTf6MFMQTFwpNhCGeDvTfbcC5YCCcAXWcnUT/GKFsvRT0Wje i/A4PMopJkfLRreM1GSZ6JY0cOL3U071M6YhVmldrL4X74OUAjfdo8mYosxe/qTe Ji0s6jg/oiXgY/hHLZzDE+AtQ6LRi0/1tpms2Qw9/7pXs8HHErQ0V2lsbGlhbSBN LiBHcm92ZSAoR1BHIFB1YmxpYyBLZXkpIDxncm92ZTAwMUB1bW4uZWR1PohbBBMR AgAbBQJPrUvRBgsJCAcDAgMVAgMDFgIBAh4BAheAAAoJEG2MPb65SVkFnNcAnRCi URO4Oy9dk5Kb6j/1yN1iElTJAKCnlwfwOt0J8jf2nJ78sNkiZNjM6bkCDQRPrUvT EAgAg52fj9/7k6BuJgmY+ynKm4rwWaRW6VjrnS5VI+luTmEkQi/w9Vz6dOmCJWlz VLsm7mEW0tvB9VwZbkrttoWnJOmqzRWNJxlRlj78jRq6yN16JZNhbPGDSbjL/+JV LzlhMetUpuTYPziuatSr7C2RxYYRd5pabgd3G/XowqSBtyD3jrViJ8CqcZlMQGfJ BRoiyBDlVEOO+aJ/KGLinfbhyz06tR50BpXFFlxBN1m+DInRLF6zRkjZUO+OyFO6 Acy0SM4htfetEI0H0+X1w/Cw8e2Ew48Wwsve+USRr+OLXqwkgBP/jTNRko6cUI5m LiXjkSXYdR0JonTzgUSh66TDBwADBQf/TR/xC5P5Vs6X5OPu3iHsKRp/dGVpL33D KBH2Ofw3ps48kQ65la/uMW+fZDvwZJukTIJL4dq5OGjxnwznj3hkZJZWLknRFDx+ ByoumskJXPENphqlnpDV+W2stNwoSY3f5gO9xTQSGeoqqLB6+Xwe4CDZo/wOuhMA nCXQeIJxiH3GsCnxkZY9e3rw9HdcE/6NbvKGZ/OOLMdorASG9eSuyQZDgKN7jXZ3 2GPbppaRjzUWuRGau1RBv9V0QMY5C0Z1c51njCHuR0cgcItad69IcgaujlC2kouu C/MW/VOA5rfGhvBBI7nr26MjhxwHUYWQw6i/KgzYwJ+X43cEBCoITIhGBBgRAgAG BQJPrUvTAAoJEG2MPb65SVkFDLoAn36dv6dBZyIrTD+lF3y1lXO/R1OHAJ0Sn0bi 0TdH/wwxUEFeq5eCMfoZTg== =91Ch -END PGP PUBLIC KEY BLOCK- [[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] VAR simulation help
http://r.789695.n4.nabble.com/file/n4657370/untitled.jpg
Hi Everyone,
I am a new comer to R circle. I am trying to simulate a VAR estimation. The
problem is given above in the image. Assume that the errors are iid and
normally distributed. Here the number of observations x1=x2=64.
I've written the code below. Not sure if it is correct.
pi=matrix(c(0.9,0.1,-0.1,0.7),nrow=2,ncol=2)
# errors
error1=rnorm(64,mean=0,sd=1)
error2=rnorm(64,mean=0,sd=1)
#variables x1 and x2
x1=sample(64)
x2=sample(64)
#VAR simulation#
obs = 64
set.seed(123)
E=matrix(c(error1,error2),nrow=2,ncol=64)
pi=matrix(c(0.9,0.1,-0.1,0.7),nrow=2,ncol=2)
X=matrix(c(x1,x2),nrow=2,ncol=64)
for (i in 2:obs) {
X[,i] = pi%*%X[,i-1]+E[,i]
}
I would appreciate if you could comment whether I've correctly simulated it
or not.
Thank you,
lphd
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Re: [R] Change default order of colors line types
You can change colors by manually creating your own palette. You can start with the Set1 color palette from RColorBrewer, then move the values. You then use this new palette in your plot: library(RColorBrewer) numGroups -5 #here is where you define the size of your color palette customPalette - brewer.pal(numGroups, Set1) #create your color palette customPalette - append(customPalette, customPalette[1], after=4) #insert the first color into the correct place on the list customPalette - customPalette[2:len(customPalette)] #drop the first color To use this in your plot, instead of using scale_color_brewer(palette=Set1), you would use scale_color_manual(values=customPalette). Linetype will need to be handled a slightly different way. Remove linetype=method from your initial qplot command, and then add a manual linetype scale that would look something like this: + scale_linetype_manual(breaks=c(m1, m2, m3, m4, m5), values=c(2,3,4,1,5)) where m1, m2, etc are the method names that you are grouping everything by. You should keep your geom_line() command the same. Note that the values in scale_linetype_manual() are rearranged to put 1 in the 4th position. I'm sure there is a more elegant way to do it, but this way should work as long as you aren't creating lots of plots with changing method names. -- Cheers, Harris Butler On Fri, Feb 1, 2013 at 1:41 PM, Soyeon Kim soyeon.sophia@gmail.com wrote: Dear R users, I'd like to change the default order of colors line types. Especially I am using ggplot2 and using color Set1. In Set1, the default color order is red, blue, green, violet,.. ect. However, I want to put red in fourth (not first). Likewise, I want to change the order of default linetype. I want to put solid line in fourth. How can I do thses? R code to draw the graph is qplot(variable, power, data = m.powers, colour = method, linetype=method, ylab = Power) + geom_line(aes(group = method), ylim = c(0,1)) + scale_colour_brewer(palette=Set1) Thank you, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] create and save a simulated dataset
Hello, I am trying to learn how to create a simulated dataset of a forest stand: I must simulate 10 stands, for each of the stands I have been creating a random number of trees. For each tree I should create a random diameter. The problem I have is that I cannot use a matrix because the length of the three items is different. I was trying with list but I have no clue on how to make a dataset out of them... tried merge Anyone has some clever suggestion? THANK you !! Giovanna Giovanna Ottaviani Aalmo Stipendiat/Ph.D. Student --- Norsk institutt for skog og landskap Pb 115, NO-1431 Ås T (+47) 64 94 9094 M(+47) 980 30 422 F(+47) 64 94 90 80 --- www.skogoglandskap.nohttp://www.skogoglandskap.no/ --- [[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] Reading .rdata files
On 13-02-02 2:23 PM, Andrew DeCarlo wrote: Hello, I am a developer working on extracting information from .rdata files using java. I have been looking through the documentation a lot trying to figure out how to read these files, but am a bit stuck at the moment. I have generated two .RDATA files from StatTransfer and have noticed that they both begin in similar ways. *File 1* RDA2 A 2 131585 66560 1026 1 4105 10 .Traceback 254 1026 1 4105 *8 - these are the only difference* SPSSTest -FileName 787 * * *File 2* RDA2 A 2 131585 66560 1026 1 4105 10 .Traceback 254 1026 1 4105 *9 **- these are the only difference* statatest -FileName 787 Because of the remarkable similarity between these two beginning segments I assume that this is some sort of header. After examining the R Internals Manual (Section 1.8) I was able to discover that the first line describes the file RDA2 means its an ascii data file opposed to XDR or word order binary and the 2 indicates that it is version 2 of R. After this first line there are supposed to be 3 integers that also describe the version. However this is pretty much where section 1.8 ends and even though I searched through the rest of the document (and a lot of Google) I couldn't find anything about what the rest of the bytes/numbers indicate. I may just be missing something (I have seen a lot of people reference the Internals Manual) but is there any documentation on the different data files (RDA, RDX, RDB) and how they are organized. Any info would be greatly appreciated. Thanks a bunch! You could try the source code: the files are written by the code in https://svn.r-project.org/R/trunk/src/main/serialize.c The comments in there describe it, and the code does what it does. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Why replacement has length zero? And How can I fix it?
Hi
for the loop section runif needs curved brackets
Try
IAP -NA
for (i in 1:Sample.Size){
if (DataSet$SES[i]0) {
IAP[i] - ifelse(runif(1)0.75, 1, 0) # High SES, higher
chance to be in
Treatment #
}
else {
IAP[i] - ifelse(runif(1)=0.25, 1, 0) # Low SES, lower chance
to be in
Treatment #
}
} # End loop #
IAP
IAP
zjiaqi19880219 wrote
Hi, all,
I am working on a project to run a simulation. I am now working on the
main body of the simulation body, I generate the data and all work well
before I identify a variable IAP, there is always an error message as
replacement has length zero, and I do not know how to fix it. Can anyone
help?
Here is the code:
library (mvtnorm)
options(digits=2)
# Set variable name #
namelist1 - c(Age, Gender, SES, ISS, IQTe, PreTe)
# Generate Data #
Sample.Size - 10 # Variable Sample Size #
CorMat - matrix(c(1.0 ,0.0 ,0.0 , 0.5, 0.2, 0.5,
0.0, 1.0 ,0.0, 0.0, 0.0, 0.0,
0.0 ,0.0 ,1.0, 0.5, 0.4, 0.5,
0.5, 0.0, 0.5, 1.0, 0.4, 0.7,
0.2, 0.0, 0.4, 0.4, 1.0, 0.9,
0.5, 0.0, 0.5, 0.7, 0.9, 1.0),ncol=6) #
Correlation Matrix #
DataSet - rmvnorm(Sample.Size, mean=c(0,0,0,0,0,0), sigma=CorMat) # Draw
Sample.Size covariate set all mean 0 #
DataSet - as.data.frame(DataSet)
names(DataSet) - namelist1
DataSet
# For Treatment IAP #
IAP - matrix(0, nrow=Sample.Size)
for (i in 1:Sample.Size){
IAP[i] - if (DataSet$SES0)
{
ifelse(runif[1]0.75, 1, 0) # High SES, higher chance to be in
Treatment
#
}
IAP[i] - if (DataSet$SES=0)
{
ifelse(runif[1]=0.25, 1, 0) # Low SES, lower chance to be in
Treatment
#
}
} # End loop #
IAP
Error in IAP[i] - if (DataSet$SES 0) { : replacement has length zero
In addition: Warning message:
In if (DataSet$SES 0) { :
the condition has length 1 and only the first element will be used
Thanks!
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Re: [R] create and save a simulated dataset
Is this a homework problem? (We don't do homework here). What is missing from your query is any specification of what you mean by random, which, by itself, is meaningless. It also might be the case that the trees within each stand are correlated in some way -- perhaps they should to be more similar in diameter within a stand than between. So I think your first task is to think more carefully about **what** you want to do and how to specify it precisely and completely before worrying about **how** to do it.. Cheers, Bert On Sat, Feb 2, 2013 at 12:10 PM, Giovanna Ottaviani g...@skogoglandskap.no wrote: Hello, I am trying to learn how to create a simulated dataset of a forest stand: I must simulate 10 stands, for each of the stands I have been creating a random number of trees. For each tree I should create a random diameter. The problem I have is that I cannot use a matrix because the length of the three items is different. I was trying with list but I have no clue on how to make a dataset out of them... tried merge Anyone has some clever suggestion? THANK you !! Giovanna Giovanna Ottaviani Aalmo Stipendiat/Ph.D. Student --- Norsk institutt for skog og landskap Pb 115, NO-1431 Ås T (+47) 64 94 9094 M(+47) 980 30 422 F(+47) 64 94 90 80 --- www.skogoglandskap.nohttp://www.skogoglandskap.no/ --- [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create and save a simulated dataset
Hello, Once you formulate the problem properly, it's not hard to solve it in R. Bert's questions should be answered before continuing. I'll make two assumptions, not present in your post. All diameters are assumed independent and to follow the same uniform distribution. (iid scenario; the two assumptions you have to decide about.) With these in mind, you can get some inspiration from the following code. It produces 10 vectors of a random numbers of trees of random diameters. min.diam - 1 # trees' diameters max.diam - 2 # max.trees - 10 # trees per stand lapply(1:10, function(i) runif(sample(max.trees, 1), min.diam, max.diam)) Hope this helps, Rui Barradas Em 02-02-2013 20:10, Giovanna Ottaviani escreveu: Hello, I am trying to learn how to create a simulated dataset of a forest stand: I must simulate 10 stands, for each of the stands I have been creating a random number of trees. For each tree I should create a random diameter. The problem I have is that I cannot use a matrix because the length of the three items is different. I was trying with list but I have no clue on how to make a dataset out of them... tried merge Anyone has some clever suggestion? THANK you !! Giovanna Giovanna Ottaviani Aalmo Stipendiat/Ph.D. Student --- Norsk institutt for skog og landskap Pb 115, NO-1431 Ås T (+47) 64 94 9094 M(+47) 980 30 422 F(+47) 64 94 90 80 --- www.skogoglandskap.nohttp://www.skogoglandskap.no/ --- [[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] Why replacement has length zero? And How can I fix it?
Hi:
No offense, but this is code is inefficient on several levels.
Firstly, a loop is unnecessary (see below); secondly, ifelse() is a
vectorized function and you're attempting to apply it to each row of
DataSet. Try this instead, given
dput(DataSet)
structure(list(Age = c(-0.552450789929175, -2.06988485889524,
-1.1290497440272, -0.268777167923272, -0.012748181465016, 0.554761979685086,
0.179401201802416, 0.0973216900180107, -0.914782595075178, -1.53708818506571
), Gender = c(-0.0705367578252014, 0.969426004829301, 0.0556317051882303,
1.33716753655214, -0.638382962845772, 0.480939398350137, -0.755053441639217,
1.25393329162734, 1.45933385954704, 0.672871276697159), SES =
c(-1.88102131807556,
0.529181612783923, 0.271835979074789, -0.656415898217512, -0.517622943602359,
0.639094817790355, -1.08205962796926, -0.193116856300625, 0.707374091321319,
1.98180909069287), ISS = c(-1.74208565286036, -1.235721877648,
0.459828219284939, -0.87781916182417, -0.429262433047679, 0.379335629940134,
0.949088546564315, 0.689758637278236, -1.14832161185134, 1.40363651773545
), IQTe = c(-0.266523341317252, -0.32477349599139, -0.75723700635683,
0.74556355929622, -0.238794158234391, 0.0906173696194765, 0.331200954821995,
-0.373009866084321, 0.417237160703897, 0.755603172535694), PreTe =
c(-0.743400287678305,
-1.1501727014007, -0.668278397501659, -0.0766957410591086, -0.232360222985187,
0.320050787942651, 0.474114337341715, -0.0181639478062117, -0.0905594374421625,
0.517614989060291)), .Names = c(Age, Gender, SES, ISS,
IQTe, PreTe), row.names = c(NA, -10L), class = data.frame)
u - runif(10)
u
[1] 0.72 0.57 0.71 0.64 0.38 0.53 0.14 0.14 0.82 0.12
# Remember two things:
# (i) ifelse() is a vectorized function;
# (ii) logical statements return either TRUE (1) or FALSE (0).
# As a result,
# less efficient:
IAP0 - ifelse(DataSet$SES 0, ifelse(u 0.75, 1, 0),
ifelse(u = 0.25, 1, 0))
# more efficient:
IAP1 - ifelse(DataSet$SES 0, u 0.75, u = 0.25) * 1
IAP1
[1] 0 0 0 0 0 0 1 1 1 0
identical(IAP0, IAP1)
[1] TRUE
# Verification:
cbind(DataSet$SES, u, IAP1)
This result conforms to what one would expect with u as the realized
uniform random vector. Moreover, the code is transparent: if SES 0,
then apply the test u 0.75, else apply the test u = 0.25). The
post-multiplication by 1 ensures that the result is numeric rather
than logical.
Dennis
On Sat, Feb 2, 2013 at 12:57 PM, soon yi soon...@ymail.com wrote:
Hi
for the loop section runif needs curved brackets
Try
IAP -NA
for (i in 1:Sample.Size){
if (DataSet$SES[i]0) {
IAP[i] - ifelse(runif(1)0.75, 1, 0) # High SES, higher
chance to be in
Treatment #
}
else {
IAP[i] - ifelse(runif(1)=0.25, 1, 0) # Low SES, lower
chance to be in
Treatment #
}
} # End loop #
IAP
IAP
zjiaqi19880219 wrote
Hi, all,
I am working on a project to run a simulation. I am now working on the
main body of the simulation body, I generate the data and all work well
before I identify a variable IAP, there is always an error message as
replacement has length zero, and I do not know how to fix it. Can anyone
help?
Here is the code:
library (mvtnorm)
options(digits=2)
# Set variable name #
namelist1 - c(Age, Gender, SES, ISS, IQTe, PreTe)
# Generate Data #
Sample.Size - 10 # Variable Sample Size #
CorMat - matrix(c(1.0 ,0.0 ,0.0 , 0.5, 0.2, 0.5,
0.0, 1.0 ,0.0, 0.0, 0.0, 0.0,
0.0 ,0.0 ,1.0, 0.5, 0.4, 0.5,
0.5, 0.0, 0.5, 1.0, 0.4, 0.7,
0.2, 0.0, 0.4, 0.4, 1.0, 0.9,
0.5, 0.0, 0.5, 0.7, 0.9, 1.0),ncol=6)
# Correlation Matrix #
DataSet - rmvnorm(Sample.Size, mean=c(0,0,0,0,0,0), sigma=CorMat) # Draw
Sample.Size covariate set all mean 0 #
DataSet - as.data.frame(DataSet)
names(DataSet) - namelist1
DataSet
# For Treatment IAP #
IAP - matrix(0, nrow=Sample.Size)
for (i in 1:Sample.Size){
IAP[i] - if (DataSet$SES0)
{
ifelse(runif[1]0.75, 1, 0) # High SES, higher chance to be in
Treatment
#
}
IAP[i] - if (DataSet$SES=0)
{
ifelse(runif[1]=0.25, 1, 0) # Low SES, lower chance to be in
Treatment
#
}
} # End loop #
IAP
Error in IAP[i] - if (DataSet$SES 0) { : replacement has length zero
In addition: Warning message:
In if (DataSet$SES 0) { :
the condition has length 1 and only the first element will be used
Thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/Why-replacement-has-length-zero-And-How-can-I-fix-it-tp4657373p4657380.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do
Re: [R] vectorisation
Hi,
Not sure this helps:
ml - data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm - apply(ml, 2, cumsum)
starts- data.frame(matrix(0,600,4))
starts1- data.frame(matrix(0,600,4))
for (i in 1:4){
starts1[,i][mm[,i]] -1
}
starts2-as.data.frame(do.call(cbind,lapply(1:4,function(i)
{starts[,i][mm[,i]]-1;starts[,i]})))
colnames(starts2)- colnames(starts1)
identical(starts1,starts2)
#[1] TRUE
#loop should be fast compared to lapply
ml - data.frame(matrix(sample(1:1e4,1e3, replace=TRUE),100,4))
mm - apply(ml, 2, cumsum)
starts- data.frame(matrix(0,1e6,4))
starts1- data.frame(matrix(0,1e6,4))
system.time({for (i in 1:4){
starts1[,i][mm[,i]] -1
}})
#user system elapsed
# 0.63 0.30 0.92
system.time(starts2-as.data.frame(do.call(cbind,lapply(1:4,function(i)
{starts[,i][mm[,i]]-1;starts[,i]}
# user system elapsed
# 0.74 0.26 1.03
colnames(starts2)- colnames(starts1)
identical(starts1,starts2)
#[1] TRUE
A.K.
- Original Message -
From: Brett Robinson brett.robin...@7dials.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Saturday, February 2, 2013 11:38 AM
Subject: [R] vectorisation
Hi
I'm trying to set up a simulation problem without resorting to (m)any loops. I
want to set entries in a data frame of zeros ('starts' in the code below) to 1
at certain points and the points have been randomly generated and stored in a
separate data.frame ('sl'), which has the same number of columns.
An example of the procedure is as follows:
ml - data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm - apply(ml, 2, cumsum)
starts- data.frame(matrix(0,600,4))
I can achieve the result I want with a loop:
for (i in 1:4){
lstarts[,i][mm[,i]] -1
}
But as I want to use a large number of columns I would like to do away with the
loop
Can anyone suggest how this might be done?
Thanks in advance
Brett Robinson
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorisation
Hello,
Try the following.
set.seed(4315)
ml - data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm - apply(ml, 2, cumsum)
s2 - starts - data.frame(matrix(0,600,4))
for (i in 1:4){
starts[,i][mm[,i]] - 1
}
s2[] - lapply(seq_len(ncol(mm)), function(i) {s2[,i][mm[,i]] - 1; s2[,i]})
identical(s2, starts) # TRUE
Note that lapply is a loop in disguise.
Hope this helps,
Rui Barradas
Em 02-02-2013 16:38, Brett Robinson escreveu:
Hi
I'm trying to set up a simulation problem without resorting to (m)any loops. I
want to set entries in a data frame of zeros ('starts' in the code below) to 1
at certain points and the points have been randomly generated and stored in a
separate data.frame ('sl'), which has the same number of columns.
An example of the procedure is as follows:
ml - data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm - apply(ml, 2, cumsum)
starts- data.frame(matrix(0,600,4))
I can achieve the result I want with a loop:
for (i in 1:4){
lstarts[,i][mm[,i]] -1
}
But as I want to use a large number of columns I would like to do away with the
loop
Can anyone suggest how this might be done?
Thanks in advance
Brett Robinson
__
This email has been scanned by the Symantec Email Security.cloud service.
For more information please visit http://www.symanteccloud.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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Filter according to the latest data
On Fri, Feb 1, 2013 at 8:05 AM, nalluri pratap pratap_s...@yahoo.co.in wrote: library(sqldf) sqldf(select * from k1 group by No having max(Date)) HAVING is only used to select groups and only works by chance in this example but if the data were to change then it would likely not work. Try this instead. It makes use of an sqlite-specific feature that guarantees that when MAX is used in a GROUP BY that the other columns will be from the same row: sqldf(select ID, No, Change, max(Date) Date from k1 group by No) ID NoChange Date 1 D 111 final 2013-01-12 2 C 123 bug fixed 2013-01-17 -- 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.
[R] Fortan to R
Dear UseRs, How can i connect my FTN95 fortran compiler with R in window 7?Thanks in Advance, Elisa [[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] Nested loop and output help
Dear R help moderator,
My name is Thomas and I am new here. I am wondering if my post violated
some of the rules of posting here. If so can you please let me know what I
have done so that I can repost? I reread the posting guidelines and don't
see anything that I violate. I have been reading a great deal online but
just can figure out how to run the loop correctly.
Thank you kindly,
Tom
On Feb 1, 2013 2:12 PM, staysafe23 staysaf...@gmail.com wrote:
Thank you very much Petr,
I believe I have fixed my inquiry to not use floating points in my cycle
as you pointed out and to use the list structure to keep my results. I am
still at a loss as to how to run the multiple loops. I have tried quite a
few different strategies but my failure seems to illustrate that my
understanding of how the loops will run is nonexistent.
I would like to simultaneously let the following 4 things vary:
z1 - rnorm(ss,mean=400, sd=70) and z2 - rnorm(ss,mean=450, sd=90) by ss
- seq(5,9,by=1) which yields 5 6 7 8 9
r - cc by cc - seq(-0.5,0.5, by=0.25) which yields -0.50 -0.25 0.00 0.25
0.50
dev1 - oo1 by oo1 - seq(-10,10, by=5) which yields -10 -5 0 5 10
dev2 - oo2 by oo2 - seq(0,20, by=5) which yields 0 5 10 15 20
I tried to run the loops that would vary each of these above conditions
with the looped code attached below and failed very badly.
Thank you Petr and all,
Best,
Thomas
###SINGLE RUN
CODE
lll - vector(mode = list, length = 10)
names(lll) - c(mat1, mat2, mat3, mat4, cut1, cut2, out3a,
out3b, out4a, out4b)
z1 - rnorm(20,mean=400, sd=70)
z2 - rnorm(20,mean=450, sd=90)
cor - runif(1,min=0.4, max=0.6)
X - z1
Y = cor*z1+(1-cor)*z2
lll[[mat1]] - cbind(X,Y)
dev1 - sample(-40:40, 1, replace=T)
lll[[cut1]] - mean(X) + dev1
dev2 - sample(12:54, 1, replace=T)
lll[[cut2]] - mean(X) + dev1 + dev2
X2 - X-lll[[cut1]]
Y2 - Y-lll[[cut2]]
c3 - cor(X2,Y2)
lll[[mat2]] -cbind(X2,Y2)
a11 - ifelse( X lll[[cut1]] Y lll[[cut2]], 1, 0)
a12 - ifelse( X lll[[cut1]] Y = lll[[cut2]], 1, 0)
a21 - ifelse( X = lll[[cut1]] Y lll[[cut2]], 1, 0)
a22 - ifelse( X = lll[[cut1]] Y = lll[[cut2]], 1, 0)
lll[[mat3]] -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
lll[[mat4]] -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
lll[[out3a]] - mcnemar.test(lll[[mat3]], correct=FALSE)
lll[[out3b]] - mcnemar.test(lll[[mat3]], correct=TRUE)
lll[[out4a]] - chisq.test(lll[[mat4]], correct = FALSE)
lll[[out4b]] - chisq.test(lll[[mat4]], correct = TRUE)
print(lll)
capture.output(print(lll), file = C:/Chi_Square_fix/temp.txt, append =
TRUE)
##LOOPED
CODE#
lll - vector(mode = list, length = 10)
names(lll) - c(mat1, mat2, mat3, mat4, cut1, cut2, out3a,
out3b, out4a, out4b)
ss - seq(5,9,by=1)
cc - seq(-0.5,0.5, by=0.25)
oo1 - seq(-10,10, by=5)
oo2 - seq(0,20, by=5)
for(i in ss) {
for (j in cc) {
for (k in oo1) {
for (l in oo2) {
z1 - rnorm(ss,mean=400, sd=70)
z2 - rnorm(ss,mean=450, sd=90)
r - cc
X - z1
Y = r*z1+(1-r)*z2
lll[[mat1]] - cbind(X,Y)
dev1 - oo1
lll[[cut1]] - mean(X) + dev1
dev2 - oo2
lll[[cut2]] - mean(X) + dev1 + dev2
X2 - X-lll[[cut1]]
Y2 - Y-lll[[cut2]]
c3 - cor(X2,Y2)
lll[[mat2]] -cbind(X2,Y2)
a11 - ifelse( X lll[[cut1]] Y lll[[cut2]], 1, 0)
a12 - ifelse( X lll[[cut1]] Y = lll[[cut2]], 1, 0)
a21 - ifelse( X = lll[[cut1]] Y lll[[cut2]], 1, 0)
a22 - ifelse( X = lll[[cut1]] Y = lll[[cut2]], 1, 0)
lll[[mat3]] -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
lll[[mat4]] -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
lll[[out3a]] - mcnemar.test(lll[[mat3]], correct=FALSE)
lll[[out3b]] - mcnemar.test(lll[[mat3]], correct=TRUE)
lll[[out4a]] - chisq.test(lll[[mat4]], correct = FALSE)
lll[[out4b]] - chisq.test(lll[[mat4]], correct = TRUE)
print(lll)
capture.output(print(lll), file = C:/Chi_Square_fix/temp.txt, append =
TRUE)
}
}
}
}
On Feb 1, 2013 2:01 AM, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
see inline
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of staysafe23
Sent: Friday, February 01, 2013 1:01 AM
To: r-help@r-project.org
Subject: [R] Nested loop and output help
Hello Everyone,
My name is Thomas and I have been using R for one week. I recently
found your site and have been able to search the archives of posts.
This has given me some great information that has allowed me to craft
an initial design to an inquiry I would like to make into the breakdown
of McNemar's test. I have read an intro to R manual and the posting
guides and hope I am not violating them with this post. If so I will
re-ask my question in the proper format.
I have succeeded in writing a loop to vary one condition of my inquiry
[R] Split xts data set into weeks
Hi I am trying to use split() to split a xts data set into weeks, but the result seems not right. original data is as following: head(xec) Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... and I did the split() as following head(split(x=xec,f=weeks)) [[1]] Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 [[2]] Open HighLow Close mean 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... the 2011-02-28 is the Monday but not being grouped with the other days in that week, which makes the second groups begin from Tuesday (2011-03-01). I want it to be as following [[1]] Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 [[2]] Open HighLow Close mean 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... Could anyone please give some advice? Thanks in advance. Seimizu Joukan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Question: write an R script with help information available to the user
Dear All, I would like to ask a question on how to incorporate into an R script help information for the user. I vaguely recall that I saw some instructions on an R manual, but am not able to figure them out. Hereunder is the basic setting: 1. I finished writing an R script, my_script.r, that is a function (the function is named my_func) for a statistical procedure. This means that the user only needs to: source(directory_of_script\my_script.r), then run, my_func(arg1, arg2, ..), to execute the procedure. 2. I would like to add some lines into my_script.r, so that a user can get information on this file and function by using ?my_func Any help or suggestion is appreciated. Best regards, Chee [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question: write an R script with help information available to the user
The normal expectations of an R user is that useful functions you want to share are in packages, which include help files. There is no way to both avoid the package development process and offer help to the user within R. Read the Writing R Extensions document for the most up-to-date information about this process. RStudio has some nice enhancements for making this process a little quicker, but not so much that you can escape reading the documentation I just mentioned. I happen to like using roxygen2 for building help files, but this just means reading even more documentation. If you are determined to use the source function instead of building a package, then you cannot take advantage of the users' familiarity with the help system. However, you could just put your instructions in comments in your R file and hope they will look there. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Chee Chen chen...@purdue.edu wrote: Dear All, I would like to ask a question on how to incorporate into an R script help information for the user. I vaguely recall that I saw some instructions on an R manual, but am not able to figure them out. Hereunder is the basic setting: 1. I finished writing an R script, my_script.r, that is a function (the function is named my_func) for a statistical procedure. This means that the user only needs to: source(directory_of_script\my_script.r), then run, my_func(arg1, arg2, ..), to execute the procedure. 2. I would like to add some lines into my_script.r, so that a user can get information on this file and function by using ?my_func Any help or suggestion is appreciated. Best regards, Chee [[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] Question: write an R script with help information available to the user
On Sat, Feb 2, 2013 at 6:31 PM, Chee Chen chen...@purdue.edu wrote: Dear All, I would like to ask a question on how to incorporate into an R script help information for the user. I vaguely recall that I saw some instructions on an R manual, but am not able to figure them out. Hereunder is the basic setting: 1. I finished writing an R script, my_script.r, that is a function (the function is named my_func) for a statistical procedure. This means that the user only needs to: source(directory_of_script\my_script.r), then run, my_func(arg1, arg2, ..), to execute the procedure. 2. I would like to add some lines into my_script.r, so that a user can get information on this file and function by using ?my_func You may wish to create a package from your script; however, if you would prefer not then a workaround would be to redefine the ? function in your script to provide some help. For example if you had this line in your script: `?` - function(...) if (match.call()[[2]] == mytopic) cat(whatever\n) else help(...) then this would work: ?mytopic whatever ?dim where the last line launches normal help. If you wish to display a file rather than cat strings to the console then use: file.show -- 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] Split xts data set into weeks
Your example of the problem is not reproducible [1]. This behavior could arise due to small discrepancies in the index values, or from specifying frequency instead of f as the second argument, our perhaps you have found a bug that only your data triggers. Any verification of what your problem is will require a reproducible example. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Seimizu Joukan saim...@gmail.com wrote: Hi I am trying to use split() to split a xts data set into weeks, but the result seems not right. original data is as following: head(xec) Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... and I did the split() as following head(split(x=xec,f=weeks)) [[1]] Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 [[2]] Open HighLow Close mean 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... the 2011-02-28 is the Monday but not being grouped with the other days in that week, which makes the second groups begin from Tuesday (2011-03-01). I want it to be as following [[1]] Open HighLow Close mean 2011-02-28 112.34 113.34 111.96 112.87 112.6275 2011-03-01 112.89 113.71 112.75 112.80 113.0375 2011-03-02 112.75 113.56 112.50 113.54 113.0875 2011-03-03 113.50 115.08 113.10 115.05 114.1825 2011-03-04 115.16 115.97 114.85 115.06 115.2600 [[2]] Open HighLow Close mean 2011-03-07 115.21 115.26 114.55 114.85 114.9675 ... Could anyone please give some advice? Thanks in advance. Seimizu Joukan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Question: write an R script with help information available to the user
A related approach which, if memory serves, was originally in S eons ago, is to define a doc attribute of any function (or object, for that matter) that you wish to document that contains text for documentation and a doc() function of the form: doc - function(obj) cat(attr(obj,doc)) used as: f- function(x) NULL attr(f,doc) - Some text\n\n doc(f) doc(f) Some text This is pretty primitive, but I suppose you could instead have the attribute point to something like an HTML file and the doc() function open it in a web browser, which is basically what R's built-in package document system does anyway. Except you wouldn't have to build a package and don't have to learn or follow R's procedures. Which means you don't get R's standardization and organization and no one but a private bunch of users will be able to use your function. But maybe that's sufficient for your needs. -- Bert On Sat, Feb 2, 2013 at 10:01 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Sat, Feb 2, 2013 at 6:31 PM, Chee Chen chen...@purdue.edu wrote: Dear All, I would like to ask a question on how to incorporate into an R script help information for the user. I vaguely recall that I saw some instructions on an R manual, but am not able to figure them out. Hereunder is the basic setting: 1. I finished writing an R script, my_script.r, that is a function (the function is named my_func) for a statistical procedure. This means that the user only needs to: source(directory_of_script\my_script.r), then run, my_func(arg1, arg2, ..), to execute the procedure. 2. I would like to add some lines into my_script.r, so that a user can get information on this file and function by using ?my_func You may wish to create a package from your script; however, if you would prefer not then a workaround would be to redefine the ? function in your script to provide some help. For example if you had this line in your script: `?` - function(...) if (match.call()[[2]] == mytopic) cat(whatever\n) else help(...) then this would work: ?mytopic whatever ?dim where the last line launches normal help. If you wish to display a file rather than cat strings to the console then use: file.show -- 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem in installing reshape2 (urgent)
Dear R users, In order to install reshape2 package, I gave the following command: install.packages(reshape2,dep=TRUE) However, it is not getting installed. The message that I am getting is: Warning: dependencies plyr, stringr, testthat are not available trying URL ' http://ftp.ctex.org/mirrors/CRAN/src/contrib/reshape2_1.2.2.tar.gz' Content type 'application/x-gzip' length 27220 bytes (26 Kb) opened URL == downloaded 26 Kb * installing *source* package reshape2 ... ** R ** data ** moving datasets to lazyload DB ** inst ** preparing package for lazy loading Error in loadNamespace(i, c(lib.loc, .libPaths())) : there is no package called 'plyr' ERROR: lazy loading failed for package reshape2 * removing /home/dell/R/i486-pc-linux-gnu-library/2.10/reshape2 Please suggest. [[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] problem in installing reshape2 (urgent)
Suggestion: follow instructions appropriate for your OS (see CRAN) to install the current version of R. (2.15.2) --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Partha Saha kuttus...@gmail.com wrote: Dear R users, In order to install reshape2 package, I gave the following command: install.packages(reshape2,dep=TRUE) However, it is not getting installed. The message that I am getting is: Warning: dependencies �plyr�, �stringr�, �testthat� are not available trying URL ' http://ftp.ctex.org/mirrors/CRAN/src/contrib/reshape2_1.2.2.tar.gz' Content type 'application/x-gzip' length 27220 bytes (26 Kb) opened URL == downloaded 26 Kb * installing *source* package �reshape2� ... ** R ** data ** moving datasets to lazyload DB ** inst ** preparing package for lazy loading Error in loadNamespace(i, c(lib.loc, .libPaths())) : there is no package called 'plyr' ERROR: lazy loading failed for package �reshape2� * removing �/home/dell/R/i486-pc-linux-gnu-library/2.10/reshape2� Please suggest. [[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.
