[R] Graphing a piece-wise defined function
Hello. I am new to R, and am typing some homework for my undergrad Analysis class. I am trying to graph the following function in R: f(x) = x^2 for x =0, and f(x) = 0 for x 0. How do I do this in R? Thanks for the help. -- View this message in context: http://n4.nabble.com/Graphing-a-piece-wise-defined-function-tp1584105p1584105.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Average regions of non-zeros
Hi Jim I was following this thread and found that your answer is perfect
there. However I could not comprehend the meaning of the expression
cumsum(x == 0). If I paste it in R window, I get following :
cumsum(x == 0)
[1] 1 2 2 2 2 3 4 4 4 4
I gone through the help page of cumsum() function I correctly understand
that this function calculates the cumulative sum. But could not understand
really the meaning of cumsum(x == 0)
Would you please explain that?
Thanks,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of jim holtman
Sent: 08 March 2010 08:32
To: Daren Tan
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] Average regions of non-zeros
Try this:
x - c(0,0,1,2,3,0,0,4,5,6)
# partition the data
x.p - split(x, cumsum(x == 0))
# now only process groups 1
x.mean - lapply(x.p, function(a){
+ if (length(a) == 1) return(NULL)
+ return(list(grp=tail(a, -1), mean=mean(tail(a, -1
+ })
# now only return the real values
x.mean[unlist(lapply(x.mean, length) != 0)]
$`2`
$`2`$grp
[1] 1 2 3
$`2`$mean
[1] 2
$`4`
$`4`$grp
[1] 4 5 6
$`4`$mean
[1] 5
On Sun, Mar 7, 2010 at 9:48 PM, Daren Tan dare...@hotmail.com wrote:
x - c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and average c(1,2,3) and c(4,5,6)
to get 2 and 5.
Thanks
_
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-guide.html
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Odp: Graphing a piece-wise defined function
Hi r-help-boun...@r-project.org napsal dne 08.03.2010 05:23:25: Hello. I am new to R, and am typing some homework for my undergrad Analysis class. I am trying to graph the following function in R: f(x) = x^2 for x =0, and f(x) = 0 for x 0. How do I do this in R? I would probably use curve function. Regards Petr Thanks for the help. -- View this message in context: http://n4.nabble.com/Graphing-a-piece-wise- defined-function-tp1584105p1584105.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] barplot with factors problem
On 03/08/2010 04:48 AM, casperyc wrote: http://n4.nabble.com/file/n1583733/100307070476876317b486a941.jpg I want to get a histogram by factors. Hi casperyc, Have a look at the third example for the barp function in the plotrix package. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] asdate parsing
Thanks Gabor - sprintf did the trick -- View this message in context: http://n4.nabble.com/as-date-parsing-tp1582868p1584218.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lattice: barchart, error bars and grouped data
Hi,
How can I, given the code snippet below, draw the error bars in the center
of each grouped bar rather than in the center of the group?
Thanks for any hints,
Joh
library(lattice)
barley[[SD]] - 5
barchart(
yield ~ variety | site,
data = barley,
groups=year,
origin=0,
lowDev=barley[[SD]],
highDev=barley[[SD]],
panel = function(
x,
y,
...,
lowDev,
highDev
){
panel.barchart(x, y, ...)
panel.segments(
as.numeric(x),
as.numeric(y) - lowDev,
as.numeric(x),
as.numeric(y) + highDev,
col = 'red', lwd = 2,
...)
}
)
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Re: [R] Redhat Linux Install
Marc Schwartz wrote: On Mar 5, 2010, at 3:45 PM, Ryan Garner wrote: I just installed R on Redhat Linux at work for the first time and have two questions. 1. I tried to install R to have png and cairo capabilities and was unsuccessful. Before running make, I ran ./configure --with-libpng=yes --with-x=no --with-cairo=yes --with-readline-yes . R installed fine, but when I run R and type capabilities() capabilities() jpeg pngtiff tcltk X11 aqua http/ftp sockets TRUE TRUE TRUE TRUEFALSEFALSE TRUE TRUE l ibxmlfifo cledit iconv NLS profmemcairo TRUEFALSE TRUE TRUE TRUE TRUEFALSE Why are png and cairo still FALSE? 2. I would also like to have X11 enabled. From reading the message board, the consesus seems to be to install xorg-dev. I'm unable to do this because I don't have root or super user priveleges. But if I'm able to log into my work servers with PuTTY and Xming and run xemacs or xvim, does this mean that X11 is already installed somewhere? If so, how do I specify this when doing ./configure? There is conflicting information here. You specified --with-x=no, yet you want X. You indicate that you installed R, yet you do not have root access. In order to compile R from source and have the functionality that you seem to want, you will need either have root access to install the required libraries or have the SysAdmin do so. In order to install R using the defaults, you need to have root access or have your SysAdmin do so. The required libraries are the 'dev' or development versions of the RPMs for each of the components such as libpng, cairo, readline and X. These contain the header files (.h) that are required to compile R from source and support these features. These issues are described in the R Installation and Administration manual: http://cran.r-project.org/doc/manuals/R-admin.html The easier option would be to have the SysAdmin simply use the available RPMs for R rather than compiling from source. I presume that by Red Hat Linux, you mean RHEL. You can point your SysAdmin to the EPEL (http://fedoraproject.org/wiki/EPEL) which provides pre-built RPMs for R, installable by using 'yum'. In addition to Marc, CRAN also provides .rpm version of R [1]. It could be that these are newer than the ones on EPEL, but I'm not sure. cheers, Paul [1] http://cran.r-project.org/bin/linux/redhat/ If you can use ssh to login to the server using PuTTY and that supports X as you indicate, then it means that the server has been configured to support 'X forwarding' and that your ssh login is using the '-X' option to request it on your end of the connection. This means that the server supports X, but may or may not have the X related development RPMs installed, which as I note, are required to compile R from source and support X. Xming, on the other hand, I believe provides its own X server implementation, which potentially brings other issues into play. I have not used it, so would defer to others on the details. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +3130 274 3113 Mon-Tue Phone: +3130 253 5773 Wed-Fri http://intamap.geo.uu.nl/~paul __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] plot a distance matrix
Dear friends, I have problems to plot a matrix from an already distance-matrix. I Just want to plot it, the distance calculation were already done, I don' have the original data to re-calculate the distance in R that will be easy. If some one can prove, here goes the data. Thank you Rosa.__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 match vector with a list ?
Thank you for answers.
My code is very slow compared with yours ;-)
#my code
system.time(r0-f0(iBig,jBig))
user system elapsed
82.489 15.060 97.544
#Holtman's code
system.time(r1-f1(iBig,jBig))
user system elapsed
0.100 0.012 0.113
#Dunlap's code
system.time(r2-f2(iBig,jBig))
user system elapsed
0.084 0.004 0.088
2010/3/5 William Dunlap wdun...@tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Carlos Petti
Sent: Friday, March 05, 2010 9:43 AM
To: r-help@r-project.org
Subject: [R] How to match vector with a list ?
Dear list,
I have a vector of characters and a list of two named elements :
i - c(a,a,b,b,b,c,c,d)
j - list(j1 = c(a,c), j2 = c(b,d))
I'm looking for a fast way to obtain a vector with names, as follows :
[1] j1 j1 j2 j2 j2 j1 j1 j2
A request with a such a nice copy-and-pastable
example in it deserves an answer.
It looks to me like you want to map the item names
in i to the group names that are the names of the list j,
which maps group names to the items in each group.
When there are lots of groups it can be faster to
first invert the list j into a mapping vector pair,
as in:
f2 - function (i, j) {
groupNames - rep(names(j), sapply(j, length)) # map to groupName
itemNames - unlist(j, use.names = FALSE) # map from itemName
groupNames[match(i, itemNames, nomatch = NA)]
}
I put your original code into a function, as this makes
testing and development easier:
f0 - function (i, j) {
match - lapply(j, function(x) {
which(i %in% x)
})
k - vector()
for (y in 1:length(match)) {
k[match[[y]]] - names(match[y])
}
k
}
With your original data these give identical results:
identical(f0(i,j), f2(i,j))
[1] TRUE
I made a list describing 1000 groups, each containing
an average of 10 members:
jBig - split(paste(N,1:1,sep=),
sample(paste(G,1:1000,sep=),size=1,replace=TRUE))
and a vector of a million items sampled from the those
member names:
iBig - sample(paste(N,1:1,sep=), replace=TRUE, size=1e6)
Then I compared the times it took f0 and f2 to compute
the result and verified that their outputs were identical:
system.time(r0-f0(iBig,jBig))
user system elapsed
100.89 10.20 111.27
system.time(r2-f2(iBig,jBig))
user system elapsed
0.140.000.14
identical(r0,r2)
[1] TRUE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
I used :
match - lapply(j, function (x) {which(i %in% x)})
k - vector()
for (y in 1:length(match)) {
k[match[[y]]] - names(match[y])}
k
[1] j1 j1 j2 j2 j2 j1 j1 j2
But, I think a better way exists ...
Thanks in advance,
Carlos
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Some hints for the R beginner
Hi Patrick, I've read it quickly and it seems to be a good resource for beginners that have just downloaded R and have no idea what to do. My guess is that it should be a good introduction to other documents such as the An introduction to R. I'll test it with the next students in my team that will have to learn R! One good thing would be to have it as PDF, it's easier to have it always close to you. Regards, Ivan Le 3/7/2010 20:30, Patrick Burns a écrit : There is now a document called Some hints for the R beginner whose purpose is to get people up and running with R as quickly as possible. Direct access to it is: http://www.burns-stat.com/pages/Tutor/hints_R_begin.html JRR Tolkien wrote a story (sans hobbits) called 'Leaf by Niggle' that has always resonated with me. I offer you an imperfect, incomplete tree (but my roof is intact). Suggestions for improvements are encouraged. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unsigned Posts; Was Setting graphical parameters
On 4 March 2010 23:47, Jim Lemon j...@bitwrit.com.au wrote: On 03/05/2010 04:11 AM, Bert Gunter wrote: Folks: Rolf's (appropriate, in my view) response below seems symptomatic of an increasing tendency of posters to hide their identities with pseudonyms and fake headers. While some of this may be due to identity paranoia (which I think is overblown for this list), I suspect that a good chunk of it is lazy students trying to beat the system by having us do their homework. The etiquette of this list has traditionally been, like the software, open: we sign our names. I would urge helpeRs to adhere to this etiquette and ignore unsigned posts. I had a working hypothesis about this, but decided to check the data before replying. Looking at the ten most recent obvious pseudonyms, all were from free email accounts like gmail or yahoo. A few of these included all or part of the name of the user anyway. As such accounts tend to be used for lots of things, the users may well be concerned about identification, even if everyone on the R help list is of the highest moral standing. 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. Personally I am just paranoid. I used my normal email address for the Debian mailing lists a few years ago, then found myself deluged with spam, and bounced emails where my email had been inserted into the from field (these generally originated from Russian or Brazilian domains - if you checked the full headers). Maybe the debian lists weren't the source and it was harvested by other means, but my full email address was viewable on some of the news list archives. I have also suffered some credit card fraud which was fortunately swiftly curtailed, but it means I am inclined to keep aspects of my life out of the online domain, and use this completely separate email address for mailing lists. It also means I can dip in and out of the lists, and not clog up a work or personal email account. R is a tool that I use for part of my job - albeit an excellent tool. I am particularly happy with the graphics that I have managed to produce. -- Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] error_hier.part
Hi everyone, BEGINNER question: I get the error below when running hier.part. Probably i´m doing something wrong. Error in glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, : object 'fit' not found In addition: Warning messages: 1: In glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, : no observations informative at iteration 1 2: In glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, : algorithm did not converge The steps i followed: - read.table to import 9 ascii files with no header - data.frame to join those nine objects (factors) - unlist to turn another imported ascii into vector (dependent variable) Note: novalues in the ascii appear as -. Importing them like that gives factors; if i change that to NaN, data frames result. I know i must be doing something wrong. Can someone please give me some clues on what that is? Thanks in advance Marco [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to match vector with a list ?
Maybe you can create a helper vector first:
helper - structure(names = unlist(j), rep(names(j), sapply(j, length)))
helper
acbd
j1 j1 j2 j2
helper[i]
aabbbccd
j1 j1 j2 j2 j2 j1 j1 j2
On Sat, Mar 6, 2010 at 1:42 AM, Carlos Petti carlos.pe...@gmail.com wrote:
Dear list,
I have a vector of characters and a list of two named elements :
i - c(a,a,b,b,b,c,c,d)
j - list(j1 = c(a,c), j2 = c(b,d))
I'm looking for a fast way to obtain a vector with names, as follows :
[1] j1 j1 j2 j2 j2 j1 j1 j2
I used :
match - lapply(j, function (x) {which(i %in% x)})
k - vector()
for (y in 1:length(match)) {
k[match[[y]]] - names(match[y])}
k
[1] j1 j1 j2 j2 j2 j1 j1 j2
But, I think a better way exists ...
Thanks in advance,
Carlos
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Lattice: barchart, error bars and grouped data
Johannes wrote: How can I, given the code snippet below, draw the error bars in the center of each grouped bar rather than in the center of the group? http://markmail.org/message/oljgimkav2qcdyre Dieter -- View this message in context: http://n4.nabble.com/Lattice-barchart-error-bars-and-grouped-data-tp1584239p1584376.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Graphing a piece-wise defined function
On Sun, 7 Mar 2010 20:23:25 -0800 (PST) thedoctor81877 thedoctor81877 @gmail.com wrote: Hello. I am new to R, and am typing some homework for my undergrad Analysis class. I am trying to graph the following function in R: f(x) = x^2 for x =0, and f(x) = 0 for x 0. How do I do this in R? f=function(x) ifelse(x = 0, x^2, 0) curve(f, -2, 4) -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Interpretation of 'swtich'
Duncan Murdoch wrote: On 07/03/2010 5:26 PM, rkevinbur...@charter.net wrote: Thatnk you. The documentation indicates as you indicated that if there is not an exact match then the next element is chosen. But it does not indicate the case that contains an exact match but there is not value to be returned (=, case). From what you indicate this is treated as if it was not a match. I think the writing is not very clear (and Brian Ripley has improved it in R-devel), but it does intend to say: - If there is an exact match with a value, then that value is returned. Re-reading it, I think my clarification is unclear. The matching occurs to the argument name. I should have written: - If there is an exact match to an argument name with an associated value, then that value is returned. Duncan Murdoch - If there is an exact match with no value, then the next value is returned. - If there is no match, then the 1st unnamed arg (or 2nd if EXPR isn't named) is returned. I think this is different from your interpretation. Duncan Murdoch Kevin Uwe Ligges lig...@statistik.tu-dortmund.de wrote: On 06.03.2010 21:49, rkevinbur...@charter.net wrote: In browsing the source I see the following construct: res- switch(type, working = , response = r, deviance = , pearson = if (is.null(object$weights)) r else r * sqrt(object$weights), partial = r) I understand that 'switch' will execute the code that is matched by its corresponding string value (in this case 'type'). What I don't understand is the empty code. Is this code saying that if the type is deviance then fill the 'res' variable with an empty value? From my naive point of view it seems that 'res' will only get a value(s) if 'type' is 'response', 'pearson', or 'partial'. Please help with my understanding. Please do read the help pages! From ?switch: If there is an exact match then that element is evaluated and returned if there is one, otherwise the next element is chosen, [...] Example: switch(A, A=1, B=, C=2) # 1 switch(B, A=1, B=, C=2) # 2 Uwe Ligges Kevin Burton rkevinbur...@charter.net __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] POSIXct type lost
It appears that I am creating a matrix where als columns are of type number, so my Date column has been converted to a number. Is there a way to show or display this number column as a Date again? -- View this message in context: http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584410.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] POSIXct type lost
I am generating a column of dates using POSIXct, but when I try to assign it to an existing dataframe - it gets stored as numbers instead of as POSIXct. Is there a way to force a column to be a specific type (POSIXct)? Thanks, Moon -- View this message in context: http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584379.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] POSIXct type lost
Add the number to the POSIXct origin. See table at end of R News 4/1. On Mon, Mar 8, 2010 at 7:11 AM, ManInMoon xmoon2...@googlemail.com wrote: It appears that I am creating a matrix where als columns are of type number, so my Date column has been converted to a number. Is there a way to show or display this number column as a Date again? -- View this message in context: http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584410.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Average regions of non-zeros
What I was looking for was the string of non-zero values and where they
'broke' at. I could have used 'rle', but I sometime find this approach just
as easy. Every place there is a zero will be TRUE which has the value 1.
'cumsum' will generate a running sum of these values. When there is a
non-zero value, you will get consecutive values of cumsum to be the same.
The is what you saw in pasting the value into the window. Notice that the
run of '2's begins with a value of zero and then includes all the non-zero
values following. By using 'split', I cn create a list of each group. If
the group is of length 1, then it only contains zero and I ignore it. If
the length is greater than 1, then we have some non-zero values and we have
to throw away the leading zero in the group (tail(a, -1)) and then take the
mean.
HTH
On Mon, Mar 8, 2010 at 3:26 AM, bogaso.christofer
bogaso.christo...@gmail.com wrote:
Hi Jim I was following this thread and found that your answer is perfect
there. However I could not comprehend the meaning of the expression
cumsum(x == 0). If I paste it in R window, I get following :
cumsum(x == 0)
[1] 1 2 2 2 2 3 4 4 4 4
I gone through the help page of cumsum() function I correctly understand
that this function calculates the cumulative sum. But could not understand
really the meaning of cumsum(x == 0)
Would you please explain that?
Thanks,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of jim holtman
Sent: 08 March 2010 08:32
To: Daren Tan
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] Average regions of non-zeros
Try this:
x - c(0,0,1,2,3,0,0,4,5,6)
# partition the data
x.p - split(x, cumsum(x == 0))
# now only process groups 1
x.mean - lapply(x.p, function(a){
+ if (length(a) == 1) return(NULL)
+ return(list(grp=tail(a, -1), mean=mean(tail(a, -1
+ })
# now only return the real values
x.mean[unlist(lapply(x.mean, length) != 0)]
$`2`
$`2`$grp
[1] 1 2 3
$`2`$mean
[1] 2
$`4`
$`4`$grp
[1] 4 5 6
$`4`$mean
[1] 5
On Sun, Mar 7, 2010 at 9:48 PM, Daren Tan dare...@hotmail.com wrote:
x - c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and average c(1,2,3) and
c(4,5,6)
to get 2 and 5.
Thanks
_
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Cincinnati, OH
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What is the problem that you are trying to solve?
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Speed up sparse matrices
Dear R, I have three matrices like this K: pp-by-pp commutation matrix, I: p-by-p diagonal matrix, X: p-by-q dense matrix, and I wish to calculate K(IoX) where `o' denotes Kronecker product. Can you give me any suggestion to speed it up when `p' and `q' are large? Thanks in advance. Feng -- Feng Li Department of Statistics Stockholm University 106 91 Stockholm, Sweden http://feng.li/ [[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] Average regions of non-zeros
Nice shot of cumsum(). Just improve it a little:
x - c(0,0,1,2,3,0,0,4,5,6)
x.groups - split(x, (x != 0) * cumsum(x == 0))[-1]
x.groups
$`2`
[1] 1 2 3
$`4`
[1] 4 5 6
lapply(x.groups, mean)
$`2`
[1] 2
$`4`
[1] 5
On Mon, Mar 8, 2010 at 11:02 AM, jim holtman jholt...@gmail.com wrote:
Try this:
x - c(0,0,1,2,3,0,0,4,5,6)
# partition the data
x.p - split(x, cumsum(x == 0))
# now only process groups 1
x.mean - lapply(x.p, function(a){
+ if (length(a) == 1) return(NULL)
+ return(list(grp=tail(a, -1), mean=mean(tail(a, -1
+ })
# now only return the real values
x.mean[unlist(lapply(x.mean, length) != 0)]
$`2`
$`2`$grp
[1] 1 2 3
$`2`$mean
[1] 2
$`4`
$`4`$grp
[1] 4 5 6
$`4`$mean
[1] 5
On Sun, Mar 7, 2010 at 9:48 PM, Daren Tan dare...@hotmail.com wrote:
x - c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and average c(1,2,3) and c(4,5,6)
to get 2 and 5.
Thanks
_
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] black cluster in salt and pepper image
Hi, on a lattice, I have binary 0/1 data. 1s are rare and may form clusters. I would like to know the size/length of largest cluster. Any help warmly welcome, Sylvain. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] arbitrary scaling
Hi all
I know I probably reinvented wheel but it was maybe simpler then search in
docs or ask help before I did my part.
I made a simple function which can scale a vector between chosen values.
Do anybody know simpler/better approach?
myscale-function(x, miny=0.5, maxy=1) {
rx - diff(range(x, na.rm=T))
minx - min(x, na.rm=T)
tga - (maxy-miny)/rx
b - miny - tga* minx
res - x*tga+b
res
}
x - c(5,30,50)
myscale(x)
[1] 0.500 0.778 1.000
Thank you
Regards
Petr
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Re: [R] scientific (statistical) foundation for Y-RANDOMIZATION in regression analysis
That sounds like a particular form of permutation test. If the
scrambling is replaced by sampling with replacement (i.e., some data
points can be sampled more than once while others can be left out),
that's the simple (or nonparametric) bootstrap. The goal is to generate
the distribution of the statistic of interest (R^2 or q^2) under the
null hypothesis that there's no relationship between the activity (or
property) and the structure.
To make the test valid, one needs to ensure that the entire model
building process is carried through for all of the sampled data,
including feature selections, etc.
Andy
From: Damjan Krstajic
Dear all,
I am a statistician doing research in QSAR, building
regression models where the dependent variable is a numerical
expression of some chemical activity and input variables are
chemical descriptors, e.g. molecular weight, number of carbon
atoms, etc.
I am building regression models and I am confronted with a
widely a technique called Y-RANDOMIZATION for which I have
difficulties in finding references in general statistical
literature regarding regression analysis. I would be grateful
if someone could point me to papers/literature in statistical
regression analysis which give scientific (statistical)
foundation for using Y-RANDOMIZATION.
Y-RANDOMIZATION is a widely used technique in QSAR community
to unsure the robustness of a QSPR (regression) model. It is
used after the best regression model is selected and to
make sure that there are no chance correlations. Here is a
short description. The dependent variable vector (Y-vector)
is randomly shuffled and a new QSPR (regression) model is
fitted using the original independent variable matrix. By
repeating this a number of times, say 100 times, one will get
hundred R2 and q2 (leave one out cross-validation R2) based
on hundred shuffled Y. It is expected that the resulting
regression models should generally have low R2 and low q2
values. However, if the majority of hundred regression models
obtained in the Y-randomization have relatively high R2 and
high q2 then it implies that an acceptable regression model
cannot be obtained for the given data set by the current
modelling method.
I cannot find any references to Y-randomization or
Y-scrambling anywhere in the literature outside
chemometrics/QSAR. Any links or references would be much appreciated.
Thanks in advance.
DK
--
Damjan Krstajic
Director
Research Centre for Cheminformatics
Belgrade, Serbia
--
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Re: [R] How can I understand this sentenc e,and express it by means of Mathema tical approach？
If your ultimate interest is in real scientific progress, I'd suggest that you ignore that sentence (and any conclusion drawn subsequent to it). Cheers, Andy From: bbslover This topic refer to independent variables reduction, as we know ,a lot of method can do with it,however, for pre-processing independent varibles, a method like the sentence below can reduce many variable, How can I understand it? what is significant correlation at 5% level, what is the criterion？ P value？or what？ Independent variables whose correlation with the response variable was not significant at 5% level were removed how can I calucate the correlation between them? thank you! -- View this message in context: http://n4.nabble.com/How-can-I-understand-this-sentence-and-ex press-it-by-means-of-Mathematical-approach- tp1584036p1584036.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] arbitrary scaling
Perhaps approx:
approx(range(x), c(0.5, 1), xout = x)$y
A one-linear, but longer, is also possible based on lm:
predict(lm(c(0.5, 1) ~ x, data.frame(x = range(x))), data.frame(x))
On Mon, Mar 8, 2010 at 9:37 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi all
I know I probably reinvented wheel but it was maybe simpler then search in
docs or ask help before I did my part.
I made a simple function which can scale a vector between chosen values.
Do anybody know simpler/better approach?
myscale-function(x, miny=0.5, maxy=1) {
rx - diff(range(x, na.rm=T))
minx - min(x, na.rm=T)
tga - (maxy-miny)/rx
b - miny - tga* minx
res - x*tga+b
res
}
x - c(5,30,50)
myscale(x)
[1] 0.500 0.778 1.000
Thank you
Regards
Petr
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Re: [R] r code to generate interaction columns
thanks Kieth. I wanted something generic code to check column data type and loop through and create the interaction columns automatically as I want to test this out as a new algorithm for data mining. Traditional regression may give misleading results with multi-collinearity and thus I wanted to take interaction terms and run them through random forests and rpart as they would need interaction terms to be manually created. Hope that clarifies. Dhruv -Original Message- From: kMan [mailto:kchambe...@gmail.com] Sent: Sunday, March 07, 2010 8:08 PM To: Sharma, Dhruv; r-help@r-project.org Subject: RE: [R] r code to generate interaction columns Dear Dhruv, You could create interaction variables manually (assuming A is your dependent variable). Just multiply the variables together. cd.int-C*D ce.int-C*E cde.int-C*D*E # what about D*E, or interactions with B? Include those in your model, such as A~B+C+D+E+cd.int+cd.int+ce.int+cde.int. Then you can compare those models to the results you get when you specify the interaction in the model formula directly using the documented syntax. In your R-console, type ?formula, or help(formula) for details. Sincerely, KeithC. -Original Message- From: Sharma, Dhruv [mailto:dhruv.sha...@penfed.org] Sent: Saturday, March 06, 2010 10:30 AM To: r-help@r-project.org Subject: [R] r code to generate interaction columns Hi, is there a way to take a dataset and extract numeric columns and create interaction columns from it automatically? For e.g. there are 5 columns of data: A,B,C,D,E. CDE are numeric. Can someone provide code to automatically create more columns such as: 1) C*D, C*E, C*D*E, (C+E)/(D+.01 (to avoid divide by zero), (D+E)/(C+.01 (to avoid divide by zero), (C+D)/(E+.01 (to avoid divide by zero)) ? I know in glm multiplying can create terms but i want the columns to be part of the data set so that i can feed this into Random forest to pick out predictive interaction terms as regression cannot reliably handle correlated interaction terms. if anyone has some simple code that can do this that would be helpful. thanks Dhruv [[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] Is there an equivalence of lm's anova for an rpart object ?
One way to do it (no p-values) is explained in the original CART book.
You basically add up all the improvement (in fit$split[, improve])
due to each splitting variable.
Andy
From: Tal Galili
Simple example:
# Classification Tree with rpart
library(rpart)
# grow tree
fit - rpart(Kyphosis ~ Age + Number + Start,
method=class, data=kyphosis)
Now I would like to know how can I measure the importance
of each of my
three explanatory variables (Age, Number, Start) in the model?
If this was a regression model, I could have looked at p
values from the
anova F test (between lm models with and without the
variable). But what
is the equivalence of using anova on lm to an rpart object ?
Any pointers, insights and references to this question will
be helpful.
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
(Hebrew) |
www.r-statistics.com (English)
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Re: [R] arbitrary scaling
Thanks
I would never deduct it out from the help page of approx.
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.03.2010 15:47:37:
Perhaps approx:
approx(range(x), c(0.5, 1), xout = x)$y
A one-linear, but longer, is also possible based on lm:
predict(lm(c(0.5, 1) ~ x, data.frame(x = range(x))), data.frame(x))
On Mon, Mar 8, 2010 at 9:37 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi all
I know I probably reinvented wheel but it was maybe simpler then
search in
docs or ask help before I did my part.
I made a simple function which can scale a vector between chosen
values.
Do anybody know simpler/better approach?
myscale-function(x, miny=0.5, maxy=1) {
rx - diff(range(x, na.rm=T))
minx - min(x, na.rm=T)
tga - (maxy-miny)/rx
b - miny - tga* minx
res - x*tga+b
res
}
x - c(5,30,50)
myscale(x)
[1] 0.500 0.778 1.000
Thank you
Regards
Petr
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[R] setClass or setValidity?
Hi, I'm reading up on S4 classes *). There seem to be at least two ways of input validation: setClass() (using the 'validity' argument) and setValidity(). Is it a matter of taste which function is used? Or should more complex validation code better be put in a setValiditity call? *) A (Not So) Short Introduction to S4 Object Oriented Programming in R V0.5.1 Christophe Genolini August 20, 2008 And, inside those validity checks, is most of the checking done with 'if' 'else' computations, or is it also common to use except()? Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ [[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] setClass or setValidity?
Sorry: there was an error in the last sentence: And, inside those validity checks, is most of the checking done with 'if' 'else' computations, or is it also common to use try()? Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ --- On Mon, 3/8/10, Albert-Jan Roskam fo...@yahoo.com wrote: From: Albert-Jan Roskam fo...@yahoo.com Subject: [R] setClass or setValidity? To: r-help@r-project.org Date: Monday, March 8, 2010, 4:14 PM Hi, I'm reading up on S4 classes *). There seem to be at least two ways of input validation: setClass() (using the 'validity' argument) and setValidity(). Is it a matter of taste which function is used? Or should more complex validation code better be put in a setValiditity call? *) A (Not So) Short Introduction to S4 Object Oriented Programming in R V0.5.1 Christophe Genolini August 20, 2008 And, inside those validity checks, is most of the checking done with 'if' 'else' computations, or is it also common to use except()? Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ [[alternative HTML version deleted]] -Inline Attachment Follows- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Data.frame issue (pls help)
Hi:
I want to obtain a particular value from a data.frame. Following is my
dataframe:
Quotes
BID ASK
Name
CT2 GOVT99.9296999.9375 CT2
TUM0 COMDTY 108.53125 108.5469TUM0
CT5 GOVT100.10156 100.1094GT5
FVM0 COMDTY 115.56250 115.5703FVM0
TYM0 COMDTY 116.93750 116.9531TYM0
If I try to run: QuoteTUM0BID = Quotes[Quotes$Name %in% TUM0, BID]
and print QuoteTUM0BID, I get 108.5312, instead of 108.53125 as an
answer. Please let me know why is it ignoring the last digit.
Additional Information.
If I run QuoteBID = Quotes[, BID], I get the whole array in which TUM0
BID is 108.53125 (a correct number).
Thanks in advance...
Rgds,
Brijesh
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[R] using sprintf to pass a variable to a RMySQL query
Hello,
I am using RmySQL and would like to iterate through a few queries.
I would like to use sprintf but I think I'm having problems mixing and
matching the sprintf syntax and the SQL regex.
I have checked my sqlcmd and it works when I wan to match %MG1% but how
do I iterate for i 1-72? Escape characters,?
thanks in advance
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id LIKE
'%MG%s%' ,i)
= Here is my vague error message
Error: unexpected input in:
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Re: [R] POSIXct type lost
What is R News 4/1? -- View this message in context: http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584464.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fit a gamma pdf using Residual Sum-of-Squares
Hi all, I would like to fit a gamma pdf to my data using the method of RSS (Residual Sum-of-Squares). Here are the data: x - c(86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174) y - c(2, 5, 10, 17, 26, 60, 94, 128, 137, 128, 77, 68, 65, 60, 51, 26, 17, 9, 5, 2, 3, 7, 3) I have typed the following code, using nls method: fit - nls(y ~ (1/((s^a)*gamma(a))*x^(a-1)*exp(-x/s)), start = c(s=3, a=75, x=86)) But I have the following message error (sorry, this is in German): Fehler in qr(.swts * attr(rhs, gradient)) : Dimensionen [Produkt 3] passen nicht zur Länge des Objektes [23] Zusätzlich: Warnmeldung: In .swts * attr(rhs, gradient) : Länge des längeren Objektes ist kein Vielfaches der Länge des kürzeren Objektes Could anyone help me with the code? I would greatly appreciate it. Sincerely yours, Vincent Laperrière. [[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] (box-) plot annotation: italic within paste?
Dear R users,
in the example below the name of the genus will be displayed in the main
titles using the variable predictor[i] and paste.
I would like to have the genus name in italic. However all my attempts
using expression and substitute failed.
Does anybody know a solution?
Thanks a lot in advance. bernd
Acrobeles -c(65.1,0.0,0.0,0.0,0.0,0.0)
Acrobeloides -c(0.0,9.8,76.7,51.1,93.9,43.9)
Alaimus-c(0.0,4.9,0.0,0.0,0.0,6.3)
Aphelenchoides -c(126.5,29.3,76.7,134.1,176.7,87.9)
x-data.frame(Acrobeles,Acrobeloides,Alaimus,Aphelenchoides)
predictor - colnames(x)
ylabel -Numerical abundance
mainlabel1 -Boxplot for
mainlabel2 -sp.
cexalabel -1.8 # axis label
cexmlabel -1.6 # main label
par(oma=c(6,6,3,3),mar = c(6, 4, 4, 2) + 0.1,mfrow=c(2,2))
for (i in 1:ncol(x)){
boxplot(x[,i],
main=paste(mainlabel,predictor[i],mainlabel2),ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
}
---
Bernd Panassiti
National Institute of Public Health the Environment (RIVM)
Laboratory for Ecological Risk Assessment (LER)
P.O. Box 1
3720 BA Bilthoven
The Netherlands
e-mail: bernd.panass...@rivm.nl
tel. +31 30 274 3647
Radboud University Nijmegen
Department of Environmental Science
b.panass...@science.ru.nl
Disclaimer RIVM
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[R] tcltk
Hi, I'm trying to install tcltk in R-2.10.1, however I get error. someone can help? thanks Vasco __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] POSIXct type lost
Try google. On Mon, Mar 8, 2010 at 8:09 AM, ManInMoon xmoon2...@googlemail.com wrote: What is R News 4/1? -- View this message in context: http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584464.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] combinations and table selection problem
Dear all, I have a table like this: a - read.csv(test.csv, header = TRUE, sep = ;) a UTM pUrb pUrb_class pAgri pAgri_class pNatFor pNatFor_class 1 NF188520.160307 NA 79.921386NA 0.00 NA 2 NF188651.965649 NA 46.657713NA 0.00 NA 3 NF189326.009581 NA 40.269204NA 0.00 NA 4 NF18943.141484 NA 0.00 NA 0.00 NA 5 NF189564.296826 NA 0.440691 NA 0.00 NA 6 NF189614.174068 NA 25.613839NA 0.00 NA 7 NF189740.985589 NA 37.680521NA 0.00 NA 8 NF189834.054325 NA 66.027334NA 0.00 NA 9 NF189920.657632 NA 79.424024NA 0.00 NA 10 NF198294.857605 NA 45.368606NA 0.00 NA ... And I executed the following code: #data classification# a$pUrb_class-cut(a$pUrb, c(-Inf,80,Inf), labels = c(0,1)) a$pAgri_class-cut(a$pAgri, c(-Inf,80,Inf), labels = c(0,1)) a$pNatFor_class-cut(a$pNatFor, c(-Inf,80,Inf), labels = c(0,1)) a UTM pUrb pUrb_class pAgri pAgri_class pNatFor pNatFor_class 1 NF188520.160307 079.9213860 0.00 0 2 NF188651.965649 046.6577130 0.00 0 3 NF189326.009581 040.2692040 0.00 0 4 NF18943.141484 0 0.00 0 0.00 0 5 NF189564.296826 0 0.440691 0 0.00 0 6 NF189614.174068 025.6138390 0.00 0 7 NF189740.985589 037.6805210 0.00 0 8 NF189834.054325 066.0273340 0.00 0 9 NF189920.657632 079.4240240 0.00 0 10 NF198294.857605 145.3686060 0.00 0 ... #obtaining the number of combinations present in the data base# library(survival) b-strata(a$pUrb_class,a$pAgri_class,a$pNatFor_class, sep=,) table(b) b a$pUrb_class=0,a$pAgri_class=0,a$pNatFor_class=0 17698 a$pUrb_class=0,a$pAgri_class=0,a$pNatFor_class=1 112 a$pUrb_class=0,a$pAgri_class=1,a$pNatFor_class=0 4360 a$pUrb_class=1,a$pAgri_class=0,a$pNatFor_class=0 160 median(table(b)) [1] 2260 In this stage I have 3 questions: 1st: how can I obtain the combinations witch are present over the median (in this case the first and the second combination)? 2nd: how can I obtain the combinations witch are present over the median and have at least one condition present (in this case only the second combination)? 3rd: how can I select/extract from the original table the rows witch comply with the 2nd question, in this case: UTM pUrb pUrb_class pAgri pAgri_class pNatFor pNatFor_class 10 NF198294.857605 145.3686060 0.00 0 ... Thanks in advance, Carlos Guerra __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] tcltk
I doubt it. Guess why? Dr. Rubén Roa-Ureta AZTI - Tecnalia / Marine Research Unit Txatxarramendi Ugartea z/g 48395 Sukarrieta (Bizkaia) SPAIN -Mensaje original- De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] En nombre de Vasco Cadavez Enviado el: lunes, 08 de marzo de 2010 15:46 Para: r-help@r-project.org Asunto: [R] tcltk Hi, I'm trying to install tcltk in R-2.10.1, however I get error. someone can help? thanks Vasco __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Data.frame issue (pls help)
Hi,
I cannot really test since your dataframe is completely distorted, but
what happens if you try: format(QuoteBID, nsmall=5)?
I think it's just a matter of printing, which uses the number of digits
from options(digits=).
See: ?options, ?format, etc.
But I'm not an expert and cannot really test for this.
HTH
Ivan
Le 3/8/2010 14:15, GULATI, BRIJESH (Global Markets FFO NY) a écrit :
Hi:
I want to obtain a particular value from a data.frame. Following is my
dataframe:
Quotes
BID ASK
Name
CT2 GOVT99.9296999.9375 CT2
TUM0 COMDTY 108.53125 108.5469TUM0
CT5 GOVT100.10156 100.1094GT5
FVM0 COMDTY 115.56250 115.5703FVM0
TYM0 COMDTY 116.93750 116.9531TYM0
If I try to run: QuoteTUM0BID = Quotes[Quotes$Name %in% TUM0, BID]
and print QuoteTUM0BID, I get 108.5312, instead of 108.53125 as an
answer. Please let me know why is it ignoring the last digit.
Additional Information.
If I run QuoteBID = Quotes[, BID], I get the whole array in which TUM0
BID is 108.53125 (a correct number).
Thanks in advance...
Rgds,
Brijesh
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--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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Re: [R] [help] deleting rows which contain more than 2 NAs or zeros
If the data is a dataframe or matrix 'd': d - d[apply(d, 1, function(v) sum( is.na(v) ) = 2 sum(v==0, na.rm=T) = 2 ), ] which can be deconstructed as follows: i1 - apply(d, 1, function(v) sum(is.na(v)) = 2 ) ## true for rows with 2 or fewer na's i2 - apply(d, 1, function(v) sum( v == 0, na.rm=T ) = 2 ##true for rows with 2 or fewer 0's i1 i2 ##logical vector, true for rows satisfying both conditions d[ i1 i2, ] ##only those rows satisfying the condition, and all columns -- View this message in context: http://n4.nabble.com/help-deleting-rows-which-contain-more-than-2-NAs-or-zeros-tp1584613p1584641.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] quickest way convert 1-col df to vector?
anything shorter than as.vector(as.matrix( df ) )? -- View this message in context: http://n4.nabble.com/quickest-way-convert-1-col-df-to-vector-tp1584646p1584646.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (box-) plot annotation: italic within paste?
Hello,
Try this way (not sure if it's the best way, but it works):
boxplot(x[,i],
main=substitute(expression(paste(a, ,italic(b),
,c)),list(a=mainlabel1,b=predictor[i],c=mainlabel2)),
ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
Best,
Miguel
On Mon, Mar 8, 2010 at 2:27 PM, Bernd Panassiti bernd.panass...@rivm.nlwrote:
Dear R users,
in the example below the name of the genus will be displayed in the main
titles using the variable predictor[i] and paste.
I would like to have the genus name in italic. However all my attempts
using expression and substitute failed.
Does anybody know a solution?
Thanks a lot in advance. bernd
Acrobeles -c(65.1,0.0,0.0,0.0,0.0,0.0)
Acrobeloides -c(0.0,9.8,76.7,51.1,93.9,43.9)
Alaimus-c(0.0,4.9,0.0,0.0,0.0,6.3)
Aphelenchoides -c(126.5,29.3,76.7,134.1,176.7,87.9)
x-data.frame(Acrobeles,Acrobeloides,Alaimus,Aphelenchoides)
predictor - colnames(x)
ylabel -Numerical abundance
mainlabel1 -Boxplot for
mainlabel2 -sp.
cexalabel -1.8 # axis label
cexmlabel -1.6 # main label
par(oma=c(6,6,3,3),mar = c(6, 4, 4, 2) + 0.1,mfrow=c(2,2))
for (i in 1:ncol(x)){
boxplot(x[,i],
main=paste(mainlabel,predictor[i],mainlabel2),ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
}
---
Bernd Panassiti
National Institute of Public Health the Environment (RIVM)
Laboratory for Ecological Risk Assessment (LER)
P.O. Box 1
3720 BA Bilthoven
The Netherlands
e-mail: bernd.panass...@rivm.nl
tel. +31 30 274 3647
Radboud University Nijmegen
Department of Environmental Science
b.panass...@science.ru.nl
Disclaimer RIVM
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Re: [R] quickest way convert 1-col df to vector?
sjaffe wrote: anything shorter than as.vector(as.matrix( df ) )? df[[1]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] quickest way convert 1-col df to vector?
D'oh -- thanks! I'm always forgetting the double-bracket extractor... -Original Message- From: Erik Iverson [mailto:er...@ccbr.umn.edu] Sent: Monday, March 08, 2010 10:50 AM To: Steve Jaffe Cc: r-help@r-project.org Subject: Re: [R] quickest way convert 1-col df to vector? sjaffe wrote: anything shorter than as.vector(as.matrix( df ) )? df[[1]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using sprintf to pass a variable to a RMySQL query
Try this:
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id LIKE
\'%%MG%d%%\'' ,i)
sqlcmd_ScaffLen
Your problem:
1. Need %% to create % when using sprintf
2. Need to use %d and not %s for integer values
3. Need to escape the quote marks.
On Mon, Mar 8, 2010 at 8:06 AM, alison waller alison.wal...@embl.de wrote:
Hello,
I am using RmySQL and would like to iterate through a few queries.
I would like to use sprintf but I think I'm having problems mixing and
matching the sprintf syntax and the SQL regex.
I have checked my sqlcmd and it works when I wan to match %MG1% but how
do I iterate for i 1-72? Escape characters,?
thanks in advance
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id
LIKE
'%MG%s%' ,i)
= Here is my vague error message
Error: unexpected input in:
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] using sprintf to pass a variable to a RMySQL query
Another possibility is to use fn$ in the gsubfn package. Just preface
any command with fn$ to enable a quasi-perl-like string interpolation.
In this example $i is replaced with 1:
library(gsubfn)
library(sqldf)
i - 1
fn$sqldf(select count(*) from CO2 where Plant like '%n$i%')
count(*)
1 14
# as seen here:
fn$identity(select count(*) from CO2 where Plant like '%n$i%')
[1] select count(*) from CO2 where Plant like '%n1%'
See http://gsubfn.googlecode.com for more.
On Mon, Mar 8, 2010 at 11:08 AM, jim holtman jholt...@gmail.com wrote:
Try this:
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id LIKE
\'%%MG%d%%\'' ,i)
sqlcmd_ScaffLen
Your problem:
1. Need %% to create % when using sprintf
2. Need to use %d and not %s for integer values
3. Need to escape the quote marks.
On Mon, Mar 8, 2010 at 8:06 AM, alison waller alison.wal...@embl.de wrote:
Hello,
I am using RmySQL and would like to iterate through a few queries.
I would like to use sprintf but I think I'm having problems mixing and
matching the sprintf syntax and the SQL regex.
I have checked my sqlcmd and it works when I wan to match %MG1% but how
do I iterate for i 1-72? Escape characters,?
thanks in advance
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id
LIKE
'%MG%s%' ,i)
= Here is my vague error message
Error: unexpected input in:
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Help with Hmisc, cut2, split and quantile
Hello, I have a set of data with two columns: Target and Actual. A http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is attached but the data looks like this: Actual Target -0.125 0.016124906 0.135 0.120799865 ... ... ... ... I want to be able to break the data into tables based on quantiles in the Target column. I can see (using cut2, and also quantile) how to get the barrier points between the different quantiles, and I can see how I would achieve this if I was just looking to split up a vector. However I am trying to break up the whole table based on those quantiles, not just the vector. The following code shows me the ranges for the deciles of the Target data: library(Hmisc) read_data=read.table(C:/Sample table.txt, head = T) table(cut2(Read_data$Target,g=10)) However I would like to be able to break the table into ten separate tables, each with both Actual and Target data, based on the Target data deciles: top_decile = ...(top decile of read_data, based on Target data) next_decile = ...and so on... bottom_decile = ... That way I could manipulate the deciles, graph them separately (and together) and so on, just as easily as I can the whole table. I'm sure this must be simple, but I can't see the way forward. I have also looked at split() and quantile() but have not been able to get them to achieve what I am after. Can anybody see a simple way foward on this? Thanks, Guy -- View this message in context: http://n4.nabble.com/Help-with-Hmisc-cut2-split-and-quantile-tp1584647p1584647.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [help] deleting rows which contain more than 2 NAs or zeros
Hello. I have just started learning how to work with R program but I have encountered a problem. I can't think up how to remove the rows which contain two (2) or more NA or Zero (0). I would be glad if you could help me because I just have some basic knowledge so far and I even haven't mastered all the basics yet as well. Thanks in advance. -- View this message in context: http://n4.nabble.com/help-deleting-rows-which-contain-more-than-2-NAs-or-zeros-tp1584613p1584613.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] compare tables
Hi!
I need some help to finish my script.
I have two tables that I combine randomly to produce a third table.
This I do for hundreds of iterations. In the output file I get all the
simulated tables after each other. It looks like this (in this case 3
iterations):
output file:
[[1]]
[,1] [,2] [,3]
[1,] GM030005 WI920024
[2,] GM930026 WI920362
[3,] GM980051 WI920007 CGCC
[4,] GM970009 WI920417
[5,] GM920089 WI920023
[6,] GM930109 WI920359
[7,] GM980007 WI920428 CGCC
[8,] GM940039 WI920430
[9,] GM990027 WI920349
[10,] GM920222 WI920410 CGCC
[11,] GM930029 WI920001 CGCC
[12,] GM990105 WI920431
[13,] GM050009 WI920430
[14,] GM920224 WI920369
[15,] GM920224 WI920352
[16,] GM960028 WI920427
[17,] GM940031 WI920004
[18,] GM930040 WI920441
[19,] GM930040 WI920441
[20,] GM050099 WI920417
[21,] GM050099 WI920423 CCCG
[22,] GM920096 WI920370
[23,] GM920034 WI920437
[24,] GM960023 WI920017
[25,] GM920031 WI920430
[26,] GM920202 WI920367 CCCG
[27,] GM990066 WI920410
[[2]]
[,1] [,2] [,3]
[1,] GM030005 WI920017
[2,] GM930026 WI920415
[3,] GM980051 WI920028 CGCC
[4,] GM970009 WI920017
[5,] GM920089 WI920028
[6,] GM930109 WI920353
[7,] GM980007 WI920009 CGCT
[8,] GM940039 WI920415
[9,] GM990027 WI920423 CCCG
[10,] GM920222 WI920423 CGCG
[11,] GM930029 WI920363 CGCC
[12,] GM990105 WI920362
[13,] GM050009 WI920365
[14,] GM920224 WI920362
[15,] GM920224 WI920410
[16,] GM960028 WI920355 CCCG
[17,] GM940031 WI920361
[18,] GM930040 WI920356
[19,] GM930040 WI920353
[20,] GM050099 WI920360
[21,] GM050099 WI920353
[22,] GM920096 WI920023
[23,] GM920034 WI920426
[24,] GM960023 WI920024
[25,] GM920031 WI920022
[26,] GM920202 WI920009 CCCG
[27,] GM990066 WI920001
[[3]]
[,1] [,2] [,3]
[1,] GM030005 WI920433
[2,] GM930026 WI920408
[3,] GM980051 WI920352 CGCC
[4,] GM970009 WI920416
[5,] GM920089 WI920022
[6,] GM930109 WI920369
[7,] GM980007 WI920415 CGCC
[8,] GM940039 WI920022
[9,] GM990027 WI920361
[10,] GM920222 WI920024 CGCC
[11,] GM930029 WI920437 CGCC
[12,] GM990105 WI920423 CCCG
[13,] GM050009 WI920416
[14,] GM920224 WI920423 CCCG
[15,] GM920224 WI920427
[16,] GM960028 WI920437
[17,] GM940031 WI920441
[18,] GM930040 WI920417
[19,] GM930040 WI920370
[20,] GM050099 WI920015
[21,] GM050099 WI920428
[22,] GM920096 WI920007
[23,] GM920034 WI920009 CCCG
[24,] GM960023 WI920410
[25,] GM920031 WI920430
[26,] GM920202 WI920015
[27,] GM990066 WI920415
Now I would like to compare one of the tables used to create the
output tables with every output table, one after the other. In detail,
I am comparing row 1 of the creator table with row 1 of the first
output table and then row 2 of the creator table with row 2 of the
first output table and so on until row 27 and each row for all
columns. Then, when the first output table is finished I go on
comparing the first creator table with the second table in the
output, row for row for all columns. I do this for all iterations.
The first creator table is called data_mc.
# apply similarity function (lettermatch) to my data
for (i in 1:(nrow(data_mc))){
for (y in 1:(ncol(data_mc))) {
creator_table - data_mc[data_mc$Status==mother,y]
output_tables - ???
output[i,y]-(lettermatch(creator_table, output_tables))
}
}
Could you please help me how I have to call up the output tables in
the way I need them (described above) for the function lettermatch?
Maybe I need to change the format of the output file?
Thank you.
Laetitia
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Re: [R] (box-) plot annotation: italic within paste?
Here's a variation on the theme:
boxplot(x[,i])
title(main =
bquote(.(mainlabel1)~~italic(.(predictor[i]))~~.(mainlabel2))
)
-Peter Ehlers
On 2010-03-08 8:46, Miguel Porto wrote:
Hello,
Try this way (not sure if it's the best way, but it works):
boxplot(x[,i],
main=substitute(expression(paste(a, ,italic(b),
,c)),list(a=mainlabel1,b=predictor[i],c=mainlabel2)),
ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
Best,
Miguel
On Mon, Mar 8, 2010 at 2:27 PM, Bernd Panassitibernd.panass...@rivm.nlwrote:
Dear R users,
in the example below the name of the genus will be displayed in the main
titles using the variable predictor[i] and paste.
I would like to have the genus name in italic. However all my attempts
using expression and substitute failed.
Does anybody know a solution?
Thanks a lot in advance. bernd
Acrobeles-c(65.1,0.0,0.0,0.0,0.0,0.0)
Acrobeloides-c(0.0,9.8,76.7,51.1,93.9,43.9)
Alaimus-c(0.0,4.9,0.0,0.0,0.0,6.3)
Aphelenchoides-c(126.5,29.3,76.7,134.1,176.7,87.9)
x-data.frame(Acrobeles,Acrobeloides,Alaimus,Aphelenchoides)
predictor- colnames(x)
ylabel-Numerical abundance
mainlabel1-Boxplot for
mainlabel2-sp.
cexalabel-1.8 # axis label
cexmlabel-1.6 # main label
par(oma=c(6,6,3,3),mar = c(6, 4, 4, 2) + 0.1,mfrow=c(2,2))
for (i in 1:ncol(x)){
boxplot(x[,i],
main=paste(mainlabel,predictor[i],mainlabel2),ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
}
---
Bernd Panassiti
National Institute of Public Health the Environment (RIVM)
Laboratory for Ecological Risk Assessment (LER)
P.O. Box 1
3720 BA Bilthoven
The Netherlands
e-mail: bernd.panass...@rivm.nl
tel. +31 30 274 3647
Radboud University Nijmegen
Department of Environmental Science
b.panass...@science.ru.nl
Disclaimer RIVM
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--
Peter Ehlers
University of Calgary
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Re: [R] fit a gamma pdf using Residual Sum-of-Squares
Thanks for making it quickly reproducible - I was able to see that message in English within a few seconds. The start has x=86, but the data is also called x. Remove x=86 from start and you get a different error. P.S. - please do include the R version information. It saves time for us, and we like it if you save us time. vincent laperriere vincent_laperri...@yahoo.fr wrote in message news:883644.16455...@web24106.mail.ird.yahoo.com... Hi all, I would like to fit a gamma pdf to my data using the method of RSS (Residual Sum-of-Squares). Here are the data: x - c(86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174) y - c(2, 5, 10, 17, 26, 60, 94, 128, 137, 128, 77, 68, 65, 60, 51, 26, 17, 9, 5, 2, 3, 7, 3) I have typed the following code, using nls method: fit - nls(y ~ (1/((s^a)*gamma(a))*x^(a-1)*exp(-x/s)), start = c(s=3, a=75, x=86)) But I have the following message error (sorry, this is in German): Fehler in qr(.swts * attr(rhs, gradient)) : Dimensionen [Produkt 3] passen nicht zur Länge des Objektes [23] Zusätzlich: Warnmeldung: In .swts * attr(rhs, gradient) : Länge des längeren Objektes ist kein Vielfaches der Länge des kürzeren Objektes Could anyone help me with the code? I would greatly appreciate it. Sincerely yours, Vincent Laperrière. [[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] black cluster in salt and pepper image
Hello, The function bwlabel() in the Bioconductor package EBImage, extracts the connected components of an image. Denoting your binary matrix by x, the following code gives you the first 10 largest clusters (in size). library(EBImage) y = bwlabel(x) sort(table(y), dec=TRUE)[1:10] See http://www.bioconductor.org/packages/release/bioc/html/EBImage.html how to download/install EBImage. Best regards, Greg --- Gregoire Pau EMBL Research Officer http://www.ebi.ac.uk/~gpau/ Sylvain Sardy wrote: Hi, on a lattice, I have binary 0/1 data. 1s are rare and may form clusters. I would like to know the size/length of largest cluster. Any help warmly welcome, Sylvain. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Is there an equivalence of lm's anova for an rpart object ?
Thanks Liaw!
I just implemented it using tapply:
tapply(fit$splits[, improve], rownames(fit$splits), sum)
If you can reference me to any other source / example and so on - it would
be great. but either way - you helped me a lot, thank you !
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Mon, Mar 8, 2010 at 4:52 PM, Liaw, Andy andy_l...@merck.com wrote:
One way to do it (no p-values) is explained in the original CART book.
You basically add up all the improvement (in fit$split[, improve])
due to each splitting variable.
Andy
From: Tal Galili
Simple example:
# Classification Tree with rpart
library(rpart)
# grow tree
fit - rpart(Kyphosis ~ Age + Number + Start,
method=class, data=kyphosis)
Now I would like to know how can I measure the importance
of each of my
three explanatory variables (Age, Number, Start) in the model?
If this was a regression model, I could have looked at p
values from the
anova F test (between lm models with and without the
variable). But what
is the equivalence of using anova on lm to an rpart object ?
Any pointers, insights and references to this question will
be helpful.
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
(Hebrew) |
www.r-statistics.com (English)
--
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Re: [R] fit a gamma pdf using Residual Sum-of-Squares
On 2010-03-08 8:24, vincent laperriere wrote: Hi all, I would like to fit a gamma pdf to my data using the method of RSS (Residual Sum-of-Squares). Here are the data: x- c(86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174) y- c(2, 5, 10, 17, 26, 60, 94, 128, 137, 128, 77, 68, 65, 60, 51, 26, 17, 9, 5, 2, 3, 7, 3) I have typed the following code, using nls method: fit- nls(y ~ (1/((s^a)*gamma(a))*x^(a-1)*exp(-x/s)), start = c(s=3, a=75, x=86)) There are a couple of problems: 1) don't include a start value for x; it's not a 'parameter'; 2) you're trying to fit a *density* function to data that's clearly not normalized. A quick check shows that your empirical curve integrates to about 4000: y[-1] * diff(x) # 3992 This almost works: fit- nls(y ~ 4000*(1/((s^a)*gamma(a))*x^(a-1)*exp(-x/s)), start = c(s=3, a=75)) but not quite; I still get an error. So let's do the right thing and plot the data and some test fits: plot(y ~ x) curve(4000 * dgamma(x, shape=75, scale=3), add=TRUE) # no good curve(4000 * dgamma(x, shape=75, scale=1), add=TRUE) # no good curve(4000 * dgamma(x, shape=75, scale=1.6), add=TRUE) # pretty good! fit- nls(y ~ 4000*(1/((s^a)*gamma(a))*x^(a-1)*exp(-x/s)), start = c(s=1.6, a=75)) coef(fit) #s a # 1.399638 86.395409 xx - seq(86, 174, length=100) yy - predict(fit, data.frame(x=xx)) plot(y ~ x) lines(yy ~ xx, col='red') -Peter Ehlers But I have the following message error (sorry, this is in German): Fehler in qr(.swts * attr(rhs, gradient)) : Dimensionen [Produkt 3] passen nicht zur L�nge des Objektes [23] Zus�tzlich: Warnmeldung: In .swts * attr(rhs, gradient) : L�nge des l�ngeren Objektes ist kein Vielfaches der L�nge des k�rzeren Objektes Could anyone help me with the code? I would greatly appreciate it. Sincerely yours, Vincent Laperri�re. [[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. -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Combinations and table selection problem (reviewed)
Dear all, I have the following dataset: t - structure(list(pUrb = c(20.160307, 51.965649, 26.009581, 3.141484, 64.296826 ), pUrb_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(0, 1), class = factor), pAgri = c(79.921386, 46.657713, 40.269204, 0, 0.440691), pAgri_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(0, 1), class = factor), pNatFor = c(0, 0, 0, 0, 0), pNatFor_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(0, 1), class = factor), pArtFor = c(0, 0, 0, 0, 24.566125), pArtFor_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(0, 1), class = factor), pMixFor = c(0, 0, 33.578923, 96.940185, 10.655666), pMixFor_class = structure(c(1L, 1L, 1L, 2L, 1L), .Label = c(0, 1), class = factor), pPioMo = c(0, 0, 0, 0, 0), pPioMo_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(0, 1), class = factor), SwiLU = c(0.565419, 0.937982, 1.071812, 0.056002, 0.831812 ), SwiLU_class = structure(c(1L, 2L, 2L, 1L, 2L), .Label = c(1, 0, 2), class = factor), NumP = c(4L, 7L, 3L, 4L, 6L ), NumP_class = structure(c(2L, 3L, 1L, 2L, 3L), .Label = c(1, 0, 2), class = factor), Roaddist = c(1615.55, 2140.09, 2308.68, 2088.06, 2000), Roaddist_class = c(NA, NA, NA, NA, NA), Roaddens = c(0, 0, 0, 0, 0), Roaddens_class = c(NA, NA, NA, NA, NA), SwiSlo = c(0, 0, 0, 0, 0), SwiSlo_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(1, 2, 0, 3), class = factor), SwiAlt = c(0, 0, 0, 0, 0), SwiAlt_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c(1, 2, 0, 3), class = factor)), .Names = c( pUrb, pUrb_class, pAgri, pAgri_class, pNatFor, pNatFor_class, pArtFor, pArtFor_class, pMixFor, pMixFor_class, pPioMo, pPioMo_class, SwiLU, SwiLU_class, NumP, NumP_class, Roaddist, Roaddist_class, Roaddens, Roaddens_class, SwiSlo, SwiSlo_class, SwiAlt, SwiAlt_class), row.names = c(NF1885, NF1886, NF1893, NF1894, NF1895), class = data.frame) #obtaining the number of combinations present in the data base# library(survival) w - strata(t$NumP_class,t$SwiLU_class,t$pMixFor_class, sep=,) table(w) w t$NumP_class=1,t$SwiLU_class=0,t$pMixFor_class=0 1 t$NumP_class=0,t$SwiLU_class=1,t$pMixFor_class=0 1 t$NumP_class=0,t$SwiLU_class=1,t$pMixFor_class=1 1 t$NumP_class=2,t$SwiLU_class=0,t$pMixFor_class=0 2 #obtaining median value# median(table(w)) [1] 1 In this stage I have 3 questions: 1st: how can I obtain the combinations witch are present over the median (in this case the fourth combination)? 2nd: how can I obtain the combinations witch are present over the median and have at least one condition 2 (in this case the fourth combination)? 3rd: how can I select/extract from the original table the rows witch comply with the 2nd question, in this case (row id: NF1895): pUrb pUrb_class pAgri pAgri_class pNatFor pNatFor_class pArtFor pArtFor_class pMixFor NF1895 64.296826 0 0.440691 0 0 0 24.56612 0 10.65567 pMixFor_class pPioMo pPioMo_classSwiLU SwiLU_class NumP NumP_class Roaddist Roaddist_class Roaddens NF1895 0 00 0.831812 06 2 2000.00 NA0 Roaddens_class SwiSlo SwiSlo_class SwiAlt SwiAlt_class NF1895 NA 01 01 Best Regards, Carlos __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to convert character variables into numeric variables directly
Here is the example. age=18:29 height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5) type=c(A, B, C, D,A, B, C, D,A, B, C, D) typec=c(0,4,2,9,0,7,2,3,0,1,2,3) typen=c(0,1,2,3,0,1,2,3,0,1,2,3) data1=data.frame(age=age,height=height, type=type, typec=typec, typen=typen) data1[,3]=as.numeric(data1[,3]) data1[,4]=as.numeric(data1[,4]) data1[,5]=as.numeric(data1[,5]) print(data1) and I got the output as: age height type typec typen 1 18 76.11 1 0 2 19 77.02 5 1 3 20 78.13 3 2 4 21 78.24 7 3 5 22 78.81 1 0 6 23 79.72 6 1 7 24 79.93 3 2 8 25 81.14 4 3 9 26 81.21 1 0 10 27 81.82 2 1 11 28 82.83 3 2 12 29 83.54 4 3 The typec is not what I expected. How can I get the direct conversion from character to numeric and get the following output? age height type typec typen 1 18 76.11 0 0 2 19 77.02 4 1 3 20 78.13 2 2 4 21 78.24 9 3 5 22 78.81 0 0 6 23 79.72 7 1 7 24 79.93 2 2 8 25 81.14 3 3 9 26 81.21 0 0 10 27 81.82 1 1 11 28 82.83 2 2 12 29 83.54 3 3 Thanks. Xumin [[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] questions
Can somebody tell me how can I calculate derivatives of some functions through r citing some examples?? by Taylor's expansion formulae how can I handle complecated functions which can not be handled properly manually by r-script? -- View this message in context: http://n4.nabble.com/questions-tp1584771p1584771.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to convert character variables into numeric variables directly
Try: as.numeric(as.character( typec)) milton On Mon, Mar 8, 2010 at 12:55 PM, Xumin Zeng xumin.z...@abbott.com wrote: Here is the example. age=18:29 height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5) type=c(A, B, C, D,A, B, C, D,A, B, C, D) typec=c(0,4,2,9,0,7,2,3,0,1,2,3) typen=c(0,1,2,3,0,1,2,3,0,1,2,3) data1=data.frame(age=age,height=height, type=type, typec=typec, typen=typen) data1[,3]=as.numeric(data1[,3]) data1[,4]=as.numeric(data1[,4]) data1[,5]=as.numeric(data1[,5]) print(data1) and I got the output as: age height type typec typen 1 18 76.11 1 0 2 19 77.02 5 1 3 20 78.13 3 2 4 21 78.24 7 3 5 22 78.81 1 0 6 23 79.72 6 1 7 24 79.93 3 2 8 25 81.14 4 3 9 26 81.21 1 0 10 27 81.82 2 1 11 28 82.83 3 2 12 29 83.54 4 3 The typec is not what I expected. How can I get the direct conversion from character to numeric and get the following output? age height type typec typen 1 18 76.11 0 0 2 19 77.02 4 1 3 20 78.13 2 2 4 21 78.24 9 3 5 22 78.81 0 0 6 23 79.72 7 1 7 24 79.93 2 2 8 25 81.14 3 3 9 26 81.21 0 0 10 27 81.82 1 1 11 28 82.83 2 2 12 29 83.54 3 3 Thanks. Xumin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to convert character variables into numeric variables directly
Thanks, it works! Xumin milton ruser milton.ru...@gmail.com 03/08/2010 01:02 PM To Xumin Zeng xumin.z...@abbott.com cc r-help r-help@r-project.org Subject Re: [R] how to convert character variables into numeric variables directly Try: as.numeric(as.character( typec)) milton On Mon, Mar 8, 2010 at 12:55 PM, Xumin Zeng xumin.z...@abbott.com wrote: Here is the example. age=18:29 height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5) type=c(A, B, C, D,A, B, C, D,A, B, C, D) typec=c(0,4,2,9,0,7,2,3,0,1,2,3) typen=c(0,1,2,3,0,1,2,3,0,1,2,3) data1=data.frame(age=age,height=height, type=type, typec=typec, typen=typen) data1[,3]=as.numeric(data1[,3]) data1[,4]=as.numeric(data1[,4]) data1[,5]=as.numeric(data1[,5]) print(data1) and I got the output as: age height type typec typen 1 18 76.11 1 0 2 19 77.02 5 1 3 20 78.13 3 2 4 21 78.24 7 3 5 22 78.81 1 0 6 23 79.72 6 1 7 24 79.93 3 2 8 25 81.14 4 3 9 26 81.21 1 0 10 27 81.82 2 1 11 28 82.83 3 2 12 29 83.54 4 3 The typec is not what I expected. How can I get the direct conversion from character to numeric and get the following output? age height type typec typen 1 18 76.11 0 0 2 19 77.02 4 1 3 20 78.13 2 2 4 21 78.24 9 3 5 22 78.81 0 0 6 23 79.72 7 1 7 24 79.93 2 2 8 25 81.14 3 3 9 26 81.21 0 0 10 27 81.82 1 1 11 28 82.83 2 2 12 29 83.54 3 3 Thanks. Xumin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Hmisc, cut2, split and quantile
On 2010-03-08 8:47, Guy Green wrote: Hello, I have a set of data with two columns: Target and Actual. A http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is attached but the data looks like this: Actual Target -0.125 0.016124906 0.135 0.120799865 ... ... ... ... I want to be able to break the data into tables based on quantiles in the Target column. I can see (using cut2, and also quantile) how to get the barrier points between the different quantiles, and I can see how I would achieve this if I was just looking to split up a vector. However I am trying to break up the whole table based on those quantiles, not just the vector. The following code shows me the ranges for the deciles of the Target data: library(Hmisc) read_data=read.table(C:/Sample table.txt, head = T) table(cut2(Read_data$Target,g=10)) However I would like to be able to break the table into ten separate tables, each with both Actual and Target data, based on the Target data deciles: top_decile = ...(top decile of read_data, based on Target data) next_decile = ...and so on... bottom_decile = ... I would just add a factor variable indicating to which decile a particular observation belongs: dat$DEC - with(dat, cut(Target, breaks=10, labels=1:10)) If you really want to have separate data frames you can then split on the decile: L - split(dat, dat$DEC) -Peter Ehlers That way I could manipulate the deciles, graph them separately (and together) and so on, just as easily as I can the whole table. I'm sure this must be simple, but I can't see the way forward. I have also looked at split() and quantile() but have not been able to get them to achieve what I am after. Can anybody see a simple way foward on this? Thanks, Guy -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Working with combinations
I had a bug in my last solution.
This one should work.
next.combin - function(oldcomb,n){
lcomb - length(oldcomb)
hole.pos - last.hole.pos(oldcomb,n)
if ((hole.pos == lcomb) oldcomb[lcomb]==n) {
return(NA)
}
newcomb-oldcomb
newcomb[hole.pos:lcomb]-oldcomb[hole.pos]+(1:(lcomb-hole.pos+1))
return(newcomb)
}
last.hole.pos - function(comb,n){
lcomb - length(comb)
if (comb[lcomb]n) {
return(lcomb)
}
diffs - comb[-1]-comb[-lcomb]
if (max(diffs)==1) {
return(lcomb)
}
diffpos - which(diffs1)
return(diffpos[length(diffpos)])
}
demo - function(n,k){
currCombin - 1:k
#this is the first possible combination k out if n.
selectedSet - NULL
while (!any(is.na(currCombin))){
# if(currCombin passes test) {
selectedSet - rbind(selectedSet,currCombin)
# }
currCombin - next.combin(currCombin,n)
}
rownames(selectedSet)-NULL
selectedSet
}
On Mar 6, 2010, at 9:50 PM, Herm Walsh wrote:
The usage here is exactly what I am looking for, thanks. However this
function seems to omit some combinations. Continuing with the example below,
given an always true condition in #***, it will only produce 7 combinations
(omitting 1,5 1,4 and 2,5).
Am I overlooking something that makes it produces all of the combinations?
From: Erich Neuwirth erich.neuwi...@univie.ac.at
To: Herm Walsh hermwa...@yahoo.com; r-help r-help@r-project.org
Sent: Sat, March 6, 2010 9:12:13 AM
Subject: Re: [R] Working with combinations
currCombin - c(1,2) #this is the first possible combination 2 out if 5.
Since the vector has length 2, we are doing 2 out of x here.
selectedSet - NULL
while (!is.na(currCombin)){
if( #*** currCombin passes test ***# ) {
selectedSet - rbind(selectedSet,currCombin)
}
currCombin - next.combin(currCombin,5) #here we say that we want 2 out of 5
}
On Mar 6, 2010, at 5:54 PM, Herm Walsh wrote:
Erich-
This approach would be great for my context. However, in the code below I
do not see how to restrict the output to the set of combinations I am
looking for. For example, suppose I am looking for the 10 two element
combinations of 1:5. Can you give me some psuedocode that shows how to do
this?
After putting the code below in a loop I do not see how to restrict the
output to containing only numbers from 1:5.
thanks very much.
From: Erich Neuwirth erich.neuwi...@univie.ac.at
To: Herm Walsh hermwa...@yahoo.com
Cc: David Winsemius dwinsem...@comcast.net; r-help@r-project.org
Sent: Wed, March 3, 2010 2:10:34 PM
Subject: Re: [R] Working with combinations
The following code takes a combination of type n over k represented by an
increasing sequence
as input an produces the lexicographically next combinations.
So you can single step through all possible combinations and apply your
filter criteria
before you produce the next combination.
next.combin - function(oldcomb,n){
lcomb - length(oldcomb)
hole.pos - last.hole.pos(oldcomb,n)
if ((hole.pos == lcomb) oldcomb[lcomb]==n) {
return(NA)
}
newcomb-oldcomb
newcomb[hole.pos]-oldcomb[hole.pos]+1
return(newcomb)
}
last.hole.pos - function(comb,n){
lcomb - length(comb)
diffs - comb[-1]-comb[-lcomb]
if (max(diffs)==1) {
return(lcomb)
}
diffpos - which(diffs1)
return(diffpos[length(diffpos)])
}
On Mar 3, 2010, at 7:35 PM, Herm Walsh wrote:
Thanks David for the thoughts. The challenge I have with this approach is
that the criteria I have is defined by a series of tests--which I do not
think I could substitute in in place of the logical indexing.
In the combinations code I was hoping there is a step where, each new
combination is added to the current list of combinations. If this were the
case, I could put my series of tests in the code right there and then store
the combination if appropriate.
However, evalutating the code--which uses recursion--I am not sure if this
approach will work. The combinations code is listed below. Is there a
simple place(s) where I could insert my tests, operating on the current
combination?
function (n, r, v = 1:n, set = TRUE, repeats.allowed = FALSE)
{
if (mode(n) != numeric || length(n) != 1 || n 1 || (n%%1) !=
0)
stop(bad value of n)
if (mode(r) != numeric || length(r) != 1 || r 1 || (r%%1) !=
0)
stop(bad value of r)
if (!is.atomic(v) || length(v) n)
stop(v is either non-atomic or too short)
if ((r n) repeats.allowed == FALSE)
stop(r n and repeats.allowed=FALSE)
if (set) {
v - unique(sort(v))
if (length(v) n)
stop(too few different elements)
}
v0 - vector(mode(v), 0)
if (repeats.allowed)
sub - function(n, r, v) {
if (r == 0)
v0
else if (r == 1)
matrix(v, n, 1)
else if (n == 1)
[R] Executable for Production Use
Hi,
A few of the developers on our Quant team are using R for data calculation and
to generate a resulting CSV file. They have R installed on their workstations.
We are interested in having this deployed to user workstations where the users
will not have R installed on their workstations. Is there a way to create an
executable that the users can just run without R installed on their workstation?
Thanks in advance for your help.
Ismail Ma
Head of IT Applications
MEAG New York Corporation
Telephone: (212) 583-4850
Fax: (646) 521-7950
E-Mail: i...@meag-ny.commailto:i...@meag-ny.com
This e-mail and any files transmitted with it are confid...{{dropped:14}}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Executable for Production Use
On Mon, Mar 8, 2010 at 6:44 PM, Ma Ismail - NewYork-MEAG-NY i...@meag-ny.com wrote: Hi, A few of the developers on our Quant team are using R for data calculation and to generate a resulting CSV file. They have R installed on their workstations. We are interested in having this deployed to user workstations where the users will not have R installed on their workstations. Is there a way to create an executable that the users can just run without R installed on their workstation? No, not for any meaningful value of the word 'installed'. If you want to run R, you need the R interpreter and all the functions and stuff it carries around with it.Maybe you could bundle this all up into one monster executable, but why bother? R install is easy and portable (stick it anywhere, it runs). Even probably a network share. Oh, you didn't say what OS you were using. Did you read the posting guide? Also, this has been asked before. Did you search the mailing lists? I've noticed a lot of financial corporates getting into R and getting free help from R-help. From now on, I'm charging $50 per email to answer questions from anyone advertising as a 'quant'. Who do I invoice? Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using sprintf to pass a variable to a RMySQL query
I always use paste()
i - 1
sqlcmd_ScaffLen - paste(SELECT scaffold.length
FROM scaffold, scaffold2contig, contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND
contig2read.read_id LIKE '%MG, i ,%', sep='')
That should create bits like
LIKE '%MG1%'
LIKE '%MG2%'
and so on.
You just have to get the nesting of the single and double quotes
correct - the SQL requires single quotes, so use double quotes for
the fixed character strings insidte paste(). That, and use sep='' to
get rid of unwanted space characters.
Using paste is also effective for constructs like
IN (3,4,5)
or
IN ('a','b','c')
though it can be necessary to nest one paste within another
-Don
At 2:06 PM +0100 3/8/10, alison waller wrote:
Hello,
I am using RmySQL and would like to iterate through a few queries.
I would like to use sprintf but I think I'm having problems mixing and
matching the sprintf syntax and the SQL regex.
I have checked my sqlcmd and it works when I wan to match %MG1% but how
do I iterate for i 1-72? Escape characters,?
thanks in advance
i-1
sqlcmd_ScaffLen-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id LIKE
'%MG%s%' ,i)
= Here is my vague error message
Error: unexpected input in:
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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062
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Re: [R] Monetary support to the R-project (Was: Re: Executable for Production Use)
On Mon, Mar 8, 2010 at 8:46 PM, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote: On Mon, Mar 8, 2010 at 6:44 PM, Ma Ismail - NewYork-MEAG-NY i...@meag-ny.com wrote: Hi, A few of the developers on our Quant team are using R for data calculation and to generate a [snip] I've noticed a lot of financial corporates getting into R and getting free help from R-help. From now on, I'm charging $50 per email to answer questions from anyone advertising as a 'quant'. Who do I invoice? For companies and others wondering how to give something back, it is possible to support R and the R Foundation either through a donation: http://www.r-project.org/ - Foundation - Donations [http://www.r-project.org/foundation/donations.html] or via a membership: http://www.r-project.org/ - Foundation - Membership [http://www.r-project.org/foundation/membership.html] or both. /Henrik Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Executable for Production Use
Ma Ismail - NewYork-MEAG-NY ima at meag-ny.com writes: Hi, A few of the developers on our Quant team are using R for data calculation andto generate a resulting CSV file. They have R installed on their workstations. We are interested in having this deployed to user workstations where the users will not have R installed on their workstations. Is there a way to create an executable that the users can just run without R installed on their workstation? Thanks in advance for your help. Ismail Ma Head of IT Applications MEAG New York Corporation Telephone: (212) 583-4850 Fax: (646) 521-7950 E-Mail: ima at meag-ny.commailto:ima at meag-ny.com Maybe I've misunderstood You, but You can manage a single R installation on a central server, and let clients hook up to it from Emacs+ESS or Eclipse+StatET. I've used both solutions, and they work like a charm... The emacs solution uses SSH and Eclipse a Java server thingy. Plz note that I'm from the Linux side of things. So my server is a Debian, and my clients are all Ubuntu. I have no clue how (or even if) these setups work on windows. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] likelihood
Hi all, Does any one know how to write the likelihood function for Poisson distribution in R when P(x=0). For normal case, it an be written as follows, n * log(lambda) - lambda * n * mean(dat) Any help is highly appreciated Ashta __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] error when using svm routine: Error in if (any(co)) { : missing value where TRUE/FALSE needed
Hi,
I met with this error message with the following data set. Do you know how
to resolve it? Thanks.
data-read.table(c://temp3//abc.csv, sep = ,, header=T)
classwt-c( 0.5806452, 0.4193548)
y-data[,1]
x-data[,2:ncol(data)]
print(y)
[1] 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1
[36] 1 1 1 1 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2
print(x)
rs2289472 rs1551398 rs7927894
1 CTAACT
2 TTAACC
3 TTAGTT
4 TTAACT
5 CCAACT
6 CTAACT
7 TTAACT
8 CTAGCT
9 CCAACT
10TTAGTT
11CTGGCC
12CTGGCC
13CTAGCC
14CTAGTT
15CTAACT
16TTAACT
17CTAGCT
18TTAACT
19TTAATT
20CTAGCT
21CTAGTT
22CTAACT
23TTAACT
24CCAACC
25CTAGCC
26CTAGCT
27CTGGCT
28CTAGCC
29CTGGCC
30TTGGCT
31CTAGCT
32TTAGTT
33CCGGCC
34TTAACT
35CTGGTT
36TTAGCT
37CTAGCC
38TTAACC
39TTAATT
40TTAATT
41CTGGCT
42CTAGTT
43TTAACT
44CTAGCC
45TTAGCC
46CTAACC
47CCAACT
48CTAACT
49CCAACC
50TTAATT
51TTGGCT
52CTAGCT
53TTAGTT
54TTAACT
55CTAACC
56CTAGCT
57CTAGCC
58CTAACC
59CTAGCC
60CTAGTT
61TTAACC
62TTGGCT
svm.fit=svm(y=as.factor(y),x=x,class.weights = classwt)
Error in if (any(co)) { : missing value where TRUE/FALSE needed
In addition: Warning messages:
1: In FUN(newX[, i], ...) : NAs introduced by coercion
2: In FUN(newX[, i], ...) : NAs introduced by coercion
3: In FUN(newX[, i], ...) : NAs introduced by coercion
Xumin
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[R] Making FTP operations with R
Dears I need to make some very basic FTP operations with R. I need to do a lot of get and issue a respective delete command too on the same connection. How can I do that? Thanks in advance Caveman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] why this function does not run correctly?
Hi,
my name is Marco Bressan i'm working to improve ADati package. I study
psicology ad Padua University (Italy). I have this problem: why bartlett.test
function running good and my anova.welch function no?
Ciao,
il mio nome è Marco Bressan e sto lavorando per migliorare il pacchetto ADati.
Studio psicologia all'università di Padova. Non capisco come mai la funzione
che ho fatto mi dia quell'errore
this is anova.welch:
anova.welch - function(x, ...) UseMethod(anova.welch)
anova.welch.default -
## this is the algoritm, I think it's ok (I copy this from Welch() inside ADati)
function (x, y = NULL, nu = c(0,0) ,...)
{
mx - tapply(x,y,mean)
s2x - tapply(x,y,var)
k - length(nu)
w - nu/s2x
Xp - sum(w * mx)/sum(w)
Fnum - sum(w * (mx - Xp)^2)/(k - 1)
Fden - 0
for (h in 1:k) {
a - 1/(nu[h] - 1)
b - (1 - (w[h]/sum(w)))^2
Fden - Fden + (a * b)
}
gl2.den - Fden * 3
Fden - Fden * ((2 * (k - 2))/(k^2 - 1)) + 1
Fw - Fnum/Fden
STATISTIC - Fw
gl1 - k - 1
gl2.num - k^2 - 1
gl2 - gl2.num/gl2.den
PARAMETER - c(gl1, gl2)
PVAL - pf(Fw, gl1, gl2, lower.tail = FALSE)
METHOD - Welch ANOVA
DNAME - NA
names(STATISTIC) - F
names(PARAMETER) - c(num df, denom df)
RVAL - list(statistic = STATISTIC, parameter = PARAMETER,
p.value = PVAL, method = METHOD)
class(RVAL) - htest
return(RVAL)
}
anova.welch.formula -
## I copy this from bartlett.test
function(formula, data, subset, na.action, ...)
{
if(missing(formula) || (length(formula) != 3L))
stop('formula' mancante o incorretta)
m - match.call(expand.dots = FALSE)
if(is.matrix(eval(m$data, parent.frame(
m$data - as.data.frame(data)
m[[1L]] - as.name(model.frame)
mf - eval(m, parent.frame())
DNAME - paste(names(mf), collapse = by )
names(mf) - NULL
y - do.call(anova.welch, as.list(mf))
y$data.name - DNAME
y
}
n1=10
## this is the test
n2=15
n3=20
y=c(rnorm((n1+n2),5,2),rnorm(n3,7,8))
A=factor(c(rep(1,n1),rep(2,n2),rep(3,n3)))
anova.welch(y,A,c(n1,n2,n3))
Welch ANOVA
data:
F = 2.3025, num df = 2.000, denom df = 27.384, p-value = 0.1191
anova.welch(y~A,nu=c(n1,n2,n3))
Errore in model.frame.default(formula = y ~ A, ... = list(nu = c(n1, n2, :
invalid type (pairlist) for variable '(...)'
Sorry for my english, tanks you if you can help me :)
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Re: [R] setClass or setValidity?
On 03/08/2010 07:18 AM, Albert-Jan Roskam wrote: Sorry: there was an error in the last sentence: And, inside those validity checks, is most of the checking done with 'if' 'else' computations, or is it also common to use try()? For me it's a matter of taste, and usual to use if... (because you know explicitly what you're trying to validate, whereas try() implies a kind of 'something might go wrong...'). I find myself using setValidity() to separate out class definition from implementation. Best, Martin Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ --- On Mon, 3/8/10, Albert-Jan Roskam fo...@yahoo.com wrote: From: Albert-Jan Roskam fo...@yahoo.com Subject: [R] setClass or setValidity? To: r-help@r-project.org Date: Monday, March 8, 2010, 4:14 PM Hi, I'm reading up on S4 classes *). There seem to be at least two ways of input validation: setClass() (using the 'validity' argument) and setValidity(). Is it a matter of taste which function is used? Or should more complex validation code better be put in a setValiditity call? *) A (Not So) Short Introduction to S4 Object Oriented Programming in R V0.5.1 Christophe Genolini August 20, 2008 And, inside those validity checks, is most of the checking done with 'if' 'else' computations, or is it also common to use except()? Cheers!! Albert-Jan ~~ In the face of ambiguity, refuse the temptation to guess. ~~ [[alternative HTML version deleted]] -Inline Attachment Follows- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] questions about Cusum
I've found the surveillance package useful for monitoring walk-in clinic visits in our county as the influenza pandemic evolved. It might serve your needs for monitoring IFI (invasive fungal infections?) in your hospital. --Chris Christopher W. Ryan, MD SUNY Upstate Medical University Clinical Campus at Binghamton 425 Robinson Street, Binghamton, NY 13904 cryanatbinghamtondotedu If you want to build a ship, don't drum up the men to gather wood, divide the work and give orders. Instead, teach them to yearn for the vast and endless sea. [Antoine de St. Exupery] Achim Zeileis wrote: On Sun, 7 Mar 2010, sdzhangping wrote: Dear friends: I have just read an article entitled Monitoring of nosocomial invasive aspergillosis and early evidence of an outbreak using cumulative sum tests (CUSUM), which is published in Clinical Microbiology and Infection. We have great need to estimate the fluctuation of incidence of IFI in our hospital. But I don't know the details of the stastical method and don't know where can I get a Cusum package. Can you give me some materials about Cusum test? An example is more appreciated. There are many techniques with the label CUSUM and I didn't look up what the specific meaning is in the article you cite. There are tests based on (mostly linear) regression models with that label, many (but not all conceivable) of these are implemented in the strucchange package. See vignette(strucchange-intro, package = strucchange) as well as citation(strucchange) for pointers to papers explaining the ideas behind it. But beyond that there are also other techniques, in particular from statistical process control. See the spc, qcc, IQCC packages among others. hth, Z Yours sincerely Ping Zhang March 7, 2010 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] page boundaries for latex printing of summary.formula objects in Hmisc
Hello,
Warning, I'm guessing only those who have used the Hmisc package's
summary.formula function with LaTeX will be able to offer much help here.
I am using the Hmisc package's summary.formula function to produce
tables for a LaTeX report. The latex function in the same package
supports longtables in LaTeX. Ideally, I would like for page breaks in
the LaTeX output to only occur at variable boundaries.
## Sample R code
## load the Hmisc package
library(Hmisc)
## create an example data.frame
test.df - data.frame(sex = gl(2, 110, labels = c(Male, Female)),
fac1 = sample(gl(11, 100, labels = paste(V1 Level, 1:11))),
fac2 = sample(gl(11, 100, labels = paste(V2 Level, 1:11))),
fac3 = sample(gl(11, 100, labels = paste(V3 Level, 1:11))),
fac4 = sample(gl(11, 100, labels = paste(V4 Level, 1:11))),
fac5 = sample(gl(11, 100, labels = paste(V5 Level, 1:11))),
fac6 = sample(gl(11, 100, labels = paste(V6 Level, 1:11
## create the summary.formula object
sf - summary.formula(sex ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6,
data = test.df,
method = reverse)
## print out the LaTeX code to the screen, not a file
latex(sf, file = , longtable = TRUE)
Notice how the LaTeX output puts the newline in the middle of factor 4,
instead of before or after.
excerpt of LaTeX ouput follows
fac4~:~V4~Level~111\%~{\scriptsize~(59)}~7\%~{\scriptsize~(41)}\tabularnewline
V4~Level~210\%~{\scriptsize~(56)}~8\%~{\scriptsize~(44)}\tabularnewline
V4~Level~3~9\%~{\scriptsize~(47)}10\%~{\scriptsize~(53)}\tabularnewline
V4~Level~4~9\%~{\scriptsize~(49)}~9\%~{\scriptsize~(51)}\tabularnewline
V4~Level~5~9\%~{\scriptsize~(51)}~9\%~{\scriptsize~(49)}\tabularnewline
V4~Level~6~9\%~{\scriptsize~(50)}~9\%~{\scriptsize~(50)}\tabularnewline
V4~Level~7~7\%~{\scriptsize~(39)}11\%~{\scriptsize~(61)}\tabularnewline
\newpage
V4~Level~8~9\%~{\scriptsize~(51)}~9\%~{\scriptsize~(49)}\tabularnewline
V4~Level~9~9\%~{\scriptsize~(51)}~9\%~{\scriptsize~(49)}\tabularnewline
V4~Level~10~9\%~{\scriptsize~(47)}10\%~{\scriptsize~(53)}\tabularnewline
V4~Level~11~9\%~{\scriptsize~(50)}~9\%~{\scriptsize~(50)}\tabularnewline
I expect this given the documentation in ?latex of lines.page, which is
set to 40 by default.
lines.page:
Applies if ‘longtable=TRUE’. No more than ‘lines.page’
lines in the body of a table will be placed on a single page.
Page breaks will only occur at ‘rgroup’ boundaries.
The problem is that variable boundaries don't in general correspond to
constants, like 40 lines.
So, rgroup sounds promising. I want the lines per variable to
correspond to be the n.rgroup values, but since my tables are dynamic,
in that the variables and number of levels in them change over time, I
can't think of a way to define n.rgroup without specifying it per
variable. My first thought was to compute it from the number of levels
per variable in the formula.
In fact, I did try this but immediately ran into some misconceptions I
had about how continuous variables are represented internally within the
latex function.
Is there any easier way to accomplish this breaking of pages on variable
boundaries using this set of functions? I suspect not, but thought I'd
ask. I think I can figure out the approach I suggested in the preceding
2 paragraphs, but just want to make sure I'm not missing something ...
Thanks a lot!
Erik Iverson
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Re: [R] aggregate for zoo or its?
Or for real power and flexibility, see the 'convert()' function in package tis. Jeff Gabor Grothendieck ggrothendi...@gmail.com writes: See ?aggregate.zoo, e.g. library(zoo) z - zoo(1:1000, as.Date(2000-01-01) + 0:999) aggregate(z, as.yearmon, mean) or replace mean with whatever summarization you want. On Sun, Mar 7, 2010 at 5:29 PM, Erin Hodgess erinm.hodg...@gmail.com wrote: Dear R People: The aggregate function works very well on regular time series. Is there a version for zoo or its that would take daily data and convert it to monthly, please? -- Jeff __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] confused by classes and methods.
Hello, I have a simple class that looks like:
setClass(statisticInfo,
representation( max = numeric,
min = numeric,
beg = numeric,
current = numeric,
avg = numeric,
obs = vector
)
)
and the following function
updateStatistic - function(statistic, newData){
statis...@obs = c(statis...@obs, newData)
statis...@max = max(newData, statis...@max, na.rm=T)
statis...@min = min(newData, statis...@min, na.rm=T)
statis...@avg = mean(statis...@obs)
statis...@current = newData
if(length(statis...@obs)==1 || is.na(statis...@beg)){
statis...@beg = newData
}
return(statistic)
}
Firstly,
I know you can use methods which seems to add some value. I looked at
http://developer.r-project.org/methodDefinition.html but I try
setMethod(update, signature(statistic=statisticInfo, newData=numeric),
function(statistic, newData){
statis...@obs = c(statis...@obs, newData)
statis...@max = max(newData, statis...@max, na.rm=T)
statis...@min = min(newData, statis...@min, na.rm=T)
statis...@avg = mean(statis...@obs)
statis...@current = newData
if(length(statis...@obs)==1 || is.na(statis...@beg)){
statis...@beg = newData
}
return(statistic)
}
)
Creating a new generic function for update in .GlobalEnv
Error in match.call(fmatch, fcall) :
unused argument(s) (statistic = statisticInfo, newData = numeric)
1: source(tca.init.R, chdir = T)
2: eval.with.vis(ei, envir)
3: eval.with.vis(expr, envir, enclos)
4: source(../../studies/tca.tradeClassifyFuncs.R)
5: eval.with.vis(ei, envir)
6: eval.with.vis(expr, envir, enclos)
7: setMethod(update, signature(statistic = statisticInfo, newData =
numeric), function(statistic, newData) {
8: isSealedMethod(f, signature, fdef, where = where)
9: getMethod(f, signature, optional = TRUE, where = where, fdef = fGen)
10: matchSignature(signature, f
I don't understand this any help would be appreciated.
Secondly, can anyone give any examples of where methods are used that makes
sense besides just checking the class inputs?
Thirdly, I've looked into passing by reference in R, and some options come
up, but in general they seem to be fairly complicated.
I would like update to work more like my update function to work without
having to return a a new object.
Something like
statList = list(new(statisticInfo))
updateStatistic(statList[[1]],3)
statList[[1]]
#this would then have the updated one and not the old one.
Anyways,
The main reason I'm asking these questions is because I can't really find a
good online resource for this. Any help would be greatly appreciated.
Thanks,
Rob
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[R] variance of discrete uniform distribution
Hi all, I am REALLY confused with the variance right now. for a discrete uniform distribution on [1,12] the mean is (1+12)/2=6.5 which is ok. y=1:12 mean(y) then var(y) gives me 13 1- on http://en.wikipedia.org/wiki/Uniform_distribution_%28discrete%29 wiki the variance is (12^2-1)/12=143/12 2- http://www.solvemymath.com/online_math_calculator/statistics/continuous_distributions/uniform/param_uniform.php here which used (12-1)^2/12=121/12 all different 3 answers!!! All I am looking for is the variance of a random variable from discrete uniform distribution. Can someone clearify that for me please? Thanks. -- View this message in context: http://n4.nabble.com/variance-of-discrete-uniform-distribution-tp1585328p1585328.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variance of discrete uniform distribution
On 9/03/2010, at 12:13 PM, casperyc wrote: Hi all, I am REALLY confused with the variance right now. You need to learn the difference (a) Between sample variance (*estimate* of population variance) and population variance. and (b) Between discrete and continuous distributions. Given that you understand those differences you will see that all three answers are correct. cheers, Rolf Turner for a discrete uniform distribution on [1,12] the mean is (1+12)/2=6.5 which is ok. y=1:12 mean(y) then var(y) gives me 13 1- on http://en.wikipedia.org/wiki/Uniform_distribution_%28discrete%29 wiki the variance is (12^2-1)/12=143/12 2- http://www.solvemymath.com/online_math_calculator/statistics/continuous_distributions/uniform/param_uniform.php ***LOOK*** at the above. Does it or does it not contain the string ``continuous_distributions''??? And doesn't your question involve the ***discrete*** uniform distribution??? R. T. here which used (12-1)^2/12=121/12 all different 3 answers!!! All I am looking for is the variance of a random variable from discrete uniform distribution. Can someone clearify that for me please? Thanks. ## Attention: This e-mail message is privileged and confidential. If you are not the intended recipient please delete the message and notify the sender. Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshal www.marshalsoftware.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] variance of discrete uniform distribution
Hi Rolf Turner , God, it directed to the wrong page. I firstly find the formula in wiki, than tried to verify the answer in R, now, given that 143/12 ((n^2-1)/12 ) is the correct answer for a discrete uniform random variable, I am still not sure what R is calculating there? why it gives me 13? Thanks! -- View this message in context: http://n4.nabble.com/variance-of-discrete-uniform-distribution-tp1585328p1585355.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] test the goodness of it for negative binomial type 2
Hi Achim Zeileis-4, That's very helpful. Thanks! -- View this message in context: http://n4.nabble.com/test-the-goodness-of-it-for-negative-binomial-type-2-tp1575892p1585357.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variance of discrete uniform distribution
On Mon, Mar 8, 2010 at 3:44 PM, casperyc caspe...@hotmail.co.uk wrote: Hi Rolf Turner , God, it directed to the wrong page. I firstly find the formula in wiki, than tried to verify the answer in R, now, given that 143/12 ((n^2-1)/12 ) is the correct answer for a discrete uniform random variable, I am still not sure what R is calculating there? why it gives me 13? Of RT's two points, you addressed (b) continuous vs. discrete, but you have yet to address (a) population estimate based on a sample. Hint: var(1:12) tries to estimate the population variance based on a sample. You are interested in the population variance. They are calculated different formulas that differ *only in the denominator*. Michael Thanks! -- View this message in context: http://n4.nabble.com/variance-of-discrete-uniform-distribution-tp1585328p1585355.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] compare tables
You can just access the data from the list:
result - lapply(output, function(.data){
lettermatch(creator, .data)
})
You can then take the result and possibly 'cbind' back into the matrix you
want.
On Mon, Mar 8, 2010 at 10:59 AM, Laetitia Schmid laetitia.sch...@gmx.chwrote:
Hi!
I need some help to finish my script.
I have two tables that I combine randomly to produce a third table.
This I do for hundreds of iterations. In the output file I get all the
simulated tables after each other. It looks like this (in this case 3
iterations):
output file:
[[1]]
[,1] [,2] [,3]
[1,] GM030005 WI920024
[2,] GM930026 WI920362
[3,] GM980051 WI920007 CGCC
[4,] GM970009 WI920417
[5,] GM920089 WI920023
[6,] GM930109 WI920359
[7,] GM980007 WI920428 CGCC
[8,] GM940039 WI920430
[9,] GM990027 WI920349
[10,] GM920222 WI920410 CGCC
[11,] GM930029 WI920001 CGCC
[12,] GM990105 WI920431
[13,] GM050009 WI920430
[14,] GM920224 WI920369
[15,] GM920224 WI920352
[16,] GM960028 WI920427
[17,] GM940031 WI920004
[18,] GM930040 WI920441
[19,] GM930040 WI920441
[20,] GM050099 WI920417
[21,] GM050099 WI920423 CCCG
[22,] GM920096 WI920370
[23,] GM920034 WI920437
[24,] GM960023 WI920017
[25,] GM920031 WI920430
[26,] GM920202 WI920367 CCCG
[27,] GM990066 WI920410
[[2]]
[,1] [,2] [,3]
[1,] GM030005 WI920017
[2,] GM930026 WI920415
[3,] GM980051 WI920028 CGCC
[4,] GM970009 WI920017
[5,] GM920089 WI920028
[6,] GM930109 WI920353
[7,] GM980007 WI920009 CGCT
[8,] GM940039 WI920415
[9,] GM990027 WI920423 CCCG
[10,] GM920222 WI920423 CGCG
[11,] GM930029 WI920363 CGCC
[12,] GM990105 WI920362
[13,] GM050009 WI920365
[14,] GM920224 WI920362
[15,] GM920224 WI920410
[16,] GM960028 WI920355 CCCG
[17,] GM940031 WI920361
[18,] GM930040 WI920356
[19,] GM930040 WI920353
[20,] GM050099 WI920360
[21,] GM050099 WI920353
[22,] GM920096 WI920023
[23,] GM920034 WI920426
[24,] GM960023 WI920024
[25,] GM920031 WI920022
[26,] GM920202 WI920009 CCCG
[27,] GM990066 WI920001
[[3]]
[,1] [,2] [,3]
[1,] GM030005 WI920433
[2,] GM930026 WI920408
[3,] GM980051 WI920352 CGCC
[4,] GM970009 WI920416
[5,] GM920089 WI920022
[6,] GM930109 WI920369
[7,] GM980007 WI920415 CGCC
[8,] GM940039 WI920022
[9,] GM990027 WI920361
[10,] GM920222 WI920024 CGCC
[11,] GM930029 WI920437 CGCC
[12,] GM990105 WI920423 CCCG
[13,] GM050009 WI920416
[14,] GM920224 WI920423 CCCG
[15,] GM920224 WI920427
[16,] GM960028 WI920437
[17,] GM940031 WI920441
[18,] GM930040 WI920417
[19,] GM930040 WI920370
[20,] GM050099 WI920015
[21,] GM050099 WI920428
[22,] GM920096 WI920007
[23,] GM920034 WI920009 CCCG
[24,] GM960023 WI920410
[25,] GM920031 WI920430
[26,] GM920202 WI920015
[27,] GM990066 WI920415
Now I would like to compare one of the tables used to create the
output tables with every output table, one after the other. In detail,
I am comparing row 1 of the creator table with row 1 of the first
output table and then row 2 of the creator table with row 2 of the
first output table and so on until row 27 and each row for all
columns. Then, when the first output table is finished I go on
comparing the first creator table with the second table in the
output, row for row for all columns. I do this for all iterations.
The first creator table is called data_mc.
# apply similarity function (lettermatch) to my data
for (i in 1:(nrow(data_mc))){
for (y in 1:(ncol(data_mc))) {
creator_table - data_mc[data_mc$Status==mother,y]
output_tables - ???
output[i,y]-(lettermatch(creator_table, output_tables))
}
}
Could you please help me how I have to call up the output tables in
the way I need them (described above) for the function lettermatch?
Maybe I need to change the format of the output file?
Thank you.
Laetitia
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What is the problem that you are trying to solve?
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Re: [R] How can I understand this sentence,and express it by means of Mathematical approach？
On 9/03/2010, at 3:48 AM, Liaw, Andy wrote: (in response to a question on the meaning of the sentence: Independent variables whose correlation with the response variable was not significant at 5% level were removed) If your ultimate interest is in real scientific progress, I'd suggest that you ignore that sentence (and any conclusion drawn subsequent to it). Surely a fortune candidate. cheers, Rolf Turner ## Attention: This e-mail message is privileged and confidential. If you are not the intended recipient please delete the message and notify the sender. Any views or opinions presented are solely those of the author. This e-mail has been scanned and cleared by MailMarshal www.marshalsoftware.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] conditioning variable in panel.xyplot?
Ah, wonderful, thank you for the code Deepayan. To recap for posterity: I
have two datafiles, d and q: each has x-y coordinates that are conditioned
by site (The actual data, for me, is maps of parent trees and their
seedlings). I wanted to superimpose the xy plots of d and q, by site,
without going to the trouble of merging the d q datasets into a single
dataset. The solution is to use the which.packet statement is
d - data.frame(site = c(rep(A,12), rep(B,12)),
x=rnorm(24),y=rnorm(24))# Create the main xy dataset
q - data.frame(site = c(rep(A,7), rep(B,7)),
x=rnorm(14),y=rnorm(14))# Create the alternate xy dataset
q.split - split(q, q$site) # Split up the alternate
dataset by site
mypanel - function(..., alt.data) {
with(alt.data[[ which.packet()[1] ]], #
which.packet passes index of the relevant data subset...
panel.xyplot(x = x, y = y, col=red)) # ... to
panel.xyplot()
panel.xyplot(...)
}
xyplot(y ~ x | site, d, alt.data = q.split, # After providing
the alternative dataset and the panel...
panel = mypanel) # ...everything prints out
properly, like magic!
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
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Re: [R] conditioning variable in panel.xyplot?
Alternatively
library(latticeExtra)
xyplot(y ~ x | site, d) +
xyplot(y ~ x | site, q, col = red)
(which is a shortcut for:)
xyplot(y ~ x | site, d) +
as.layer(xyplot(y ~ x | site, q, col = red))
On 9 March 2010 11:17, Seth W Bigelow sbige...@fs.fed.us wrote:
Ah, wonderful, thank you for the code Deepayan. To recap for posterity: I
have two datafiles, d and q: each has x-y coordinates that are conditioned
by site (The actual data, for me, is maps of parent trees and their
seedlings). I wanted to superimpose the xy plots of d and q, by site,
without going to the trouble of merging the d q datasets into a single
dataset. The solution is to use the which.packet statement is
d - data.frame(site = c(rep(A,12), rep(B,12)),
x=rnorm(24),y=rnorm(24)) # Create the main xy dataset
q - data.frame(site = c(rep(A,7), rep(B,7)),
x=rnorm(14),y=rnorm(14)) # Create the alternate xy dataset
q.split - split(q, q$site) # Split up the alternate
dataset by site
mypanel - function(..., alt.data) {
with(alt.data[[ which.packet()[1] ]], #
which.packet passes index of the relevant data subset...
panel.xyplot(x = x, y = y, col=red)) # ... to
panel.xyplot()
panel.xyplot(...)
}
xyplot(y ~ x | site, d, alt.data = q.split, # After providing
the alternative dataset and the panel...
panel = mypanel) # ...everything prints out
properly, like magic!
Dr. Seth W. Bigelow
Biologist, USDA-FS Pacific Southwest Research Station
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Fenner School of Environment and Society [Bldg 48a]
The Australian National University
Canberra ACT 0200 Australia
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[R] varComb in gls/lme
Dear R-help members,
I have a question regarding how to use varComb function to specify a
variance function for the weights in the gls. I need to fit a
linear model with heteroscedasticity. The variance function is
exp(c0+nu0*W +nu1*W^2) where W is a covariate. Initially I want to use
varFunc to define my own variance function following the instruction in
the Pinheiro and Bates (2000), but I could not make it work. Then I used
varComb in gls with weights=varComb(varExp(form=~W),
varExp(form=~I(W^2). But the estimated variance parameters seems to
have a large discrepancy from the true values
(I used the simulated data). This makes me wonder if it is a right way
to model variance function exp(c0+nu0*W +nu1*W^2) using varComb. The
codes and outputs are copied below.
Any suggestions and help are very apprecited
library(nlme)
simulate.pilot = function(m, mn, sigma, alpha0, alpha1, c0, nu0, nu1) {
pilot.dat=data.frame(W=rnorm(m, mean=mn, sd=sigma))
pilot.dat=transform(pilot.dat, Y=rnorm(m, mean=alpha0 + alpha1*W,
sd=sqrt(exp(c0+nu0*W+nu1*W^2
pilot.dat
}
mn=3.3
sigma=sqrt(0.5)
alpha0=0.1
alpha1=3
m=200
n=200
c0=-2.413; nu0=-0.2; nu1=0.3
simu.dat=simulate.pilot(m, mn, sigma, alpha0, alpha1, c0, nu0, nu1)
fit1=try(gls(Y~W, data=simu.dat, weights=varComb(varExp(form=~W),
varExp(form=~I(W^2)
c0.hat= log(fit1$sigma^2)
nu0.hat=2*fit1$modelStruct$varStruct$A[1]
nu1.hat=2*fit1$modelStruct$varStruct$B[1]
c0.hat
[1] -1.570104
nu0.hat
[1] -0.787264
nu1.hat
[1] 0.4057129
Thanks
Xuesong
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Re: [R] A slight trap in read.table/read.csv.
Rolf Turner r.tur...@auckland.ac.nz wrote: I solved the problem by putting in a colClasses argument in my call to read.csv(). But I really think that the read functions are being too clever by half here. If field entries are surrounded by quotes, shouldn't they be left as character? Even if they are all F's and T's? Furthermore using F's and T's to represent TRUE's and FALSE's is bad practice anyway. Since FALSE and TRUE are reserved words it would make sense for the read function to assume that a field is logical if it consists entirely of these words. But T's and F's I don't think so. I would argue that this behaviour should be changed. I can see no downside to such a change. I agree with you, Rolf, that this is horrid behavior. It is such automatic devices that have made people hate (e.g.) Microsoft Word with a passion. Yet, in R this is a designed-in bug (e.g., feature) that probably can't be changed without making some legacy code not work. But at least, T and F could be removed soon as synonms for TRUE and FALSE. We have seen that _ was removed as an assignment operator, and the world did not crumble. The use of T and F is no less error-prone, and possibly more. The only immediate solution to this accretion of overly clever behavior would be for someone to write new functions (say, Read.csv) that didn't do all those conversions behind the scenes. I'm not about to do that. Are you? Best of luck! -- Mike Prager, NOAA, Beaufort, NC * Opinions expressed are personal and not represented otherwise. * Any use of tradenames does not constitute a NOAA endorsement. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] A slight trap in read.table/read.csv.
On 9/03/2010, at 11:17 AM, Mike Prager wrote:
Rolf Turner r.tur...@auckland.ac.nz wrote:
I solved the problem by putting in a colClasses argument in my
call to read.csv(). But I really think that the read functions
are being too clever by half here. If field entries are surrounded
by quotes, shouldn't they be left as character? Even if they are
all F's and T's?
Furthermore using F's and T's to represent TRUE's and FALSE's is
bad practice anyway. Since FALSE and TRUE are reserved words it
would make sense for the read function to assume that a field is
logical if it consists entirely of these words. But T's and F's
I don't think so.
I would argue that this behaviour should be changed. I can see no
downside to such a change.
I agree with you, Rolf, that this is horrid behavior. It is such
automatic devices that have made people hate (e.g.) Microsoft
Word with a passion.
Yet, in R this is a designed-in bug (e.g., feature) that
probably can't be changed without making some legacy code not
work. But at least, T and F could be removed soon as synonms for
TRUE and FALSE. We have seen that _ was removed as an
assignment operator, and the world did not crumble. The use of T
and F is no less error-prone, and possibly more.
I would definitely support the removal of the use of T
and F for TRUE and FALSE. Some code would break, but
it would be easy to trace the source of the problem and
easy to fix.
The only immediate solution to this accretion of overly clever
behavior would be for someone to write new functions (say,
Read.csv) that didn't do all those conversions behind the
scenes. I'm not about to do that. Are you?
NFL!!!
cheers,
Rolf
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Re: [R] why this function does not run correctly?
On 2010-03-08 14:45, Marco Bressan wrote:
Hi,
my name is Marco Bressan i'm working to improve ADati package. I study
psicology ad Padua University (Italy). I have this problem: why bartlett.test
function running good and my anova.welch function no?
Ciao,
il mio nome � Marco Bressan e sto lavorando per migliorare il pacchetto ADati.
Studio psicologia all'universit� di Padova. Non capisco come mai la funzione
che ho fatto mi dia quell'errore
this is anova.welch:
anova.welch- function(x, ...) UseMethod(anova.welch)
anova.welch.default-##
this is the algoritm, I think it's ok (I copy this from Welch() inside ADati)
function (x, y = NULL, nu = c(0,0) ,...)
{
mx- tapply(x,y,mean)
s2x- tapply(x,y,var)
k- length(nu)
w- nu/s2x
Xp- sum(w * mx)/sum(w)
Fnum- sum(w * (mx - Xp)^2)/(k - 1)
Fden- 0
for (h in 1:k) {
a- 1/(nu[h] - 1)
b- (1 - (w[h]/sum(w)))^2
Fden- Fden + (a * b)
}
gl2.den- Fden * 3
Fden- Fden * ((2 * (k - 2))/(k^2 - 1)) + 1
Fw- Fnum/Fden
STATISTIC- Fw
gl1- k - 1
gl2.num- k^2 - 1
gl2- gl2.num/gl2.den
PARAMETER- c(gl1, gl2)
PVAL- pf(Fw, gl1, gl2, lower.tail = FALSE)
METHOD- Welch ANOVA
DNAME- NA
names(STATISTIC)- F
names(PARAMETER)- c(num df, denom df)
RVAL- list(statistic = STATISTIC, parameter = PARAMETER,
p.value = PVAL, method = METHOD)
class(RVAL)- htest
return(RVAL)
}
anova.welch.formula-
## I copy this from bartlett.test
function(formula, data, subset, na.action, ...)
{
if(missing(formula) || (length(formula) != 3L))
stop('formula' mancante o incorretta)
m- match.call(expand.dots = FALSE)
if(is.matrix(eval(m$data, parent.frame(
m$data- as.data.frame(data)
m[[1L]]- as.name(model.frame)
mf- eval(m, parent.frame())
DNAME- paste(names(mf), collapse = by )
names(mf)- NULL
y- do.call(anova.welch, as.list(mf))
y$data.name- DNAME
y
}
bartlett.test doesn't have a 'nu' argument. Try this:
anova.welch.formula -
function(formula, data, subset, na.action, nu = c(0,0), ...)
{
if(missing(formula) || (length(formula) != 3L))
stop('formula' mancante o incorretta)
m - match.call(expand.dots = FALSE)
m[[nu]] - NULL ## needed so that eval(m,) will work
if(is.matrix(eval(m$data, parent.frame(
m$data - as.data.frame(data)
m[[1L]] - as.name(model.frame)
mf - eval(m, parent.frame())
DNAME - paste(names(mf), collapse = by )
names(mf) - NULL
y - do.call(anova.welch, c(as.list(mf), list(nu)))
## include 'nu' in the parameters passed to
##anova.welch.default
y$data.name - DNAME
y
}
(You might also find the code for lm() instructive.)
I assume that you're aware of the oneway.test() function in
package:stats. So why re-invent the wheel?
-Peter Ehlers
n1=10
## this is the test
n2=15
n3=20
y=c(rnorm((n1+n2),5,2),rnorm(n3,7,8))
A=factor(c(rep(1,n1),rep(2,n2),rep(3,n3)))
anova.welch(y,A,c(n1,n2,n3))
Welch ANOVA
data:
F = 2.3025, num df = 2.000, denom df = 27.384, p-value = 0.1191
anova.welch(y~A,nu=c(n1,n2,n3))
Errore in model.frame.default(formula = y ~ A, ... = list(nu = c(n1, n2, :
invalid type (pairlist) for variable '(...)'
Sorry for my english, tanks you if you can help me :)
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and provide commented, minimal, self-contained, reproducible code.
--
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University of Calgary
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Re: [R] A slight trap in read.table/read.csv.
Ditching T/F for TRUE/FALSE would get my vote, too.
-Peter Ehlers
On 2010-03-08 17:44, Rolf Turner wrote:
On 9/03/2010, at 11:17 AM, Mike Prager wrote:
Rolf Turnerr.tur...@auckland.ac.nz wrote:
I solved the problem by putting in a colClasses argument in my
call to read.csv(). But I really think that the read functions
are being too clever by half here. If field entries are surrounded
by quotes, shouldn't they be left as character? Even if they are
all F's and T's?
Furthermore using F's and T's to represent TRUE's and FALSE's is
bad practice anyway. Since FALSE and TRUE are reserved words it
would make sense for the read function to assume that a field is
logical if it consists entirely of these words. But T's and F's
I don't think so.
I would argue that this behaviour should be changed. I can see no
downside to such a change.
I agree with you, Rolf, that this is horrid behavior. It is such
automatic devices that have made people hate (e.g.) Microsoft
Word with a passion.
Yet, in R this is a designed-in bug (e.g., feature) that
probably can't be changed without making some legacy code not
work. But at least, T and F could be removed soon as synonms for
TRUE and FALSE. We have seen that _ was removed as an
assignment operator, and the world did not crumble. The use of T
and F is no less error-prone, and possibly more.
I would definitely support the removal of the use of T
and F for TRUE and FALSE. Some code would break, but
it would be easy to trace the source of the problem and
easy to fix.
The only immediate solution to this accretion of overly clever
behavior would be for someone to write new functions (say,
Read.csv) that didn't do all those conversions behind the
scenes. I'm not about to do that. Are you?
NFL!!!
cheers,
Rolf
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
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--
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University of Calgary
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Re: [R] Help with Hmisc, cut2, split and quantile
try as.numeric(read_data$DEC) this should turn it into a numeric variable that you can work with hth David Freedman CDC, Atlanta Guy Green wrote: Hi Peter others, Thanks (Peter) - that gets me really close to what I was hoping for. The one problem I have is that the cut approach breaks the data into intervals based on the absolute value of the Target data, rather than their frequency. In other words, if the data ranged from 0 to 50, the data would be separated into 0-5, 5-10 and so on, regardless of the frequency within those categories. However I want to get the data into deciles. The code that does this (incorporating Peter's) is: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC - with(read_data, cut(Target, breaks=10, labels=1:10)) L - split(read_data, read_data$DEC) This means that I can get separate data frames, such as L$'10', which comes out tidy, but only containing 2 data items (the sample has 63 rows, so each decile should have 6+ data items): ActualTarget DEC 9 0.572 0.3778386 10 31 0.2990.3546606 10 If I try to adjust this to get deciles using cut2(), I can break the data into deciles as follows: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC - with(read_data, cut2(read_data$Target, g=10), labels=1:10) L - split(read_data, read_data$DEC) However this time, while the data is broken into even data frames, the labels for the separate data frames are unuseable, e.g.: $`[ 0.26477, 0.37784]` ActualTarget DEC 6 0.243 0.2650960[ 0.26477, 0.37784] 9 0.572 0.3778386[ 0.26477, 0.37784] 10 -0.049 0.3212681[ 0.26477, 0.37784] 15 0.780 0.2778518[ 0.26477, 0.37784] 31 0.299 0.3546606[ 0.26477, 0.37784] 33 0.105 0.2647676[ 0.26477, 0.37784] Could anyone suggest a way of rearranging this to make the labels useable again? Sample data is reattached http://n4.nabble.com/file/n1585427/Sample_table.txt Sample_table.txt . Thanks, Guy Peter Ehlers wrote: On 2010-03-08 8:47, Guy Green wrote: Hello, I have a set of data with two columns: Target and Actual. A http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is attached but the data looks like this: Actual Target -0.125 0.016124906 0.135 0.120799865 ... ... ... ... I want to be able to break the data into tables based on quantiles in the Target column. I can see (using cut2, and also quantile) how to get the barrier points between the different quantiles, and I can see how I would achieve this if I was just looking to split up a vector. However I am trying to break up the whole table based on those quantiles, not just the vector. However I would like to be able to break the table into ten separate tables, each with both Actual and Target data, based on the Target data deciles: top_decile = ...(top decile of read_data, based on Target data) next_decile = ...and so on... bottom_decile = ... I would just add a factor variable indicating to which decile a particular observation belongs: dat$DEC - with(dat, cut(Target, breaks=10, labels=1:10)) If you really want to have separate data frames you can then split on the decile: L - split(dat, dat$DEC) -Peter Ehlers -- Peter Ehlers University of Calgary -- View this message in context: http://n4.nabble.com/Help-with-Hmisc-cut2-split-and-quantile-tp1584647p1585503.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RGtk2:::gdkColorToString throws an error
Dear Michael, thanks. I have installed gtk+ 2-12.9 revision 2 from http://gladewin32.sourceforge.net/, it still doesn't work. I tried gtk2-runtime-2.16.6-2010-02-24-ash.exe from http://gtk-win.sourceforge.net/home/index.php/en/Downloads, it did not solve the issue neither. Could you please give me more hints? My OS is windows vista. Regards Ronggui On 26 October 2009 23:27, Michael Lawrence mflaw...@fhcrc.org wrote: Hi and sorry for the late reply. The Gdk library is part of the GTK+ bundle (GTK+, Gdk and GdkPixbuf are distributed together and have synchronized versions). So you'll just need GTK+ 2.12 or higher. Michael On Tue, Oct 20, 2009 at 8:16 AM, Ronggui Huang ronggui.hu...@gmail.com wrote: Dear all, I try to use RGtk2:::gdkColorToString, but it throws an error: Error in .RGtkCall(S_gdk_color_to_string, object, PACKAGE = RGtk2) : gdk_color_to_string exists only in Gdk = 2.12.0 I know what it means, but don't know to solve this problem because I don't know where I can download the referred gdk library. Any information? Thank you. -- HUANG Ronggui, Wincent Doctoral Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html -- Wincent Ronggui HUANG Doctoral Candidate Dept of Public and Social Administration City University of Hong Kong http://asrr.r-forge.r-project.org/rghuang.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ctree - party package multivariate response variables
Hi, I have a problem with ctree of party package. I have data on distribution of more than one species (about 50 species) and I would like identify the relation of this multivariate object (species distribution) with a number of explanatory variables. rs is the name of my dataframe containing the species (columns from 2 to 51) and the explanatory variables (columns 52 and 53). Rows are my sampling sites. I wrote: species-rs[,2:51] v1-rs[,52] v2-rs[53] tree-ctree(species~v1+v2) It does not work , but when I use the same formula for the univariate case (i.e. a single column - e.g. the total number of species in each samplig sites) it works. I know that ctree can handle multivariate response variables, but I cannot figure out how to do that. Someone can help me? Thank you Valeriano __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Hmisc, cut2, split and quantile
Hi Peter others, Thanks (Peter) - that gets me really close to what I was hoping for. The one problem I have is that the cut approach breaks the data into intervals based on the absolute value of the Target data, rather than their frequency. In other words, if the data ranged from 0 to 50, the data would be separated into 0-5, 5-10 and so on, regardless of the frequency within those categories. However I want to get the data into deciles. The code that does this (incorporating Peter's) is: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC - with(read_data, cut(Target, breaks=10, labels=1:10)) L - split(read_data, read_data$DEC) This means that I can get separate data frames, such as L$'10', which comes out tidy, but only containing 2 data items (the sample has 63 rows, so each decile should have 6+ data items): ActualTarget DEC 9 0.572 0.3778386 10 31 0.2990.3546606 10 If I try to adjust this to get deciles using cut2(), I can break the data into deciles as follows: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC - with(read_data, cut2(read_data$Target, g=10), labels=1:10) L - split(read_data, read_data$DEC) However this time, while the data is broken into even data frames, the labels for the separate data frames are unuseable, e.g.: $`[ 0.26477, 0.37784]` ActualTarget DEC 6 0.243 0.2650960[ 0.26477, 0.37784] 9 0.572 0.3778386[ 0.26477, 0.37784] 10 -0.049 0.3212681[ 0.26477, 0.37784] 15 0.780 0.2778518[ 0.26477, 0.37784] 31 0.299 0.3546606[ 0.26477, 0.37784] 33 0.105 0.2647676[ 0.26477, 0.37784] Could anyone suggest a way of rearranging this to make the labels useable again? Sample data is reattached http://n4.nabble.com/file/n1585427/Sample_table.txt Sample_table.txt . Thanks, Guy Peter Ehlers wrote: On 2010-03-08 8:47, Guy Green wrote: Hello, I have a set of data with two columns: Target and Actual. A http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is attached but the data looks like this: Actual Target -0.125 0.016124906 0.1350.120799865 ... ... ... ... I want to be able to break the data into tables based on quantiles in the Target column. I can see (using cut2, and also quantile) how to get the barrier points between the different quantiles, and I can see how I would achieve this if I was just looking to split up a vector. However I am trying to break up the whole table based on those quantiles, not just the vector. However I would like to be able to break the table into ten separate tables, each with both Actual and Target data, based on the Target data deciles: top_decile = ...(top decile of read_data, based on Target data) next_decile = ...and so on... bottom_decile = ... I would just add a factor variable indicating to which decile a particular observation belongs: dat$DEC - with(dat, cut(Target, breaks=10, labels=1:10)) If you really want to have separate data frames you can then split on the decile: L - split(dat, dat$DEC) -Peter Ehlers -- Peter Ehlers University of Calgary -- View this message in context: http://n4.nabble.com/Help-with-Hmisc-cut2-split-and-quantile-tp1584647p1585427.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Making FTP operations with R
R does provide support for basic FTP requests. Not for DELETE requests. And not for communication on the same connection. I think your best approach is to use the RCurl package (http://www.omegahat.org/RCurl). D. Orvalho Augusto wrote: Dears I need to make some very basic FTP operations with R. I need to do a lot of get and issue a respective delete command too on the same connection. How can I do that? Thanks in advance Caveman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Help with Hmisc, cut2, split and quantile
On 2010-03-08 18:00, Guy Green wrote: Hi Peter others, Thanks (Peter) - that gets me really close to what I was hoping for. The one problem I have is that the cut approach breaks the data into intervals based on the absolute value of the Target data, rather than their frequency. In other words, if the data ranged from 0 to 50, the data would be separated into 0-5, 5-10 and so on, regardless of the frequency within those categories. However I want to get the data into deciles. The code that does this (incorporating Peter's) is: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC- with(read_data, cut(Target, breaks=10, labels=1:10)) L- split(read_data, read_data$DEC) This means that I can get separate data frames, such as L$'10', which comes out tidy, but only containing 2 data items (the sample has 63 rows, so each decile should have 6+ data items): ActualTarget DEC 9 0.572 0.3778386 10 31 0.2990.3546606 10 If I try to adjust this to get deciles using cut2(), I can break the data into deciles as follows: read_data=read.table(C:/Sample table.txt, head = T) read_data$DEC- with(read_data, cut2(read_data$Target, g=10), labels=1:10) L- split(read_data, read_data$DEC) However this time, while the data is broken into even data frames, the labels for the separate data frames are unuseable, e.g.: $`[ 0.26477, 0.37784]` ActualTarget DEC 6 0.243 0.2650960[ 0.26477, 0.37784] 9 0.572 0.3778386[ 0.26477, 0.37784] 10 -0.049 0.3212681[ 0.26477, 0.37784] 15 0.780 0.2778518[ 0.26477, 0.37784] 31 0.299 0.3546606[ 0.26477, 0.37784] 33 0.105 0.2647676[ 0.26477, 0.37784] Could anyone suggest a way of rearranging this to make the labels useable again? Sample data is reattached http://n4.nabble.com/file/n1585427/Sample_table.txt Sample_table.txt . I think that the easiest way would be to relabel the levels of DEC: read_data$DEC - factor(read_data$DEC, labels = 1:10) or, since I would prefer letters as factor levels: read_data$DEC - factor(read_data$DEC, labels = LETTERS[1:10]) Another way would be to use cut2() with onlycuts=TRUE to get the breaks and then use these with cut() as in my original post: brks - cut2(read_data$Target, g=10, onlycuts=TRUE) read_data$DEC- with(read_data, cut(Target, breaks=brks, labels=1:10)) But I still don't see why you want a list of separate data frames. For most analyses, it's more convenient to just use the factor variable to subset the data as needed. -Peter Ehlers Thanks, Guy Peter Ehlers wrote: On 2010-03-08 8:47, Guy Green wrote: Hello, I have a set of data with two columns: Target and Actual. A http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is attached but the data looks like this: Actual Target -0.125 0.016124906 0.135 0.120799865 ... ... ... ... I want to be able to break the data into tables based on quantiles in the Target column. I can see (using cut2, and also quantile) how to get the barrier points between the different quantiles, and I can see how I would achieve this if I was just looking to split up a vector. However I am trying to break up the whole table based on those quantiles, not just the vector. However I would like to be able to break the table into ten separate tables, each with both Actual and Target data, based on the Target data deciles: top_decile = ...(top decile of read_data, based on Target data) next_decile = ...and so on... bottom_decile = ... I would just add a factor variable indicating to which decile a particular observation belongs: dat$DEC- with(dat, cut(Target, breaks=10, labels=1:10)) If you really want to have separate data frames you can then split on the decile: L- split(dat, dat$DEC) -Peter Ehlers -- Peter Ehlers University of Calgary -- Peter Ehlers University of Calgary __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help! I turned my data into junk!
It's probably binary data - which implies that you can only read it with the application that created it. What is the filename extension? Ravi -- View this message in context: http://n4.nabble.com/Help-I-turned-my-data-into-junk-tp1585481p1585585.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: how to convert character variables into numeric variables directly
Hi r-help-boun...@r-project.org napsal dne 08.03.2010 18:55:10: Here is the example. age=18:29 height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5) type=c(A, B, C, D,A, B, C, D,A, B, C, D) typec=c(0,4,2,9,0,7,2,3,0,1,2,3) typen=c(0,1,2,3,0,1,2,3,0,1,2,3) data1=data.frame(age=age,height=height, type=type, typec=typec, typen=typen) With data1=data.frame(age=age,height=height, type=type, typec=typec, typen=typen, stringsAsFactors=F) your conversion shall work Regards Petr data1[,3]=as.numeric(data1[,3]) data1[,4]=as.numeric(data1[,4]) data1[,5]=as.numeric(data1[,5]) print(data1) and I got the output as: age height type typec typen 1 18 76.11 1 0 2 19 77.02 5 1 3 20 78.13 3 2 4 21 78.24 7 3 5 22 78.81 1 0 6 23 79.72 6 1 7 24 79.93 3 2 8 25 81.14 4 3 9 26 81.21 1 0 10 27 81.82 2 1 11 28 82.83 3 2 12 29 83.54 4 3 The typec is not what I expected. How can I get the direct conversion from character to numeric and get the following output? age height type typec typen 1 18 76.11 0 0 2 19 77.02 4 1 3 20 78.13 2 2 4 21 78.24 9 3 5 22 78.81 0 0 6 23 79.72 7 1 7 24 79.93 2 2 8 25 81.14 3 3 9 26 81.21 0 0 10 27 81.82 1 1 11 28 82.83 2 2 12 29 83.54 3 3 Thanks. Xumin [[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.
