Re: [R] maps does not work
I think the problem is that tempdir() has been set somewhere invalid.
However, does help() work at all, since that uses the same mechanism?
Here are some diagnostic hints:
Start an R session, printout tempdir() and see if it looks right. Search
for the directory on your file system and see if you can create a file in
it. If it is on a remote file system there could well be permission
issues.
See ?tempdir and set (or unset) environment variables to change tempdir()
to somewhere else you know is valid.
On Fri, 12 Oct 2007, Lukas Gudmundsson wrote:
2007/10/11, Ray Brownrigg [EMAIL PROTECTED]:
On Thu, 11 Oct 2007, Lukas Gudmundsson wrote:
Hello,
I am trying to draw geographical maps with the maps package. However
if I try to access the data following error occurs:
require(maps)
map()
Fehler in zip.file.extract(file, Rdata.zip) :
'destination' existiert nicht
# Error in zip.file.extract(file, Rdata.zip) : 'destination' does not
exist
traceback()
4: zip.file.extract(file, Rdata.zip)
3: data(list = dbname)
2: maptype(database)
1: map()
To check if I have principle problems with my system I tried:
require(mapdata)
map('worldHires', col=1:10)
# Works
I am running R 2.6.0 and a current version of maps. But I had a
similar problem with R 2.5.1
Is this a known issue?
thanks
Lukas
You don't say which OS you are using and I can't reproduce this on Windows so
it is hard to give explicit recommendations, but what if you try running:
R --vanilla
then:
library(maps); map()
Alternately, try:
map(world)
I suspect you have something in your workspace or your environment that is
clashing.
Ray
First at all thanks a lot for the reply and sorry for the incomplete
informations:
OS: Windows XP, SP2, version 2002
The computer is connected to a network with several data servers
(uni). R is installed locally at the default location. I have an user
and an admin account for this computer. As the problem I have seams
not to be common a short history of what I have done so far (without
success):
1. Installed the package using the admin login at the default location
2. Installed the package from the usr account into R_LIBS_USER (C:\RLIBS)
3. I have tried to dig into the error using options(error=recover). I
did not get to far but
file == C:/RLIBS/maps/data/worldMapEnv.r.
I had a look at the zip file in C:/RLIBS/maps/data. It contains
worldMapEnv.r (but as I do not really understand the package
architecture I can not interpret its single line contents (worldMapEnv
- R_MAP_DATA_DIR))
4. If I am trying to run any thing like example(lm) I am getting the
same error in zip.file.extract (but so far i did not care)
5. I tried to call R from the DOS box using R --vanilla but that did
not work either
6. I reinstalled maps form several CRAN mirrors
7. I tried to run maps on my private computer at home and that did
work... which increases the confusion.
again thanks for any help!
Lukas
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[R] random forest mtry and mse
I have been using random forest on a data set with 226 sites and 36 explanatory variables (continuous and categorical). When I use tune.randomforest to determine the best value to use in mtry there is a fairly consistent and steady decrease in MSE, with the optimum of mtry usually equal to 1. Why would that occur, and what does it signify? What I would assume is that most of my explanatory variables have little to no explanatory power. Does that sound about right? Thanks for any insight that anyone may have. dave __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] accessing ylim set by xyplot
At 22:48 12/10/07, you wrote:
Hello,
I would like to know if there is a clever way to avoid the problem
illustrated below within the xyplot function.
x - seq(1:10)
y - seq(1:10)
pr1 - xyplot(x ~ y)
u - seq(1:12)
v - seq(1:12)
pr2 - xyplot(u ~ v, col = red, more = FALSE)
prts - list(pr1, pr2)
for(i in prts) print(i, more = TRUE)
I realize that one possibility is to explicitly set the same values for
xlim and ylim for the plot, but that doesn't work very well for me
because I am accessing data via a function in a simulation loop in which
the ylims vary substantially depending on the variable being plotted.
What I would like to be able to do is to access the limits that xyplot
automatically sets when it initially creates a plot and then use those
values to estimate a ylim for succeeding rounds of the simulation. For
instance, if in the first round of a simulation the ylim set by xyplot
for a variable is c(0,20), I could then use that value as the ylim in
the following rounds for the same variable (other variables will be
assigned their limits accordingly). At the end of the program I will
have plots for 20 or so variables with each plot having several 'lines'
for each round of the simulation.
In short, is there a way to access the automatically determined ylim?
Kind regards,
Jeff
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Try using the prepanel function (untested for your example)
xyplot(x,y,
prepanel = function(y ...){list(ylim = range(y))}
)
you may have to use pretty to improve the values from the range
Regards
Duncan Mackay
Dept of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email [EMAIL PROTECTED]
home: [EMAIL PROTECTED]
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Re: [R] Anybody has ever met the problem to add a legend to a figure generated by image()
have you tried the function image.plot in the package fields? On Fri, 12 Oct 2007, zhijie zhang wrote: Dear friends, Anybody has ever met the problem to add a legend to a figure generated by image()? I have three variables,x,y and z. x and y are the coordinates, and z is the third values. we can use image(x, y, z,...) to generate a figure according to the z-values, but the problem is the figure legend. How can the legend be added to a figure generated by image()? Note that filled.contour() can add the figure legend automatically, but there are some problems sometime. I want to know the specific method for adding the legend to the figure generated by image(). Thanks very much. -- With Kind Regards, oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [***] Zhi Jie,Zhang ,PHD Tel:86-21-54237149 Dept. of Epidemiology,School of Public Health,Fudan University Address:No. 138 Yi Xue Yuan Road,Shanghai,China Postcode:200032 Email:[EMAIL PROTECTED] Website: www.statABC.com [***] oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] collapsing a data frame
Here is one way. Not sure what you wanted done with some of the other
variables, so I just chose the first one; you could do max/min:
z - by(h, h$BROOD, function(x){
+ # take first value of elements you don't want to change
+ data.frame(BROOD=x$BROOD[1], TICKS.mean=mean(x$TICKS),
TICKS.sd=sd(x$TICKS),
+ HEIGHT=x$HEIGHT[1], YEAR=x$YEAR[1], LOCATION=x$LOCATION[1])
+ })
do.call('rbind', z)
BROOD TICKS.mean TICKS.sd HEIGHT YEAR LOCATION
501 501 0 0.00465 95 32
502 502 0 NA472 95 36
503 503 1 1.732051475 95 37
On 10/12/07, Ben Bolker [EMAIL PROTECTED] wrote:
Trying to find a quick/slick/easily interpretable way to
collapse a data set.
Suppose I have a data set that looks like this:
h - structure(list(INDEX = structure(1:6, .Label = c(1, 2, 3,
4, 5, 6), class = factor), TICKS = c(0, 0, 0, 0, 0, 3
), BROOD = structure(c(1L, 1L, 2L, 3L, 3L, 3L), .Label = c(501,
502, 503), class = factor), HEIGHT = c(465, 465, 472, 475,
475, 475), YEAR = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c(95,
96, 97), class = factor), LOCATION = structure(c(1L, 1L,
2L, 3L, 3L, 3L), .Label = c(32, 36, 37), class = factor)), .Names =
c(INDEX,
TICKS, BROOD, HEIGHT, YEAR, LOCATION), row.names = c(NA,
6L), class = data.frame)
i.e.,
h
INDEX TICKS BROOD HEIGHT YEAR LOCATION
1 1 0 501465 95 32
2 2 0 501465 95 32
3 3 0 502472 95 36
4 4 0 503475 95 37
5 5 0 503475 95 37
6 6 3 503475 95 37
I want a data set that looks like this:
BROOD TICKS.mean HEIGHT YEAR LOCATION
501 0 465 95 32
502 0 472 95 36
503 1 475 95 37
(for example). I.e., I want to collapse it to a dataset by brood,
taking the mean of TICKS and reducing each of
the other variables (would be nice to allow multiple summary
statistics, e.g. TICKS.mean and TICKS.sd ...)
In some ways, this is the opposite of a database join/merge
operation -- I want to collapse the data frame back down.
If I had the unmerged (i.e., the brood table) handy I could
use it.
I know I can construct this table a bit at a time,
using tapply() or by() or aggregate() to get the means.
Here's a solution that takes the first element of each factor
and the mean of each numeric variable. I can imagine there
are more general/flexible solutions. (One might want to
specify more than one summary function, or specify that
factors that vary within group should be dropped.)
vtype = sapply(h,class) ## variable types [numeric or factor]
vtypes = unique(vtype) ## possible types
v2 = lapply(vtypes,function(z) which(vtype==z)) ## which are which?
cfuns = list(factor=function(z)z[1],numeric=mean)## functions to apply
m = mapply(function(w,f) { aggregate(h[w],list(h$BROOD),f) },
v2,cfuns,SIMPLIFY=FALSE)
data.frame(m[[1]],m[[2]][-1])
My question is whether this is re-inventing the wheel. Is there
some function or package that performs this task?
cheers
Ben Bolker
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http://www.nabble.com/collapsing-a-data-frame-tf4614195.html#a13177053
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Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] Basic plot question: Figure 1.1 Pinheiro Bates
On 10/12/07, David Afshartous [EMAIL PROTECTED] wrote: All, Sorry for overly simplistic question, but I can't seem to remember how to create the basic plot shown in Figure 1.1 of Pinheiro Bates (2004; p.4). The y-axis delineates a factor (Rail) while the x-axis displays the distribution of a continuous variable (time) according to each level of the factor. Didn't see it in archives but perhaps I'm not searching on correct key words, any help appreciated, David library(nlme) Rail Grouped Data: travel ~ 1 | Rail Rail travel 1 1 55 2 1 53 3 1 54 4 2 26 5 2 37 6 2 32 7 3 78 8 3 91 9 3 85 104 92 114100 124 96 135 49 145 51 155 50 166 80 176 85 186 83 That plot can be reproduced by library(lattice) data(Rail, package = nlme) dotplot(Rail ~ travel, Rail) However, this relies on the Rail$Rail factor being ordered by increasing mean travel time, which is fine for the plot but may get in the way of other uses of the data. In a way we only want to assign an order to the levels of the Rail factor for the purposes of the plot. As Deepayan Sarkar mentions in his useR!2007 presentation (http://user2007.org/program/presentations/sarkar.pd) he has done a considerable amount of upgrading of the lattice package, relative to the original design of Trellis graphics, in part to make the nlme plots easier to create. The version of the Rail data in the MEMSS package removes the ordering on the levels of the Rail factor so it defaults to the original order. Compare data(Rail, package = MEMSS) dotplot(Rail ~ travel, Rail) dotplot(reorder(Rail, travel) ~ travel, Rail) The latter plot is the way I would recommend constructing that plot now. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Automating binning for chisq.test()
On Fri, 2007-10-12 at 11:16 -0600, D. R. Evans wrote: The standard chisq.test() and fisher.test() functions, when applied to two distributions (to determine whether the same underlying distribution applies to both) requires one to pre-bin the distributions. Is there a library function (either built-in or in a package) that acts more like the ks.test() function, in that one can simply pass the two distributions and have it do the necessary binning as well as the actual statistical test? (Yes, you can accuse me of laziness: I just don't fancy trying to figure out a routine that would make sure that there more than 5 samples in each of the expected bins before applying the chi-squared test. It seems too much like re-inventing an elementary wheel that must have been invented by someone else.) You might want to review the following article: Chi-squared and Fisher-Irwin tests of two-by-two tables with small sample recommendations Ian Campbell Stat in Med 26:3661-3675; 2007 http://www3.interscience.wiley.com/cgi-bin/abstract/114125487/ABSTRACT Frank Harrell has offered some comments here (bottom of page): http://biostat.mc.vanderbilt.edu/twiki/bin/view/Main/DataAnalysisDisc 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.
Re: [R] Automating binning for chisq.test()
On 10/12/2007 1:16 PM, D. R. Evans wrote: The standard chisq.test() and fisher.test() functions, when applied to two distributions (to determine whether the same underlying distribution applies to both) requires one to pre-bin the distributions. Is there a library function (either built-in or in a package) that acts more like the ks.test() function, in that one can simply pass the two distributions and have it do the necessary binning as well as the actual statistical test? (Yes, you can accuse me of laziness: I just don't fancy trying to figure out a routine that would make sure that there more than 5 samples in each of the expected bins before applying the chi-squared test. It seems too much like re-inventing an elementary wheel that must have been invented by someone else.) If you have a quantile function q() for the distribution, a sample size of N, and want expected counts of 5 in each bin, just calculate the cutpoints as nbins - floor(N/5) cutpoints - c(-Inf, q( (1:(nbins-1)/nbins)), Inf) Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] collapsing a data frame
Trying to find a quick/slick/easily interpretable way to
collapse a data set.
Suppose I have a data set that looks like this:
h - structure(list(INDEX = structure(1:6, .Label = c(1, 2, 3,
4, 5, 6), class = factor), TICKS = c(0, 0, 0, 0, 0, 3
), BROOD = structure(c(1L, 1L, 2L, 3L, 3L, 3L), .Label = c(501,
502, 503), class = factor), HEIGHT = c(465, 465, 472, 475,
475, 475), YEAR = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c(95,
96, 97), class = factor), LOCATION = structure(c(1L, 1L,
2L, 3L, 3L, 3L), .Label = c(32, 36, 37), class = factor)), .Names =
c(INDEX,
TICKS, BROOD, HEIGHT, YEAR, LOCATION), row.names = c(NA,
6L), class = data.frame)
i.e.,
h
INDEX TICKS BROOD HEIGHT YEAR LOCATION
1 1 0 501465 95 32
2 2 0 501465 95 32
3 3 0 502472 95 36
4 4 0 503475 95 37
5 5 0 503475 95 37
6 6 3 503475 95 37
I want a data set that looks like this:
BROOD TICKS.mean HEIGHT YEAR LOCATION
501 0 465 95 32
502 0 472 95 36
503 1 475 95 37
(for example). I.e., I want to collapse it to a dataset by brood,
taking the mean of TICKS and reducing each of
the other variables (would be nice to allow multiple summary
statistics, e.g. TICKS.mean and TICKS.sd ...)
In some ways, this is the opposite of a database join/merge
operation -- I want to collapse the data frame back down.
If I had the unmerged (i.e., the brood table) handy I could
use it.
I know I can construct this table a bit at a time,
using tapply() or by() or aggregate() to get the means.
Here's a solution that takes the first element of each factor
and the mean of each numeric variable. I can imagine there
are more general/flexible solutions. (One might want to
specify more than one summary function, or specify that
factors that vary within group should be dropped.)
vtype = sapply(h,class) ## variable types [numeric or factor]
vtypes = unique(vtype) ## possible types
v2 = lapply(vtypes,function(z) which(vtype==z)) ## which are which?
cfuns = list(factor=function(z)z[1],numeric=mean)## functions to apply
m = mapply(function(w,f) { aggregate(h[w],list(h$BROOD),f) },
v2,cfuns,SIMPLIFY=FALSE)
data.frame(m[[1]],m[[2]][-1])
My question is whether this is re-inventing the wheel. Is there
some function or package that performs this task?
cheers
Ben Bolker
--
View this message in context:
http://www.nabble.com/collapsing-a-data-frame-tf4614195.html#a13177053
Sent from the R help mailing list archive at Nabble.com.
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[R] Batch-Plot Histograms
Hi everyone,
anybodyâs got an idea why the following script doesnât produce
batch-histograms?
Iâm using Windows XP and R 2.5.1.
Hereâs the script:
matrix-read.csv(C:\\Stadtwerke_mit_Umlage.csv, header=TRUE,sep=;,dec=.)
Stadtwerke-colnames(matrix)
Bereich_blau-66.67
Schleifen-dim(matrix)[2]
Plan-c(35.46,1.75,63.49,13.52,39.38,21.83)
Median-apply(matrix,2,median)
breaks-round((0.01)*dim(matrix[1]))
for(i in 1:Schleifen){
hist(matrix[,i],main=paste(Risiko-Chancenprofil der Stadtwerke ,
Stadtwerke[i]),col=blue,xlab=EBIT in Mio. â¬,
ylab=Häufigkeit,breaks[1])
abline(v=Median[i],lwd=2,col=grey,lty=3)
axis(side=1,at=Median[i],labels=Median,col=grey)
hist_data-hist(matrix[,i],plot=F,breaks[1])
Quantil_unten-quantile(matrix[,i],((100-Bereich_blau)/2)/100)
Quantil_oben-quantile(matrix[,i], (100-(100-Bereich_blau)/2)/100)
#Plotten der roten unterer Teil
for(j in 1:length(hist_data$breaks)){
if (hist_data$breaks[j]Quantil_unten)
hv-j
}
for(k in 1:hv){
rect(hist_data$breaks[k],0,hist_data$breaks[k+1],hist_data$counts[k],col=red)
}
for(l in 1:length(hist_data$breaks)){
if (hist_data$breaks[l]Quantil_oben)
hv2-l+1
}
for(m in hv2:length(hist_data$breaks)){
rect(hist_data$breaks[m],0,hist_data$breaks[m+1],hist_data$counts[m],col=red)
}
dev.copy(device = postscript,file=paste(C:\\,Stadtwerke[i],.ps))
dev.off()
}
It seems, that the error is in line âdev.copyâ⦠but I donât understand
why. It doesnât matter if Iâm using device = jpeg or postscript. Maybe I
must add a waiting time of 2 seconds or something like that?
Thanks a lot,
Thomas
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Re: [R] calculate impulse responses
Perhaps the VAR and irf functions in the vars library will allow you to do what you want. good luck, spencer On 10/12/07, Martin Ivanov [EMAIL PROTECTED] wrote: Dear R users, I need perform structural analysis on a no intercept VAR model. Unfortunately the functions irf.VAR and dfev that come with the MSBVAR package only work with objects output by the reduced.form.var function, which seems to only evaluate VAR models with intercept. Is there a way to suppress the estimation of intercept term in reduced.form.var? Do I need to modify the code, and if I do, how? Is there a way to apply irf and dfev to objects of type systemfit? Regards, Martin Ivanov - ëÒÁÊÎÁ ÃÅÌ - äÁ ÏÃÅÌÅÅÛ! www.survivor.btv.bg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] embedding plots
Hello! I am trying to embed a plot of a curve(say x^2) on a matrix that I am viewing using the image(matrix) command. I was wondering if someone could give me some idea of how to do this. Thanks, J. [[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] Basic plot question: Figure 1.1 Pinheiro Bates
On Friday 12 October 2007 17:31:22 David Afshartous wrote: DA All, DA Sorry for overly simplistic question, but I can't seem to remember how to DA create the basic plot shown in Figure 1.1 of Pinheiro Bates (2004; p.4). DA The y-axis delineates a factor (Rail) while the x-axis displays the DA distribution of a continuous variable (time) according to each level of the DA factor. Didn't see it in archives but perhaps I'm not searching on correct DA key words, any help appreciated, DA David library(lattice) xyplot(Rail~travel,data=Rail) signature.asc Description: This is a digitally signed message part. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Disregard my previous question
I did ?par and it looks switching to mfrow will fix my problem so
thanks but disregard my previous message and
I apologize for not checking ?par first. I never thought it would be
that simple.
This is not an offer (or solicitation of an offer) to bu...{{dropped:22}}
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[R] hist () x-axis
Hello all together, I (R Beginner) think the answer for my question is easy but anyway I have not found the solution. I am using hist for a frequency histogramm. But the divisions of the xaxis is to large How can I change the division (I am not sure if this is the right name for what I mean) of the xaxis, so that it shows not only 0.3 and 0.4 but also 0.35? Thanks in advance for your help. Regards Birgit Birgit Lemcke Institut für Systematische Botanik Zollikerstrasse 107 CH-8008 Zürich Switzerland Ph: +41 (0)44 634 8351 [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with two libraries
Hi, I'm running sessionInfo() R version 2.6.0 (2007-10-03) i386-apple-darwin8.10.1 locale: en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base I have two libraries: (1) /Users/mk/R/i386-apple-darwin8.10.1-library/2.6/ and (2) /Library/Frameworks/R.framework/Versions/2.6/Resources/library/ When I try to load a recently-installed package (plotAndPlayGTK) which got installed in (1) it tries to load RGtk2 which is in (2) and it fails. I would like advice on how to consolidate these two in (2) and make (2) the default library (if that's adivsable) and how to make the upgrade to future versions of R as trouble-free as possible. _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 know whether Vector B is a subset of Vector A
On Fri, 2007-10-12 at 18:47 +0800, Samuel wrote: Hi, I'm quite fresh to R, and a layman of English as well. I hope I can make you understood. Now I have two vectors A and B. Is there any quick way to know whether B is a subset of A? and If B is a subset of A, can I know easily which elements in A (the index of A) equals to B's elements accordingly? For example, a-1:20 b=c(2,5,9,7,4,8,3) Question 1: we know b is a subset of a, but how does R know that? Question 2: since we know b is a subset of a, then which a's elements equals to b? I do it like this: test=0 for (i in 1:length(b)) test[i]=which(a==b[i]) test Is there any easier way to know these indexes? Thanks a lot for you helps in advance You can use %in% to determine whether or not the elements of 'b' are in 'a': a - 1:20 b - c(2, 5, 9, 7, 4, 8, 3) b %in% a [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE To get the indices of b's elements in 'a', you can do: which(a %in% b) [1] 2 3 4 5 7 8 9 If you want to know whether or not ALL of the elements of 'b' are in 'a', you add the use of all() to the first example: all(b %in% a) [1] TRUE See ?%in% and ?all 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.
Re: [R] how to know whether Vector B is a subset of Vector A
Samuel wrote: Hi, I'm quite fresh to R, and a layman of English as well. I hope I can make you understood. Now I have two vectors A and B. Is there any quick way to know whether B is a subset of A? and If B is a subset of A, can I know easily which elements in A (the index of A) equals to B's elements accordingly? For example, a-1:20 b=c(2,5,9,7,4,8,3) Question 1: we know b is a subset of a, but how does R know that? Question 2: since we know b is a subset of a, then which a's elements equals to b? I do it like this: test=0 for (i in 1:length(b)) test[i]=which(a==b[i]) test Is there any easier way to know these indexes? Thanks a lot for you helps in advance a-1:20 b-c(2,5,9,7,4,8,3) !length(setdiff(b,a)) [1] TRUE a %in% b [1] FALSE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE FALSE FALSE FALSE [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE which(a %in% b) [1] 2 3 4 5 7 8 9 -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Anybody has ever met the problem to add a legend to a figure generated by image()?
Dear friends, Anybody has ever met the problem to add a legend to a figure generated by image()? I have three variables,x,y and z. x and y are the coordinates, and z is the third values. we can use image(x, y, z,...) to generate a figure according to the z-values, but the problem is the figure legend. How can the legend be added to a figure generated by image()? Note that filled.contour() can add the figure legend automatically, but there are some problems sometime. I want to know the specific method for adding the legend to the figure generated by image(). Thanks very much. -- With Kind Regards, oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [***] Zhi Jie,Zhang ,PHD Tel:86-21-54237149 Dept. of Epidemiology,School of Public Health,Fudan University Address:No. 138 Yi Xue Yuan Road,Shanghai,China Postcode:200032 Email:[EMAIL PROTECTED] Website: www.statABC.com [***] oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Q-type factor analysis
Hallo! Is there a package in R that does Q-type factor analysis? I know how to do principal component analysis, but haven't found any application of Q-type factor analysis. Thx, Julia -- Pt! Schon vom neuen GMX MultiMessenger gehört? Der kanns mit allen: http://www.gmx.net/de/go/multimessenger __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] irfs from a no intercept VAR
Dear R users, I need perform structural analysis on a no intercept VAR model. Unfortunately the functions irf.VAR and dfev that come with the MSBVAR package only work with objects output by the reduced.form.var function, which seems to only evaluate VAR models with intercept. Is there a way to suppress the estimation of intercept term in reduced.form.var? Do I need to modify the code, and if I do, how? Is there a way to apply irf and dfev to objects of type systemfit? Regards, Martin Ivanov - Крайна цел - Да оцелееш! www.survivor.btv.bg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Basic plot question: Figure 1.1 Pinheiro Bates
All, Sorry for overly simplistic question, but I can't seem to remember how to create the basic plot shown in Figure 1.1 of Pinheiro Bates (2004; p.4). The y-axis delineates a factor (Rail) while the x-axis displays the distribution of a continuous variable (time) according to each level of the factor. Didn't see it in archives but perhaps I'm not searching on correct key words, any help appreciated, David library(nlme) Rail Grouped Data: travel ~ 1 | Rail Rail travel 1 1 55 2 1 53 3 1 54 4 2 26 5 2 37 6 2 32 7 3 78 8 3 91 9 3 85 104 92 114100 124 96 135 49 145 51 155 50 166 80 176 85 186 83 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 know whether Vector B is a subset of Vector A
Hi, I'm quite fresh to R, and a layman of English as well. I hope I can make you understood. Now I have two vectors A and B. Is there any quick way to know whether B is a subset of A? and If B is a subset of A, can I know easily which elements in A (the index of A) equals to B's elements accordingly? For example, a-1:20 b=c(2,5,9,7,4,8,3) Question 1: we know b is a subset of a, but how does R know that? Question 2: since we know b is a subset of a, then which a's elements equals to b? I do it like this: test=0 for (i in 1:length(b)) test[i]=which(a==b[i]) test Is there any easier way to know these indexes? Thanks a lot for you helps in advance -- Samuel Wu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] random forest mtry and mse
Dave, I have been using random forest on a data set with 226 sites and 36 explanatory variables (continuous and categorical). When I use tune.randomforest to determine the best value to use in mtry there is a fairly consistent and steady decrease in MSE, with the optimum of mtry usually equal to 1. Why would that occur, and what does it signify? What I would assume is that most of my explanatory variables have little to no explanatory power. Does that sound about right? I'm not sure that it means anything (I've seen this happen too). Essentially, this would indicate that, for this particular dataset, the random forest model needs the trees to be as uncorrelated as possible. If it were to like mtry = # predictors, this would indicate that bagging was the optimal model. There is the no free lunch theorem and this would apply to possible random forest sub-models; without information related to the specifics of the problem at hand, there is no reason to believe that any one model is uniformly best across problems. Did you have any subject-specific reason to think larger values of mtry were optimal? What was the difference in performance across all of the candidate values of mtry? I don't usually see a huge effect of altering mtry (change in accuracy or Rsquared = 5% in classification and regression models, respectively) relative the variation in resampling. Max __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Differencing data by groups
Colleagues,
I am analyzing data collected during oceanographic cruises. We have
conducted many cruises over the last decade. On each cruise we visit ~50
stations. At each station (termed EventNum)we lower an instrument that
measures depth, temperature, salinity and oxygen every few seconds as it is
lowered through the water. Data from all EventNums on all cruises are
stacked, generating a data file that might look like
CruiseID Year Season EventNum Region Depth SalinityTemp Oxygen
1CF0101 2001 Spring 849 Lower 0.056 21.1753 15.5626 8.12344
2CF0101 2001 Spring 849 Lower 0.103 21.3264 15.5511 8.11873
3CF0101 2001 Spring 849 Lower 0.131 21.4780 15.5492 8.10568
4CF0101 2001 Spring 849 Lower 0.169 21.5296 15.5492 8.10633
15000 CF0103 2001 Fall 929Mid 14.449 17.692 22.901 7.743
15001 CF0103 2001 Fall 929Mid 14.478 17.691 22.900 7.743
15002 CF0103 2001 Fall 929Mid 14.527 17.692 22.899 7.743
15003 CF0103 2001 Fall 929Mid 14.595 17.692 22.900 7.743
15004 CF0103 2001 Fall 929Mid 14.654 17.694 22.902 7.742
15005 CF0103 2001 Fall 929Mid 14.722 17.695 22.903 7.742
15006 CF0103 2001 Fall 929Mid 14.790 17.695 22.903 7.742
15007 CF0103 2001 Fall 929Mid 14.839 17.694 22.903 7.742
15008 CF0103 2001 Fall 929Mid 14.887 17.695 22.902 7.742
15009 CF0103 2001 Fall 929Mid 14.946 17.694 22.902 7.742
15010 CF0103 2001 Fall 929Mid 14.985 17.693 22.902 7.743
..
In total there are ~350k rows of data
I can calculate simple EventNum averages using
meanobs-aggregate(ctd[,7:9],by=list(EventNum=EventNum),mean)
But, I want to calculate a depth-weighted mean for salinity, temp and oxygen
for each EventNum. The depth-weighted mean is calculated as
Xbar = Sum{depth_interval*temp}/Sum(depth_intervals)
Where the depth interval is the differenced depths (i.e., 0.103-0.056 for
the second row of data).
I cannot seem to get diff to work the way I want it to - advice appreciated
Tom
Thomas J. Miller, Ph.D.
Professor
Chesapeake Biological Laboratory
University of Maryland Center for Environmental Science
Solomons, MD 20688
Tel: (410) 326-7276
Fax: (410) 326-7318
WWW: hjort.cbl.umces.edu
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[R] Plotting question
I am constructing plots ( regular not lattice ) and my initial command
is
par(mar=c(3,4,2,2), mfcol=c(5,2))
and then I create 10 plots on the page. It looks great but the plots on
the page go in the order
16
27
38
49
510
Where the numbers denote decile breakdowns.
Is there an easy way to make them go from left to right so
12
34
56
78
910
I could try to stay with the first way and reorder the deciles but that
would be tricky and beyond
my current knowledge base. Thanks.
This is not an offer (or solicitation of an offer) to bu...{{dropped:22}}
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Re: [R] Plotting question
Have you looked at layout() ?
Hadley
On 10/12/07, Leeds, Mark (IED) [EMAIL PROTECTED] wrote:
I am constructing plots ( regular not lattice ) and my initial command
is
par(mar=c(3,4,2,2), mfcol=c(5,2))
and then I create 10 plots on the page. It looks great but the plots on
the page go in the order
16
27
38
49
510
Where the numbers denote decile breakdowns.
Is there an easy way to make them go from left to right so
12
34
56
78
910
I could try to stay with the first way and reorder the deciles but that
would be tricky and beyond
my current knowledge base. Thanks.
This is not an offer (or solicitation of an offer) to bu...{{dropped:22}}
__
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--
http://had.co.nz/
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Re: [R] variance explained by each term in a GAM
Dear Prof. Wood, Just another quick question. I am doing model selection following Wood and Augustin (2002). One of the criteria for retaining a term is to see if removing it causes an increase in the GCV score. When doing this, do I also need to fix the smooth parameters? Thanks, Julian Burgos Fisheries Acoustics Research Lab School of Aquatic and Fishery Science University of Washington 1122 NE Boat Street Seattle, WA 98105 Simon Wood wrote: I think that your approach is reasonable, except that you should use the same smoothing parameters throughout. i.e the reduced models should use the same smoothing parameters as the full model. Otherwise you get in trouble if x1 and x2 are correlated, since the smoothing parameters will then tend to change alot when terms are dropped as one smooth tries to `do the work' of the other. Here's an example, (which is modifiable to illustrate the problem with not fixing the sp's) ## simulate some data set.seed(0) n-400 x1 - runif(n, 0, 1) ## to see problem with not fixing smoothing parameters ## remove the `##' from the next line, and the `sp' ## arguments from the `gam' calls generating b1 and b2. x2 - runif(n, 0, 1) ## *.1 + x1 f1 - function(x) exp(2 * x) f2 - function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 f - f1(x1) + f2(x2) e - rnorm(n, 0, 2) y - f + e ## fit full and reduced models... b - gam(y~s(x1)+s(x2)) b1 - gam(y~s(x1),sp=b$sp[1]) b2 - gam(y~s(x2),sp=b$sp[2]) b0 - gam(y~1) ## calculate proportions deviance explained... (deviance(b1)-deviance(b))/deviance(b0) ## prop explained by s(x2) (deviance(b2)-deviance(b))/deviance(b0) ## prop explained by s(x1) On Monday 08 October 2007 20:19, Julian M Burgos wrote: Hello fellow R's, I do apologize if this is a basic question. I'm doing some GAMs using the mgcv package, and I am wondering what is the most appropriate way to determine how much of the variability in the dependent variable is explained by each term in the model. The information provided by summary.gam() relates to the significance of each term (F, p-value) and to the wiggliness of the fitted smooth (edf), but (as far as I understand) there is no information on the proportion of variance explained. One alternative may be to fit alternative models without each term, and calculate the reduction in deviance. For example: m1=gam(y~s(x1) + s(x2)) # Full model m2=gam(y~s(x2)) m3=gam(y~s(x1)) ddev1=deviance(m1)-deviance(m2) ddev2=deviance(m1)-deviance(m3) Here, ddev1 would measure the relative proportion of the variability in y explained by x1, and ddev2 would do the same for x2. Does this sound like an appropriate approach? Julian Julian Burgos FAR lab University of Washington __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] wireframe shade=T colorkey
On 10/11/2007 6:32 PM, Deepayan Sarkar wrote: On 10/11/07, Karim Rahim [EMAIL PROTECTED] wrote: Thank you for your reply. In this graphics context, I'm not sure what the incident or reflected light rays are. May I ask for an example of using a colour key with the volcano data using the colours that appear when you select the shade option? It is simple to have the colour key appear using drape. Perhaps it is not so simple to have a colour key using shade colours or different colours. Once again, may I ask for an example of setting these colour key and/or colour options? I'm not really sure what you want. The goal of the colorkey is to associate a given z-value (or height) with a specific color. drape=TRUE does this, e.g., wireframe(volcano, drape = TRUE, colorkey = TRUE) Now, with shade=TRUE, e.g., wireframe(volcano, shade = TRUE) the SAME Z-VALUE CAN HAVE DIFFERENT COLORS depending on the orientation of the facet with respect to the viewing direction and the light source. So, a colorkey DOES NOT MAKE SENSE. But the colorkey could display what the color would be at some particular orientation, with height allowed to vary, or perhaps with fixed values for irradiance and reflective angle. With the standard palette this should be readable because the height determines the hue, and (non-colorblind) people are good at recognizing hue even when saturation and brightness vary. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] wireframe shade=T colorkey
On 10/12/07, Duncan Murdoch [EMAIL PROTECTED] wrote: On 10/11/2007 6:32 PM, Deepayan Sarkar wrote: On 10/11/07, Karim Rahim [EMAIL PROTECTED] wrote: Thank you for your reply. In this graphics context, I'm not sure what the incident or reflected light rays are. May I ask for an example of using a colour key with the volcano data using the colours that appear when you select the shade option? It is simple to have the colour key appear using drape. Perhaps it is not so simple to have a colour key using shade colours or different colours. Once again, may I ask for an example of setting these colour key and/or colour options? I'm not really sure what you want. The goal of the colorkey is to associate a given z-value (or height) with a specific color. drape=TRUE does this, e.g., wireframe(volcano, drape = TRUE, colorkey = TRUE) Now, with shade=TRUE, e.g., wireframe(volcano, shade = TRUE) the SAME Z-VALUE CAN HAVE DIFFERENT COLORS depending on the orientation of the facet with respect to the viewing direction and the light source. So, a colorkey DOES NOT MAKE SENSE. But the colorkey could display what the color would be at some particular orientation, with height allowed to vary, or perhaps with fixed values for irradiance and reflective angle. With the standard palette this should be readable because the height determines the hue, and (non-colorblind) people are good at recognizing hue even when saturation and brightness vary. Yes, and since it's easy to specify your own colors and breakpoints, you could do shade.col.fun - trellis.par.get(shade.colors)$palette shade.colors - shade.col.fun(0.5, 0.5, seq(0, 1, length = 100)) wireframe(volcano, shade = TRUE, colorkey = list(col = shade.colors, at = do.breaks(range(volcano), 100))) While this works for the standard color palette, it doesn't do anything useful for the standard black and white palette, which ignores height. Given the lack of generality, I would be reluctant to have something this happen automatically. -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] collapsing a data frame
On 10/12/07, Ben Bolker bolker at ufl.edu wrote: Trying to find a quick/slick/easily interpretable way to collapse a data set. Another alternative for SQL fans is the sqldf package. I used the MySQL driver here since SQLite does not support standard deviation. sqldf(select BROOD, avg(TICKS) as 'TICKS.mean', stddev_samp(TICKS) as 'TICKS.sd', HEIGHT, YEAR, LOCATION from h group by BROOD, HEIGHT, YEAR, LOCATION, drv=MySQL) BROOD TICKS.mean TICKS.sd HEIGHT YEAR LOCATION 1 501 0 0.00465 95 32 2 502 0 NA472 95 36 3 503 1 1.73475 95 37 Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Q-type factor analysis
On 10/12/07, Julia Kröpfl [EMAIL PROTECTED] wrote: Is there a package in R that does Q-type factor analysis? I know how to do principal component analysis, but haven't found any application of Q-type factor analysis. Q-mode factor analysis is not a separate type of factor analysis but (in old-fashioned psychological slang) analyzing of rows rather than the columns of data matrix. So you can transpose your data (with t() if it's a matrix) and do an ordinary factor analysis or PCA. Kenn [[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] Addition operation based on specific columns and rows of twodata frames
My guess is that there's an easier way but this gives what you want.
newY.df-aggregate(Y.df$Counts, list(Y.df[,1],Y.df[,2]), FUN=sum)
names(newY.df)-names(X.df)
temp.df-merge(newY.df, X.df,
by=intersect(names(X.df),names(newY.df)),all=TRUE)
almost.df-aggregate(temp.df$Counts, list(temp.df[,1],temp.df[,2]),
FUN=sum)
temp-order(almost.df$Group.1)
final.df-almost.df[temp,]
names(final.df)-names(X.df)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Luke Neraas
Sent: Friday, October 12, 2007 5:15 PM
To: r-help
Subject: [R] Addition operation based on specific columns and rows of
twodata frames
#Hello,
# I have a question about the addition of values in specific columns and
rows of a Data frame.
# Below I have created two data frames, X.df and Y.df.
## creation of X.df data frame
X- matrix(0,16,3)
X.df-data.frame(X)
X.df[,1] - c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
X.df[,2] - c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4)
names(X.df)[1]-L(A)a(i)
names(X.df)[2]-L(B)a(j)
names(X.df)[3]-Counts
X.df
## creation of Y.df data frame
Y- matrix(0,7,3)
Y.df-data.frame(Y)
Y.df[,1]-c(1,3,4,2,2,1,2)
Y.df[,2]-c(2,4,2,4,1,2,3)
Y.df[,3]-c(1,2,1,0.5,1,1,1)
names(Y.df)[1]-L(A)a(i)
names(Y.df)[2]-L(B)a(j)
names(Y.df)[3]-Counts
Y.df
# I would like to add the value in the Counts column of Y.df to the
Counts Column of X.df, based on matching # the row value from the
first two columns of each data.frame. I have tried the merge function
but that # doesn't give me the addition operation.
# here is what i would like the resulting operation to end up with in
the X.df data frame
X.df$Counts - c(0,2,0,0,1,0,1,0.5,0,0,0,2,0,1,0,0)
X.df
# Any help or ideas would be greatly appreciated
# Thanks in advance
# Luke Neraas
# [EMAIL PROTECTED]
# University of Alaska Fairbanks
# School of Fisheries and Ocean Sciences # 11120 Glacier Highway # UAF
Fisheries Division # Juneau, AK 99801
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Re: [R] use 'lapply' to creat 2 new columns based on old ones in a data frame
runner said the following on 10/12/2007 4:46 PM:
There is a dataset 'm', which has 3 columns: 'index', 'old1' and 'old2';
I want to create 2 new columns: 'new1' and 'new2' on this condition:
if 'index'==i, then 'new1'='old1'+add[i].
'add' is a vector of numbers to be added to old columns, e.g. add=c(10,20,30
...)
Like this:
index old1old2new1new2
1 5 6 15 16
2 5 6 2526
3 5 6 35 36
3 50 60 80 90
Since the actual dataset is huge, I use 'lapply'. I am able to add 1 column:
do.call(rbind, lapply( 1:nrow(m),
function(i) {m$new1[i]=m[i,2]+add[m[i,1]];
return (m[i,])}
))
but don't know how to do for 2 columns at the same time, sth. like this
simply doesn't work:
do.call(rbind,lapply(1:nrow(m),
function(i){ m$new1[i]=m[i,2]+add[m[i,1]];
m$new2[i]=m[i,3]+add[m[i,1]];
return (m[i,])}
))
Could you please tell me how? or any other better approach?
No need for lapply.
x$new1 - x$old1 + add[x$index]
x$new2 - x$old2 + add[x$index]
To see how this works, try:
add - c(10, 20, 30)
index - c(1, 2, 1, 3, 1, 2, 3)
add[index]
but be careful if 'add' is length 3 and 'index' has a 4 in it you will get
index - c(index, 4)
add[index] ## produces an 'NA'
I hope I understood your question correctly. It's happy hour on the U.S.
east coast.
HTH,
--sundar
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Re: [R] Differencing data by groups
What problem are you actually having with 'diff'? Now if you are using
'diff', you will get a vector that is shorter by one than the
original. Now do you want to do do something like:
Xbar = Sum{c(Depth[1], diff(Depth))*temp}/Sum(c(Depth[1], diff(Depth))
On 10/12/07, Thomas Miller [EMAIL PROTECTED] wrote:
Colleagues,
I am analyzing data collected during oceanographic cruises. We have
conducted many cruises over the last decade. On each cruise we visit ~50
stations. At each station (termed EventNum)we lower an instrument that
measures depth, temperature, salinity and oxygen every few seconds as it is
lowered through the water. Data from all EventNums on all cruises are
stacked, generating a data file that might look like
CruiseID Year Season EventNum Region Depth SalinityTemp Oxygen
1CF0101 2001 Spring 849 Lower 0.056 21.1753 15.5626 8.12344
2CF0101 2001 Spring 849 Lower 0.103 21.3264 15.5511 8.11873
3CF0101 2001 Spring 849 Lower 0.131 21.4780 15.5492 8.10568
4CF0101 2001 Spring 849 Lower 0.169 21.5296 15.5492 8.10633
15000 CF0103 2001 Fall 929Mid 14.449 17.692 22.901 7.743
15001 CF0103 2001 Fall 929Mid 14.478 17.691 22.900 7.743
15002 CF0103 2001 Fall 929Mid 14.527 17.692 22.899 7.743
15003 CF0103 2001 Fall 929Mid 14.595 17.692 22.900 7.743
15004 CF0103 2001 Fall 929Mid 14.654 17.694 22.902 7.742
15005 CF0103 2001 Fall 929Mid 14.722 17.695 22.903 7.742
15006 CF0103 2001 Fall 929Mid 14.790 17.695 22.903 7.742
15007 CF0103 2001 Fall 929Mid 14.839 17.694 22.903 7.742
15008 CF0103 2001 Fall 929Mid 14.887 17.695 22.902 7.742
15009 CF0103 2001 Fall 929Mid 14.946 17.694 22.902 7.742
15010 CF0103 2001 Fall 929Mid 14.985 17.693 22.902 7.743
..
In total there are ~350k rows of data
I can calculate simple EventNum averages using
meanobs-aggregate(ctd[,7:9],by=list(EventNum=EventNum),mean)
But, I want to calculate a depth-weighted mean for salinity, temp and oxygen
for each EventNum. The depth-weighted mean is calculated as
Xbar = Sum{depth_interval*temp}/Sum(depth_intervals)
Where the depth interval is the differenced depths (i.e., 0.103-0.056 for
the second row of data).
I cannot seem to get diff to work the way I want it to - advice appreciated
Tom
Thomas J. Miller, Ph.D.
Professor
Chesapeake Biological Laboratory
University of Maryland Center for Environmental Science
Solomons, MD 20688
Tel: (410) 326-7276
Fax: (410) 326-7318
WWW: hjort.cbl.umces.edu
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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[R] TLCA Admin Email Connectivity Re: Returned mail: Data format error
Due to an intermittent Broadband connection, I cannot check email as often as I'd like. Please be assured that all your emails are being received I will respond to those that require attention as my ISP Connection allows. CIncinnati Bell cannot advise as to when this will be fixed, so I ask that you please be patient in awaiting for a response. I am doing everything I am able to in order to respond in a timely fashion. Thank You! If you need to renew your current membership, or would like to join, you may do so on-line at www.tlca.org. Please click on JOIN in the upper right hand corner. All technical questions can be directed to www.ih8mud.com The TLCA on-line store is currently off-line. If you are reaching a site that is requesting a password, you are at an incorrect site. If you have recently joined the TLCA, please allow 4 - 6 weeks for processing. If you joined on-line, your pay pal receipt is your confirmation, Thank you! -- Thanks! Jennifer Lorincz TLCA Administrator 800.655.3810 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] collapsing a data frame
On 10/12/07, Ben Bolker [EMAIL PROTECTED] wrote:
Trying to find a quick/slick/easily interpretable way to
collapse a data set.
Suppose I have a data set that looks like this:
h - structure(list(INDEX = structure(1:6, .Label = c(1, 2, 3,
4, 5, 6), class = factor), TICKS = c(0, 0, 0, 0, 0, 3
), BROOD = structure(c(1L, 1L, 2L, 3L, 3L, 3L), .Label = c(501,
502, 503), class = factor), HEIGHT = c(465, 465, 472, 475,
475, 475), YEAR = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c(95,
96, 97), class = factor), LOCATION = structure(c(1L, 1L,
2L, 3L, 3L, 3L), .Label = c(32, 36, 37), class = factor)), .Names =
c(INDEX,
TICKS, BROOD, HEIGHT, YEAR, LOCATION), row.names = c(NA,
6L), class = data.frame)
i.e.,
h
INDEX TICKS BROOD HEIGHT YEAR LOCATION
1 1 0 501465 95 32
2 2 0 501465 95 32
3 3 0 502472 95 36
4 4 0 503475 95 37
5 5 0 503475 95 37
6 6 3 503475 95 37
I want a data set that looks like this:
BROOD TICKS.mean HEIGHT YEAR LOCATION
501 0 465 95 32
502 0 472 95 36
503 1 475 95 37
(for example). I.e., I want to collapse it to a dataset by brood,
taking the mean of TICKS and reducing each of
the other variables (would be nice to allow multiple summary
statistics, e.g. TICKS.mean and TICKS.sd ...)
In some ways, this is the opposite of a database join/merge
operation -- I want to collapse the data frame back down.
If I had the unmerged (i.e., the brood table) handy I could
use it.
I know I can construct this table a bit at a time,
using tapply() or by() or aggregate() to get the means.
Here's a solution that takes the first element of each factor
and the mean of each numeric variable. I can imagine there
are more general/flexible solutions. (One might want to
specify more than one summary function, or specify that
factors that vary within group should be dropped.)
vtype = sapply(h,class) ## variable types [numeric or factor]
vtypes = unique(vtype) ## possible types
v2 = lapply(vtypes,function(z) which(vtype==z)) ## which are which?
cfuns = list(factor=function(z)z[1],numeric=mean)## functions to apply
m = mapply(function(w,f) { aggregate(h[w],list(h$BROOD),f) },
v2,cfuns,SIMPLIFY=FALSE)
data.frame(m[[1]],m[[2]][-1])
My question is whether this is re-inventing the wheel. Is there
some function or package that performs this task?
Maybe the reshape package? http://had.co.nz/reshape
hm - melt(h, m = TICKS)
cast(hm, BROOD + HEIGHT + YEAR + LOCATION ~ ., mean)
cast(hm, BROOD + HEIGHT + LOCATION ~ YEAR, mean)
cast(hm, BROOD ~ HEIGHT ~ YEAR, mean)
You should be able to create just about any data structure you need,
and if you can't let me know.
Hadley
--
http://had.co.nz/
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Re: [R] collapsing a data frame
Here's a solution that takes the first element of each factor
and the mean of each numeric variable. I can imagine there
are more general/flexible solutions. (One might want to
specify more than one summary function, or specify that
factors that vary within group should be dropped.)
vtype = sapply(h,class) ## variable types [numeric or factor]
vtypes = unique(vtype) ## possible types
v2 = lapply(vtypes,function(z) which(vtype==z)) ## which are which?
cfuns = list(factor=function(z)z[1],numeric=mean)## functions to apply
m = mapply(function(w,f) { aggregate(h[w],list(h$BROOD),f) },
v2,cfuns,SIMPLIFY=FALSE)
data.frame(m[[1]],m[[2]][-1])
My question is whether this is re-inventing the wheel. Is there
some function or package that performs this task?
Maybe the reshape package? http://had.co.nz/reshape
hm - melt(h, m = TICKS)
cast(hm, BROOD + HEIGHT + YEAR + LOCATION ~ ., mean)
cast(hm, BROOD + HEIGHT + LOCATION ~ YEAR, mean)
cast(hm, BROOD ~ HEIGHT ~ YEAR, mean)
You should be able to create just about any data structure you need,
and if you can't let me know.
Oh, and you can easily use multiple summary functions too:
cast(hm, BROOD + HEIGHT + YEAR + LOCATION ~ ., c(mean, sd, length))
Hadley
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Re: [R] Image analysis and image questions
See the EBImage package on Bioconductor.org. It builds on top of ImageMagick. /Henrik On 10/12/07, Bio7 [EMAIL PROTECTED] wrote: Dear R users, in my application i can transfer images to R with the help of Rserve. The images come from a java application. When i plot a greyscale image (values 0-255) with images (imageMatrix...as grey) the image is created with inverse colours. My first question is how can i plot the image with the default grey values. In the application the greys are (ImageJ): white=0,black=255. I would be also interested in operations on images inside R to test some image functionality. So which packages exist to do some image operations and how can i create and plot images with them (with available data structures like a list or a matrix)? Another point is the creation of coloured images inside R. How could i create and plot coloured images inside R and which data structure do i need? In addition to this question for coloured images - are there any bit arithmetics available to seperate the colour space or do i have to transfer the rgb values seperately? Any help is appreciated With kind regards Marcel -- View this message in context: http://www.nabble.com/Image-analysis-and-image-questions-tf4611810.html#a13170210 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.
