Re: [R] Euclidean Distance in 3 Dimensions
This function unfortunately does not work in 3d space.
Thoughts?
On Wed, Aug 20, 2014 at 4:57 PM, Don McKenzie d...@u.washington.edu wrote:
?dist
from the help
dist {stats}R Documentation
Distance Matrix Computation
Description
This function computes and returns the distance matrix computed by using
the specified distance measure to compute the distances between the rows of
a data matrix.
Is this what you want? Computing on a matrix whose rows are your x, y,
and z values?
On Aug 20, 2014, at 1:12 PM, Patzelt, Edward patz...@g.harvard.edu
wrote:
R Community -
I am attempting to write a function that will calculate the distance
between points in 3 dimensional space for unique regions (e.g. localized
brain regions such as the frontal lobe).
For example I'm looking to compare each point in region 45 to every other
region in 45 to establish if they are a distance of 8 or more apart. I
can
do this linearly comparing each distance to the previous but this is not
comparing all points.
structure(list(Cluster.Index = c(46L, 46L, 46L, 46L, 46L, 45L,
45L, 45L, 45L, 45L, 44L, 44L, 44L, 44L, 44L, 43L, 43L, 43L, 43L,
43L), Value = c(8.21, 7.96, 7.85, 7.83, 7.8, 5.38, 4.56, 4.5,
4, 3.99, 5.42, 4.82, 4.21, 4.18, 3.91, 4.79, 4.27, 3.24, 3.06,
3.04), x = c(33L, 38L, 37L, 36L, 38L, 47L, 42L, 43L, 44L, 42L,
50L, 41L, 39L, 41L, 44L, 46L, 45L, 45L, 41L, 46L), y = c(15L,
12L, 12L, 13L, 13L, 91L, 84L, 84L, 95L, 96L, 69L, 70L, 65L, 65L,
59L, 41L, 40L, 46L, 44L, 47L), z = c(41L, 38L, 41L, 39L, 33L,
39L, 40L, 42L, 44L, 45L, 34L, 36L, 30L, 35L, 39L, 53L, 47L, 61L,
52L, 57L), X = c(NA, 6.557438524302, 3.16227766016838, 2.44948974278318,
6.32455532033676, 78.7464284904401, 8.66025403784439, 2.23606797749979,
11.2249721603218, 2.44948974278318, 30.2324329156619, 9.2736184954957,
8.06225774829855, 5.3851648071345, 7.81024967590665, 22.8910462845192,
6.16441400296898, 15.2315462117278, 10.0498756211209, 7.68114574786861
)), .Names = c(Cluster.Index, Value, x, y, z, X), row.names =
c(NA,
20L), class = data.frame)
mainDat - data.frame()
for(i in 2:nrow(dat)){
tempDist - (sqrt((dat$x[i] - dat$x[i-1])^2 + (dat$y[i] - dat$y[i-1])^2 +
(dat$z[i] - dat$z[i-1])^2))
dat$X[i] - c(tempDist)
if(dat$Cluster.Index[i] != dat$Cluster.Index[i-1]){
mainDat - rbind(mainDat, dat[i,])
}
if((dat$Cluster.Index[i] == dat$Cluster.Index[i-1])) {
if(tempDist 8){
mainDat - rbind(mainDat, dat[i,])
}
}
}
--
*Edward H Patzelt | Clinical Science PhD StudentPsychology | Harvard
University *
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Don McKenzie
Research Ecologist
Pacific Wildland Fire Sciences Lab
US Forest Service
Affiliate Professor
School of Environmental and Forest Sciences
University of Washington
d...@uw.edu
--
*Edward H Patzelt | Clinical Science PhD StudentPsychology | Harvard
University *
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Re: [R] Euclidean Distance in 3 Dimensions
Your first description is correct with slight modification compare point 1
to all the other points in that Cluster.Index and see if any of euclidean
distances are greater than 8; do this for each point (i.e. point 2, point
3) in that specific Cluster.Index (i.e. 45)
On Thu, Aug 21, 2014 at 3:35 PM, David L Carlson dcarl...@tamu.edu wrote:
The dist() function works just fine in 2d or 3d or 100d. Your
description of what you want to accomplish is not clear. Your code compares
rows 1 and 2, then 2 and 3, then 3 and 4, and so on. You are comparing only
adjacent points, but your description makes it sound like you want to
compare point 1 to all the other points and see if they are in the same
group and over 8 or in another group. If you type the following command you
will see that your dat$X is just the diagonal of the distance matrix: 1
with 2, 2 with 3, 3 with 4 etc:
dist(dat[, 3:5])
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
*From:* r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
*On Behalf Of *Don McKenzie
*Sent:* Thursday, August 21, 2014 1:44 PM
*To:* Patzelt, Edward
*Cc:* R-help@r-project.org
*Subject:* Re: [R] Euclidean Distance in 3 Dimensions
Ugh sorry. I misread your message obviously. Cc�ing back to the list (as
is the protocol)
I�m surprised no one else has replied. I�m a lightweight compared to
others on the list. It looks as if the dist() function has compiled code,
which suggests that there is some gnarly linear algebra underneath to
speed it up even in 2D. Not for the faint-of-heart to hack.
Others? �dist3D�?
On Aug 21, 2014, at 11:34 AM, Patzelt, Edward patz...@g.harvard.edu
wrote:
This function unfortunately does not work in 3d space.
Thoughts?
On Wed, Aug 20, 2014 at 4:57 PM, Don McKenzie d...@u.washington.edu
wrote:
?dist
from the help
dist {stats}R Documentation
Distance Matrix Computation
Description
This function computes and returns the distance matrix computed by using
the specified distance measure to compute the distances between the rows of
a data matrix.
Is this what you want? Computing on a matrix whose rows are your x, y,
and z values?
On Aug 20, 2014, at 1:12 PM, Patzelt, Edward patz...@g.harvard.edu
wrote:
R Community -
I am attempting to write a function that will calculate the distance
between points in 3 dimensional space for unique regions (e.g.
localized
brain regions such as the frontal lobe).
For example I'm looking to compare each point in region 45 to every
other
region in 45 to establish if they are a distance of 8 or more apart. I
can
do this linearly comparing each distance to the previous but this is
not
comparing all points.
structure(list(Cluster.Index = c(46L, 46L, 46L, 46L, 46L, 45L,
45L, 45L, 45L, 45L, 44L, 44L, 44L, 44L, 44L, 43L, 43L, 43L, 43L,
43L), Value = c(8.21, 7.96, 7.85, 7.83, 7.8, 5.38, 4.56, 4.5,
4, 3.99, 5.42, 4.82, 4.21, 4.18, 3.91, 4.79, 4.27, 3.24, 3.06,
3.04), x = c(33L, 38L, 37L, 36L, 38L, 47L, 42L, 43L, 44L, 42L,
50L, 41L, 39L, 41L, 44L, 46L, 45L, 45L, 41L, 46L), y = c(15L,
12L, 12L, 13L, 13L, 91L, 84L, 84L, 95L, 96L, 69L, 70L, 65L, 65L,
59L, 41L, 40L, 46L, 44L, 47L), z = c(41L, 38L, 41L, 39L, 33L,
39L, 40L, 42L, 44L, 45L, 34L, 36L, 30L, 35L, 39L, 53L, 47L, 61L,
52L, 57L), X = c(NA, 6.557438524302, 3.16227766016838,
2.44948974278318,
6.32455532033676, 78.7464284904401, 8.66025403784439, 2.23606797749979,
11.2249721603218, 2.44948974278318, 30.2324329156619, 9.2736184954957,
8.06225774829855, 5.3851648071345, 7.81024967590665, 22.8910462845192,
6.16441400296898, 15.2315462117278, 10.0498756211209, 7.68114574786861
)), .Names = c(Cluster.Index, Value, x, y, z, X),
row.names =
c(NA,
20L), class = data.frame)
mainDat - data.frame()
for(i in 2:nrow(dat)){
tempDist - (sqrt((dat$x[i] - dat$x[i-1])^2 + (dat$y[i] -
dat$y[i-1])^2 +
(dat$z[i] - dat$z[i-1])^2))
dat$X[i] - c(tempDist)
if(dat$Cluster.Index[i] != dat$Cluster.Index[i-1]){
mainDat - rbind(mainDat, dat[i,])
}
if((dat$Cluster.Index[i] == dat$Cluster.Index[i-1])) {
if(tempDist 8){
mainDat - rbind(mainDat, dat[i,])
}
}
}
--
*Edward H Patzelt | Clinical Science PhD StudentPsychology | Harvard
University *
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Don McKenzie
Research Ecologist
Pacific Wildland Fire Sciences Lab
US Forest Service
Affiliate Professor
School of Environmental and Forest Sciences
University of Washington
[R] Subsetting data for split-sample validation, then repeating 1000x
Hi all, I'm doing some within-dataset model validation and would like to subset a dataset 70/30 and fit a model to 70% of the data (the training data), then validate it by predicting the remaining 30% (the testing data), and I would like to do this split-sample validation 1000 times and average the correlation coefficient and r2 between the training and testing data. I have the following working for a single iteration, and would like to know how to use either the replicate() or for-loop functions to average the 1000 'r2' and 'cor' outputs. -- # create 70% training sample A.samp - sample(1:nrow(A),floor(0.7*nrow(A)), replace = TRUE) # Fit model (I'm modeling native plant richness, 'nat.r') A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family = poisson, data = A[A.samp,]) # Use the model to predict the remaining 30% of the data A.pred - predict(A.model, newdata = A[-A.samp,], type = response) # Correlation between predicted 30% and actual 30% cor - cor(A[-A.samp,]$nat.r, A.pred, method = pearson) # r2 between predicted and observed lm.A - lm(A.pred ~ A[-A.samp,]$nat.r) r2 - summary(lm.A)$r.squared # print values r2 cor -- Thanks for your time! Cheers, Angela -- Angela E. Boag Ph.D. Student, Environmental Studies CAFOR Project Researcher University of Colorado, Boulder Mobile: 720-212-6505 [[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] loading saved files with objects in same names
Have you tried the 'envir' argument to load()? E.g., envA - new.environment() load(A.RData, envir=envA) envB - new.environment() load(B.RData, envir=envB) plot(A$object, B$object) Bill Dunlap TIBCO Software wdunlap tibco.com An alternative that I have been advocating is using attach(A.RData) etc. It does something similar as the above, but more conveniently: It loads the objects into a new environment *and* attaches that environment to your search() path, so you can access them directly, but attach() will never accidentally destroy existing R objects in your global environment ( = search()[[1]] ). Martin On Mon, Aug 18, 2014 at 5:30 PM, Jinsong Zhao jsz...@yeah.net wrote: Hi there, I have several saved data files (e.g., A.RData, B.RData and C.RData). In each file, there are some objects with same names but different contents. Now, I need to compare those objects through plotting. However, I can't find a way to load them into a workspace. The only thing I can do is to rename them and then save and load again. Is there a convenient to load those objects? Thanks a lot in advance. Best regards, Jinsong __ R-help@r-project.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 on installing R packages in a Citrix
Hi
We are currently trying to migrate 3 users of R to a citrix based
environment, but are coming across major issues trying to install the packages
to the relevant image. Can someone please contact me around how the install
should be done - as we can find no supporting documentation or help on this
matter.
I already sent this request to R-packages for help, but was told this was the
correct Forum for info of this type.
Many thanks
Jon-Paul Cameron
Systems Support Manager
Jon-Paul Cameron
Systems Support Manager
Direct tel: + 44 (0)20 7680 8046
Email: j.came...@hermes.co.ukmailto:j.came...@hermes.co.uk
Hermes Fund Managers
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Switchboard: +44 (0)20 7702 0888
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Re: [R] Help on installing R packages in a Citrix
On Aug 22, 2014, at 12:36 AM, Jon-Paul Cameron wrote: Hi We are currently trying to migrate 3 users of R to a citrix based environment, Windows?, Linux? Citrix isn't usually thought of as an OS is it? but are coming across major issues trying to install the packages to the relevant image. Can someone please contact me around how the install should be done - as we can find no supporting documentation or help on this matter. http://cran.us.r-project.org/manuals.html I already sent this request to R-packages for help, but was told this was the correct Forum for info of this type. Have you installed R? Which OS? Which version? -- David Winsemius Alameda, CA, USA __ R-help@r-project.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] no visible binding for global variable and with() vs. within()
Rolf Turner r.tur...@auckland.ac.nz
on Mon, 18 Aug 2014 08:47:36 +1200 writes:
On 17/08/14 23:05, Duncan Murdoch wrote:
On 16/08/2014, 9:36 PM, Daniel Braithwaite wrote:
R CMD check does not object to this code when checking a
package:
foo1 - function (bar) { with(bar, { x }) }
but produces a warning:
foo2: no visible binding for global variable 'x'
in response to this:
foo2 - function (bar) { within(bar, { x }) }
Is this an R bug, or at least, an inadvertent
inconsistency? Here is sessionInfo() from my machine,
right after starting an interactive session:
I'm not sure, but I suspect it's an intentional
inconsistency. The code that checks for use of globals
can't do anything in with() or within() code, so bugs can
slip by if you use those. I think with() had been around
for a long time and was in wide use when that test was
added, but within() is newer, and it was less disruptive
to warn about it, so the warning has been left in. (I
don't remember whether the test came before or after
within() was introduced.)
So if you want to avoid the warning, don't use within().
Or you could have a file, say melvin.R, in the R
directory of your package, containing the line:
utils::globalVariables(x)
Yes, but that would be a quite bad idea, IMHO:
The checking code {from package 'codetools' BTW}
would no longer warn you about any accidental global 'x'
variable in any of your functions in your package.
After all, these codetools checks *are* very helpful in
detecting typos and thinkos.
Consequently, I'd strongly advise to only use
globalVariables(.) on *rare* variable names.
Martin Maechler,
ETH Zurich
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Re: [R] loading saved files with objects in same names
On 20/08/2014, 8:58 AM, Barry Rowlingson wrote:
On Tue, Aug 19, 2014 at 1:30 AM, Jinsong Zhao jsz...@yeah.net wrote:
Hi there,
I have several saved data files (e.g., A.RData, B.RData and C.RData). In
each file, there are some objects with same names but different
contents. Now, I need to compare those objects through plotting.
However, I can't find a way to load them into a workspace. The only
thing I can do is to rename them and then save and load again.
Is there a convenient to load those objects?
Thanks a lot in advance.
The technique of loading into an environment already mentioned can be
cleaned up and put into a function.
First lets save a thing called x into two files with different values:
x=first
save(x,file=f.RData))
x=second
save(x,file=s.RData)
This little function wraps the loading:
getFrom=function(file, name){e=new.env();load(file,env=e);e[[name]]}
So now I can get 'x' from the first file - the value is returned from
`getFrom` so I can assign it to anything:
x1 = getFrom(f.RData,x)
x1
[1] first
x2 = getFrom(s.RData,x)
x2
[1] second
And I can even loop over RData files and read in all the `x`s into a vector:
sapply(c(f.RData,s.RData),function(f){getFrom(f,x)})
f.RData s.RData
first second
(on second thoughts, possibly 'loadFrom' is a better name)
That's a nice little function. You could also have lsFrom, that lists
the objects stored in the file, along the same lines:
lsFrom - function(file, all.names = FALSE, pattern) {
e - new.env()
load(file, envir = e)
ls(e, all.names = all.names, pattern = pattern)
}
Duncan Murdoch
Note that the solution of simply using
attach(f.RData)
makes the use of ls() very natural,
and for Rstudio and similar gui users, even automatically part of
their GUI.
Martin Maechler
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[R] knitr and stopifnot replacement.
I have a daily generated report in which I put a check using stopifnot. Unfortunately I didn’t check the effect very well, turns out that if the condition checked fails this is not shown in the knitr output (I only get an error much later due to a missing object). So my question is, what is the correct way to use assertions in a filename.Rnw file. I want to perform a check, and if it fails everything should stop with a message I can generate. Thanks, Best, Bart __ R-help@r-project.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] knitr and stopifnot replacement.
On 22/08/2014, 6:02 AM, Bart Kastermans wrote: I have a daily generated report in which I put a check using stopifnot. Unfortunately I didn’t check the effect very well, turns out that if the condition checked fails this is not shown in the knitr output (I only get an error much later due to a missing object). So my question is, what is the correct way to use assertions in a filename.Rnw file. I want to perform a check, and if it fails everything should stop with a message I can generate. One of the differences between knitr and Sweave is that knitr handles errors, and Sweave doesn't. I expect there's some knitr option to tell it to quit in case of error (a hook?), but you'll have to check the documentation to find it. 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] knitr and stopifnot replacement.
On 22 Aug 2014, at 12:39, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 22/08/2014, 6:02 AM, Bart Kastermans wrote: I have a daily generated report in which I put a check using stopifnot. Unfortunately I didn’t check the effect very well, turns out that if the condition checked fails this is not shown in the knitr output (I only get an error much later due to a missing object). So my question is, what is the correct way to use assertions in a filename.Rnw file. I want to perform a check, and if it fails everything should stop with a message I can generate. One of the differences between knitr and Sweave is that knitr handles errors, and Sweave doesn't. I expect there's some knitr option to tell it to quit in case of error (a hook?), but you'll have to check the documentation to find it. Thanks; that helped me to find the exact right option: filelist, error=FALSE= # check all files exist before continuing stopifnot(all(sapply(files, file.exists))) @ 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] Help on installing R packages in a Citrix
We are currently trying to migrate 3 users of R to a citrix based
environment, but are coming across major issues trying to install the
packages to the relevant image.
Why can't you open a virtualised OS instance, install and start R in the normal
way, install the packages normally in R using install.packages and then save
the image for re-use?
S
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Re: [R] Help on installing R packages in a Citrix
On Fri, Aug 22, 2014 at 2:03 PM, S Ellison s.elli...@lgcgroup.com wrote: We are currently trying to migrate 3 users of R to a citrix based environment, but are coming across major issues trying to install the packages to the relevant image. Please be more precised in your issue. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with analysis of variance
Hi, I have an observational dataset which consists of eight annual observations (year) for children (id) recording the rate of unemployment in the neighbourhood in which they lived (rate). I know if children move home so the data also has an identifier for spells in the same neighbourhood (spell). I want to decompose the overall variation in children's experience of area unemployment, given by the sum of (rate - mean rate)^2, into a) the component within a residential spell, sum of (rate - spell mean of rate)^2, b) the component between spells, sum of (spell mean), and c) the component between children, sum of (rate - mean rate for child). I think I can do this longhand using the calculations below: mobility - structure(list(year = c(2002L, 2003L, 2004L, 2005L, 2006L, 2007L,2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L), rate = c(13.08962, 14.27165, 4.496403, 3.89839, 4.60199, 5.138746, 5.251025, 4.874652, 5.880996, 5.813953, 6.204044, 6.93802, 6.866853, 7.614808, 4.405841, 4.826733, 4.760742, 3.762136, 4.60199, 5.138746, 5.251025, 4.405841, 4.826733, 4.760742, 3.762136, 4.60199, 5.138746, 5.251025, 4.405841, 5.789474, 5.889423, 4.61211, 4.642526, 6.838906, 9.683488), spell = c(1L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L)), .Names = c(year, rate, spell, id), row.names = c(NA, -35L), class = data.frame) mobility$id - factor(mobility$id) mobility$spell - factor(mobility$spell) mobility$spellmean - ave(mobility$rate,mobility$id,mobility$spell,FUN=mean) mobility$personmean - ave(mobility$rate,mobility$id,FUN=mean) mobility$totalmean - mean(mobility$rate,na.rm=TRUE) N - dim(mobility)[1] # observation deviation from overall mean sum(((mobility$rate-mobility$totalmean)^2)/N) 5.159846 # observation deviation from spell mean sum(((mobility$rate-mobility$spellmean)^2)/N) 2.039461 # deviation of spell mean from person mean sum(((mobility$spellmean-mobility$personmean)^2)/N) 2.13787 # deviation of person mean from overall mean sum(((mobility$personmean-mobility$totalmean)^2)/N) 0.982515 I think this is correct because the sum of the three components of variation sums to the total: 2.039461+2.13787+0.982515 = 5.159846 Can someone show me how to use the analysis of variance functions in R to get the same result. Thanks. Andrew McCulloch Leeds Metropolitan University From 22 September 2014 Leeds Metropolitan University will become Leeds Beckett University. Find out more at http://www.leedsbeckett.ac.uk To view the terms under which this email is distributed, please go to:- http://www.leedsmet.ac.uk/email-disclaimer.htm [[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] print vectors with consecutive numbers
Hi all, I have a matrix with consecutive and non-consecutive numbers in columns. For example, the first 2 columns have consecutive numbers. I want R to print only columns with consecutive numbers. Here is the matrix and how I did using conditional statement: ## mat=matrix(data=c(9,2,3,4,5,6,10,13,15,17,19,22, 25,27,29,31,34,37,39,41),ncol=5) mat difference = diff(mat)==1 difference y1=difference[1,] y2=difference[2,] y3=difference[3,] y=(y1|y2|y3) y if (y==TRUE) mat else 0 ## However, R still print all 5 columns, not the first 2 columns I wanted. I got the Warning message: In if (y == TRUE) mat else 0 : the condition has length 1 and only the first element will be used How can I change the code to get only the first 2 columns with consecutive numbers printed? I am new to R. Thanks in advance for your help. James [[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] print vectors with consecutive numbers
Hi James, Try mat[, apply(mat, 2, function(x) any(diff(x) == 1))] HTH, Jorge.- On Fri, Aug 22, 2014 at 10:18 PM, James Wei zwei0...@hotmail.com wrote: Hi all, I have a matrix with consecutive and non-consecutive numbers in columns. For example, the first 2 columns have consecutive numbers. I want R to print only columns with consecutive numbers. Here is the matrix and how I did using conditional statement: ## mat=matrix(data=c(9,2,3,4,5,6,10,13,15,17,19,22, 25,27,29,31,34,37,39,41),ncol=5) mat difference = diff(mat)==1 difference y1=difference[1,] y2=difference[2,] y3=difference[3,] y=(y1|y2|y3) y if (y==TRUE) mat else 0 ## However, R still print all 5 columns, not the first 2 columns I wanted. I got the Warning message: In if (y == TRUE) mat else 0 : the condition has length 1 and only the first element will be used How can I change the code to get only the first 2 columns with consecutive numbers printed? I am new to R. Thanks in advance for your help. James [[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 analysis of variance
Andrew, I plotted your data first. Then I made id, spell, year into factors and did the ANOVA. From the graph, it looks like the two ids who had more than one spell show variability in rate. The ANOVA table agrees by showing high significance for the id:spell interaction. The numbers in this ANOVA table are not similar to your numbers. Rich mobility$id - factor(mobility$id) mobility$spell - factor(mobility$spell) mobility$year - factor(mobility$year) xyplot(rate ~ year | id, group=spell, data=mobility, pch=19, layout=c(1,5), scales=list(alternating=FALSE), between=list(y=1), strip=FALSE, strip.left=strip.custom(strip.names=c(TRUE,TRUE))) mobility.aov - aov(rate ~ year + id/spell, data=mobility) anova(mobility.aov) On Fri, Aug 22, 2014 at 7:51 AM, McCulloch, Andrew a.mccull...@leedsmet.ac.uk wrote: Hi, I have an observational dataset which consists of eight annual observations (year) for children (id) recording the rate of unemployment in the neighbourhood in which they lived (rate). I know if children move home so the data also has an identifier for spells in the same neighbourhood (spell). I want to decompose the overall variation in children's experience of area unemployment, given by the sum of (rate - mean rate)^2, into a) the component within a residential spell, sum of (rate - spell mean of rate)^2, b) the component between spells, sum of (spell mean), and c) the component between children, sum of (rate - mean rate for child). I think I can do this longhand using the calculations below: mobility - structure(list(year = c(2002L, 2003L, 2004L, 2005L, 2006L, 2007L,2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L, 2002L, 2003L, 2004L, 2005L, 2006L, 2007L, 2008L), rate = c(13.08962, 14.27165, 4.496403, 3.89839, 4.60199, 5.138746, 5.251025, 4.874652, 5.880996, 5.813953, 6.204044, 6.93802, 6.866853, 7.614808, 4.405841, 4.826733, 4.760742, 3.762136, 4.60199, 5.138746, 5.251025, 4.405841, 4.826733, 4.760742, 3.762136, 4.60199, 5.138746, 5.251025, 4.405841, 5.789474, 5.889423, 4.61211, 4.642526, 6.838906, 9.683488), spell = c(1L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L), id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L)), .Names = c(year, rate, spell, id), row.names = c(NA, -35L), class = data.frame) mobility$id - factor(mobility$id) mobility$spell - factor(mobility$spell) mobility$spellmean - ave(mobility$rate,mobility$id,mobility$spell,FUN=mean) mobility$personmean - ave(mobility$rate,mobility$id,FUN=mean) mobility$totalmean - mean(mobility$rate,na.rm=TRUE) N - dim(mobility)[1] # observation deviation from overall mean sum(((mobility$rate-mobility$totalmean)^2)/N) 5.159846 # observation deviation from spell mean sum(((mobility$rate-mobility$spellmean)^2)/N) 2.039461 # deviation of spell mean from person mean sum(((mobility$spellmean-mobility$personmean)^2)/N) 2.13787 # deviation of person mean from overall mean sum(((mobility$personmean-mobility$totalmean)^2)/N) 0.982515 I think this is correct because the sum of the three components of variation sums to the total: 2.039461+2.13787+0.982515 = 5.159846 Can someone show me how to use the analysis of variance functions in R to get the same result. Thanks. Andrew McCulloch Leeds Metropolitan University From 22 September 2014 Leeds Metropolitan University will become Leeds Beckett University. Find out more at http://www.leedsbeckett.ac.uk To view the terms under which this email is distributed, please go to:- http://www.leedsmet.ac.uk/email-disclaimer.htm [[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] knitr and stopifnot replacement.
Yep, that is exactly the answer. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Web: http://yihui.name On Fri, Aug 22, 2014 at 6:35 AM, Bart Kastermans kaste...@kasterma.net wrote: On 22 Aug 2014, at 12:39, Duncan Murdoch murdoch.dun...@gmail.com wrote: On 22/08/2014, 6:02 AM, Bart Kastermans wrote: I have a daily generated report in which I put a check using stopifnot. Unfortunately I didn’t check the effect very well, turns out that if the condition checked fails this is not shown in the knitr output (I only get an error much later due to a missing object). So my question is, what is the correct way to use assertions in a filename.Rnw file. I want to perform a check, and if it fails everything should stop with a message I can generate. One of the differences between knitr and Sweave is that knitr handles errors, and Sweave doesn't. I expect there's some knitr option to tell it to quit in case of error (a hook?), but you'll have to check the documentation to find it. Thanks; that helped me to find the exact right option: filelist, error=FALSE= # check all files exist before continuing stopifnot(all(sapply(files, file.exists))) @ __ R-help@r-project.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] Euclidean Distance in 3 Dimensions
This function unfortunately does not work in 3d space.
[I think 'this' is refering to the 'dist' function.]
Can you show how it is not working for you? I.e., what does it
produce compared to what you want for a given input?
dist() does work on a 3-column (or n-column) matrix or data.frame,
which is how R generally represents 3 dimensional (or n dimensional)
data. E.g.,
d - data.frame(One=1:3, Two=c(3,5,8), Three=c(4,8,16))
d
One Two Three
1 1 3 4
2 2 5 8
3 3 816
dist(d)
1 2
2 4.582576
3 13.152946 8.602325
as.matrix(dist(d)) # the matrix format makes further compuations easier
12 3
1 0.00 4.582576 13.152946
2 4.582576 0.00 8.602325
3 13.152946 8.602325 0.00
which(as.matrix(dist(d))8, arr.ind=TRUE)
row col
3 3 1
3 3 2
1 1 3
2 2 3
sqrt(sum((d[,2] - d[,3])^2)) # the 2,3 or 3,2 element, by hand
[1] 8.602325
I think it would help if you restated your problem. I found the original
description confusing. A small example, with the expected output, would
be very helpful.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Aug 21, 2014 at 11:34 AM, Patzelt, Edward patz...@g.harvard.edu wrote:
This function unfortunately does not work in 3d space.
Thoughts?
On Wed, Aug 20, 2014 at 4:57 PM, Don McKenzie d...@u.washington.edu wrote:
?dist
from the help
dist {stats}R Documentation
Distance Matrix Computation
Description
This function computes and returns the distance matrix computed by using
the specified distance measure to compute the distances between the rows of
a data matrix.
Is this what you want? Computing on a matrix whose rows are your x, y,
and z values?
On Aug 20, 2014, at 1:12 PM, Patzelt, Edward patz...@g.harvard.edu
wrote:
R Community -
I am attempting to write a function that will calculate the distance
between points in 3 dimensional space for unique regions (e.g. localized
brain regions such as the frontal lobe).
For example I'm looking to compare each point in region 45 to every other
region in 45 to establish if they are a distance of 8 or more apart. I
can
do this linearly comparing each distance to the previous but this is not
comparing all points.
structure(list(Cluster.Index = c(46L, 46L, 46L, 46L, 46L, 45L,
45L, 45L, 45L, 45L, 44L, 44L, 44L, 44L, 44L, 43L, 43L, 43L, 43L,
43L), Value = c(8.21, 7.96, 7.85, 7.83, 7.8, 5.38, 4.56, 4.5,
4, 3.99, 5.42, 4.82, 4.21, 4.18, 3.91, 4.79, 4.27, 3.24, 3.06,
3.04), x = c(33L, 38L, 37L, 36L, 38L, 47L, 42L, 43L, 44L, 42L,
50L, 41L, 39L, 41L, 44L, 46L, 45L, 45L, 41L, 46L), y = c(15L,
12L, 12L, 13L, 13L, 91L, 84L, 84L, 95L, 96L, 69L, 70L, 65L, 65L,
59L, 41L, 40L, 46L, 44L, 47L), z = c(41L, 38L, 41L, 39L, 33L,
39L, 40L, 42L, 44L, 45L, 34L, 36L, 30L, 35L, 39L, 53L, 47L, 61L,
52L, 57L), X = c(NA, 6.557438524302, 3.16227766016838, 2.44948974278318,
6.32455532033676, 78.7464284904401, 8.66025403784439, 2.23606797749979,
11.2249721603218, 2.44948974278318, 30.2324329156619, 9.2736184954957,
8.06225774829855, 5.3851648071345, 7.81024967590665, 22.8910462845192,
6.16441400296898, 15.2315462117278, 10.0498756211209, 7.68114574786861
)), .Names = c(Cluster.Index, Value, x, y, z, X), row.names =
c(NA,
20L), class = data.frame)
mainDat - data.frame()
for(i in 2:nrow(dat)){
tempDist - (sqrt((dat$x[i] - dat$x[i-1])^2 + (dat$y[i] - dat$y[i-1])^2 +
(dat$z[i] - dat$z[i-1])^2))
dat$X[i] - c(tempDist)
if(dat$Cluster.Index[i] != dat$Cluster.Index[i-1]){
mainDat - rbind(mainDat, dat[i,])
}
if((dat$Cluster.Index[i] == dat$Cluster.Index[i-1])) {
if(tempDist 8){
mainDat - rbind(mainDat, dat[i,])
}
}
}
--
*Edward H Patzelt | Clinical Science PhD StudentPsychology | Harvard
University *
[[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.
Don McKenzie
Research Ecologist
Pacific Wildland Fire Sciences Lab
US Forest Service
Affiliate Professor
School of Environmental and Forest Sciences
University of Washington
d...@uw.edu
--
*Edward H Patzelt | Clinical Science PhD StudentPsychology | Harvard
University *
[[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
Re: [R] Help on installing R packages in a Citrix
On Aug 22, 2014, at 7:37 AM, Chirag Patel wrote: Hi David I have installed R Version 2.15.1 on the image but having problems install rcom and rscproxy. Those are commercial packages and not maintained or supported by R-Core (That is also a rather old version of R and would have gotten the advice to update before further comments were offered if the above sentence were not applicable.) Since the error message says the packages are nota available for R 2.15.1 I would go to the author's website and see what versions those packages are available for. Standard message to users: read the error messages for meaning. -- David. I'm following this guide: http://homepage.univie.ac.at/erich.neuwirth/php/rcomwiki/doku.php?id=wiki:how_to_install#installation_of_r_r_d_com_server_and_rexcel Download the statconn DCOM server and execute the program you downloaded Start R as administrator (on Windows 7 you need to right-click the R icon and click the corresponding item) In R, run the following commands (you must start R as administrator to do this) install.packages(c(rscproxy,rcom),repos=http://rcom.univie.ac.at/download,lib=.Library) library(rcom) comRegisterRegistry() i get the following message: image001.png Please can you help me? Many Thanks Chirag Patel -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: 22 August 2014 09:29 To: Jon-Paul Cameron Cc: 'r-help@R-project.org'; Scott Waters; #BST - Citrix Support Subject: Re: [R] Help on installing R packages in a Citrix On Aug 22, 2014, at 12:36 AM, Jon-Paul Cameron wrote: Hi We are currently trying to migrate 3 users of R to a citrix based environment, Windows?, Linux? Citrix isn't usually thought of as an OS is it? but are coming across major issues trying to install the packages to the relevant image. Can someone please contact me around how the install should be done - as we can find no supporting documentation or help on this matter. http://cran.us.r-project.org/manuals.html I already sent this request to R-packages for help, but was told this was the correct Forum for info of this type. Have you installed R? Which OS? Which version? -- David Winsemius Alameda, CA, USA __ David Winsemius Alameda, CA, USA __ R-help@r-project.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] print vectors with consecutive numbers
Hi Jorge, Thanks so much, it is working perfectly. There are so many for me to learn. Cheers. James From: jorgeivanve...@gmail.com Date: Fri, 22 Aug 2014 22:28:40 +1000 Subject: Re: [R] print vectors with consecutive numbers To: zwei0...@hotmail.com CC: r-help@r-project.org Hi James, Try mat[, apply(mat, 2, function(x) any(diff(x) == 1))] HTH, Jorge.- On Fri, Aug 22, 2014 at 10:18 PM, James Wei zwei0...@hotmail.com wrote: Hi all, I have a matrix with consecutive and non-consecutive numbers in columns. For example, the first 2 columns have consecutive numbers. I want R to print only columns with consecutive numbers. Here is the matrix and how I did using conditional statement: ## mat=matrix(data=c(9,2,3,4,5,6,10,13,15,17,19,22, 25,27,29,31,34,37,39,41),ncol=5) mat difference = diff(mat)==1 difference y1=difference[1,] y2=difference[2,] y3=difference[3,] y=(y1|y2|y3) y if (y==TRUE) mat else 0 ## However, R still print all 5 columns, not the first 2 columns I wanted. I got the Warning message: In if (y == TRUE) mat else 0 : the condition has length 1 and only the first element will be used How can I change the code to get only the first 2 columns with consecutive numbers printed? I am new to R. Thanks in advance for your help. James [[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] print vectors with consecutive numbers
Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Fri, Aug 22, 2014 at 7:16 AM, James Wei zwei0...@hotmail.com wrote: Hi Jorge, Thanks so much, it is working perfectly. There are so many for me to learn. Have you done an R tutorial? e.g. http://cran.r-project.org/doc/manuals/R-intro.pdf (This ships with R. There are many others around the web. Just search). If not, please do not post further until you have done so. You cannot complain of ignorance when you have not made an honest effort to learn. And if you have already, post away. Cheers, Bert Cheers. James From: jorgeivanve...@gmail.com Date: Fri, 22 Aug 2014 22:28:40 +1000 Subject: Re: [R] print vectors with consecutive numbers To: zwei0...@hotmail.com CC: r-help@r-project.org Hi James, Try mat[, apply(mat, 2, function(x) any(diff(x) == 1))] HTH, Jorge.- On Fri, Aug 22, 2014 at 10:18 PM, James Wei zwei0...@hotmail.com wrote: Hi all, I have a matrix with consecutive and non-consecutive numbers in columns. For example, the first 2 columns have consecutive numbers. I want R to print only columns with consecutive numbers. Here is the matrix and how I did using conditional statement: ## mat=matrix(data=c(9,2,3,4,5,6,10,13,15,17,19,22, 25,27,29,31,34,37,39,41),ncol=5) mat difference = diff(mat)==1 difference y1=difference[1,] y2=difference[2,] y3=difference[3,] y=(y1|y2|y3) y if (y==TRUE) mat else 0 ## However, R still print all 5 columns, not the first 2 columns I wanted. I got the Warning message: In if (y == TRUE) mat else 0 : the condition has length 1 and only the first element will be used How can I change the code to get only the first 2 columns with consecutive numbers printed? I am new to R. Thanks in advance for your help. James [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.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 on installing R packages in a Citrix
On Aug 22, 2014, at 9:13 AM, David Winsemius wrote: On Aug 22, 2014, at 7:37 AM, Chirag Patel wrote: Hi David I have installed R Version 2.15.1 on the image but having problems install rcom and rscproxy. Those are commercial packages and not maintained or supported by R-Core I need to clarify that statement: http://homepage.univie.ac.at/erich.neuwirth/php/rcomwiki/doku.php?id=wiki:how_to_install rcom is placed on CRAN but authored by the people who distribute RExcel, so technically it's not a commercial product itself but it is made available to support a commercial product. You will notice in its requirements statconnDCOM: And it's wiki says rcom does not comply with CRAN open source requirements anymore: http://homepage.univie.ac.at/erich.neuwirth/php/rcomwiki/doku.php?id=start -- David. (That is also a rather old version of R and would have gotten the advice to update before further comments were offered if the above sentence were not applicable.) Since the error message says the packages are nota available for R 2.15.1 I would go to the author's website and see what versions those packages are available for. Standard message to users: read the error messages for meaning. -- David. I'm following this guide: http://homepage.univie.ac.at/erich.neuwirth/php/rcomwiki/doku.php?id=wiki:how_to_install#installation_of_r_r_d_com_server_and_rexcel Download the statconn DCOM server and execute the program you downloaded Start R as administrator (on Windows 7 you need to right-click the R icon and click the corresponding item) In R, run the following commands (you must start R as administrator to do this) install.packages(c(rscproxy,rcom),repos=http://rcom.univie.ac.at/download,lib=.Library) library(rcom) comRegisterRegistry() i get the following message: image001.png Please can you help me? Many Thanks Chirag Patel -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: 22 August 2014 09:29 To: Jon-Paul Cameron Cc: 'r-help@R-project.org'; Scott Waters; #BST - Citrix Support Subject: Re: [R] Help on installing R packages in a Citrix On Aug 22, 2014, at 12:36 AM, Jon-Paul Cameron wrote: Hi We are currently trying to migrate 3 users of R to a citrix based environment, Windows?, Linux? Citrix isn't usually thought of as an OS is it? but are coming across major issues trying to install the packages to the relevant image. Can someone please contact me around how the install should be done - as we can find no supporting documentation or help on this matter. http://cran.us.r-project.org/manuals.html I already sent this request to R-packages for help, but was told this was the correct Forum for info of this type. Have you installed R? Which OS? Which version? -- David Winsemius Alameda, CA, USA __ David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.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] loading saved files with objects in same names
On 2014/8/22 1:02, Martin Maechler wrote: Have you tried the 'envir' argument to load()? E.g., envA - new.environment() load(A.RData, envir=envA) envB - new.environment() load(B.RData, envir=envB) plot(A$object, B$object) Bill Dunlap TIBCO Software wdunlap tibco.com An alternative that I have been advocating is using attach(A.RData) etc. It does something similar as the above, but more conveniently: It loads the objects into a new environment *and* attaches that environment to your search() path, so you can access them directly, but attach() will never accidentally destroy existing R objects in your global environment ( = search()[[1]] ). Martin Thanks a lot. I try your method, and I got: attach(D2.1.RData) The following objects are masked _by_ .GlobalEnv: coda.jags.1, df.1, jags.1, Mean, N In this case, how to access the masked objects? Best, Jinsong On Mon, Aug 18, 2014 at 5:30 PM, Jinsong Zhao jsz...@yeah.net wrote: Hi there, I have several saved data files (e.g., A.RData, B.RData and C.RData). In each file, there are some objects with same names but different contents. Now, I need to compare those objects through plotting. However, I can't find a way to load them into a workspace. The only thing I can do is to rename them and then save and load again. Is there a convenient to load those objects? Thanks a lot in advance. Best regards, Jinsong __ R-help@r-project.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] filled.contour key axis
Hi Jim, all,
Thx, I was hoping for percentage scores, such that R puts numbers with %
there.
And by automatically labeling I meant giving the axis a title, sorry for
mixing that up.
I havn't found an option in the parameters for filled.contour.
Haiko
Von: Jim Lemon [j...@bitwrit.com.au]
Gesendet: Montag, 18. August 2014 14:41
An: r-help@r-project.org
Cc: Lietz, Haiko
Betreff: Re: [R] filled.contour key axis
On Mon, 18 Aug 2014 08:51:57 AM Lietz, Haiko wrote:
Hi all,
Using filled.contour...
foo - matrix(seq(0.1, 0.9, 0.1), ncol = 3)
filled.contour(foo)
how can I set the key axis to give percentages?
And is there a way to automatically label the key axis except for
placing
text there?
Thanks
Haiko
Hi Haiko,
Try this:
filled.contour(foo,
plot.axes={axis(1);
axis(2,at=seq(0,1,by=0.2),labels=seq(0,100,by=20))},
key.axes=
{axis(4,at=seq(0,1,by=0.2),labels=seq(0,100,by=20))})
Jim
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] no visible binding for global variable and with() vs. within()
I'm dealing with these type of false NOTEs as:
foo1 - function (bar) {
# To please R CMD check
x - NULL; rm(list=x)
with(bar, {
x })
}
Of course, that may one day break with more clever code inspections.
My $.02
/Henrik
On Fri, Aug 22, 2014 at 2:40 AM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
Rolf Turner r.tur...@auckland.ac.nz
on Mon, 18 Aug 2014 08:47:36 +1200 writes:
On 17/08/14 23:05, Duncan Murdoch wrote:
On 16/08/2014, 9:36 PM, Daniel Braithwaite wrote:
R CMD check does not object to this code when checking a
package:
foo1 - function (bar) { with(bar, { x }) }
but produces a warning:
foo2: no visible binding for global variable 'x'
in response to this:
foo2 - function (bar) { within(bar, { x }) }
Is this an R bug, or at least, an inadvertent
inconsistency? Here is sessionInfo() from my machine,
right after starting an interactive session:
I'm not sure, but I suspect it's an intentional
inconsistency. The code that checks for use of globals
can't do anything in with() or within() code, so bugs can
slip by if you use those. I think with() had been around
for a long time and was in wide use when that test was
added, but within() is newer, and it was less disruptive
to warn about it, so the warning has been left in. (I
don't remember whether the test came before or after
within() was introduced.)
So if you want to avoid the warning, don't use within().
Or you could have a file, say melvin.R, in the R
directory of your package, containing the line:
utils::globalVariables(x)
Yes, but that would be a quite bad idea, IMHO:
The checking code {from package 'codetools' BTW}
would no longer warn you about any accidental global 'x'
variable in any of your functions in your package.
After all, these codetools checks *are* very helpful in
detecting typos and thinkos.
Consequently, I'd strongly advise to only use
globalVariables(.) on *rare* variable names.
Martin Maechler,
ETH Zurich
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Re: [R] filled.contour key axis
You seem to be thinking of Excel. R does not dress up numbers for you... if you
want them represented in a particular way, you need to format them into a
character string. There are plenty of options for doing that.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On August 22, 2014 10:36:23 AM PDT, Lietz, Haiko haiko.li...@gesis.org
wrote:
Hi Jim, all,
Thx, I was hoping for percentage scores, such that R puts numbers with
% there.
And by automatically labeling I meant giving the axis a title, sorry
for mixing that up.
I havn't found an option in the parameters for filled.contour.
Haiko
Von: Jim Lemon [j...@bitwrit.com.au]
Gesendet: Montag, 18. August 2014 14:41
An: r-help@r-project.org
Cc: Lietz, Haiko
Betreff: Re: [R] filled.contour key axis
On Mon, 18 Aug 2014 08:51:57 AM Lietz, Haiko wrote:
Hi all,
Using filled.contour...
foo - matrix(seq(0.1, 0.9, 0.1), ncol = 3)
filled.contour(foo)
how can I set the key axis to give percentages?
And is there a way to automatically label the key axis except for
placing
text there?
Thanks
Haiko
Hi Haiko,
Try this:
filled.contour(foo,
plot.axes={axis(1);
axis(2,at=seq(0,1,by=0.2),labels=seq(0,100,by=20))},
key.axes=
{axis(4,at=seq(0,1,by=0.2),labels=seq(0,100,by=20))})
Jim
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting data for split-sample validation, then repeating 1000x
You can use replicate() or a for (i in 1:1000){} loop to do your replications,
but you have other issues first.
1. You are sampling with replacement which makes no sense at all. Your 70%
sample will contain some observations multiple times and will use less than 70%
of the data most of the time.
2. You compute r using cor() and r.squared using summary.lm(). Why? Once you
have computed r, r*r or r^2 is equal to r.squared for the simple linear model
you are using.
# To split your data, you need to sample without replacement, e.g.
train - sample.int(nrow(A), floor(nrow(A)*.7))
test - (1:nrow(A))[-train]
# Now run your analysis on A[train,] and test it on A[test,]
# Fit model (I'm modeling native plant richness, 'nat.r')
A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family =
poisson, data = A[train,])
# Correlation between predicted 30% and actual 30%
cor - cor(Atest$nat.r, predict(A.model, newdata = A[test,], type = response))
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Angela Boag
Sent: Thursday, August 21, 2014 4:46 PM
To: r-help@r-project.org
Subject: [R] Subsetting data for split-sample validation, then repeating 1000x
Hi all,
I'm doing some within-dataset model validation and would like to subset a
dataset 70/30 and fit a model to 70% of the data (the training data), then
validate it by predicting the remaining 30% (the testing data), and I would
like to do this split-sample validation 1000 times and average the
correlation coefficient and r2 between the training and testing data.
I have the following working for a single iteration, and would like to know
how to use either the replicate() or for-loop functions to average the 1000
'r2' and 'cor' outputs.
--
# create 70% training sample
A.samp - sample(1:nrow(A),floor(0.7*nrow(A)), replace = TRUE)
# Fit model (I'm modeling native plant richness, 'nat.r')
A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family =
poisson, data = A[A.samp,])
# Use the model to predict the remaining 30% of the data
A.pred - predict(A.model, newdata = A[-A.samp,], type = response)
# Correlation between predicted 30% and actual 30%
cor - cor(A[-A.samp,]$nat.r, A.pred, method = pearson)
# r2 between predicted and observed
lm.A - lm(A.pred ~ A[-A.samp,]$nat.r)
r2 - summary(lm.A)$r.squared
# print values
r2
cor
--
Thanks for your time!
Cheers,
Angela
--
Angela E. Boag
Ph.D. Student, Environmental Studies
CAFOR Project Researcher
University of Colorado, Boulder
Mobile: 720-212-6505
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subsetting data for split-sample validation, then repeating 1000x
Combine your code into a function:
Plant - function() {
train - sample.int(nrow(A), floor(nrow(A)*.7))
test - (1:nrow(A))[-train]
A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family =
poisson, data = A[train,])
cor(Atest$nat.r, predict(A.model, newdata = A[test,], type = response))
}
Test the function. It should return a single correlation and no errors or
warnings.
Plant()
If not, debug and run it again. When it works:
Out - replicate(1000, Plant())
Out should be a vector with 1000 correlation values.
hist(Out) # for a histogram of the correlation values
David C
From: Angela Boag [mailto:angela.b...@colorado.edu]
Sent: Friday, August 22, 2014 4:01 PM
To: David L Carlson
Subject: Re: [R] Subsetting data for split-sample validation, then repeating
1000x
Hi David,
Thanks for the feedback. I actually sampled without replacement initially but
it's been a while since I looked at this code and just changed it because I
thought it made more sense logically, but you've reassured me that my original
hunch was right.
The real issue I'm having is how to use either the replicate() or for(i in
1:1000){} loop code to get the average r value of 1000 repetitions as my
output. I'm not familiar with either tool, so any suggestions on what that code
would look like would be very helpful.
Thanks!
Angela
--
Angela E. Boag
Ph.D. Student, Environmental Studies
CAFOR Project Researcher
University of Colorado, Boulder
Mobile: 720-212-6505
On Fri, Aug 22, 2014 at 2:46 PM, David L Carlson dcarl...@tamu.edu wrote:
You can use replicate() or a for (i in 1:1000){} loop to do your replications,
but you have other issues first.
1. You are sampling with replacement which makes no sense at all. Your 70%
sample will contain some observations multiple times and will use less than 70%
of the data most of the time.
2. You compute r using cor() and r.squared using summary.lm(). Why? Once you
have computed r, r*r or r^2 is equal to r.squared for the simple linear model
you are using.
# To split your data, you need to sample without replacement, e.g.
train - sample.int(nrow(A), floor(nrow(A)*.7))
test - (1:nrow(A))[-train]
# Now run your analysis on A[train,] and test it on A[test,]
# Fit model (I'm modeling native plant richness, 'nat.r')
A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family =
poisson, data = A[train,])
# Correlation between predicted 30% and actual 30%
cor - cor(Atest$nat.r, predict(A.model, newdata = A[test,], type = response))
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Angela Boag
Sent: Thursday, August 21, 2014 4:46 PM
To: r-help@r-project.org
Subject: [R] Subsetting data for split-sample validation, then repeating 1000x
Hi all,
I'm doing some within-dataset model validation and would like to subset a
dataset 70/30 and fit a model to 70% of the data (the training data), then
validate it by predicting the remaining 30% (the testing data), and I would
like to do this split-sample validation 1000 times and average the
correlation coefficient and r2 between the training and testing data.
I have the following working for a single iteration, and would like to know
how to use either the replicate() or for-loop functions to average the 1000
'r2' and 'cor' outputs.
--
# create 70% training sample
A.samp - sample(1:nrow(A),floor(0.7*nrow(A)), replace = TRUE)
# Fit model (I'm modeling native plant richness, 'nat.r')
A.model - glmmadmb(nat.r ~ isl.sz + nr.mead, random = ~ 1 | site, family =
poisson, data = A[A.samp,])
# Use the model to predict the remaining 30% of the data
A.pred - predict(A.model, newdata = A[-A.samp,], type = response)
# Correlation between predicted 30% and actual 30%
cor - cor(A[-A.samp,]$nat.r, A.pred, method = pearson)
# r2 between predicted and observed
lm.A - lm(A.pred ~ A[-A.samp,]$nat.r)
r2 - summary(lm.A)$r.squared
# print values
r2
cor
--
Thanks for your time!
Cheers,
Angela
--
Angela E. Boag
Ph.D. Student, Environmental Studies
CAFOR Project Researcher
University of Colorado, Boulder
Mobile: 720-212-6505
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] loading saved files with objects in same names
On 22/08/2014, 1:14 PM, Jinsong Zhao wrote: On 2014/8/22 1:02, Martin Maechler wrote: Have you tried the 'envir' argument to load()? E.g., envA - new.environment() load(A.RData, envir=envA) envB - new.environment() load(B.RData, envir=envB) plot(A$object, B$object) Bill Dunlap TIBCO Software wdunlap tibco.com An alternative that I have been advocating is using attach(A.RData) etc. It does something similar as the above, but more conveniently: It loads the objects into a new environment *and* attaches that environment to your search() path, so you can access them directly, but attach() will never accidentally destroy existing R objects in your global environment ( = search()[[1]] ). Martin Thanks a lot. I try your method, and I got: attach(D2.1.RData) The following objects are masked _by_ .GlobalEnv: coda.jags.1, df.1, jags.1, Mean, N In this case, how to access the masked objects? Don't ever use attach(), and this won't be a problem. Martin gave you bad advice. 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] BNF description for R
Thank you,that help a lot. At 2014-08-17 19:09:00,Duncan Murdoch murdoch.dun...@gmail.com wrote: On 16/08/2014, 7:32 PM, kevin2059 wrote: HOW can I get a completely BNF description for R? I try to write a parser for R now. The R grammar is defined in the src/main/gram.y file (in Bison format). You can get that file from https://svn.r-project.org/R/trunk/src/main/gram.y. 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] filled.contour key axis
On Fri, 22 Aug 2014 05:36:23 PM Lietz, Haiko wrote:
Hi Jim, all,
Thx, I was hoping for percentage scores, such that R puts numbers
with %
there.
And by automatically labeling I meant giving the axis a title, sorry for
mixing that up.
I havn't found an option in the parameters for filled.contour.
Haiko
Hi Haiko,
I think what Jeff meant was:
filled.contour(foo,
plot.axes={axis(1);
axis(2,at=seq(0,1,by=0.2),
labels=paste(seq(0,100,by=20),%))},
key.axes=
{axis(4,at=seq(0,1,by=0.2),
labels=paste(seq(0,100,by=20),%))})
Even Excel can't read your mind, although it sometimes thinks it can.
Jim
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