[R] Time series
A simplified ask is: when I use the time series plot function, R treats each time on 9/19/13 as an individual day, when clearly it isn't. Thank you ahead of time. Keith ts(chum) with(chum,plot.ts(Time,PA)) Chum- Time PA 9/18/13 18:29 16 9/19/13 13:29 14 9/19/13 16:29 13.2 9/19/13 17:29 13.1 9/19/13 18:29 13 9/20/13 18:29 12 9/21/13 18:29 10 9/22/13 18:29 9 9/23/13 18:29 7 9/24/13 18:29 5 9/25/13 18:29 3 9/26/13 18:29 2 9/27/13 18:29 1 M. Keith Cox, Ph.D. Principal MKConsulting 17105 Glacier Hwy Juneau, AK 99801 U.S. 907.957.4606 [[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] Time series
On Tue, 1 Apr 2014, Marlin Keith Cox wrote: A simplified ask is: when I use the time series plot function, R treats each time on 9/19/13 as an individual day, when clearly it isn't. The ts class can handle only regular time series. See the zoo or xts packages for dealing with time series that have an irregular time index. Here, you could use POSIXct or chron time stamps. Thank you ahead of time. Keith ts(chum) with(chum,plot.ts(Time,PA)) Chum- Time PA 9/18/13 18:29 16 9/19/13 13:29 14 9/19/13 16:29 13.2 9/19/13 17:29 13.1 9/19/13 18:29 13 9/20/13 18:29 12 9/21/13 18:29 10 9/22/13 18:29 9 9/23/13 18:29 7 9/24/13 18:29 5 9/25/13 18:29 3 9/26/13 18:29 2 9/27/13 18:29 1 M. Keith Cox, Ph.D. Principal MKConsulting 17105 Glacier Hwy Juneau, AK 99801 U.S. 907.957.4606 [[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] Time series
Hello Keith, Your example is clearly not reproducible and wrong (Chum and chum). Please use dput() to attach sample dataset. At first glance, you work with hourly data. Package such as xts might be more useful. Regards, Pascal On Wed, Apr 2, 2014 at 3:16 PM, Marlin Keith Cox marlink...@gmail.com wrote: A simplified ask is: when I use the time series plot function, R treats each time on 9/19/13 as an individual day, when clearly it isn't. Thank you ahead of time. Keith ts(chum) with(chum,plot.ts(Time,PA)) Chum- Time PA 9/18/13 18:29 16 9/19/13 13:29 14 9/19/13 16:29 13.2 9/19/13 17:29 13.1 9/19/13 18:29 13 9/20/13 18:29 12 9/21/13 18:29 10 9/22/13 18:29 9 9/23/13 18:29 7 9/24/13 18:29 5 9/25/13 18:29 3 9/26/13 18:29 2 9/27/13 18:29 1 M. Keith Cox, Ph.D. Principal MKConsulting 17105 Glacier Hwy Juneau, AK 99801 U.S. 907.957.4606 [[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. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ 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] pre-allocation not always a timesaver
I don't think you got a response to this one;
x - array(dim=(c(j, n)))
for (i in 1:n) {
x[,i] - rnorm(j)
}
Note that array() allocates a logical array by default, which means
that in your first iteration (i==1) it has to be coerced to a double
array before assigning the value of rnorm(). That takes time. It also
takes time to garbage collect the stray logical array afterward.
Using,
x - array(NA_real_, dim=(c(j, n)))
for (i in 1:n) {
x[,i] - rnorm(j)
}
avoids this.
For updating list elements, you can avoid repetitive overhead from $-
and $ by replacing:
a$myx - array(dim=c(j, n))
for (i in 1:n) {
a$myx[,i] - rnorm(j)
}
a$myx
with
myx - array(NA_real, dim=c(j, n))
for (i in 1:n) {
myx[,i] - rnorm(j)
}
a$myx - myx
myx
Similarly for S4 slots and @- and @.
/Henrik
On Thu, Feb 27, 2014 at 7:53 PM, Ross Boylan r...@biostat.ucsf.edu wrote:
The R Inferno advises that if you are building up results in pieces it's
best to pre-allocate the result object and fill it in. In some testing,
I see a benefit with this strategy for regular variables. However, when
the results are held by a class, the opposite seems to be the case.
Comments? Explanations?
Possibly for classes any update causes the entire object to be
replaced--perhaps to trigger the validation machinery?--and so
preallocation simply means on average a bigger object is being
manipulated.
Here is some test code, with CPU seconds given in the comments. I tried
everything twice in case there was some first-time overhead such as
growing total memory in the image. When the 2 times differed noticeably
I reported both values.
# class definitions
refbase - setRefClass(refBase, fields = list(dispatch=ANY, myx=ANY),
methods = list( initialize = function(x0=NULL, ...) {
usingMethods(foo)
dispatch - foo
myx - x0
}
# some irrelevant methods edited out
))
myclass - setClass(simple, representation=list(myx=ANY))
### Method 1: regular variables
pre - function(n, j=1000) {
x - array(dim=(c(j, n)))
for (i in 1:n) {
x[,i] - rnorm(j)
}
x
}
system.time(pre(1000)) #0.3s
nopre - function(n, j=1000) {
x - numeric(0)
for (i in 1:n)
x - c(x, rnorm(j))
x
}
system.time(nopre(1000)) # 2.0s, 2.7s
# Method 2: with ref class
pre2 - function(n, j=1000) {
a - refbase(x0=numeric(0))
a$myx - array(dim=c(j, n))
for (i in 1:n) {
a$myx[,i] - rnorm(j)
}
a$myx
}
system.time(pre2(1000)) # 4.0 s
nopre2 - function(n, j=1000) {
a - refbase(x0=numeric(0))
for (i in 1:n)
a$myx - c(a$myx, rnorm(j))
a$myx
}
system.time(nopre2(1000)) # 2.9s, 4.3
# Method 3: with regular class
pre3 - function(n, j=1000) {
a - myclass()
a@myx - array(dim=c(j, n))
for (i in 1:n) {
a@myx[,i] - rnorm(j)
}
a@myx
}
system.time(pre3(1000)) # 7.3 s
nopre3 - function(n, j=1000) {
a - myclass(myx=numeric(0))
for (i in 1:n)
a@myx - c(a@myx, rnorm(j))
a@myx
}
system.time(nopre3(1000)) # 4.2s
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Difficulty coding time-forced functions in deSolve
On 4/2/2014 5:51 AM, Aimee Kopolow wrote: Any pointers as to how I can code a function that relies on solutions from previous time steps? Such a system would be called a delay differential equation (DDE). It can be solved with the dede function, see ?dede for details. However if you want to model something like this: Explicitly: I want to introduce vaccination 7 days after the proportion of I2/N2 reaches 0.01. Than this is called root finding, that can be combined with events, see example EVENTS triggered by a root function in ?events. More can be found in the papers listed at: http://desolve.r-forge.r-project.org ... or you may consider to ask the R-sig-dynamic models mailing list: https://stat.ethz.ch/mailman/listinfo/r-sig-dynamic-models Hope it helps Thomas Petzoldt Dr. Thomas Petzoldt Technische Universitaet Dresden Faculty of Environmental Sciences Institute of Hydrobiology 01062 Dresden, Germany http://tu-dresden.de/Members/thomas.petzoldt __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A vector of normal distributed values with a sum-to-zero constraint
Hi Marc
I think that we could help you better if we knew in which context you need
sample from a sum constrained normal distribution. However this is more a
question on probability theory than on how to do it in R.
The proposal so far has been linear transformation of multivariate normal
distribution (Marc, Rui), mixture of normal and reflected normal distribution
(Boris, try that with e.g. mu = 2), normal distribution mixed with single point
with positive mass (Jlucke), degenerated normal distribution (Greg).
What you in fact want to do is to draw samples from a conditional distribution.
The condition is the sum constraint so if we have x = (x1, x2, ..., xn) then
sum_{i=1}^n xi = 0 or x1 + x2 + ... x{n-1} = xn so you want to draw samples
from P(x given that x is normal distributed and sum(x)=0). The sum constraint
gives in fact what is called distributions on the simplex. Google for normal
distribution simplex and you will get almost 2 mill hits. The second shows how
to sample using Gibbs sampling
(http://dobigeon.perso.enseeiht.fr/papers/Dobigeon_TechReport_2007b.pdf).
However you can probably just use other distributions given sum constraint
since you say that you only need the sample as initial values for a MCMC
algorithm. Many methods are available from compositional statistics (google
for that, Aitchison 1986 is the pioneer).
At least two packages are available for R:compositions with the latest
version from 2013 and can be found in the archives and robComposition still
maintained.
Hope that helps, it is help for yourself to find a solution.
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730 5135
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Marc Marí Dell'Olmo
Sent: 1. april 2014 16:57
To: Boris Steipe
Cc: r-help@r-project.org
Subject: Re: [R] A vector of normal distributed values with a sum-to-zero
constraint
Boris is right. I need this vector to include as initial values of a
MCMC process (with openbugs) and If I use this last approach sum(x)
could be a large (or extreme) value and can cause problems.
The other approach x - c(x, -x) has the problem that only vectors
with even values are obtained.
Thank you!
2014-04-01 16:25 GMT+02:00 Boris Steipe boris.ste...@utoronto.ca:
But the result is not Normal. Consider:
set.seed(112358)
N - 100
x - rnorm(N-1)
sum(x)
[1] 1.759446 !!!
i.e. you have an outlier at 1.7 sigma, and for larger N...
set.seed(112358)
N - 1
x - rnorm(N-1)
sum(x)
[1] -91.19731
B.
On 2014-04-01, at 10:14 AM, jlu...@ria.buffalo.edu wrote:
The sum-to-zero constraint imposes a loss of one degree of freedom. Of
N samples, only (N-1) can be random. Thus the solution is
N - 100
x - rnorm(N-1)
x - c(x, -sum(x))
sum(x)
[1] -7.199102e-17
Boris Steipe boris.ste...@utoronto.ca
Sent by: r-help-boun...@r-project.org
04/01/2014 09:29 AM
To
Marc Marí Dell'Olmo marceivi...@gmail.com,
cc
r-help@r-project.org r-help@r-project.org
Subject
Re: [R] A vector of normal distributed values with a sum-to-zero
constraint
Make a copy with opposite sign. This is Normal, symmetric, but no longer
random.
set.seed(112358)
x - rnorm(5000, 0, 0.5)
x - c(x, -x)
sum(x)
hist(x)
B.
On 2014-04-01, at 8:56 AM, Marc Marí Dell'Olmo wrote:
Dear all,
Anyone knows how to generate a vector of Normal distributed values
(for example N(0,0.5)), but with a sum-to-zero constraint??
The sum would be exactly zero, without decimals.
I made some attempts:
l - 100
aux - rnorm(l,0,0.5)
s - sum(aux)/l
aux2 - aux-s
sum(aux2)
[1] -0.0006131392
aux[1]- -sum(aux[2:l])
sum(aux)
[1] -0.03530422
but the sum is not exactly zero and not all parameters are N(0,0.5)
distributed...
Perhaps is obvious but I can't find the way to do it..
Thank you very much!
Marc
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and provide commented, minimal, self-contained,
Re: [R] A vector of normal distributed values with a sum-to-zero constraint
On 01 Apr 2014, at 17:22 , Rui Barradas ruipbarra...@sapo.pt wrote: Hello, One way is to use ?scale. ...except that the sd will be less than 0.5 (not obvious at n=1e6, though). However, if you want - normal distribution - symmetry - constant marginal variance of sigma^2 - fixed sum = 0 I don't see any way more straightforward than generating n normal variates and subtracting the mean. The only snag is that the variance of a residual is sigma^2(1-1/n), so generate the original data with a variance of sigma^2/(1-1/n) = n/(n-1) sigma^2 I.e. x - rnorm(n,0,0.5*sqrt(n/(n-1))) x - x - mean(x) All of this applies within the boundaries of numerical precision. You're not going to beat the FPU: n - 1e6 x - rnorm(n,0,0.5*sqrt(n/(n-1))) x - x - mean(x) sum(x) [1] -1.625718e-11 mean(x) [1] -1.452682e-17 The problem of getting a sum or mean of _exactly_ 0 is just not well-defined, since sums and averages depend on the summation order: sum(x) [1] -1.625718e-11 sum(sample(x)) [1] -1.624851e-11 sum(sort(x)) [1] -1.508771e-11 sum(rev(sort(x))) [1] -1.599831e-11 set.seed(4867) l - 100 aux - rnorm(l, 0, 0.5) aux - scale(aux, scale = FALSE) sum(aux) hist(aux, prob = TRUE) curve(dnorm(x, 0, 0.5), from = -2, to = 2, add = TRUE) Hope this helps, Rui Barradas Em 01-04-2014 16:01, jlu...@ria.buffalo.edu escreveu: Then what's wrong with centering your initial values around the mean? Marc Marí Dell'Olmo marceivi...@gmail.com 04/01/2014 10:56 AM To Boris Steipe boris.ste...@utoronto.ca, cc jlu...@ria.buffalo.edu, r-help@r-project.org r-help@r-project.org Subject Re: [R] A vector of normal distributed values with a sum-to-zero constraint Boris is right. I need this vector to include as initial values of a MCMC process (with openbugs) and If I use this last approach sum(x) could be a large (or extreme) value and can cause problems. The other approach x - c(x, -x) has the problem that only vectors with even values are obtained. Thank you! 2014-04-01 16:25 GMT+02:00 Boris Steipe boris.ste...@utoronto.ca: But the result is not Normal. Consider: set.seed(112358) N - 100 x - rnorm(N-1) sum(x) [1] 1.759446 !!! i.e. you have an outlier at 1.7 sigma, and for larger N... set.seed(112358) N - 1 x - rnorm(N-1) sum(x) [1] -91.19731 B. On 2014-04-01, at 10:14 AM, jlu...@ria.buffalo.edu wrote: The sum-to-zero constraint imposes a loss of one degree of freedom. Of N samples, only (N-1) can be random. Thus the solution is N - 100 x - rnorm(N-1) x - c(x, -sum(x)) sum(x) [1] -7.199102e-17 Boris Steipe boris.ste...@utoronto.ca Sent by: r-help-boun...@r-project.org 04/01/2014 09:29 AM To Marc Marí Dell'Olmo marceivi...@gmail.com, cc r-help@r-project.org r-help@r-project.org Subject Re: [R] A vector of normal distributed values with a sum-to-zero constraint Make a copy with opposite sign. This is Normal, symmetric, but no longer random. set.seed(112358) x - rnorm(5000, 0, 0.5) x - c(x, -x) sum(x) hist(x) B. On 2014-04-01, at 8:56 AM, Marc Marí Dell'Olmo wrote: Dear all, Anyone knows how to generate a vector of Normal distributed values (for example N(0,0.5)), but with a sum-to-zero constraint?? The sum would be exactly zero, without decimals. I made some attempts: l - 100 aux - rnorm(l,0,0.5) s - sum(aux)/l aux2 - aux-s sum(aux2) [1] -0.0006131392 aux[1]- -sum(aux[2:l]) sum(aux) [1] -0.03530422 but the sum is not exactly zero and not all parameters are N(0,0.5) distributed... Perhaps is obvious but I can't find the way to do it.. Thank you very much! Marc __ 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. [[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
[R] Strange sprintf Behavior
All, I'm getting this: sprintf(%.17f, 0.8) [1] 0.80004 Where does the `4` at the end come from? Shouldn't it be zero at the end? Maybe I'm missing something. sessionInfo() R version 3.0.2 (2013-09-25) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C LC_TIME=en_US.utf8 [4] LC_COLLATE=en_US.utf8 LC_MONETARY=en_US.utf8 LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C LC_ADDRESS=C [10] LC_TELEPHONE=CLC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base Thanks, M __ 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] mzR and Rcpp version bug
Good morning everyone, I'm having the following error when I try to load the *mzR* libary. Does someone have a clue where to search for a solution. I tried to re-install *mzR *and *Rcpp* also, but with no effect. - library(mzR) Lade nötiges Paket: Rcpp Error : .onLoad in loadNamespace() für 'mzR' fehlgeschlagen, Details: Aufruf: value[[3L]](cond) Fehler: failed to load module Ramp from package mzR kann Vektor der Größe 13.2 GB nicht allozieren Zusätzlich: Warnmeldungen: 1: In fun(libname, pkgname) : mzR has been built against a different Rcpp version than is installed on your system. This might lead to errors when loading mzR. If you encounter such issues, please send a report, including the output of sessionInfo() to the Bioc mailing list -- http://www.bioconductor.org/help/mailing-list. 2: In Module(m, pkg, mustStart = TRUE) : Reached total allocation of 8183Mb: see help(memory.size) 3: In Module(m, pkg, mustStart = TRUE) : Reached total allocation of 8183Mb: see help(memory.size) 4: In Module(m, pkg, mustStart = TRUE) : Reached total allocation of 8183Mb: see help(memory.size) 5: In Module(m, pkg, mustStart = TRUE) : Reached total allocation of 8183Mb: see help(memory.size) Fehler: Laden von Paket oder Namensraum für 'mzR' fehlgeschlagen sessionInfo() R version 3.0.3 (2014-03-06) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252 LC_MONETARY=German_Germany.1252 LC_NUMERIC=C [5] LC_TIME=German_Germany.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] Rcpp_0.11.1 loaded via a namespace (and not attached): [1] Biobase_2.20.1 BiocGenerics_0.6.0 parallel_3.0.3 - Best Regards, László-András Zsurzsa Master of Informatics Student, Technical University Munich, Germany [[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] Strange sprintf Behavior
On Apr 2, 2014, at 6:32 AM, Michael Smith my.r.h...@gmail.com wrote: All, I'm getting this: sprintf(%.17f, 0.8) [1] 0.80004 Where does the `4` at the end come from? Shouldn't it be zero at the end? Maybe I'm missing something. Hi, First, please start a new thread when posting, do not just reply to an existing thread and change the subject line. Your post gets lost in the archive and is improperly linked to other posts. Second, see the Most Frequently Asked Question: http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f Regards, 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.
[R] gradientForest input data structure
Dear All, Following on from my last post (randomForest warning: The response has five or fewer unique values. Are you sure you want to do regression?) which presented two problems whilst trying to conduct a gradientForest regression, the warning I got was not an issue as Andy kindly pointed out, but I still have the second problem relating to the data structure of my input data and I would really appreciate your help on this. I think this is simply a data structure issue and nothing specific to gradientForest. I am a relative beginner to R (also not a mathematician) and have tried to figure out how the data is structured to get the analysis to work but to no avail. I can run the analysis with the data provided within the gradientForest package according to the instructions but when I try it with my own data it doesnt work and it does not consider all the response variables (please see output in the previous post below). So my understanding is that gradientForest regression requires a set of response variables and a set of predictor variables which then need to be combined. The structure of the predictor variables according to the example data accompanying the gradientForest package is: load(GZ.phys.site.Rdata) str(Phys_site) 'data.frame': 197 obs. of 28 variables: $ BATHY : num -16.7 -26.6 -32.8 -32.5 -29.7 ... $ SLOPE : num 0.505 0.784 0.12 0.332 0.467 ... $ ASPECT : num 234 116 192 172 230 ... $ BSTRESS: num 0.218 0.248 0.322 0.374 0.425 ... $ CRBNT : num 98.5 98 98.9 98.3 97.9 ... $ GRAVEL : num 39.3 39.2 30 42.4 38.8 ... $ SAND : num 59.7 59.8 63.9 54.9 62.7 ... $ MUD: num 3.16e-07 2.76e-02 5.17 9.22e-01 2.38 ... $ NO3_AV : num 0.24 0.3 0.24 0.26 0.3 0.24 0.25 0.24 0.23 0.25 ... $ NO3_SR : num 0.33 0.39 0.29 0.16 0.2 0.33 0.31 0.35 0.39 0.19 ... $ PO4_AV : num 0.15 0.15 0.15 0.15 0.16 0.16 0.16 0.16 0.15 0.15 ... $ PO4_SR : num 0.08 0.08 0.07 0.05 0.08 0.07 0.07 0.08 0.08 0.06 ... $ O2_AV : num 4.42 4.44 4.39 4.35 4.33 4.38 4.34 4.37 4.4 4.34 ... $ O2_SR : num 0.4 0.49 0.28 0.26 0.24 0.32 0.27 0.36 0.43 0.24 ... $ S_AV : num 34.9 34.9 35 34.9 34.9 ... $ S_SR : num 1.47 1.29 1.64 1.57 1.58 1.8 1.94 1.81 1.7 1.83 ... $ T_AV : num 28.2 28 28.3 28.6 28.5 ... $ T_SR : num 2.19 2.79 1.8 1.99 2.12 2.03 2.12 2.18 2.15 2.23 ... $ Si_AV : num 2.33 2.67 2.25 1.26 1.21 2.39 2.34 2.46 2.6 1.6 ... $ Si_SR : num 4.3 4.96 3.59 2.59 2.64 3.92 3.56 4.32 4.88 2.97 ... $ CHLA_AV: num 0.499 0.499 0.455 0.594 0.594 ... $ CHLA_SR: num 0.55 0.55 0.669 1.258 1.258 ... $ K490_AV: num 0.0726 0.0726 0.0672 0.075 0.075 ... $ K490_SR: num 0.0489 0.0489 0.0594 0.0732 0.0732 ... $ SST_AV : num 27 27 26.9 26.9 26.9 ... $ SST_SR : num 4.85 4.85 4.81 4.81 4.81 ... $ BIR_AV : num 0.1735 0.0688 0.2397 0.4476 0.5169 ... $ BIR_SR : num 0.1563 0.0885 0.2249 0.2667 0.256 ... Which seems to correspond to the structure of my predictor variables, so I dont think this is the problem: str(enviro) 'data.frame': 14 obs. of 8 variables: $ Temperature : num 24.8 24.4 24.3 23 24.6 24.6 24.8 24.9 24.3 24.5 ... $ Turbidity : num 0.047 0.046 0.052 0.058 0.049 0.047 0.047 0.049 0.049 0.051 ... $ Chlorophyll : num 0.24 0.23 0.29 0.26 0.25 0.23 0.23 0.28 0.3 0.29 ... $ Waveheight : num 2.14 2.13 2.12 2.12 2.12 2.12 2.11 2.12 2.11 2.12 ... $ nLw551 : num 0.231 0.228 0.228 0.236 0.226 ... $ nLw667 : num 1e-04 8e-04 1e-03 1e-03 1e-03 1e-04 1e-04 1e-03 1e-03 1e-04 ... $ Sediment.nlw551.667.: num 0.231 0.229 0.229 0.237 0.227 ... $ Depth : num 4.8 4.1 5 4 6.2 7.7 10.1 4.3 5.1 7.9 ... BUT my set of response variables seems to be in the wrong structure and this is I think the problem and where I need help. This is the structure of the example data provided with gradientForest: load(GZ.sps.mat.Rdata) str(Sp_mat) num [1:197, 1:110] 1.04 -2.11 -3.43 -2.36 -1.15 ... - attr(*, dimnames)=List of 2 ..$ : chr [1:197] 1 2 3 4 ... ..$ : chr [1:110] A1010102 A1010113 A1010206 A1010209 ... And this is the structure that my response variables are currently in (essentially a matrix created from Excel with rows indicating sites (14 of them) and coloumns indicating species (100 hundred of them) abbundances occuring at these sites (Header = TRUE): # data structure of biological data str(biological) 'data.frame': 14 obs. of 100 variables: $ a : num 0 0 0 0 0 0 0 0 0 0 ... $ b : num 0 0 0 0 257 ... $ c : int 0 0 0 0 0 0 441 0 0 0 ... $ d : num 179 0 1430 0 0 ... $ e : num 100 0 601 0 123 ... $ f : num 0 0 3 0 1.5 0 0 0 0 4.5 ... $ g : num 0 0 0 0 0 0 0 0 0 0 ... $ h : int 0 0 0 0 0 0 0 0 1 0 ... $ i : num 0 0 0 0 0 0 0 0 0 3.85 ... $ j : num 0 0 0 27.6 3.6 ... $ k : num 0 0 0 0 0 0 0 0 0 1.8 ... $ l : num 0 0 0 0 0 0 0 0 0 0 ... $ m : num 0 0 0 0 0 0 0 0 0 0 ... $ n : num 0 0 0 0 0 0 0 1.1 0 0 ... $ o : num 0 0 0 0 0 0.2 0 0 0 0 ... $ p : num 0 0.15 0
[R] Survey
Dear R-Users, I was using survey for the past years and now I am experiencing some problems with scripts that was working in the past. We are working with big data bases so I can't put all variables that I will use in the svydesign. Script working: data(api) dclus1-svydesign(id=apiclus1$dnum, weights=apiclus1$pw, fpc=apiclus1$fpc) svyby(apiclus1$api99,apiclus1$stype, dclus1, svymean) by V1 se E E 607.7917 22.81660 H H 595.7143 41.76400 M M 608.6000 32.56064 svyby(apiclus1$api99,~apiclus1$stype, dclus1, svymean) apiclus1$stype V1 se E E 607.7917 22.81660 H H 595.7143 41.76400 M M 608.6000 32.56064 dclus1-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) svyby(~api99, ~stype, dclus1, svymean) stypeapi99 se.api99 E E 607.7917 22.81660 H H 595.7143 41.76400 M M 608.6000 32.56064 sessionInfo() R version 2.12.1 (2010-12-16) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=Portuguese_Brazil.1252 LC_CTYPE=Portuguese_Brazil.1252 LC_MONETARY=Portuguese_Brazil.1252 [4] LC_NUMERIC=C LC_TIME=Portuguese_Brazil.1252 attached base packages: [1] grDevices datasets splines graphics stats tcltk utils methods base other attached packages: [1] gdata_2.8.2 survey_3.26 foreign_0.8-42 SOAR_0.99-8 svSocket_0.9-51 TinnR_1.0.3 R2HTML_2.2 [8] Hmisc_3.8-3 survival_2.36-2 loaded via a namespace (and not attached): [1] cluster_1.14.1 grid_2.12.1 gtools_2.6.2lattice_0.19-33 svMisc_0.9-63 tools_2.12.1 script not working: data(api) dclus1-svydesign(id=apiclus1$dnum, weights=apiclus1$pw, fpc=apiclus1$fpc) svyby(apiclus1$api99,apiclus1$stype, dclus1, svymean) Erro em svyby.default(apiclus1$api99, apiclus1$stype, dclus1, svymean) : objeto 'byfactor' não encontrado svyby(apiclus1$api99,~apiclus1$stype, dclus1, svymean) Erro em svyby.default(apiclus1$api99, ~apiclus1$stype, dclus1, svymean) : objeto 'byfactor' não encontrado dclus1-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) svyby(~api99, ~stype, dclus1, svymean) stypeapi99 se E E 607.7917 22.81660 H H 595.7143 41.76400 M M 608.6000 32.56064 sessionInfo() R version 3.0.3 (2014-03-06) Platform: x86_64-w64-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=Portuguese_Brazil.1252 LC_CTYPE=Portuguese_Brazil.1252 LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C LC_TIME=Portuguese_Brazil.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] survey_3.29-5 loaded via a namespace (and not attached): [1] tools_3.0.3 [[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] Strange sprintf Behavior
It is poor netiquette to reply to a thread with a different subject. Please start a new thread for a new subject. As for your question, see FAQ 7.31. This is standard floating point numerical limitations at work. --- 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 April 2, 2014 4:32:26 AM PDT, Michael Smith my.r.h...@gmail.com wrote: All, I'm getting this: sprintf(%.17f, 0.8) [1] 0.80004 Where does the `4` at the end come from? Shouldn't it be zero at the end? Maybe I'm missing something. sessionInfo() R version 3.0.2 (2013-09-25) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C LC_TIME=en_US.utf8 [4] LC_COLLATE=en_US.utf8 LC_MONETARY=en_US.utf8 LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C LC_ADDRESS=C [10] LC_TELEPHONE=CLC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base Thanks, M __ 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] ggplot: add points selectively to curves
I'm working on an example of plotting predicted probabilities from a proportional odds model. The steps below generate a data frame, plotdat, that I want to plot with ggplot. library(MASS) data(Arthritis, package=vcd) arth.polr - polr(Improved ~ Sex + Treatment + Age, data=Arthritis, Hess=TRUE) # get predicted probs for categories of Improve arth.fitp - cbind(Arthritis, predict(arth.polr, type=probs)) head(arth.fitp) # reshape probs to long library(reshape2) plotdat - melt(arth.fitp, id.vars = c(Sex, Treatment, Age, Improved), measure.vars=c(None, Some, Marked), variable.name = Level, value.name = Probability) ## view first few rows head(plotdat) head(plotdat) Sex Treatment Age Improved Level Probability 1 Male Treated 27 Some None 0.7326185 2 Male Treated 29 None None 0.7174048 3 Male Treated 30 None None 0.7096042 4 Male Treated 32 Marked None 0.6936286 5 Male Treated 46 Marked None 0.5702499 6 Male Treated 58 Marked None 0.4563432 In the plot step, I am plotting Probability vs. Age, stratified by Level, and faceted by Sex and Treatment. My question concerns the use of geom_point(). The call below plots 3 points for each case, one on each Level curve. ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + geom_point(color=black, size=1.5) + facet_grid(Sex ~ Treatment, labeller = function(x, y) sprintf(%s = %s, x, y) ) Instead, I want to plot only one point for each case, for the value of Level that corresponds to the value of Improved in this data set. Somehow, this involves something like an aes() argument to geom_point(), with Level indexed by Improved, or some such. How can I do this? -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] CORDIF test
Hi, I search on your website for a definition of the CORDIF test, but it wasnt successful. Im analyzing an article that use that test and its not really documented on the net. The article refers to your website, so I pretend that you will be able to give me a brief explanation of this test. Here is the cote that talk about this test in my article : ' To compare these regressions and to see whicheither body height or LLLis best related to performance (Pearson correlation coefficients comparison), a CORDIF test (R software [www.r-project.org], multilevel package, ver- sion 2.12.1) was performed. Does it use parametric or non-parametric values ? Is it a test to compare 2 groups only or it can be used for a comparison of more than two groups ? Why is it so hard to find information on that test on the net ? Thanks for your time Have a nice day Elizabeth Caron Physical therapist student, Laval University, Qc, Canada [[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] Insert DateTime from R into MongoDB
First time poster, for forgive if I'm not following the etiquette:
How does one properly insert a DateTime, from R, into mongoDB and have it
recognized as a DateTime in mongoDB in lieu of being recognized as a string?
I've tried a zillion things, and NOTHING works. I'm using:
library('RMongo')
require(rmongodb)
require(RJSONIO)
With a variety of commands I found online, and it always passes from R into
mognoDB as a string no matter what I do.
--
View this message in context:
http://r.789695.n4.nabble.com/Insert-DateTime-from-R-into-MongoDB-tp4688030.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Help with MANOVA
I'm trying to run a MANOVA on shoreline habitat data. The data was collected in an odd way. For each reach of shoreline, data was collected for each habitat variable (shade, veg and IWM). Each variable was described as having a certain percentage of a particular category of the variable. For example, (not from my data) Shade (the habitat variable) at Reach 1 was comprised of 50% No Shade (one of the categories), 25% 6%-25% shade, and 25% Greater than 75% shade. Therefore each of the three variables are described by amounts of each of the categories and the percentages always equal 100 when added together. So my question is, how do I run a MANOVA on this to determine if the reaches are statistically different in R? Thanks in advance for any advice! Reach No Shade 1-5% Shade 6-25% Shade 26-75%Shade 75% Shade 25% Veg 26-50% Veg 51-75%Veg 75% Veg No IWM 1-10% IMW 11-50%IWM 50% IWM MS01 0.5609 0.2631 0.0103 0.1658 0 0 0.0294 0 0.9706 0.6693 0.2281 0.1026 0 MS02 0.2916 0.5672 0.0623 0.0789 0 0.0222 0 0 0.9778 0.1312 0.4787 0.1011 0.289 MS03 0.477 0.0411 0.0609 0.184 0.2371 0.0411 0 0 0.9589 0.477 0 0.225 0.298 MS04 0.7158 0.2037 0 0.0805 0 0.2569 0 0 0.7431 0.4589 0.2037 0.3374 0 MS05 0.4938 0.5062 0 0 0 0 0 0 1 0.558 0.442 0 0 [[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] R2WinBUGS expected collection operator c error
Hi all, I am currently in a Bayesian Modeling course and am trying to implement an analog representation model using R, WinBUGS, and R2WinBUGS. I'm currently stuck banging my head against an expected collection operator c error. I am working off of a template code, and I swear I haven't moved any collection operators from the original code. Here is what my script looks like in R: # clears workspace: rm(list=ls(all=TRUE)) # sets working directories: setwd(C:\\Users\\Nick\\Desktop\\modeling) library(R2WinBUGS) bugsdir = C:/Program Files/WinBUGS14 # read the data g = matrix(scan(g.txt, sep=,), ncol=144, nrow=3, byrow=T) q = matrix(scan(q.txt, sep=,), ncol=144, nrow=3, byrow=T) S = 3 # number of subjects Q = 144 # number of questions for each child N = 9 # maximum numbers of item list data = list(g, q, S, Q, N) # to be passed on to WinBUGS myinits = list( list(sigma = rep(1,S), pitmp=rep(1/N, N))) # parameters to be monitored: parameters = c(pp,ppb,sigma) # NB. even with only 1000 iterations, the sampling can take a long time! # The following command calls WinBUGS with specific options. # For a detailed description see Sturtz, Ligges, Gelman (2005). samples = bugs(data, inits=myinits, parameters, model.file =NumberConcept_2_data.txt, n.chains=1, n.iter=1000, n.burnin=100, n.thin=1, DIC=T, bugs.directory=bugsdir, codaPkg=F, debug=T) # Now the values for the monitored parameters are in the samples object, # ready for inspection. samples$summary ### and here is the log file file from WinBUGS after R2WinBUGS passes it over: ### display(log) check(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/NumberConcept_2_data.txt) model is syntactically correct data(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/data.txt) expected collection operator c compile(1) inits(1,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/inits1.txt) command #Bugs:inits cannot be executed (is greyed out) gen.inits() command #Bugs:gen.inits cannot be executed (is greyed out) thin.updater(1) update(100) command #Bugs:update cannot be executed (is greyed out) set(pp) command #Bugs:set cannot be executed (is greyed out) set(ppb) command #Bugs:set cannot be executed (is greyed out) set(sigma) command #Bugs:set cannot be executed (is greyed out) set(deviance) command #Bugs:set cannot be executed (is greyed out) dic.set() command #Bugs:dic.set cannot be executed (is greyed out) update(900) command #Bugs:update cannot be executed (is greyed out) coda(*,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/coda) command #Bugs:coda cannot be executed (is greyed out) stats(*) command #Bugs:stats cannot be executed (is greyed out) dic.stats() DIC history(*,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/history.odc) command #Bugs:history cannot be executed (is greyed out) save(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/log.odc) save(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/log.txt) # WinBUGS also opens up a data window that looks like this: # list(g= structure(.Data= c(1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00,
Re: [R] CORDIF test
google is your friend! google r cordif test -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Wed, Apr 2, 2014 at 6:09 AM, Elizabeth Caron-Gamache babeth_...@icloud.com wrote: Hi, I search on your website for a definition of the CORDIF test, but it wasn’t successful. I’m analyzing an article that use that test and it’s not really documented on the net. The article refers to your website, so I pretend that you will be able to give me a brief explanation of this test. Here is the cote that talk about this test in my article : ‘' To compare these regressions and to see which—either body height or LLL—is best related to performance (Pearson correlation coefficients comparison), a CORDIF test (R software [www.r-project.org], multilevel package, ver- sion 2.12.1) was performed. Does it use parametric or non-parametric values ? Is it a test to compare 2 groups only or it can be used for a comparison of more than two groups ? Why is it so hard to find information on that test on the net ? Thanks for your time Have a nice day Elizabeth Caron Physical therapist student, Laval University, Qc, Canada [[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] CORDIF test
On Apr 2, 2014, at 8:09 AM, Elizabeth Caron-Gamache babeth_...@icloud.com wrote: Hi, I search on your website for a definition of the CORDIF test, but it wasn’t successful. I’m analyzing an article that use that test and it’s not really documented on the net. The article refers to your website, so I pretend that you will be able to give me a brief explanation of this test. Here is the cote that talk about this test in my article : ‘' To compare these regressions and to see which—either body height or LLL—is best related to performance (Pearson correlation coefficients comparison), a CORDIF test (R software [www.r-project.org], multilevel package, ver- sion 2.12.1) was performed. Does it use parametric or non-parametric values ? Is it a test to compare 2 groups only or it can be used for a comparison of more than two groups ? Why is it so hard to find information on that test on the net ? Thanks for your time Have a nice day Elizabeth Caron Physical therapist student, Laval University, Qc, Canada Thanks for including the citation, which indicates that the CORDIF test is part of the 'multilevel' package, which is on CRAN: http://cran.r-project.org/web/packages/multilevel/index.html The reason that it is likely difficult is that 'cordif' is an abbreviation for correlation difference, not the proper name for a test. If you review the provided documentation for the package: http://cran.r-project.org/web/packages/multilevel/multilevel.pdf you will see that there is a description of the cordif() function and a reference given: Cohen, J. Cohen, P. (1983). Applied multiple regression/correlation analysis for the behavioral sciences (2nd Ed.). Hillsdale, NJ: Lawrence Erlbaum Associates. I would review the package documentation and reference and if you have further questions, contact the authors of the paper. Regards, 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] CORDIF test
Hi Elizabeth,
In addition to the helpful suggestions you have already received, I
would add that for simple functions like this it can be very
instructive to just look at the function definition. Loading the
multilevel package and typing 'cordif' will show you
function (rvalue1, rvalue2, n1, n2)
{
zvalue1 - 0.5 * ((log(1 + rvalue1)) - (log(1 - rvalue1)))
zvalue2 - 0.5 * ((log(1 + rvalue2)) - (log(1 - rvalue2)))
zest - (zvalue1 - zvalue2)/sqrt(1/(n1 - 3) + 1/(n2 - 3))
out - list(`z value` = zest)
return(out)
}
best.
Ista
On Wed, Apr 2, 2014 at 9:09 AM, Elizabeth Caron-Gamache
babeth_...@icloud.com wrote:
Hi,
I search on your website for a definition of the CORDIF test, but it wasn’t
successful. I’m analyzing an article that use that test and it’s not really
documented on the net. The article refers to your website, so I pretend that
you will be able to give me a brief explanation of this test. Here is the
cote that talk about this test in my article :
‘' To compare these regressions and to see which—either body height or LLL—is
best related to performance (Pearson correlation coefficients comparison), a
CORDIF test (R software [www.r-project.org], multilevel package, ver- sion
2.12.1) was performed.
Does it use parametric or non-parametric values ?
Is it a test to compare 2 groups only or it can be used for a comparison of
more than two groups ?
Why is it so hard to find information on that test on the net ?
Thanks for your time
Have a nice day
Elizabeth Caron
Physical therapist student, Laval University, Qc, Canada
[[alternative HTML version deleted]]
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Re: [R] Removing White spaces with NA
Hi, May be this helps: dat - data.frame(Col1=c(A, , B,C, ,,D), stringsAsFactors=FALSE) is.na(dat) - dat=='' dat$Col1 #[1] A NA B C NA NA D A.K. Hi All, I have a table and a column with values as below Col1 A B C D I need to replace the Empty cells with the value NA as below Col1 A NA B C NA NA D I tried a code, which was not working. Table.name$column.name - gsub(,NA, table.name$column.name) Can anyone help me with this ? Thanks and regards, Praveen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Interpreting effect of ordered categorical predictor
(I posted this question in http://stackoverflow.com/questions/22781965/interpreting-effect-of-ordered-categorical-predictor without answer... I try here) Thanks a lot Marc My question is very similar to this one (https://stat.ethz.ch/pipermail/r-help/2012-March/305357.html) but I fail to understand it fully. I would like to visualize the effect of an ordered categorical predictor after a glm. First I generate some dummy data: ## data.frame with continuous values and 6 factors datagenerate - data.frame(measure=c(rnorm(20, 10, 2), rnorm(30, 15, 2), rnorm(20, 20, 2), rnorm(20, 25, 2), rnorm(20, 30, 2), rnorm(20, 35, 2)), factor=c(rep(A, 20), rep(B, 30), rep(C, 20), rep(D, 20), rep(E, 20), rep(F, 20)), stringsAsFactors=FALSE) nbfactors - length(levels(datagenerate$factor)) Now I apply a glm with an unordered category: ## First factors are unordered datagenerate$factor - as.factor(datagenerate$factor) essaiglm - glm(measure ~ factor, datagenerate, family=gaussian()) coef_unordered - coef(summary(essaiglm))[,1] plot(1:nbfactors, c(0, coef_unordered[2:nbfactors]), type=h, bty=n, las=1, xlab=Factors, ylab=Effect) All is ok. But I would like to do the same with ordered category: ## Now factors are ordered datagenerate$factor - ordered(datagenerate$factor, levels=c(A, B, C, D, E, F)) essaiglm - glm(measure ~ factor, datagenerate, family=gaussian()) coef_ordered - coef(summary(essaiglm))[,1] ## I am not sure about this line. How the ordered factors are coded ? x - ((0:(nbfactors-1))-(nbfactors-1)/2)/(nbfactors-1) y - x*coef_ordered[factor.L]+x^2*coef_ordered[factor.Q]+ x^3*coef_ordered[factor.C]+x^4*coef_ordered[factor^4]+ x^5*coef_ordered[factor^5] y - y-min(y) plot(1:nbfactors, y, type=h, bty=n, las=1, xlab=Factors, ylab=Effect) The result is highly dependent on the coding of the levels. Based on several tries, I propose x - ((0:(nbfactors-1))-(nbfactors-1)/2)/(nbfactors-1) But I am not sure. If someone has the answer, I will be very grateful. Thanks a lot Marc __ 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] 127.0.0.1:22381/doc/html/index.html hep.start won't start
Dear All, Sorry to bother you. I narrowed my problem to the function help.start() which causes the freezing of R by mill endlessly after M-x R waiting for 127.0.0.1:22381/doc/html/index.html to load. I am using Aquamacs 3.0a (GNU Emacs 24.3.50.2)with R3.0.3, newly installed: sessionInfo() --- R version 3.0.3 (2014-03-06) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] compiler_3.0.3 tools_3.0.3 - The FIREWALL is responsible, deactivating will allow help.start()? Bur I don't want to work without the firewall!! Settingthe firewall for Firefox and Aquamacs to allow incoming connections (translated from German)will also not help. Thanks for considering. Christian __ 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] looping in R
I ran the following loop on my SNP data and got an error message as
indicated
for (i in genenames){
+ current - fst1[which(fst1$Gene == i),]
+ num - nrow(current)
+ fst - max(current$fst)
+ position - mean(current$pos)
+ nposition - mean(current$newpos)
+ numhigh - nrow(current[which(current$fst threshold),])
+ colors - mean(current$colors)
+ output - matrix(NA,nrow=1,ncol=8)
+ numthresh - paste(# SNPs Fst = , threshold, sep=)
+ colnames(output) - c(gene, gene_old, pos, newpos, # Snps,
numthresh, Max.Fst, colors)
+ output[1,1] - i
+ output[1,2] - as.character(current[1, gene_old])
+ output[1,3] - position
+ output[1,4] - nposition
+ output[1,5] - num
+ output[1,6] - numhigh
+ output[1,7] - fst
+ output[1,8] - colors
+ maxfstgene - rbind(maxfstgene, output)
+ }
Error in output[1, 2] - as.character(current[1, gene_old]) :
replacement has length zero
In addition: Warning message:
In mean.default(current$pos) :
argument is not numeric or logical: returning NA
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Re: [R] looping in R
You desperately need to read the Posting Guide (mentioned in the footer of this
email) which warns you not to post in HTML format, and learn how to make a
reproducible example (e.g.
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example).
The problem lies in some interaction between your data and code, and without
both we cannot help you.
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On April 2, 2014 12:15:51 PM PDT, Abugri James jabu...@uds.edu.gh wrote:
I ran the following loop on my SNP data and got an error message as
indicated
for (i in genenames){
+ current - fst1[which(fst1$Gene == i),]
+ num - nrow(current)
+ fst - max(current$fst)
+ position - mean(current$pos)
+ nposition - mean(current$newpos)
+ numhigh - nrow(current[which(current$fst threshold),])
+ colors - mean(current$colors)
+ output - matrix(NA,nrow=1,ncol=8)
+ numthresh - paste(# SNPs Fst = , threshold, sep=)
+ colnames(output) - c(gene, gene_old, pos, newpos, #
Snps,
numthresh, Max.Fst, colors)
+ output[1,1] - i
+ output[1,2] - as.character(current[1, gene_old])
+ output[1,3] - position
+ output[1,4] - nposition
+ output[1,5] - num
+ output[1,6] - numhigh
+ output[1,7] - fst
+ output[1,8] - colors
+ maxfstgene - rbind(maxfstgene, output)
+ }
Error in output[1, 2] - as.character(current[1, gene_old]) :
replacement has length zero
In addition: Warning message:
In mean.default(current$pos) :
argument is not numeric or logical: returning NA
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Re: [R] looping in R
On 02/04/2014, 3:15 PM, Abugri James wrote:
I ran the following loop on my SNP data and got an error message as
indicated
I would assume that the error message is accurate:
as.character(current[1, gene_old]) has length zero. You'll need to
debug why that happened.
Duncan Murdoch
for (i in genenames){
+ current - fst1[which(fst1$Gene == i),]
+ num - nrow(current)
+ fst - max(current$fst)
+ position - mean(current$pos)
+ nposition - mean(current$newpos)
+ numhigh - nrow(current[which(current$fst threshold),])
+ colors - mean(current$colors)
+ output - matrix(NA,nrow=1,ncol=8)
+ numthresh - paste(# SNPs Fst = , threshold, sep=)
+ colnames(output) - c(gene, gene_old, pos, newpos, # Snps,
numthresh, Max.Fst, colors)
+ output[1,1] - i
+ output[1,2] - as.character(current[1, gene_old])
+ output[1,3] - position
+ output[1,4] - nposition
+ output[1,5] - num
+ output[1,6] - numhigh
+ output[1,7] - fst
+ output[1,8] - colors
+ maxfstgene - rbind(maxfstgene, output)
+ }
Error in output[1, 2] - as.character(current[1, gene_old]) :
replacement has length zero
In addition: Warning message:
In mean.default(current$pos) :
argument is not numeric or logical: returning NA
--
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[R] ASA Conference on Statistical Practice - call for short courses
R users, The 2015 American Statistical Association Conference on Statistical Practice is currently accepting proposals for short courses and tutorials. The conference will be held February 19-21 in New Orleans, Louisiana, USA. Two R-related short courses were offered at the last conference, both of which were well received: Elegant R Graphics with ggplot2 An Introduction to R for Data Analysts I am sure there would be interest in a course on interactive graphics, e.g., using R packages shiny or googleVis. Courses and tutorials need not be limited to R. They may touch on any one of the four conference themes: Communication, Impact, and Career Development Data Modeling and Analysis Big Data Prediction and Analytics Software, Programming, and Graphics The deadline for proposal submissions is May 13, 2014. If you are interested, you may submit your abstract on the website: http://www.amstat.org/meetings/csp/2015/courses.cfm Thank you. Jean V. Adams on behalf of the ASA-CSP 2015 Steering Committee `·.,, (((º `·.,, (((º `·.,, (((º Jean V. Adams Statistician U.S. Geological Survey Great Lakes Science Center 223 East Steinfest Road Antigo, WI 54409 USA 715-627-4317, ext. 3125 (Office) 715-216-8014 (Cell) 715-623-6773 (FAX) http://www.glsc.usgs.gov (GLSC web site) http://profile.usgs.gov/jvadams (My home page) [[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] Icelandic Characters in Mac
Dear R community I have few students that use Mac. When creating graphs they inform me that when they use Icelandic characters in title or xlab they get some wrong results. In stead of Það er sætt they get .a. er s.tt period in stead of the Icelandic character. Any Icelandic Mac user that has a solution to this? . The Icelandic characters follows. Æ æ ó Ó Ð ð,í Í á Á é É ý Ý http://en.wikipedia.org/wiki/Icelandic_alphabet Thanks Stefan Jonsson [[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] -fopenmp
Hi everybody, just wanted to know if it is possible to use the openmp library in Fortran code to be used within R. I tried this simple thing: subroutine test !$OMP parallel write(*,*) 'hello' !$OMP end parallel end subroutine test and I compiled in the following way: R CMD SHLIB test.f90 -fopenmp but is seems not working. The program correctly print me out four times 'hello' if I compile outside R simply using gfortran with the flag -fopenmp. If someone can help me I would be very grateful. Thanks in advance and regards Filippo Monari __ 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] R2WinBUGS expected collection operator c error
This is probably a BUGS rather than an R problem, hence better for a BUGS mailing list. But if we should help, then we need at least the modelfile. Best, Uwe Ligges On 02.04.2014 19:10, zumacrume wrote: Hi all, I am currently in a Bayesian Modeling course and am trying to implement an analog representation model using R, WinBUGS, and R2WinBUGS. I'm currently stuck banging my head against an expected collection operator c error. I am working off of a template code, and I swear I haven't moved any collection operators from the original code. Here is what my script looks like in R: # clears workspace: rm(list=ls(all=TRUE)) # sets working directories: setwd(C:\\Users\\Nick\\Desktop\\modeling) library(R2WinBUGS) bugsdir = C:/Program Files/WinBUGS14 # read the data g = matrix(scan(g.txt, sep=,), ncol=144, nrow=3, byrow=T) q = matrix(scan(q.txt, sep=,), ncol=144, nrow=3, byrow=T) S = 3 # number of subjects Q = 144 # number of questions for each child N = 9 # maximum numbers of item list data = list(g, q, S, Q, N) # to be passed on to WinBUGS myinits = list( list(sigma = rep(1,S), pitmp=rep(1/N, N))) # parameters to be monitored: parameters = c(pp,ppb,sigma) # NB. even with only 1000 iterations, the sampling can take a long time! # The following command calls WinBUGS with specific options. # For a detailed description see Sturtz, Ligges, Gelman (2005). samples = bugs(data, inits=myinits, parameters, model.file =NumberConcept_2_data.txt, n.chains=1, n.iter=1000, n.burnin=100, n.thin=1, DIC=T, bugs.directory=bugsdir, codaPkg=F, debug=T) # Now the values for the monitored parameters are in the samples object, # ready for inspection. samples$summary ### and here is the log file file from WinBUGS after R2WinBUGS passes it over: ### display(log) check(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/NumberConcept_2_data.txt) model is syntactically correct data(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/data.txt) expected collection operator c compile(1) inits(1,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/inits1.txt) command #Bugs:inits cannot be executed (is greyed out) gen.inits() command #Bugs:gen.inits cannot be executed (is greyed out) thin.updater(1) update(100) command #Bugs:update cannot be executed (is greyed out) set(pp) command #Bugs:set cannot be executed (is greyed out) set(ppb) command #Bugs:set cannot be executed (is greyed out) set(sigma) command #Bugs:set cannot be executed (is greyed out) set(deviance) command #Bugs:set cannot be executed (is greyed out) dic.set() command #Bugs:dic.set cannot be executed (is greyed out) update(900) command #Bugs:update cannot be executed (is greyed out) coda(*,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/coda) command #Bugs:coda cannot be executed (is greyed out) stats(*) command #Bugs:stats cannot be executed (is greyed out) dic.stats() DIC history(*,C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/history.odc) command #Bugs:history cannot be executed (is greyed out) save(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/log.odc) save(C:/Users/Nick/AppData/Local/Temp/RtmpIRZ1dh/log.txt) # WinBUGS also opens up a data window that looks like this: # list(g= structure(.Data= c(1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 1.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 2.0E+00, 1.0E+00, 2.0E+00,
Re: [R] plotting several columns of matrix in one graph
Hi, May be this helps: library(xts) library(xtsExtra) data(sample_matrix) plot(as.xts(sample_matrix),screens=1) #or library(zoo) plot(as.zoo(sample_matrix), plot.type=single,col=1:ncol(sample_matrix)) You may also check ?matplot A.K. Hi everyone, I have started using R and although I am used to some other languages, I am struggling doing a plot that contains several lines which each correspond to a column of the Matrix which all my data. I tried to google it but unfortunately, it haven't found anything which helped me and also the description didn't really give me a hint what to do. Let's say I have Matrix calles Data_Set which consists of 6 columns and let's say 100 rows. in the first column, I have the date, which is also the x-axis of my plot. The next five column contain the time series, for each of them I want I line drawn in the plot. I have installed the lattice package and I tried several things using the xyplot command, but it didn't work. thanks so much for your help. __ 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] Subset error on atomic vectors why?
I'm getting this error: Error in MOPrice$Date : $ operator is invalid for atomic vectors The cost is: subset(MOPrice, as.Date(MOPrice$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) The date column looks like: 2013-12-31 2013-12-31 2013-12-31 2013-12-31 2013-12-31 -- View this message in context: http://r.789695.n4.nabble.com/Subset-error-on-atomic-vectors-why-tp4688051.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Survey
On Thu, Apr 3, 2014 at 2:37 AM, Leandro Marino leandromar...@leandromarino.com.br wrote: Dear R-Users, I was using survey for the past years and now I am experiencing some problems with scripts that was working in the past. We are working with big data bases so I can't put all variables that I will use in the svydesign. This was never supposed to work. The variables shouldn't take up any more room in the survey design object than they do anywhere else, but in any case the right solution when you have enough memory for the variables in a given analysis but not for all the variables in your dataset is to use the database-backed designs and put the data in something like SQLite. -thomas -- Thomas Lumley Professor of Biostatistics University of Auckland [[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] Icelandic Characters in Mac
On Apr 2, 2014, at 2:18 PM, Stefán Hrafn Jónsson wrote: Dear R community I have few students that use Mac. When creating graphs they inform me that when they use Icelandic characters in title or xlab they get some wrong results. In stead of fia› er sætt they get .a. er s.tt period in stead of the Icelandic character. Any Icelandic Mac user that has a solution to this? . The Icelandic characters follows. Æ æ ó Ó ‹ ›,í Í á Á é É ‡ † http://en.wikipedia.org/wiki/Icelandic_alphabet The default screen device is named quartz(). I'm not an Icelander, but in a US locale with the standard fonts I get faithful reproduction of those characters on a Mac running 10.7.5 with R 3.0.2 They need to be looking at: ?quartz ?quartzFonts sans for a sans-serif font, serif for a serif font mono for a monospaced font. And they need to know what fonts would be appropriate to assign to the various families,s or they need to specify font families in their plotting calls. There may be a need to review how their locale settings are being specified. -- 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] Subset error on atomic vectors why?
On Apr 2, 2014, at 3:35 PM, jcrosbie wrote: I'm getting this error: Error in MOPrice$Date : $ operator is invalid for atomic vectors The cost is: subset(MOPrice, as.Date(MOPrice$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) The date column looks like: 2013-12-31 2013-12-31 2013-12-31 2013-12-31 2013-12-31 What does str(MOPrice) return? View this message in context: http://r.789695.n4.nabble.com/Subset-error-on-atomic-vectors-why-tp4688051.html Sent from the R help mailing list archive at Nabble.com. R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Subset error on atomic vectors why?
Hi, It is not mentioned whether your dataset is a matrix of data.frame. Also, please use ?dput() to show the dataset. I get similar errors with matrix. MOPrice - data.frame(Date=c(2013-12-31,2013-12-31,2013-12-31,2013-11-28),stringsAsFactors=FALSE) subset(MOPrice, as.Date(Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) #Date #4 2013-11-28 MOPrice1 - data.frame(Date=c(2013-12-31,2013-12-31,2013-12-31),stringsAsFactors=FALSE) subset(MOPrice1, as.Date(Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) #[1] Date #0 rows (or 0-length row.names) MOPrice2 - data.frame(Date=c(2013-12-31,2013-12-31,2013-12-31)) subset(MOPrice2, as.Date(MOPrice2$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) #[1] Date #0 rows (or 0-length row.names) subset(MOPrice1, as.Date(MOPrice1$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) #[1] Date #0 rows (or 0-length row.names) MOPrice3 - as.matrix(MOPrice1) subset(MOPrice3, as.Date(MOPrice3$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) #Error in MOPrice3$Date : $ operator is invalid for atomic vectors A.K. On Wednesday, April 2, 2014 6:49 PM, jcrosbie ja...@crosb.ie wrote: I'm getting this error: Error in MOPrice$Date : $ operator is invalid for atomic vectors The cost is: subset(MOPrice, as.Date(MOPrice$Date,%Y-%m-%d)==as.Date(2013-11-28,%Y-%m-%d)) The date column looks like: 2013-12-31 2013-12-31 2013-12-31 2013-12-31 2013-12-31 -- View this message in context: http://r.789695.n4.nabble.com/Subset-error-on-atomic-vectors-why-tp4688051.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot: add points selectively to curves
Thanks, Dennis Not quite. I need to have the same lines as in the original, using plotdat, but then add the points from your plotdat2 gg - ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + #geom_point(color=black, size=1.5) + # not these points facet_grid(Sex ~ Treatment, # scales = free, labeller = function(x, y) sprintf(%s = %s, x, y) ) # want these points, added to the above plot plotdat2 - subset(plotdat, as.character(Improved) == as.character(Level)) On 02/04/2014 7:09 PM, Dennis Murphy wrote: Hi Michael: Does this work? plotdat2 - subset(plotdat, as.character(Improved) == as.character(Level)) ggplot(plotdat2, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + geom_point(color=black, size=1.5) + facet_grid(Sex ~ Treatment, labeller = function(x, y) sprintf(%s = %s, x, y) ) Dennis On Wed, Apr 2, 2014 at 7:43 AM, Michael Friendly frien...@yorku.ca wrote: I'm working on an example of plotting predicted probabilities from a proportional odds model. The steps below generate a data frame, plotdat, that I want to plot with ggplot. library(MASS) data(Arthritis, package=vcd) arth.polr - polr(Improved ~ Sex + Treatment + Age, data=Arthritis, Hess=TRUE) # get predicted probs for categories of Improve arth.fitp - cbind(Arthritis, predict(arth.polr, type=probs)) head(arth.fitp) # reshape probs to long library(reshape2) plotdat - melt(arth.fitp, id.vars = c(Sex, Treatment, Age, Improved), measure.vars=c(None, Some, Marked), variable.name = Level, value.name = Probability) ## view first few rows head(plotdat) head(plotdat) Sex Treatment Age Improved Level Probability 1 Male Treated 27 Some None 0.7326185 2 Male Treated 29 None None 0.7174048 3 Male Treated 30 None None 0.7096042 4 Male Treated 32 Marked None 0.6936286 5 Male Treated 46 Marked None 0.5702499 6 Male Treated 58 Marked None 0.4563432 In the plot step, I am plotting Probability vs. Age, stratified by Level, and faceted by Sex and Treatment. My question concerns the use of geom_point(). The call below plots 3 points for each case, one on each Level curve. ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + geom_point(color=black, size=1.5) + facet_grid(Sex ~ Treatment, labeller = function(x, y) sprintf(%s = %s, x, y) ) Instead, I want to plot only one point for each case, for the value of Level that corresponds to the value of Improved in this data set. Somehow, this involves something like an aes() argument to geom_point(), with Level indexed by Improved, or some such. How can I do this? -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele StreetWeb: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] ggplot: add points selectively to curves
Hi, May be this helps: gg - ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + facet_grid(Sex ~ Treatment, # scales = free, labeller = function(x, y) sprintf(%s = %s, x, y) ) plotdat2 - subset(plotdat,as.character(Improved)==as.character(Level)) gg+geom_point(data= plotdat2, aes(x=Age, y=Probability),color=black,size=1.5) A.K. On Wednesday, April 2, 2014 9:22 PM, Michael Friendly frien...@yorku.ca wrote: Thanks, Dennis Not quite. I need to have the same lines as in the original, using plotdat, but then add the points from your plotdat2 gg - ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + # geom_point(color=black, size=1.5) + # not these points facet_grid(Sex ~ Treatment, # scales = free, labeller = function(x, y) sprintf(%s = %s, x, y) ) # want these points, added to the above plot plotdat2 - subset(plotdat, as.character(Improved) == as.character(Level)) On 02/04/2014 7:09 PM, Dennis Murphy wrote: Hi Michael: Does this work? plotdat2 - subset(plotdat, as.character(Improved) == as.character(Level)) ggplot(plotdat2, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + geom_point(color=black, size=1.5) + facet_grid(Sex ~ Treatment, labeller = function(x, y) sprintf(%s = %s, x, y) ) Dennis On Wed, Apr 2, 2014 at 7:43 AM, Michael Friendly frien...@yorku.ca wrote: I'm working on an example of plotting predicted probabilities from a proportional odds model. The steps below generate a data frame, plotdat, that I want to plot with ggplot. library(MASS) data(Arthritis, package=vcd) arth.polr - polr(Improved ~ Sex + Treatment + Age, data=Arthritis, Hess=TRUE) # get predicted probs for categories of Improve arth.fitp - cbind(Arthritis, predict(arth.polr, type=probs)) head(arth.fitp) # reshape probs to long library(reshape2) plotdat - melt(arth.fitp, id.vars = c(Sex, Treatment, Age, Improved), measure.vars=c(None, Some, Marked), variable.name = Level, value.name = Probability) ## view first few rows head(plotdat) head(plotdat) Sex Treatment Age Improved Level Probability 1 Male Treated 27 Some None 0.7326185 2 Male Treated 29 None None 0.7174048 3 Male Treated 30 None None 0.7096042 4 Male Treated 32 Marked None 0.6936286 5 Male Treated 46 Marked None 0.5702499 6 Male Treated 58 Marked None 0.4563432 In the plot step, I am plotting Probability vs. Age, stratified by Level, and faceted by Sex and Treatment. My question concerns the use of geom_point(). The call below plots 3 points for each case, one on each Level curve. ggplot(plotdat, aes(x = Age, y = Probability, color = Level)) + geom_line(size=2.5) + theme_bw() + xlim(10,80) + geom_point(color=black, size=1.5) + facet_grid(Sex ~ Treatment, labeller = function(x, y) sprintf(%s = %s, x, y) ) Instead, I want to plot only one point for each case, for the value of Level that corresponds to the value of Improved in this data set. Somehow, this involves something like an aes() argument to geom_point(), with Level indexed by Improved, or some such. How can I do this? -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. Chair, Quantitative Methods York University Voice: 416 736-2100 x66249 Fax: 416 736-5814 4700 Keele Street Web: http://www.datavis.ca Toronto, ONT M3J 1P3 CANADA __ 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] plotting several columns of matrix in one graph
HI, It is better to show a reproducible example using ?dput(). May be this helps: #vector vec1 - seq(as.Date(2002-01-01), as.Date(2009-12-31),by=1 day) #Assuming that length of the vector is the same as ?nrow of matrix. set.seed(532) mat1 - matrix(cumsum(rnorm(length(vec1)*5)),ncol=5, dimnames=list(NULL,LETTERS[1:5])) library(xts) library(xtsExtra) plot(as.xts(mat1,order.by=vec1),major.format=%Y,screens=1, main=Time series plot, auto.legend=TRUE,auto.grid=FALSE,col=5:1) You should also check ?legend() A.K. thank you so much. it tried the 2nd way and it is working perfectly. could you also tell me how I can select a vector which is then used for the x axis? let's say I want some Dates 2002 to 2009, currently I have a vector called date which contains of every single day from 2002 to 2009, is there a way to get this on the x axis (just the year)? in Addition, is there some command to add a legend, titles, Change Color etc? From what I find on the Internet, I just don't understand how that is working. thanks again for the help! On Wednesday, April 2, 2014 6:41 PM, arun smartpink...@yahoo.com wrote: Hi, May be this helps: library(xts) library(xtsExtra) data(sample_matrix) plot(as.xts(sample_matrix),screens=1) #or library(zoo) plot(as.zoo(sample_matrix), plot.type=single,col=1:ncol(sample_matrix)) You may also check ?matplot A.K. Hi everyone, I have started using R and although I am used to some other languages, I am struggling doing a plot that contains several lines which each correspond to a column of the Matrix which all my data. I tried to google it but unfortunately, it haven't found anything which helped me and also the description didn't really give me a hint what to do. Let's say I have Matrix calles Data_Set which consists of 6 columns and let's say 100 rows. in the first column, I have the date, which is also the x-axis of my plot. The next five column contain the time series, for each of them I want I line drawn in the plot. I have installed the lattice package and I tried several things using the xyplot command, but it didn't work. thanks so much for your help. __ 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] -fopenmp
On 02/04/2014 22:30, Filippo wrote: Hi everybody, just wanted to know if it is possible to use the openmp library in Fortran code to be used within R. Yes, on a supported platform. But 1) The posting guide tells you this is not the correct list for questions about compiled code. 2) 'Writing R Extensions' tells you how to do this. 3) ?SHLIB tells you about valid inputs, which do not include arbitrary flags. I tried this simple thing: subroutine test !$OMP parallel write(*,*) 'hello' !$OMP end parallel end subroutine test and I compiled in the following way: R CMD SHLIB test.f90 -fopenmp but is seems not working. The program correctly print me out four times 'hello' if I compile outside R simply using gfortran with the flag -fopenmp. If someone can help me I would be very grateful. Thanks in advance and regards Filippo Monari __ 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. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@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] Strange sprintf Behavior
All, Apologies for the thread issue, and many thanks for the pointers to the FAQs. Thanks, M On 04/02/2014 10:14 PM, Jeff Newmiller wrote: It is poor netiquette to reply to a thread with a different subject. Please start a new thread for a new subject. As for your question, see FAQ 7.31. This is standard floating point numerical limitations at work. --- 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 April 2, 2014 4:32:26 AM PDT, Michael Smith my.r.h...@gmail.com wrote: All, I'm getting this: sprintf(%.17f, 0.8) [1] 0.80004 Where does the `4` at the end come from? Shouldn't it be zero at the end? Maybe I'm missing something. sessionInfo() R version 3.0.2 (2013-09-25) Platform: x86_64-redhat-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C LC_TIME=en_US.utf8 [4] LC_COLLATE=en_US.utf8 LC_MONETARY=en_US.utf8 LC_MESSAGES=en_US.utf8 [7] LC_PAPER=en_US.utf8 LC_NAME=C LC_ADDRESS=C [10] LC_TELEPHONE=CLC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base Thanks, M __ 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] 'rms' package error
Hi everyone, I am attempting to use the R package 'rms' http://biostat.mc.vanderbilt.edu/wiki/Main/Rrms to implement a PH weibull model, using the pphsm() function. However, I get the following error, f.ph - pphsm(f) Warning message: In pphsm(f) : at present, pphsm does not return the correct covariance matrix I tried simply running the example on page 117 of the manual, i.e. set.seed(1) S - Surv(runif(100)) x - runif(100) dd - datadist(x); options(datadist='dd') f - psm(S ~ x, dist=exponential) summary(f) # effects on log(T) scale f.ph - pphsm(f) ## Not run: summary(f.ph) But I still got the above error message. I have looked through the R help archives, and it appears that this question has been asked before in 2011, but there were no replies. http://r.789695.n4.nabble.com/HELP-td3494640.html Does anyone know how to get this function to work? Or if there is an alternative package that can implement a Weibull PH model? Cheers, Lucy [[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] 'rms' package error
Hello Lucy, If you carefully read, it is is not an error message, but a warning message. It tells you that for the moment, if I am not mistaken, pphsm does not return the correct covariance matrix, for any fitting. Regards, Pascal On Thu, Apr 3, 2014 at 11:29 AM, Lucy Leigh lucy.le...@newcastle.edu.au wrote: Hi everyone, I am attempting to use the R package 'rms' http://biostat.mc.vanderbilt.edu/wiki/Main/Rrms to implement a PH weibull model, using the pphsm() function. However, I get the following error, f.ph - pphsm(f) Warning message: In pphsm(f) : at present, pphsm does not return the correct covariance matrix I tried simply running the example on page 117 of the manual, i.e. set.seed(1) S - Surv(runif(100)) x - runif(100) dd - datadist(x); options(datadist='dd') f - psm(S ~ x, dist=exponential) summary(f) # effects on log(T) scale f.ph - pphsm(f) ## Not run: summary(f.ph) But I still got the above error message. I have looked through the R help archives, and it appears that this question has been asked before in 2011, but there were no replies. http://r.789695.n4.nabble.com/HELP-td3494640.html Does anyone know how to get this function to work? Or if there is an alternative package that can implement a Weibull PH model? Cheers, Lucy [[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. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ 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] figure margins too large
Dear R experts, I tried to plot some figures in R using postscript(), but it always shows that the fugures margin is too large. I don't know how to change it. The following is my example: postscript(All.eps,width=3.27,height=1.416,pointsize=12,family=Arial) par(mar=c(5.1,4.5,4.1,2.1));boxplot(All~Nameall,ylab= expression(Size~ (log [10]~mm ^2)), boxwex=0.3, main=All species,col=c(red,yellow,blue),ylim=c(0,4.0)) Error: plot.new() : figure margins too large When I run the boxplot in R, it shows well, but once I run it in the postscript, it fails. Can someone help me on it? Note: my operational system is windows 7. Many thanks in advance Yichun [[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.
