Re: [R] foreign package read.spss() and NA levels
Hi Uwe,
Thank you so much for your reply, I am sorry to waste your time, but after
another look, the error seemed to be caused with how I had created the '.sav'
file I was using. It was converting one of the factor levels which wasn't NA to
the factor level "NA", but the problem was in the dataset as I said.
Many Thanks,
Justin
Justin Allen
Housing Consultant, BRE<https://bregroup.com/>
T: 07807122647
From: Uwe Ligges
Sent: 04 June 2021 10:57
To: Allen, Justin ; R-help@r-project.org
Cc: Whiteley, Jonathon ; Foster, Helen
Subject: Re: [R] foreign package read.spss() and NA levels
On 27.05.2021 12:27, Allen, Justin wrote:
> Hi All,
>
> Wanted to report what may be a bug or possibly an oversight, but I am unsure,
> in the "foreign" packages in the read.spss() command,
> https://cran.r-project.org/web/packages/foreign/index.html. When running the
> following code,
>
> input <- read.spss("[.sav file location]", to.data.frame = TRUE)
> str(input)
>
> The read.spss() seems to be applying addNA() to factors so NA is being set as
> a level, and there seems to be no way to get read.spss() to bring factors in
> without doing this. This seems to be a recent change as read.spss() was not
> doing this as of a few months ago. None of the arguments in read.spss() seem
> to also stop this behaviour. I am currently on the most recent version of
> both R and the package, as of 27/05/21, and am using RStudio Version 1.4.1106.
Within R (I do not use RStudio) and even with the most recent R-devel, I see
(sav <- system.file("files", "electric.sav", package = "foreign"))
dat <- read.spss(file=sav, to.data.frame=TRUE)
table(dat$DAYOFWK)
# SUNDAY MONDAY TUESDAY WEDNSDAY THURSDAY FRIDAY SATURDAY
# 19 11 19 17 15 13 16
table(dat$DAYOFWK, useNA="always")
# SUNDAY MONDAY TUESDAY WEDNSDAY THURSDAY FRIDAY SATURDAY
# 19 11 19 17 15 13 16 130
So exactly what you expected?
If you rather use
dat <- read.spss(file=sav, to.data.frame=TRUE, use.missings=FALSE)
table(dat$DAYOFWK, useNA="always")
you see the NA values are converted to a factor level called "MISSING".
If it is different on your end, pelase try in plain R, tell us the
version of R / foreign and show an example data file where this happens.
Best,
Uwe Ligges
>
> Any thoughts?
>
> Many Thanks,
>
> Justin Allen
>
> p.s. your continued maintenance and additions to R and its packages have been
> infinitely useful in my work and life and thank for that.
>
> Justin Allen
> Housing Consultant, BRE<https://bregroup.com/>
> T: 07807122647
>
>
> Follow BRE on Twitter: @BRE_Group<http://twitter.com/BRE_Group>
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[R] foreign package read.spss() and NA levels
Hi All,
Wanted to report what may be a bug or possibly an oversight, but I am unsure,
in the "foreign" packages in the read.spss() command,
https://cran.r-project.org/web/packages/foreign/index.html. When running the
following code,
input <- read.spss("[.sav file location]", to.data.frame = TRUE)
str(input)
The read.spss() seems to be applying addNA() to factors so NA is being set as a
level, and there seems to be no way to get read.spss() to bring factors in
without doing this. This seems to be a recent change as read.spss() was not
doing this as of a few months ago. None of the arguments in read.spss() seem to
also stop this behaviour. I am currently on the most recent version of both R
and the package, as of 27/05/21, and am using RStudio Version 1.4.1106.
Any thoughts?
Many Thanks,
Justin Allen
p.s. your continued maintenance and additions to R and its packages have been
infinitely useful in my work and life and thank for that.
Justin Allen
Housing Consultant, BRE<https://bregroup.com/>
T: 07807122647
Follow BRE on Twitter: @BRE_Group<http://twitter.com/BRE_Group>
Privileged and confidential information and/or copyright material may be
contained in this e-mail. If you are not the intended addressee you may not
copy or deliver it to anyone else or use it in any unauthorised manner. To do
so is prohibited and may be unlawful. If you have received this e-mail by
mistake, please advise the sender immediately by return e-mail and destroy all
copies. Thank you.
Building Research Establishment Ltd, Registered under number 3319324 in England
and Wales. VAT Registration No GB 689 9499 27
www.bregroup.com<http://www.bregroup.com>
BRE Global Limited, Registered under number 8961297 in England and Wales.
www.breglobal.com<http://www.breglobal.com>
Building Research Establishment and BRE Global are subsidiaries of the BRE
Trust.
BRE Trust is a company limited by guarantee, Registered under number 3282856 in
England and Wales, and registered as a charity in England (no. 1092193) and in
Scotland (no. SC039320). www.bretrust.org.uk<http://www.bretrust.org.uk>
Registered Offices: Bucknalls Lane, Garston, Watford, Hertfordshire WD25 9XX -
Travelling to BRE: see
www.bregroup.com/contact/directions/<http://www.bregroup.com/contact/directions/>
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[R] Opportunities for Developing R Packages (Research-Based, Open-Source)
Dear R package community, I am uncertain whether this is appropriate for this mailing list. Please let me know. If not, would you be so kind as to point me in a better direction? I am a mathematics major with a well-developed R experience. I have graduated two years ago and have been working in business operations in a cryptocurrency startup. I am rather rusty and I wish to venture back into statistical research and R-package development. My question is: For those researchers who are interested in developing tools and algorithms for their new-founded research, be it in medical statistics or data visualisation or machine learning, I was wondering whether is there a possibility for collaboration. This will help me extend my experience and possibly open more avenues for me to enter research. I am quite aware of statistical concepts and can read research papers (I've done a research internship in experimental design, linear algebra and data compression, particle filters and bayes analyses). I do not expect to be paid and am willing to commit to a project. Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] igraph problem
Run this code tree<-graph_from_literal(1-+2:3,3-+5,1-+4); graph.bfs(tree,root=1, neimode="out",father=TRUE,order=TRUE,unreachable = FALSE) I do not understand why the father values will give NA 1 1 3 1 rather than NA 1 1 1 3 The reason I am doing this is to obtain the values(by vertex names) or some index of each individual branch in tree. Does anyone have any ideas on how to do this? Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419 <07938%20674419>(UK) or +60125056192 <+60%2012-505%206192>(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] rmutil parameters for Pareto distribution
In https://en.wikipedia.org/wiki/Pareto_distribution, it is clear what the parameters are for the pareto distribution: *xmin *the scale parameter and *a* the shape parameter. I am using rmutil to generate random deviates from a pareto distribution. It says in the documentation that the probabilty density of the pareto distribution The Pareto distribution has density f(y) = s (1 + y/(m (s-1)))^(-s-1)/(m (s-1)) where m is the mean parameter of the distribution and s is the dispersion Through my experimentation of using rpareto function from the library using m as the scale parameter *xmin* value and s as the shape parameter* a* , I found that the deviates generated are not all larger than *xmin*. This leads me to believe that m and s are not the shape and scale parameter respectively. What is m and s? Could it be defined as the mean and variance respectively as shown on the wikipedia link? Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Kernel Density Estimation: Generate a sample from Epanechnikov Kernel
Below are samples from a kernel density estimated "data" with gaussian kernel. I really like this solution of estimation of a kernel because it is nice and elegant. fit<-density(data) rnorm(N, sample(data, size = N, replace = TRUE), fit$bw) #samples from kernel density estimation I am however interested in generating a kernel density estimate with an Epanechnikov kernel fit<-density(data,kernel = "epanechnikov") #is there a quick way to compute the samples and INCORPORATING THE BANDWIDTH of the #kernel density estimate Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Hmisc::latex and the use of \phantom
Dear all,
Is it possible to avoid the use of \phantom with latex function ? when I run
latex(latex=FALSE,…) I get an error message.
When I use Format(big.mark= ‘’ ‘’) the result appear correct in R console but
not in Latex code. Is it a way to combine numprint with latex ? col.just=
‘n{#}{#}’ donc work
Sincerly.
Provenance : Courrier pour
Windows 10
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Re: [R] Using apply on a three dimensional matrix and passing multiple arguments to user defined function
Dear Liusfo and Jean,
Thank you both for your help and suggestions, which both work.
As Jean mentioned, there is no real speed up using the apply family, which I
thought there would be, so I will stick with the for loop for clarity.
Liusfo, the reason I used c(1,2) in the apply function (i.e. mask_data <-
apply(data,c(1,2),mask,y=lsmask)) was because this is what you would do if you
wanted to sum all the two-dimensional vectors over time (for instance to
produce an average of a field over a year).
i.e. to get the sum you would do sum_data <- apply(data,c(1,2),sum)
I thought it would extend to using my mask function.
Anyway, thanks again for both of your help.
Cheers,
Justin
--
Justin Peter
Research Fellow
International Centre for Applied Climate Sciences,
University of Southern Queensland
West St, Toowoomba, QLD, 4350
Australia
Email: justin.pe...@usq.edu.au<mailto:justin.pe...@usq.edu.au>
Ph: +61 (0) 7 4631 1181
Fax: +61 (0) 7 4631 5581
Mob: +61 (0)474 774 107
-Original Message-
From: Luisfo <luisf...@yahoo.es<mailto:luisfo%20%3cluisf...@yahoo.es%3e>>
To: "Adams, Jean"
<jvad...@usgs.gov<mailto:%22Adams,%20jean%22%20%3cjvad...@usgs.gov%3e>>, Justin
Peter
<justin.pe...@usq.edu.au<mailto:justin%20peter%20%3cjustin.pe...@usq.edu.au%3e>>
CC: r-help@r-project.org
<r-help@r-project.org<mailto:%22r-h...@r-project.org%22%20%3cr-h...@r-project.org%3e>>
Subject: Re: [R] Using apply on a three dimensional matrix and passing multiple
arguments to user defined function
Date: Wed, 7 Sep 2016 16:05:48 +0200
Hi,
Jean's example with lapply works fine.
However, if you still want to use apply, I think this works.
One observation first. You were passing c(1,2) as second argument to apply, in
your code. And that is what makes you have lots of NAs as a result, since your
function is being applied twice, by rows and columns (first and second
dimensions) respectively.
Use:
masked_data <- apply(data,3,mask,y=lsmask)
# but now masked_data has dim(nlon*nlat,ntime), so change it
dim(masked_data) <- dim(data)
The apply goes over the third dimension (second parameter '3'), so it takes
every nlot*nlat matrix as first argument for function mask.
I think it should work.
Regards,
Luisfo Chiroque
PhD Student | PhD Candidate
IMDEA Networks Institute
http://fourier.networks.imdea.org/people/~luis_nunez/<http://fourier.networks.imdea.org/people/%7Eluis_nunez/>
On 09/07/2016 03:17 PM, Adams, Jean wrote:
Justin,
I don't think you can get the apply() function to return an array. You
could use lapply() instead, and then use simplify2array() to convert the
list of matrices to an array. Also, in your mask() function you don't need
the which() and you should return the x. See my example with toy data
below.
# toy data
nlon <- 2
nlat <- 4
ntime <- 3
data <- array(1:(nlon*nlat*ntime), dim=c(nlon, nlat, ntime))
lsmask <- array(sample(0:1, size=nlon*nlat, replace=TRUE), dim=c(nlon,
nlat))
# newly defined function
mask <- function(x, y) {
x[y==0] <- NA
x
}
# doit
data2 <- simplify2array(lapply(1:ntime, function(i) mask(data[, , i],
lsmask)))
You may prefer to stick with the for() loop approach (for clarity or
simplicity or ...) When I ramped up the toy data to much larger
dimensions, the lapply() approach was only slightly faster than the for()
loop approach on my PC.
data3 <- data
data3[ , , i] <- mask(data3[ , , i], lsmask)
Jean
On Tue, Sep 6, 2016 at 11:33 PM, Justin Peter
<justin.pe...@usq.edu.au><mailto:justin.pe...@usq.edu.au>
wrote:
Dear R-user,
I have a three-dimensional matrix of atmospheric data. The first two
dimensions are spatial (lon and lat) and the third is time, such that
dim(data) <- c(nlon,nlat,ntime)
I wish to apply a land sea mask data which is a matrix of "0" and "1" if
dim(nlon,nlat)
dim(lsmask) <- c(nlon,nlat)
I wish to set all of the elements in the two-dimensional array of
data[,,ntime] for every 1:length(ntime).
I could do this in a loop:
for (i in 1:ntime){
data[,,i][which(lsmask == 0)] <- NA
}
I would like to do this using apply, but I need to pass two variables to
the function in apply (data and lsmask), where data is a two-dimensional
array.
I tried:
mask <- function(x,y) {x[which(y==0)] <- NA}
masked_data <- apply(data,c(1,2),mask,y=lsmask)
but I get back a vector of dim(nlon,nlat) populated with NA.
Any clues as to what I am missing?
Thanks in advance for you help.
Kind regards,
Justin
--
Justin Peter
Research Fellow
International Centre for Applied Climate Sciences,
University of Southern Queensland
West St, Toowoomba, QLD, 4350
Australia
Email:
justin.pe...@usq.edu.au<mailto:justin.pe...@usq.edu.au><mailto:justin.pe...@usq.edu.au><mailto:justin.pe...@usq.edu.au>
Ph: +61 (0) 7 4631 1181
Fax: +61 (0) 7 4631 5581
Mob: +61 (0)474 774 107
[R] Using apply on a three dimensional matrix and passing multiple arguments to user defined function
Dear R-user,
I have a three-dimensional matrix of atmospheric data. The first two dimensions
are spatial (lon and lat) and the third is time, such that
dim(data) <- c(nlon,nlat,ntime)
I wish to apply a land sea mask data which is a matrix of "0" and "1" if
dim(nlon,nlat)
dim(lsmask) <- c(nlon,nlat)
I wish to set all of the elements in the two-dimensional array of data[,,ntime]
for every 1:length(ntime).
I could do this in a loop:
for (i in 1:ntime){
data[,,i][which(lsmask == 0)] <- NA
}
I would like to do this using apply, but I need to pass two variables to the
function in apply (data and lsmask), where data is a two-dimensional array.
I tried:
mask <- function(x,y) {x[which(y==0)] <- NA}
masked_data <- apply(data,c(1,2),mask,y=lsmask)
but I get back a vector of dim(nlon,nlat) populated with NA.
Any clues as to what I am missing?
Thanks in advance for you help.
Kind regards,
Justin
--
Justin Peter
Research Fellow
International Centre for Applied Climate Sciences,
University of Southern Queensland
West St, Toowoomba, QLD, 4350
Australia
Email: justin.pe...@usq.edu.au<mailto:justin.pe...@usq.edu.au>
Ph: +61 (0) 7 4631 1181
Fax: +61 (0) 7 4631 5581
Mob: +61 (0)474 774 107
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Re: [R] Estimated Effects Not Balanced
Hi, Thanks Richard, That was me playing with too many examples and having too many variables just lying around. Thanks for the tip though. On 22 August 2016 at 23:32, Bert Gunter <bgunter.4...@gmail.com> wrote: > Thanks, Rich. I didn't notice that! > > -- Bert > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) > > > On Mon, Aug 22, 2016 at 1:43 PM, Richard M. Heiberger <r...@temple.edu> > wrote: > > The problem is that you have 12 observations and 1+2+10=13 degrees of > freedom. > > There should be 1 + 2 + 8 = 11 degrees of freedom. > > Probably one of your variables is masked by something else in you > workspace. > > Protect yourself by using a data.frame > > > >> tmp <- data.frame(A=factor(c(1,1,1,1,1,1,2,2,2,2,2,2)), > > + B=factor(c(1,1,2,2,3,3,1,1,2,2,3,3)), > > + y=rnorm(12)) > >> mod <- aov(y ~ A+B, data=tmp) > >> summary(mod) > > Df Sum Sq Mean Sq F value Pr(>F) > > A 1 1.553 1.553 1.334 0.281 > > B2 3.158 1.579 1.357 0.311 > > Residuals8 9.311 1.164 > > > > On Mon, Aug 22, 2016 at 11:15 AM, Justin Thong <justinthon...@gmail.com> > wrote: > >> Something does not make sense in R. It has to do with the question of > >> balance and unbalance. > >> > >> *A<-factor(c(1,1,1,1,1,1,2,2,2,2,2,2))* > >> *B<-factor(c(1,1,2,2,3,3,1,1,2,2,3,3))* > >> *y<-rnorm(12)* > >> *mod<-aov(y~A+B)* > >> > >> I was under the impression that the design is balanced ie order does not > >> effect the sums of squares. However, when I compute the anova R reports > >> that the Estimated Effects are Unbalanced. I thought that when all > >> combinations of levels of A and B have equal replications then the > design > >> is called balanced. But, R tends to think that when not all levels of A > and > >> levels of B have equal replication, then the "Estimated Effects are > >> unbalanced" Is this the same as the design being unbalanced? Because > >> for the example below, where the error occured, the order does not > matter > >> (which make me think that the design is balanced). > >> > >> > >> *Call:* > >> * aov(formula = y ~ A + B)* > >> > >> *Terms:* > >> *A B Residuals* > >> *Sum of Squares 0.872572 0.025604 16.805706* > >> *Deg. of Freedom 1 210* > >> > >> *Residual standard error: 1.296368* > >> *Estimated effects may be unbalanced* > >> -- > >> Yours sincerely, > >> Justin > >> > >> *I check my email at 9AM and 4PM everyday* > >> *If you have an EMERGENCY, contact me at +447938674419(UK) or > >> +60125056192(Malaysia)* > >> > >> [[alternative HTML version deleted]] > >> > >> __ > >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > >> 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 -- To UNSUBSCRIBE and more, see > > 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. > -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Estimated Effects Not Balanced
Something does not make sense in R. It has to do with the question of balance and unbalance. *A<-factor(c(1,1,1,1,1,1,2,2,2,2,2,2))* *B<-factor(c(1,1,2,2,3,3,1,1,2,2,3,3))* *y<-rnorm(12)* *mod<-aov(y~A+B)* I was under the impression that the design is balanced ie order does not effect the sums of squares. However, when I compute the anova R reports that the Estimated Effects are Unbalanced. I thought that when all combinations of levels of A and B have equal replications then the design is called balanced. But, R tends to think that when not all levels of A and levels of B have equal replication, then the "Estimated Effects are unbalanced" Is this the same as the design being unbalanced? Because for the example below, where the error occured, the order does not matter (which make me think that the design is balanced). *Call:* * aov(formula = y ~ A + B)* *Terms:* *A B Residuals* *Sum of Squares 0.872572 0.025604 16.805706* *Deg. of Freedom 1 210* *Residual standard error: 1.296368* *Estimated effects may be unbalanced* -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Intercept in Model Matrix (Parameters not what I expected)
I have something which has been bugging me and I have even asked this on cross validated but I did not get a response. Let's construct a simple example. Below is the code. A<-gl(2,4) #factor of 2 levels B<-gl(4,2) #factor of 4 levels df<-data.frame(y,A,B) As you can see, B is nested within A. The peculiar result I am interested in the output of the model matrix when I fit for a nested model . *How does R decide what is included inside the intercept?* Since we are using dummy coding, the coefficients of the model is interpreted as the difference between a particular level and the reference level/the intercept for an single factor model. I understand for model ~A, A1 becomes the intercept and that for model ~A+B, A1 and B1 (both) become the intercept. *I do not get why when we use a nested model, A1:B2 appears as a column inside the model matrix. Why isn't the first parameter of the interaction subspace A1:B1 or A2:B1? *I think I am missing the concept. I think the intercept is A1. *Hence, Why do we not compare the levels of A1:B1 and A1(intercept) or A2:B1 and A1(intercept)?* #nested model > mod<-aov(y~A+A:B) > model.matrix(mod) (Intercept) A2 A1:B2 A2:B2 A1:B3 A2:B3 A1:B4 A2:B4 1 1 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 3 1 0 1 0 0 0 0 0 4 1 0 1 0 0 0 0 0 5 1 1 0 0 0 1 0 0 6 1 1 0 0 0 1 0 0 7 1 1 0 0 0 0 0 1 8 1 1 0 0 0 0 0 1 -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] RJDBC - Java connection to Oracle database crashing on execution
Hi,
I have a script that starts with some code to open a connection to an Oracle
database, however the code is crashing R/RStudio as soon as it runs. The code
does run successfully on another machine.
The script opens by loading the required RJDBC package:
library("RJDBC", lib.loc="C:/Program Files/R/R-3.3.1/library")
After, this I run the code below that points to the required ojdbc7.jar file,
that should start the connection, however this crashes R, closing it down a few
seconds later:
drv = JDBC("oracle.jdbc.OracleDriver",
classPath="C:/Program Files/R/ojdbc7.jar", " ")
The same thing happens in Rstudio - there is no error message, the program
simply crashes stating that "R encountered a fatal error . The session was
terminated".
Again, the code executes exactly as written on a another machine.I have the
latest versions of all required software/packages, so I'm not sure what is
causing the crash.
Thanks.
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[R] What is "args" in this function?
Hi again I need help
*R-code*
debug(model.matrix)
model.matrix(~S)
*model.matrix code*
ans <- .External2(C_modelmatrix, t, data) #t =terms(object) , data="data
frame of object"
*modelframe C-code*
SEXP modelframe(SEXP call, SEXP op, SEXP args, SEXP rho)
{
SEXP terms, data, names, variables, varnames, dots, dotnames, na_action;
SEXP ans, row_names, subset, tmp;
char buf[256];
int i, j, nr, nc;
int nvars, ndots, nactualdots;
const void *vmax = vmaxget();
args = CDR(args);
terms = CAR(args); args = CDR(args);
row_names = CAR(args); args = CDR(args);
variables = CAR(args); args = CDR(args);
varnames = CAR(args); args = CDR(args);
dots = CAR(args); args = CDR(args);
dotnames = CAR(args); args = CDR(args);
subset = CAR(args); args = CDR(args);
na_action = CAR(args);
. . . .
I am sorry I virtually have no experience in C.
Can someone explain to me what "args" is at the point when it enters the
function? I know CAR points to the first element of an object, and CDR
points to the complement of the first element of an object.
Does "args" represent the list of t and data?
or
Does "args" represent the thrid argument in .External2 which is data?
or
something else
I am guessing this whole process of playing CAR and CDR is just a way of
extracting variables from "args" until everything thing in "args" is
assigned to.
For instance, if args=(1,2,3,4,5,6) then below correspond in square
brackets
args = CDR(args); [(1,2,3,4,5,6)]
terms = CAR(args) ;[(1)] args = CDR(args);[(2,3,4,5,6)]
row_names = CAR(args);[(2)] args = CDR(args);[(3,4,5,6)]
variables = CAR(args);[(3)] args = CDR(args);[(4,5,6)]
varnames = CAR(args);[(4)] args = CDR(args);[(5,6)]
etc
Is this correct?
I am sorry if I am asking too many questions on C. Please advise if I am
posting inappropriately.
--
Yours sincerely,
Justin
*I check my email at 9AM and 4PM everyday*
*If you have an EMERGENCY, contact me at +447938674419(UK) or
+60125056192(Malaysia)*
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[R] Ways to understand C code (like debug function)
Hi I need some advice. Note: I do not know anything from C apart from my 2 days of research. I am currently trying to make meaning of the modelmatrix function (written in C) and called from R function model.matrix() via .External2. In trying to view the source code (in R) for model.matrix(), I have been reasonably succesful thanks to the debug command. This command was good because I was able to check line-by-line what the code was doing and obtain an output within my R console. Furthermore, checking the values of each of my variables while sequentially moving through the lines was also very useful. However, just by looking at the R source code, it is insufficient in understanding most of the computation. I have to look within the C code. In particular, within model.matrix(), a .External2 call is executed to a C function named modelmatrix. I downloaded the source from the website and can view the function modelmatrix(in model.c) in a text editor. I am now finding a way to play with the code so I understand whats going on and I don't know what's the best way to do this. I* was wondering whether there is an equivalent way as the debug function to check C code line by line. ie each line of code are typed in, and an output is obtained*. I know a package "inline" allows you to build C functions and use them in R. But I can't find anything which does what I want. * If this is not possible, is there an alternative good, easy way to run through and understand the commands in C that anyone knows about.* -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Reference for aov()
Hi Peter, Thank you for your good answer. I am sorry for the late reply. *An ortogonalized model matrix generates a decomposition of the model space into orthogonal subspaces corresponding to the terms of the model. Projections onto each of the subspaces are easily worked out. E.g., for a two-way analysis (Y~row+col+row:col) you can decompose the model effect as a row effect, a column effect, and an interaction effect. This allows straightforward calculation of the sums of squares of the ANOVA table. As you are probably well aware, if the design is unbalanced, the results will depend on the order of terms -- col + row + col:row gives a different result.* It may be a stupid question. How are projections of each sums of squares easily worked out and how does the sums of squares follow easily? Does it matter that certain parameters of the model are not estimated. R appears to just give a sums of squares despite some of the parameters being non-estimable. Thank you On 14 July 2016 at 09:50, peter dalgaard <pda...@gmail.com> wrote: > I am not aware of a detailed documentation of this beyond the actual > source code. > However, the principles are fairly straightforward, except that the rules > for constructing the design matrix from the model formula can be a bit > arcane at times. > > The two main tools are the design matrix constructor (model.matrix) and a > Gram-Schmidt type ortogonalization of its columns (the latter is called a > QR decomposition in R, which it is, but there are several algorithms for > QR, and the linear models codes depend on the QR algorithm being based on > orthogonalization - so LINPACK works and LAPACK doesn't). > > An ortogonalized model matrix generates a decomposition of the model space > into orthogonal subspaces corresponding to the terms of the model. > Projections onto each of the subspaces are easily worked out. E.g., for a > two-way analysis (Y~row+col+row:col) you can decompose the model effect as > a row effect, a column effect, and an interaction effect. This allows > straightforward calculation of the sums of squares of the ANOVA table. As > you are probably well aware, if the design is unbalanced, the results will > depend on the order of terms -- col + row + col:row gives a different > result. > > What aov() does is that it first decomposes the observations according to > the Error() term, forming the error strata, then fits the systematic part > of the model to each stratum in turn. In the nice cases, each term of the > model will be estimable in exactly one stratum, and part of the aov() logic > is to detect and remove unestimable terms. E.g., if you have a balanced two > way layout, say individual x treatment, the variable gender is a subfactor > of individual, so Y ~ gender * treatment + Error(individual/treatment), the > gender effect is estimated in the individual stratum, whereas treatment and > gender:treatment are estimated in the individual:treatment stratum. > > It should be noted that it is very hard to interpret the results of aov() > unless the Error() part of the model corresponds to a balanced experimental > design. Or put more sharply: The model implied by the decomposition into > error strata becomes nonsensical otherwise. If you do have a balanced > design, the error strata reduce to simple combinations of means and > observation, so the aov() algorithm is quite inefficient, but to my > knowledge nobody has bothered to try and do better. > > -pd > > > On 13 Jul 2016, at 18:18 , Justin Thong <justinthon...@gmail.com> wrote: > > > > Hi > > > > *I have been looking for a reference to explain how R uses the aov > > command(at a deeper level)*. More specifically, how R reads the formulae > > and R computes the sums of squares. I am not interested in understanding > > what the difference of Type 1,2,3 sum of squares are. I am more > interested > > in finding out about how R computes ~x1:x2:x3 or how R computes ~A:x1 > > emphasizing sequential nature of the way it computes, and models even > more > > complicated than this. > > > > Yours sincerely, > > Justin > > > > *I check my email at 9AM and 4PM everyday* > > *If you have an EMERGENCY, contact me at +447938674419(UK) or > > +60125056192(Malaysia)* > > > > [[alternative HTML version deleted]] > > > > __ > > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > -- > Peter Dalgaard, Professor, > Center for Statistics
[R] Linear Dependance of Model Matrix and How Fitted/ Sums of Squares Follow
0 116 4 0 0 4 4 0 117 4 0 0 4 0 4 118 4 0 0 4 0 4 119 4 0 0 4 0 4 120 4 0 0 4 0 4 121 4 0 0 0 4 4 122 4 0 0 0 4 4 123 4 0 0 0 4 4 124 4 0 0 0 4 4 125 0 4 4 4 0 0 126 0 4 4 4 0 0 127 0 4 4 4 0 0 128 0 4 4 4 0 0 129 0 4 4 0 4 0 130 0 4 4 0 4 0 131 0 4 4 0 4 0 132 0 4 4 0 4 0 133 0 4 4 0 0 4 134 0 4 4 0 0 4 135 0 4 4 0 0 4 136 0 4 4 0 0 4 137 0 4 0 4 4 0 138 0 4 0 4 4 0 139 0 4 0 4 4 0 140 0 4 0 4 4 0 141 0 4 0 4 0 4 142 0 4 0 4 0 4 143 0 4 0 4 0 4 144 0 4 0 4 0 4 145 0 4 0 0 4 4 146 0 4 0 0 4 4 147 0 4 0 0 4 4 148 0 4 0 0 4 4 149 0 0 4 4 4 0 150 0 0 4 4 4 0 151 0 0 4 4 4 0 152 0 0 4 4 4 0 153 0 0 4 4 0 4 154 0 0 4 4 0 4 155 0 0 4 4 0 4 156 0 0 4 4 0 4 157 0 0 4 0 4 4 158 0 0 4 0 4 4 159 0 0 4 0 4 4 160 0 0 4 0 4 4 161 0 0 0 4 4 4 162 0 0 0 4 4 4 163 0 0 0 4 4 4 164 0 0 0 4 4 4 *F<- factor(c(rep(1,3),rep(2,3)))* *G<- factor(c(rep(1,2),rep(2,2),rep(3,2)))* *H<-F<- factor(c(rep(1,3),rep(2,3)))* *y<-rnorm(6,2)* *test3<-aov(y~F*G)* *model.matrix(test3)* (Intercept) F2 G2 G3 F2:G2 F2:G3 1 1 0 0 0 0 0 2 1 0 0 0 0 0 3 1 0 1 0 0 0 4 1 1 1 0 1 0 5 1 1 0 1 0 1 6 1 1 0 1 0 1 attr(,"assign") [1] 0 1 2 2 3 3 attr(,"contrasts") attr(,"contrasts")$F [1] "contr.treatment" attr(,"contrasts")$G [1] "contr.treatment" *alias(test3)* Model : y ~ F * G Complete : (Intercept) F2 G2 G3 F2:G2 0 1 0 -1 F2:G3 0 0 0 1 *summary(test3)* Df Sum Sq Mean Sq F value Pr(>F) F1 0.0479 0.0479 0.059 0.830 G2 0.9762 0.4881 0.604 0.624 Residuals 2 1.6175 0.8087 -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419 <%2B447938674419>(UK) or +60125056192 <%2B60125056192>(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Soft Question: Where to find this reference.
I notice a lot of r documentation refer to this reference below. I can't seem to find it anywhere. Does anyone have a link to point to where I can either view it or buy it? *Chambers, J. M., Freeny, A and Heiberger, R. M. (1992) Analysis of variance; designed experiments* -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Missing rows anova
Hi Michael, Thank you for the reply. I am sorry I forgot to print out the anova table to make my question clear. DfSum Sq Mean Sq F value Pr(>F) S 20.199.630e-060.8180.444 x110.0002562.560e-04 21.751 9.44e-06 *** ID 47 0.0035247.498e-056.370 3.35e-15 *** Resid102 0.0012011.177e-05 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 There is *no* unique value for ID for each combination of S and x1. For example, S=1 and x1=0 can equal to either B or C or D or E or F . Perhaps you mean that for each combination of S and x1 have different values. If that is the case, I think maybe it makes sense. *What I think? * Anova has this thing where it fits the terms of 1st order first ( a formula term including no interactions) before it fits a 2nd order term ( a formula term including 1 interaction) and so on. First Order--> Second Order--> Third Order--> etc Therefore, it is known that the true fitting formula is not S+x1+S:x1+ID but it is S+x1+ID+S:x1. Hence, it appears that ID is fitted before S:x1 but since ID is a more refined factor than S:x1, it can be said that S:x1 is already included in the fit of ID so R recognizes the linear dependance and excludes the term S:x1. In other words, S:x1 is linearly dependant to ID. And so, the row S:x1 disappears because it is considered within ID. Does this makes sense? On 19 July 2016 at 16:19, Michael Dewey <li...@dewey.myzen.co.uk> wrote: > Presumably it disappears because there is a unique value of ID for eac > combination of S*x1 so they are indistinguishable. > > > On 19/07/2016 12:53, Justin Thong wrote: > >> Why does the S:x1 column disappear (presumably S:x1 goes into ID but I >> dont >> know why)? S is a factor, x1 is a covariate and ID is a factor. >> >> rich.side<-aov(y~S*x1+ID) >> summary(rich.side) >> >> Below is the model frame >> >> model.frame(~S*x1+ID) >> >> S x1 ID >> 1 1 12 A >> 2 1 12 A >> 3 1 12 A >> 4 1 12 A >> 5 1 0 B >> 6 1 0 B >> 7 1 0 B >> 8 1 0 B >> 9 1 0 C >> 10 1 0 C >> 11 1 0 C >> 12 1 0 C >> 13 1 0 D >> 14 1 0 D >> 15 1 0 D >> 16 1 0 D >> 17 1 0 E >> 18 1 0 E >> 19 1 0 E >> 20 1 0 E >> 21 1 0 F >> 22 1 0 F >> 23 1 0 F >> 24 1 0 F >> 25 2 6 AB >> 26 2 6 AB >> 27 2 6 AB >> 28 2 6 AB >> 29 2 6 AC >> 30 2 6 AC >> 31 2 6 AC >> 32 2 6 AC >> 33 2 6 AD >> 34 2 6 AD >> 35 2 6 AD >> 36 2 6 AD >> 37 2 6 AE >> 38 2 6 AE >> 39 2 6 AE >> 40 2 6 AE >> 41 2 6 AF >> 42 2 6 AF >> 43 2 6 AF >> 44 2 6 AF >> 45 2 0 BC >> 46 2 0 BC >> 47 2 0 BC >> 48 2 0 BC >> 49 2 0 BD >> 50 2 0 BD >> 51 2 0 BD >> 52 2 0 BD >> 53 2 0 BE >> 54 2 0 BE >> 55 2 0 BE >> 56 2 0 BE >> 57 2 0 BF >> 58 2 0 BF >> 59 2 0 BF >> 60 2 0 BF >> 61 2 0 CD >> 62 2 0 CD >> 63 2 0 CD >> 64 2 0 CD >> 65 2 0 CE >> 66 2 0 CE >> 67 2 0 CE >> 68 2 0 CE >> 69 2 0 CF >> 70 2 0 CF >> 71 2 0 CF >> 72 2 0 CF >> 73 2 0 DE >> 74 2 0 DE >> 75 2 0 DE >> 76 2 0 DE >> 77 2 0 DF >> 78 2 0 DF >> 79 2 0 DF >> 80 2 0 DF >> 81 2 0 EF >> 82 2 0 EF >> 83 2 0 EF >> 84 2 0 EF >> 85 3 4 ABC >> 86 3 4 ABC >> 87 3 4 ABC >> 88 3 4 ABC >> 89 3 4 ABD >> 90 3 4 ABD >> 91 3 4 ABD >> 92 3 4 ABD >> 93 3 4 ABE >> 94 3 4 ABE >> 95 3 4 ABE >> 96 3 4 ABE >> 97 3 4 ABF >> 98 3 4 ABF >> 99 3 4 ABF >> 100 3 4 ABF >> 101 3 4 ACD >> 102 3 4 ACD >> 103 3 4 ACD >> 104 3 4 ACD >> 105 3 4 ACE >> 106 3 4 ACE >> 107 3 4 ACE >> 108 3 4 ACE >>
[R] Missing rows anova
Why does the S:x1 column disappear (presumably S:x1 goes into ID but I dont know why)? S is a factor, x1 is a covariate and ID is a factor. rich.side<-aov(y~S*x1+ID) summary(rich.side) Below is the model frame model.frame(~S*x1+ID) S x1 ID 1 1 12 A 2 1 12 A 3 1 12 A 4 1 12 A 5 1 0 B 6 1 0 B 7 1 0 B 8 1 0 B 9 1 0 C 10 1 0 C 11 1 0 C 12 1 0 C 13 1 0 D 14 1 0 D 15 1 0 D 16 1 0 D 17 1 0 E 18 1 0 E 19 1 0 E 20 1 0 E 21 1 0 F 22 1 0 F 23 1 0 F 24 1 0 F 25 2 6 AB 26 2 6 AB 27 2 6 AB 28 2 6 AB 29 2 6 AC 30 2 6 AC 31 2 6 AC 32 2 6 AC 33 2 6 AD 34 2 6 AD 35 2 6 AD 36 2 6 AD 37 2 6 AE 38 2 6 AE 39 2 6 AE 40 2 6 AE 41 2 6 AF 42 2 6 AF 43 2 6 AF 44 2 6 AF 45 2 0 BC 46 2 0 BC 47 2 0 BC 48 2 0 BC 49 2 0 BD 50 2 0 BD 51 2 0 BD 52 2 0 BD 53 2 0 BE 54 2 0 BE 55 2 0 BE 56 2 0 BE 57 2 0 BF 58 2 0 BF 59 2 0 BF 60 2 0 BF 61 2 0 CD 62 2 0 CD 63 2 0 CD 64 2 0 CD 65 2 0 CE 66 2 0 CE 67 2 0 CE 68 2 0 CE 69 2 0 CF 70 2 0 CF 71 2 0 CF 72 2 0 CF 73 2 0 DE 74 2 0 DE 75 2 0 DE 76 2 0 DE 77 2 0 DF 78 2 0 DF 79 2 0 DF 80 2 0 DF 81 2 0 EF 82 2 0 EF 83 2 0 EF 84 2 0 EF 85 3 4 ABC 86 3 4 ABC 87 3 4 ABC 88 3 4 ABC 89 3 4 ABD 90 3 4 ABD 91 3 4 ABD 92 3 4 ABD 93 3 4 ABE 94 3 4 ABE 95 3 4 ABE 96 3 4 ABE 97 3 4 ABF 98 3 4 ABF 99 3 4 ABF 100 3 4 ABF 101 3 4 ACD 102 3 4 ACD 103 3 4 ACD 104 3 4 ACD 105 3 4 ACE 106 3 4 ACE 107 3 4 ACE 108 3 4 ACE 109 3 4 ACF 110 3 4 ACF 111 3 4 ACF 112 3 4 ACF 113 3 4 ADE 114 3 4 ADE 115 3 4 ADE 116 3 4 ADE 117 3 4 ADF 118 3 4 ADF 119 3 4 ADF 120 3 4 ADF 121 3 4 AEF 122 3 4 AEF 123 3 4 AEF 124 3 4 AEF 125 3 0 BCD 126 3 0 BCD 127 3 0 BCD 128 3 0 BCD 129 3 0 BCE 130 3 0 BCE 131 3 0 BCE 132 3 0 BCE 133 3 0 BCF 134 3 0 BCF 135 3 0 BCF 136 3 0 BCF 137 3 0 BDE 138 3 0 BDE 139 3 0 BDE 140 3 0 BDE 141 3 0 BDF 142 3 0 BDF 143 3 0 BDF 144 3 0 BDF 145 3 0 BEF 146 3 0 BEF 147 3 0 BEF 148 3 0 BEF 149 3 0 CDE 150 3 0 CDE 151 3 0 CDE 152 3 0 CDE 153 3 0 CDF 154 3 0 CDF 155 3 0 CDF 156 3 0 CDF 157 3 0 CEF 158 3 0 CEF 159 3 0 CEF 160 3 0 CEF 161 3 0 DEF 162 3 0 DEF 163 3 0 DEF 164 3 0 DEF -- Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Reference for aov()
Hi *I have been looking for a reference to explain how R uses the aov command(at a deeper level)*. More specifically, how R reads the formulae and R computes the sums of squares. I am not interested in understanding what the difference of Type 1,2,3 sum of squares are. I am more interested in finding out about how R computes ~x1:x2:x3 or how R computes ~A:x1 emphasizing sequential nature of the way it computes, and models even more complicated than this. Yours sincerely, Justin *I check my email at 9AM and 4PM everyday* *If you have an EMERGENCY, contact me at +447938674419(UK) or +60125056192(Malaysia)* [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] General copula model with heterogeneous marginals
I am looking to model the sum of a number of random variables with
arbitrary gamma distributions and an empirical dependence structure
that I obtain from data. Basically I observe all of the individual
pieces but I want to model their sum, as opposed to many copula
questions which observe a single outcome of a multivariate process and
seek to fit possible marginal and covariance structure.
It has been years since I coded in R, but this is what I have thus far:
library(copula)
library(scatterplot3d)
library(psych)
set.seed(1)
myCop<-
normalCopula(param=c(.1,.1,.1,.1,.1,.2,.2,.2,.2,.2,.2,.2,.4,.4,.4,.4,.4,.5,.5,.5,.5), dim=7,
dispstr="un")
myMvd<-mvdc(copula=myCop, margins=rep("gamma",7),
paramMargins=list(list(shape=3,scale=4),
list(shape=2, scale=5),
list(shape=2, scale=5),
list(shape=2, scale=5),
list(shape=2, scale=5),
list(shape=3, scale=5),
list(shape=3, scale=5)))
simulation<- rMvdc(2,myMvd)
colnames(simulation)<-c("P1","P2","P3","P4","P5","P6","P7")
total =
simulation[,1]+simulation[,2]+simulation[,3]+simulation[,4]+simulation[,5]+simulation[,6]+simulation[,7]
As you can see, I have forced 7 gamma distributions with a placeholder
covariance matrix input. The problem is that I am looking to
generalize this to the order of ~150 different marginals with
potentially differing distributions and parameters.
Ultimately I will have the following input:
• matrix of 150 marginal distributions with family and parameters
• 150x150 covariance matrix
And what I need to produce is the following:
An empirical CDF/PDF of the sum of realizations from 5-10 of the
underlying marginal distributions. To be more clear, assume each
marginal distribution is a person's response to a treatment, and I
need to calculate the cumulative treatment effect for a sub-group of
the population of 150. So, I will have a vector of 0s and 1s to
identify which members of the population are grouped together for a
trial. Then I will have a separate vector for the next group. Each
group vector will have dim=150 but have between 5 and 10 1s with the
rest 0s. I need a different empirical CDF for each vector.
Any help?
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[R] Limiting state probability for Markov chain
Dear All,
I am trying to determine the liming state probability .
my_fun-function(A,b){
for (j in 1:3){
x-A;
while ((sum(x[j,]) ==1) )
{
x - x%*%x;
print (x);
if ( b%*%x==b)
{
break;
}}}
}
A-rbind(c(.5,.3,.2), c(.3,.3,.4),c(.1,.5,.4))
b - matrix(data=c(1,0,0), nrow=1, ncol=3, byrow=FALSE)
my_fun(A,b)
I got the following warning
1: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
2: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
3: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
4: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
5: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
6: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
7: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
8: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
9: In if (b %*% x == b) { :
the condition has length 1 and only the first element will be used
your help will be appreciate
Best Regard
Ushize Rutikanga Justin
Student at African Institute for Mathematical Sciences (AIMS) South Africa
E-mail:justinush...@aims.ac.za
Tel:+27717029144
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[R] Awaiting feedback/moderation (Re: Non-convergence in boot.stepAIC function with a logit model)
Hi I posted a question to this list, and received an email indicating that it is awaiting moderation. However I have not received any feedback yet. If the question is not appropriate, I’d like to know where or how I should post differently, if possible. Thanks and Regards Justin Michell __ 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] Non-convergence in boot.stepAIC function with a logit model
Hi all I am getting warning when I try to perform a bootstrap selection procedure on variables (using boot.stepAIC function in the bootStepAIC package). I had previously established which variables were collinear and kept the one which had the lowest AIC following univariate regression on each predictor. I obtain a candidate list of variables that are not correlated at the end of this procedure. I then revisit those variables that were excluded at each step using bootstrapping. I have referred to other list questions (such as http://stackoverflow.com/questions/8596160/why-am-i-getting-algorithm-did-not-converge-and-fitted-prob-numerically-0-or) and I see that this is a common problem with logit models, but convergence fails only in a bootstrapping context. For the first set of previously excluded variables, I added individually each variable to candidate list of variables and then performed bootstrapping, and then added more than one variable to see if the algorithm would converge. Sometimes it did other times not (as indicated by dc). I suppose the presence of multicollinearity affects this process? From the next group on of excluded variables I only really considered adding variables separately one at a time and then checked if there was an improvement in AIC. If one of the previously excluded variables is in the candidate list, then i take that variable out and add the previously excluded one and see if there is an improvement in AIC. From this reasoning I end up adding two new variables to the list. They are not correlated with any of the variables in candidate list, nor are they correlated. My question is, is this a valid way to come up with my best set of predictors? Is there a way I can monitor more closely what is going on, i.e. if multicollinearity in is mathematically causing the algorithm not to converge for some variables? Here is my workflow using the boot.stepAIC function in the the forward stepwise direction (the forward direction seems to be more robust w.r.t convergence): (if reproducible code is required I can happily provide it - via dropbox for the data) #kept altitude (15454.23) (but not in candidate list) and excluded: # meanTemp (14422.72), minTemp (14435.72), bio1 (14767.88), bio6 (didn't converge (dc)), bio8 (15050.46), bio10 (14285.46), bio11 (14655.82), bio18 (15445.24), bio10+bio11 (dc), # bio10 + bio11 + # bio18 (dc), bio10+bio18 (dc), meanTemp + bio10+bio11 (14160.33), minTemp + bio10+bio11 (14204.41), meanTemp + minTemp + bio10+bio11 (14135.49), meanTemp+minTemp + bio10+bio11+bio1(dc), # bio10+bio11+bio1(dc) fit.1 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) bootGLM.1 - boot.stepAIC(fit.1, spatialVars, direction = backward, alpha = 0.05, B = 1000) #15445.24 # add bio10 (lowest AIC - 14285.46) # (backward drection dc) # kept bio2 and excluded: bio7 (dc), -bio2 + bio7 (14710.04), -bi02 - bio10 + bio7 (15676.62) fit.2 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) #15302.59 bootGLM.2 - boot.stepAIC(fit.2, spatialVars, direction = forward, alpha = 0.05, B = 1000) # keep bio2 in candidate list (+bio10) # kept bio5 and excluded: altitude (dc), -bio5 + altitude (15659.26), -bio5 +maxTemp (15637.91) fit.3 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10, weights = Examind, data = spatialVars, family=binomial) bootGLM.3 - boot.stepAIC(fit.3, spatialVars, direction = forward, alpha = 0.05, B = 1000) # keep bio5 in candidate list (+bio10) # kept bio17 (not in candidate list) (bio17 (14178.88)) and excluded: bio12 (dc), bio14 (14168.77), bio16 (14287.42), bio19 (14248.65), rain (14287.45), bio17+bio12(14162.3) fit.4 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10 + rain, weights = Examind, data = spatialVars, family=binomial) bootGLM.4 - boot.stepAIC(fit.4, spatialVars, direction = forward, alpha = 0.05, B = 1000) # add bio14 to candiate list (+bio10) # keptp bio15 (14168.77) (not included in candidate list) and excluded: bio17 (14161.03) fit.5 - glm(Pos/Examind ~ bio13 + bio15 + bio2 + bio3 + DstTClW + bio4 + NDVI + bio5 + bio9 + bio10 + bio14, weights = Examind, data = spatialVars, family=binomial) bootGLM.5 - boot.stepAIC(fit.5, spatialVars, direction = forward, alpha = 0.05, B = 1000) Thanks very much (for any help, advice or thoughts) Justin [[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] Maintaining data order in factanal with missing data
Hi Petr,
Thanks for the quick response. Unfortunately I cannot share the data I am
working with, however please find attached a suitable R workspace with
generated data. It has the appropriate variable names, only the data has
been changed.
The last function in the list (init.dfs()) I call to subset the overall data
set into the three waves, then conduct the factor analysis on each (1 factor
CFA); it's just in a function to ease re-typing in a new workspace.
Thanks,
Justin
-Original Message-
From: PIKAL Petr [mailto:petr.pi...@precheza.cz]
Sent: Friday, 26 July 2013 7:35 PM
To: Justin Delahunty; r-help@r-project.org
Subject: RE: [R] Maintaining data order in factanal with missing data
Hi
You provided functions, so far so good. But without data it would be quite
difficult to understand what the functions do and where could be the issue.
I suspect combination of complete cases selection together with subset and
factor behaviour. But I can be completely out of target too.
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of s00123...@myacu.edu.au
Sent: Friday, July 26, 2013 9:35 AM
To: r-help@r-project.org
Subject: [R] Maintaining data order in factanal with missing data
Hi,
I'm new to R, so sorry if this is a simple answer. I'm currently
trying to collapse some ordinal variables into a composite; the
program ideally should take a data frame as input, perform a factor
analysis, compute factor scores, sds, etc., and return the rescaled
scores and loadings. The difficulty I'm having is that my data set
contains a number of NA, which I am excluding from the analysis using
complete.cases(), and thus the incomplete cases are skipped. These
functions are for a longitudinal data set with repeated waves of data,
so the final rescaled scores from each wave need to be saved as
variables grouped by a unique ID (DMID). The functions I'm trying to
implement are as follows:
weighted.sd-function(x,w){
sum.w-sum(w)
sum.w2-sum(w^2)
mean.w-sum(x*w)/sum(w)
x.sd.w-sqrt((sum.w/(sum.w^2-sum.w2))*sum(w*(x-mean.w)^2))
return(x.sd.w)
}
re.scale-function(f.scores, raw.data, loadings){
fz.scores-(f.scores+mean(f.scores))/(sd(f.scores))
means-apply(raw.data,1,weighted.mean,w=loadings)
sds-apply(raw.data,1,weighted.sd,w=loadings)
grand.mean-mean(means)
grand.sd-mean(sds)
final.scores-((fz.scores*grand.sd)+grand.mean)
return(final.scores)
}
get.scores-function(data){
fact-
factanal(data[complete.cases(data),],factors=1,scores=regression)
f.scores-fact$scores[,1]
f.loads-fact$loadings[,1]
rescaled.scores-re.scale(f.scores,
data[complete.cases(data),], f.loads)
output.list-list(rescaled.scores,
f.loads)
names(output.list)-
c(rescaled.scores,
factor.loadings)
return(output.list)
}
init.dfs-function(){
ab.1.df-subset(ab.df,,select=c(dmid,g5oab2:g5ovb1))
ab.2.df-subset(ab.df,,select=c(dmid,w2oab3:w2ovb1))
ab.3.df-subset(ab.df,,select=c(dmid,
w3oab3, w3oab4, w3oab7, w3oab8, w3ovb1))
ab.1.fa-get.scores(ab.1.df[-1])
ab.2.fa-get.scores(ab.2.df[-1])
ab.3.fa-get.scores(ab.3.df[-1])
}
Thanks for your help,
Justin
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Re: [R] Maintaining data order in factanal with missing data
Hi Petr, So sorry, I accidentally attached the complete data set rather than the one with missing values. I've attached the correct file to this email. RE: init.dfs() being local, I hadn't even thought of that. I've been away from OOP for close to 15 years now, so it might be time to revise! The problem I have is that with missing values the list of factor scores returned (ab.w1.fa$factor.scores) does not map onto the originating data frame (ab.w1.df) as it no longer includes the cases which had missing values. So while the original data set for ab.w1.df contains 154 ordered cases, the factor analysis contains only 150. I am seeking a way to map the values derived from the factor analysis (ab.w1.fa$factor.scores) back to their original ordered position, so that these factor score variables may be merged back into the master data frame (ab.df). A unique ID for each case is available ($dmid) which I had thought to use when merging the new variables, however I don't know how to implement this. Thanks for your help, Justin -Original Message- From: PIKAL Petr [mailto:petr.pi...@precheza.cz] Sent: Friday, 26 July 2013 10:59 PM To: Justin Delahunty; Justin Delahunty; r-help@r-project.org Subject: RE: [R] Maintaining data order in factanal with missing data Hi Well, the function init.dfs does nothing as all data frames created inside it does not propagate to global environment and there is nothing what the function returns. Tha last line (when used outside a function) gives warnings but there is no sign of error. When head(ab.1.df) dmid g5oab2 g53 g54 g55 g5ovb1 11 1.418932 1.805227 2.791152 3.624116 3.425586 22 2.293907 1.187830 1.611237 1.748526 3.816533 33 2.836536 2.679523 1.279639 2.674986 2.452395 44 1.872259 3.278359 1.785872 2.458315 1.146480 55 1.467195 1.180747 3.564127 3.007682 2.109506 66 3.098512 3.151974 3.969379 3.750571 1.497358 head(ab.2.df) dmid w2oab3 w22 w23 w24 w2ovb1 11 4.831362 5.522764 7.809366 6.969172 7.398385 22 6.706346 4.101742 1.434697 5.266775 5.357641 33 3.653806 2.666885 1.209326 5.125556 4.963374 44 7.221255 7.649152 6.540398 6.648506 2.576081 55 1.848023 5.044314 2.761881 3.307220 1.454234 66 7.606429 4.911766 2.034813 2.638573 2.818834 head(ab.3.df) dmid w3oab3 w3oab4 w3oab7 w3oab8 w3ovb1 11 5.835609 6.108220 6.587721 2.451461 2.785467 22 4.973198 1.196815 6.388056 1.110877 4.226463 33 3.800367 6.697287 5.235345 6.666829 6.319073 44 1.093141 1.43 2.269252 3.194978 4.916342 55 1.975060 7.204516 4.825435 1.775874 3.484027 66 3.273361 2.243805 5.326547 5.720892 6.118723 str(ab.1.fa) List of 2 $ rescaled.scores: Named num [1:154] 3.43 3.83 2.43 1.1 2.08 ... ..- attr(*, names)= chr [1:154] 1 2 3 4 ... $ factor.loadings: Named num [1:5] -0.0106 -0.0227 -0.1093 -0.0912 0.9975 ..- attr(*, names)= chr [1:5] g5oab2 g53 g54 g55 ... str(ab.2.fa) List of 2 $ rescaled.scores: Named num [1:154] 6.34 5.24 5.3 1.91 2.16 ... ..- attr(*, names)= chr [1:154] 1 2 3 4 ... $ factor.loadings: Named num [1:5] -0.2042 0.0063 -0.2287 -0.0119 0.7138 ..- attr(*, names)= chr [1:5] w2oab3 w22 w23 w24 ... str(ab.3.fa) List of 2 $ rescaled.scores: Named num [1:154] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN ... ..- attr(*, names)= chr [1:154] 1 2 3 4 ... $ factor.loadings: Named num [1:5] -0.1172 0.0128 -0.0968 0.106 0.9975 ..- attr(*, names)= chr [1:5] w3oab3 w3oab4 w3oab7 w3oab8 ... Anyway I have no idea what you consider wrong? Regards Petr -Original Message- From: Justin Delahunty [mailto:a...@genius.net.au] Sent: Friday, July 26, 2013 2:22 PM To: PIKAL Petr; 'Justin Delahunty'; r-help@r-project.org Subject: RE: [R] Maintaining data order in factanal with missing data Hi Petr, Thanks for the quick response. Unfortunately I cannot share the data I am working with, however please find attached a suitable R workspace with generated data. It has the appropriate variable names, only the data has been changed. The last function in the list (init.dfs()) I call to subset the overall data set into the three waves, then conduct the factor analysis on each (1 factor CFA); it's just in a function to ease re-typing in a new workspace. Thanks, Justin -Original Message- From: PIKAL Petr [mailto:petr.pi...@precheza.cz] Sent: Friday, 26 July 2013 7:35 PM To: Justin Delahunty; r-help@r-project.org Subject: RE: [R] Maintaining data order in factanal with missing data Hi You provided functions, so far so good. But without data it would be quite difficult to understand what the functions do and where could be the issue. I suspect combination of complete cases selection together with subset and factor behaviour. But I can be completely out of target too. Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org
[R] R CMD check package mypkg-Ex.R failed
prepare_Rd: f.Rd:25-31: Dropping empty section \value prepare_Rd: f.Rd:38-40: Dropping empty section \note prepare_Rd: f.Rd:35-37: Dropping empty section \author prepare_Rd: f.Rd:32-34: Dropping empty section \references prepare_Rd: f.Rd:44-46: Dropping empty section \seealso checkRd: (5) f.Rd:0-60: Must have a \description prepare_Rd: mypkg-package.Rd:33: All text must be in a section prepare_Rd: mypkg-package.Rd:34: All text must be in a section * checking Rd metadata ... OK * checking Rd cross-references ... WARNING Unknown package(s) âpkgâ in Rd xrefs * checking for missing documentation entries ... OK * checking for code/documentation mismatches ... OK * checking Rd \usage sections ... OK * checking Rd contents ... OK * checking for unstated dependencies in examples ... NOTE Warning: parse error in file âmypkg-Ex.Râ: 11: unexpected symbol 42: 43: ~~ simple examples              ^ * checking examples ... ERROR Running examples in âmypkg-Ex.Râ failed The error most likely occurred in: ### Name: mypkg-package ### Title: What the package does (short line) ~~ package title ~~ ### Aliases: mypkg-package mypkg ### Keywords: package ### ** Examples ~~ simple examples of the most important functions ~~ Error: unexpected symbol in ~~ simple examples Execution halted My question: How to correct the above error? Thanks, Justin [[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] Getting a vector result from a computed index
Greetings,
I confess to being new to R, which I am exploring for some simulation
modeling purposes. I am a long time programmer in PHP.
I am having some trouble getting over the steep learning curve with
some very basic things which are probably just little a-ha things in
R. I have hunted and hunted through the manual and cannot figure this
one out, so I am appealing for help.
I have the following program:
results - replicate(1,0)
for (year in 2000:2050) {
print(year);
for (i in 1:1) {
x=results[i];
if (x == 0) {
prev = results[i-1];
next = results[i+1];
prob=0.1;
if (i=1) {
if (prev==1) { prob=prob+0.4; }
}
if (i1) {
if (next==1) { prob=prob+0.4; }
}
y=runif(1,0,1);
if (yprob) { x=1; }
results[i]=x;
}
}
}
No matter how I try this (and I've tried a number of variations), I
get an error similar to
Error in next = results[i + 1] : invalid (NULL) left side of assignment
I need to be able to loop through the results vector and compare the
previous/next items in the list to the current one. Is that possible?
What am I missing?
Cordially,
Justin Long
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[R] two phases sampling
Dear all, I have a question about estimation in two phases sampling. I' have a first sample of household from a complex sampling S1, a second sample is drawned from S2. from S2, I compute an estimator y2, for households of S1 not in S2 I set y2=0. I have an estimator y1 on S1 My indicator is y=y1+y2. So the variance of y is v(y)=v(y1)+v(y2)+cov(y1,y2). Sampling theory indicate how to compute v(y1) and v(y2) but how can I compute cov(y1,y2) ? Can the survey package help me for that ? Justin BEM BP 1917 Yaoundé Tél (237) 76043774 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do anova() and Anova(type=III) handle incomplete designs?
Thanks for your response, John. That was helpful. I was using Type III from Anova() as a comparison to some results I had obtained JMP, which I've lost access to and have moved on to R, and I was confused by the error. Given that I do have a continuous covariate, the analyses are not likely comparable, considering your response. I am still confused about interpretation of interactions within an anova() with an incomplete design, as mine is. Is the interaction term still informative? - Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com On Sat, Jun 16, 2012 at 9:20 PM, John Fox j...@mcmaster.ca wrote: Dear Justin, anova() and Anova() are entirely different functions; the former is part of the standard R distribution and the second part of the car package. By default, Anova() produces an error for type-III tests conducted on rank-deficient models because the hypotheses tested aren't generally sensible. From ?Anova: singular.ok defaults to TRUE for type-II tests, and FALSE for type-III tests (where the tests for models with aliased coefficients will not be straightforwardly interpretable); if FALSE, a model with aliased coefficients produces an error. and The designations type-II and type-III are borrowed from SAS, but the definitions used here do not correspond precisely to those employed by SAS. Type-II tests are calculated according to the principle of marginality, testing each term after all others, except ignoring the term's higher-order relatives; so-called type-III tests violate marginality, testing each term in the model after all of the others. This definition of Type-II tests corresponds to the tests produced by SAS for analysis-of-variance models, where all of the predictors are factors, but not more generally (i.e., when there are quantitative predictors). Be very careful in formulating the model for type-III tests, or the hypotheses tested will not make sense. I hope this helps, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 15 Jun 2012 15:01:27 -0400 Justin Montemarano jmont...@kent.edu wrote: Hello all: I am confused about the output from a lm() model with an incomplete design/missing level. I have two categorical predictors and a continuous covariate (day) that I am using to model larval mass (l.mass): leaf.species has three levels - map, syc, and oak cond.time has two levels - 30 and 150. There are no response values for Map-150, so that entire, two-way, level is missing. When running anova() on the model with Type I SS, the full factorial design does not return errors; however, using package:car Anova() and Type III SS, I receive an singularity error unless I used the argument 'singular.ok = T' (it is defaulted to F). So, why don't I receive an error with anova() when I do with Anova(type = III)? How do anova() and Anova() handle incomplete designs, and how can interactions of variables with missing levels be interpreted? I realize these are fairly broad questions, but any insight would be helpful. Thanks, all. Below is code to illustrate my question(s): lmMass - lm(log(l.mass) ~ day*leaf.species + cond.time, data = growth.data) #lm() without cond.time interactions lmMassInt - lm(log(l.mass) ~ day*leaf.species*cond.time, data = growth.data) #lm() with cond.time interactions anova(lmMass); anova(lmMassInt) #ANOVA summary of both models with Type I SS Analysis of Variance Table Response: log(l.mass) Df Sum Sq Mean Sq F valuePr(F) day1 51.373 51.373 75.7451 2.073e-15 leaf.species 2 0.340 0.170 0.25060.7786 cond.time 1 0.161 0.161 0.23690.6271 day:leaf.species 2 1.296 0.648 0.95510.3867 Residuals179 121.404 0.678 Analysis of Variance Table Response: log(l.mass) Df Sum Sq Mean Sq F value Pr(F) day 1 51.373 51.373 76.5651 1.693e-15 leaf.species 2 0.340 0.170 0.2533 0.77654 cond.time1 0.161 0.161 0.2394 0.62523 day:leaf.species 2 1.296 0.648 0.9655 0.38281 day:cond.time1 0.080 0.080 0.1198 0.72965 leaf.species:cond.time 1 1.318 1.318 1.9642 0.16282 day:leaf.species:cond.time 1 1.915 1.915 2.8539 0.09293 Residuals 176 118.091 0.671 Anova(lmMass, type = 'III'); Anova(lmMassInt, type = 'III') #ANOVA summary of both models with Type III SS Anova Table (Type III tests) Response: log(l.mass
[R] How do anova() and Anova(type=III) handle incomplete designs?
Hello all: I am confused about the output from a lm() model with an incomplete design/missing level. I have two categorical predictors and a continuous covariate (day) that I am using to model larval mass (l.mass): leaf.species has three levels - map, syc, and oak cond.time has two levels - 30 and 150. There are no response values for Map-150, so that entire, two-way, level is missing. When running anova() on the model with Type I SS, the full factorial design does not return errors; however, using package:car Anova() and Type III SS, I receive an singularity error unless I used the argument 'singular.ok = T' (it is defaulted to F). So, why don't I receive an error with anova() when I do with Anova(type = III)? How do anova() and Anova() handle incomplete designs, and how can interactions of variables with missing levels be interpreted? I realize these are fairly broad questions, but any insight would be helpful. Thanks, all. Below is code to illustrate my question(s): lmMass - lm(log(l.mass) ~ day*leaf.species + cond.time, data = growth.data) #lm() without cond.time interactions lmMassInt - lm(log(l.mass) ~ day*leaf.species*cond.time, data = growth.data) #lm() with cond.time interactions anova(lmMass); anova(lmMassInt) #ANOVA summary of both models with Type I SS Analysis of Variance Table Response: log(l.mass) Df Sum Sq Mean Sq F valuePr(F) day1 51.373 51.373 75.7451 2.073e-15 leaf.species 2 0.340 0.170 0.25060.7786 cond.time 1 0.161 0.161 0.23690.6271 day:leaf.species 2 1.296 0.648 0.95510.3867 Residuals179 121.404 0.678 Analysis of Variance Table Response: log(l.mass) Df Sum Sq Mean Sq F value Pr(F) day 1 51.373 51.373 76.5651 1.693e-15 leaf.species 2 0.340 0.170 0.2533 0.77654 cond.time1 0.161 0.161 0.2394 0.62523 day:leaf.species 2 1.296 0.648 0.9655 0.38281 day:cond.time1 0.080 0.080 0.1198 0.72965 leaf.species:cond.time 1 1.318 1.318 1.9642 0.16282 day:leaf.species:cond.time 1 1.915 1.915 2.8539 0.09293 Residuals 176 118.091 0.671 Anova(lmMass, type = 'III'); Anova(lmMassInt, type = 'III') #ANOVA summary of both models with Type III SS Anova Table (Type III tests) Response: log(l.mass) Sum Sq Df F value Pr(F) (Intercept) 39.789 1 58.6653 1.13e-12 day3.278 1 4.8336 0.02919 leaf.species 0.934 2 0.6888 0.50352 cond.time 0.168 1 0.2472 0.61968 day:leaf.species 1.296 2 0.9551 0.38672 Residuals121.404 179 Error in Anova.III.lm(mod, error, singular.ok = singular.ok, ...) : there are aliased coefficients in the model Anova(lmMassInt, type = 'III', singular.ok = T) #Given the error in Anova() above, set singular.ok = T Anova Table (Type III tests) Response: log(l.mass) Sum Sq Df F value Pr(F) (Intercept) 39.789 1 59.3004 9.402e-13 day 3.278 1 4.8860 0.02837 leaf.species 1.356 2 1.0103 0.36623 cond.time0.124 1 0.1843 0.66822 day:leaf.species 2.783 2 2.0738 0.12877 day:cond.time0.805 1 1.1994 0.27493 leaf.species:cond.time 0.568 1 0.8462 0.35888 day:leaf.species:cond.time 1.915 1 2.8539 0.09293 Residuals 118.091 176 - Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com http://www.montegraphia.com/ -- Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Violation of sample independence in Pearson's product-moment correlation
Hi all: There was a concern raised by reviewers of a manuscript of mine over the proper execution of a Pearson's correlation. In brief, this was undertaken in order to determine the relationship between the extent of wheel running (y axis) and ethanol intake (x axis) across three, separate 10 day periods in 7 animals. In the paper, the correlational plots for each 10 day-period had 70 data points: One point for each day and each animal across 10 days of experimentation. The reviewers, however, appropriately pointed out that this is a violation of the assumption of sample independence for Pearson's test, and I should have had only 7 points, which would reflect the means of my two variables for each individual animal across 10 days. Is this appropriate or is there a means of accounting for repeated sampling with a correlation test? - Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] ANN: bigml package for R bigml_0.1.tar.gz
The bigml package is an R wrapper for the BigML API: https://bigml.com/developers It contains straightforward methods for most of the relevant API end points, as well as some fancier methods that allow for direct translations from R dataframes directly into datasets appropriate for BigML. Excerpt from an upcoming blog post, which describes the package in more detail: Today BigML releases the bigml package for R. R is already well known for its capabilities in statistics and data analysis, and we use it internally for a number of different day-to-day tasks. The bigml package enables the R community to easily take advantage of our highly scalable cloud based machine learning infrastructure, while using familiar R data structures and workflows. Apologies for sending this e-mail a bit late. Best, -Justin -- blog: http://www.scwn.net aim: iujjd twitter: jjdonald [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Deleting observations from baseline that don't appear in follow up
Hello all, I'm almost embarrassed to post this , it seems so easy. Suppose I have a baseline and follow up survey but some people are missing in the follow up: baseline-data.frame(id=c(3,5,7,9,12), data= runif(5)) follow.up-data.frame(id=c(3,7,9,12), data= runif(4)) baseline id data 1 3 0.66771988 2 5 0.28794744 3 7 0.01892821 4 9 0.64863175 5 12 0.86485882 follow.up id data 1 3 0.8237210 2 7 0.8140544 3 9 0.8803674 4 12 0.8031520 Here, in follow up we are missing person #5. I need to delete him from the baseline, so that I have an equal number of rows once again. Obviously baseline-baseline[-2,] won't cut it here, since in my data set I have thousands of people. Thanks in advance Justin [[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] system command to a specific shell (bash)
I need to run a bash command, but when you call system() the default shell
is sh (see my sessionInfo below).
I found the shell command (
http://www.stat.ucl.ac.be/ISdidactique/Rhelp/library/base/html/shell.html)
but it seems to be disappeared in current versions of R?
I am running all this from R CMD BATCH with system calls to other R
scripts.
For a little more info, I'm generating sphinx documents (a python
documentation library) through R and need to use a python virtual
environment.
So I need to call system('source bin/activate'), but source isn't a
recognized command in the sh shell...
Any help is appreciated,
Justin
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8LC_PAPER=C
LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] graphics grDevices utils datasets stats grid methods
base
other attached packages:
[1] ggplot2_0.9.0 reshape2_1.2.1 plyr_1.7.1
loaded via a namespace (and not attached):
[1] colorspace_1.1-1 dichromat_1.2-4digest_0.5.1 MASS_7.3-16
memoise_0.1munsell_0.3
[7] proto_0.3-9.2 RColorBrewer_1.0-5 scales_0.2.0 stringr_0.6
tools_2.15.0
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Re: [R] system command to a specific shell (bash)
Thanks Jeff, but I'm running a python program that expects certain
functionality that bash provides and sh doesn't... I can just stop using
github checkouts and use system packages though and fix this.
I'm mostly wondering where the shell command went in base R... it sounds
like it completely solves this issue but doesn't exist in my R
On Mon, Apr 16, 2012 at 10:58 AM, Jeff Newmiller
jdnew...@dcn.davis.ca.uswrote:
You could make a hash bang bash script that sources the file and then
proceeds to do whatever you want. Bourne shell should have no problems
invoking another shell.
---
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.
Justin Haynes jto...@gmail.com wrote:
I need to run a bash command, but when you call system() the default
shell
is sh (see my sessionInfo below).
I found the shell command (
http://www.stat.ucl.ac.be/ISdidactique/Rhelp/library/base/html/shell.html
)
but it seems to be disappeared in current versions of R?
I am running all this from R CMD BATCH with system calls to other R
scripts.
For a little more info, I'm generating sphinx documents (a python
documentation library) through R and need to use a python virtual
environment.
So I need to call system('source bin/activate'), but source isn't a
recognized command in the sh shell...
Any help is appreciated,
Justin
sessionInfo()
R version 2.15.0 (2012-03-30)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8LC_PAPER=C
LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] graphics grDevices utils datasets stats grid methods
base
other attached packages:
[1] ggplot2_0.9.0 reshape2_1.2.1 plyr_1.7.1
loaded via a namespace (and not attached):
[1] colorspace_1.1-1 dichromat_1.2-4digest_0.5.1
MASS_7.3-16
memoise_0.1munsell_0.3
[7] proto_0.3-9.2 RColorBrewer_1.0-5 scales_0.2.0
stringr_0.6
tools_2.15.0
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] A little exercise in R!
Since I thought this was a cool question, I posted it to StackOverflow.
Vincent Zookynd's answer is amazing and really exercises the power of R.
http://stackoverflow.com/questions/10150161/ordering-117-by-perfect-square-pairs/10150797#10150797
On Fri, Apr 13, 2012 at 10:06 PM, Bert Gunter gunter.ber...@gene.comwrote:
... and a moment's more consideration immediately shows it cannot be
done for n = 18, since 16,17, and 18 cannot all be at an end.
-- Bert
On Fri, Apr 13, 2012 at 9:59 PM, Bert Gunter bgun...@gene.com wrote:
Folks:
IMHO this is exactly the **wrong** way t go about this. These are
mathematical exercises that should employ mathematical thinking, not
brute force checking of cases.
Consider, for example, the 1 to 17 sequence given by Ted. Then 17
**must** be one end of the sequence and 16 the other. (Why?) Hence,
starting from the 17 end, the values ** must** be 17 8 1 ...
Proceeding in this way, it takes only a couple of minutes to solve.
The more interesting point which I think the question was really
about, is can this always be done? I haven't given this any thought,
but there may be an easy proof or counterexample. If the answer to
this latter is no, then perhaps even more interesting is to
characterize the set of numbers where it can/cannot be done.
But this is all way off topic, no?
Cheers,
Bert
On Fri, Apr 13, 2012 at 6:26 PM, Philippe Grosjean
phgrosj...@sciviews.org wrote:
Hi all,
I got another solution, and it would apply probably for the ugliest one
:-(
I made it general enough so that it works for any series from 1 to n (n
not
too large, please... tested up to 30).
Hint for a better algorithm: inspect the object 'friends' in my code:
there
is a nice pattern appearing there!!!
Best,
Philippe
..¡}))
) ) ) ) )
( ( ( ( (Prof. Philippe Grosjean
) ) ) ) )
( ( ( ( (Numerical Ecology of Aquatic Systems
) ) ) ) ) Mons University, Belgium
( ( ( ( (
..
findSerie - function (n, tmax = 500) {
## Check arguments
n - as.integer(n)
if (length(n) != 1 || is.na(n) || n 1)
stop('n' must be a single positive integer)
tmax - as.integer(tmax)
if (length(tmax) != 1 || is.na(tmax) || tmax 1)
stop('tmax' must be a single positive integer)
## Suite of our numbers to be sorted
nbrs - 1:n
## Trivial cases: only one or two numbers
if (n == 1) return(1)
if (n == 2) stop(The pair does not sum to a square number)
## Compute all possible pairs
omat - outer(rep(1, n), nbrs)
## Which pairs sum to a square number?
friends - sqrt(omat + nbrs) %% 1 .Machine$double.eps
diag(friends) - FALSE # Eliminate pairs of same numbers
## Get a list of possible neighbours
neigb - apply(friends, 1, function(x) nbrs[x])
## Nbr of neighbours for each number
nf - sapply(neigb, length)
## Are there numbers without neighbours?
## then, problem impossible to solve..
if (any(!nf))
stop(Impossible to solve:\n,
paste(nbrs[!nf], collapse = , ),
sum to square with nobody else!)
## Are there numbers that can have only one neighbour?
## Must be placed at one extreme
toEnds - nbrs[nf == 1]
## I must have two of them maximum!
l - length(toEnds)
if (l 2)
stop(Impossible to solve:\n,
More than two numbers form only one pair:\n,
paste(toEnds, collapse = , ))
## The other numbers can appear in the middle of the suite
inMiddle - nbrs[!nbrs %in% toEnds]
generateSerie - function (neigb, toEnds, inMiddle) {
## Allow to generate serie by picking candidates randomly
if (length(toEnds) 1) toEnds - sample(toEnds)
if (length(inMiddle) 1) inMiddle - sample(inMiddle)
## Choose a number to start with
res - rep(NA, n)
## Three cases: 0, 1, or 2 numbers that must be at an extreme
## Following code works in all cases
res[1] - toEnds[1]
res[n] - toEnds[2]
## List of already taken numbers
taken - toEnds
## Is there one number in res[1]? Otherwise, fill it now...
if (is.na(res[1])) {
taken - inMiddle[1]
res[1] - taken
}
## For each number in the middle, choose one acceptable neighbour
for (ii in 2:(n-1)) {
prev - res[ii - 1]
allpossible - neigb[[prev]]
candidate - allpossible[!(allpossible %in% taken)]
if (!length(candidate)) break # We fail to construct the serie
## Take randomly one possible candidate
if (length(candidate) 1) take - sample(candidate, 1) else
take - candidate
res[ii] - take
taken - c(taken, take)
}
## If we manage to go to the end, check last pair...
if (length(taken) == (n - 1)) {
take - nbrs[!(nbrs %in% taken)]
res[n] - take
taken
Re: [R] A little exercise in R!
I thought this was kinda cool! Here's my solution, its not robust or
probably efficient
I'd to hear improvements or other solutions!
Justin
sq.test - function(a, b) {
## test for number pairs that sum to squares.
sqrt(sum(a, b)) == floor(sqrt(sum(a, b)))
}
ok.pairs - function(n, vec) {
## given n as a member of vec,
## which other members of vec satisfiy sq.test
vec - vec[vec!=n]
vec[sapply(vec, sq.test, b=n)]
}
grow.seq - function(y) {
## given a starting point (y) and a pairs list (pl)
## grow the squaring sequence.
ly - length(y)
if(ly == y[1]) return(y)
## this line is the one that breaks down on other number sets...
y - c(y, max(pl[[y[ly]]][!pl[[y[ly]]] %in% y]))
y - grow.seq(y)
return(y)
}
## start vector
x - 1:17
## get list of possible pairs
pl - lapply(x, ok.pairs, vec=x)
## pick start at max since few combinations there.
y - max(x)
grow.seq(y)
On Fri, Apr 13, 2012 at 2:34 PM, Ted Harding ted.hard...@wlandres.netwrote:
Greetings all!
A recent news item got me thinking that a problem stated
therein could provide a teasing little exercise in R
programming.
http://www.bbc.co.uk/news/uk-england-cambridgeshire-17680326
Cambridge University hosts first European 'maths Olympiad'
for girls
The first European girls-only mathematical Olympiad
competition is being hosted by Cambridge University.
[...]
Olympiad co-director, Dr Ceri Fiddes, said competition questions
encouraged clever thinking rather than regurgitating a taught
syllabus.
[...]
A lot of Olympiad questions in the competition are about
proving things, Dr Fiddes said.
If you have a puzzle, it's not good enough to give one answer.
You have to prove that it's the only possible answer.
[...]
In the Olympiad it's about starting with a problem that anybody
could understand, then coming up with that clever idea that
enables you to solve it, she said.
For example, take the numbers one up to 17.
Can you write them out in a line so that every pair of numbers
that are next to each other, adds up to give a square number?
Well, that's the challenge: Write (from scratch) an R program
that solves this problem. And make it neat.
NOTE: If there should happen to be some R package that can solve
this kind of problem already, without you having to think much,
then its use is illegitimate! (I.e. will be deemed regurgitation).
Over to you.
With best wishes,
Ted.
-
E-Mail: (Ted Harding) ted.hard...@wlandres.net
Date: 13-Apr-2012 Time: 22:33:43
This message was sent by XFMail
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Re: [R] Remove carriage return in writing tab-delimited file.
take a look at ?paste paste(yourmatrix, sep='\t', collapse='') On Wed, Apr 4, 2012 at 2:58 PM, kickout plant.breeding.cr...@gmail.com wrote: Having problems with the write.table function. I can write a tab delimited file just fine, but for each line in my matrix its inputs a carriage return when i dont want it to. For example my matrix might be: ID V1 V2 V3 FARY1004 1 2 3 FARY2067 2 3 1 FARY4587 2 2 2 And I want the written File to be: FARY1004 1 2 3FARY2067 2 3 1FARY4587 2 2 2 TIA -- View this message in context: http://r.789695.n4.nabble.com/Remove-carriage-return-in-writing-tab-delimited-file-tp4533322p4533322.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] sampling rows from a list
## recreating your data mydata-list(matrix(1:9, nrow=3, byrow=T), matrix(10:15, nrow=2, byrow=T), matrix(16:30, nrow=5, byrow=T)) ## get the shortest matrix in your list n - min(unlist(lapply(mydata, nrow))) ## subset the list into random samples of length n out - lapply(mydata, function(x, n) x[sample(1:nrow(x), n),], n=n) ## this structure is still a list though... ## converting directly to an array: out.array - array(unlist(out), dim=c(dim(out[[1]]), length(out))) not totally sure about what structure you're wanting in the last step, so if i missed i apologize... Hope that helps, Justin On Mon, Apr 2, 2012 at 11:24 AM, Bcampbell99 briand.campb...@ec.gc.ca wrote: Hi: I'm sure this seems like a rudimentary question, but I am not well versed with R syntax for lists. I have a ragged array from which I've removed records (entire rows) with missing data. The functions I used to remove the missing cases resulted in the generation of an R list class object, that looks something like this; mydata [[1]] [,1] [,2] [,3] [1,] 1 2 3 [2,] 4 5 6 [3,] 7 8 9 [[2]] [,1] [,2] [,3] [1,] 10 11 12 [2,] 13 14 15 [[3]] [,1] [,2] [,3] [1,] 16 17 18 [2,] 19 20 21 [3,] 22 23 24 [4,] 25 26 27 [5,] 28 29 30 Part1 What I would like to do is draw an equal number of random row samples from[[1]],[[2]] and [[3]] (to preserve the structure of [,1][,2],[,3]. Part2 Then I would like to cocerce the list object into something like an array. Help scripting out part 1 or 2 would be much appreciated. Brian Campbell -- View this message in context: http://r.789695.n4.nabble.com/sampling-rows-from-a-list-tp4526831p4526831.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] list assignment syntax?
You can also take a look at http://stackoverflow.com/questions/7519790/assign-multiple-new-variables-in-a-single-line-in-r which has some additional solutions. On Fri, Mar 30, 2012 at 4:49 PM, Peter Ehlers ehl...@ucalgary.ca wrote: On 2012-03-30 15:40, ivo welch wrote: Dear R wizards: is there a clean way to assign to elements in a list? what I would like to do, in pseudo R+perl notation is f- function(a,b) list(a+b,a-b) (c,d)- f(1,2) and have c be assigned 1+2 and d be assigned 1-2. right now, I use the clunky x- f(1,2 c- x[[1]] d- x[[2]] rm(x) which seems awful. is there a nicer syntax? regards, /iaw Ivo Welch (ivo.we...@brown.edu, ivo.we...@gmail.com) I must be missing something. Why not just assign to a vector instead of a list? f- function(a,b) c(a+b,a-b) If it's imperative that f return a list, then you could use (c, d) - unlist(f(a, b)) to get vector (c, d). Peter Ehlers __ 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] scanning data into r
What have you tried? What type of file are you trying to import from? What do you want your data to look like in R? take a look at ?read.table and ?readLines On Wed, Mar 28, 2012 at 11:23 AM, joel.green joel.gr...@live.co.uk wrote: Hey I am having trouble importing data into R, my data field looks like this 21 TEST DATA 32 year:2012 33 34 5 36 I require the the number at the start of each line however the text is not needed, i am struggling to get R to import the data with out changing the file itself? how do i import the data, i have tried using comment.char= , however this didnt work, any help would be much appreciated thanks -- View this message in context: http://r.789695.n4.nabble.com/scanning-data-into-r-tp4513182p4513182.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. [[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] Why does this work? plyr within-subset normalization
To those without access to nabble, the code in reference is: relative - ddply(ranktable, .(Timestamp), function(x) data.frame(relative = x[,5]/max(x[,5]))) I may be misunderstanding your question, but: ddply splits your data.frame, ranktable, by the column Timestamp into many smaller data.frames, one for each unique Timestamp value. Those new small data.frames are sent one at a time to the function you specify. So, when you call max(x[,5]) you're taking the max of the data.frame sent to the function rather than the max of the larger ranktable data.frame. On Wed, Mar 28, 2012 at 10:18 AM, z2.0 zack.abraham...@gmail.com wrote: Working code that normalize each row's value against the subset's maximum. Does the invocation of max() somehow instruct R to 'step back' and evaluate the subset? Thanks, Zack -- View this message in context: http://r.789695.n4.nabble.com/Why-does-this-work-plyr-within-subset-normalization-tp4512989p4512989.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] how to match exact phrase using gsub (or similar function)
In most regexs the carrot( ^ ) signifies the start of a line and the
dollar sign ( $ ) signifies the end.
gsub('^S S', 'S', a)
gsub('^S S', 'S', '3421 BIGS St')
you can use logical or inside your pattern too:
gsub('^S S|S S$| S S ', 'S', a)
the S S condition is difficult.
gsub('^S S|S S$| S S ', 'S', 'foo S S bar')
gives the wrong output. as does:
gsub('^S S | S S$| S S ', ' S ', 'foo S S bar')
gsub('^S S | S S$| S S ', ' S ', a)
so you might have to catch that with a second gsub.
gsub(' S S ', ' S ', 'foo S S bar')
On Wed, Mar 28, 2012 at 12:32 PM, Markus Weisner r...@themarkus.com wrote:
trying to switch out addresses that have double directions, such as the
following example:
a = S S Main St Interstate 95
a = gsub(pattern=S S , replacement=S , a)
… the problem is that I don't want to affect instances where this might be
a correct address such as the following:
3421 BIGS St
what I want to say is switch out only if this is either of the following
situations
[beginning of char]S S
S S
S S[end of char]
Is there anyway of making gsub or a similar function make the replacements
I want? Thanks in advance for your help.
~Markus
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Re: [R] how to match exact phrase using gsub (or similar function)
wow! and here I thought I was starting to know most things about regexes...
On Wed, Mar 28, 2012 at 1:34 PM, William Dunlap wdun...@tibco.com wrote:
You can use the \ and \ patterns (backslashing the backslashes) to
mean start and end of word, respectively. E.g.,
addresses - c(S S Main St Interstate 95, 3421 BIGS St)
gsub(\\S S\\, S, addresses)
[1] S Main St Interstate 95 3421 BIGS St
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Justin Haynes
Sent: Wednesday, March 28, 2012 1:24 PM
To: Markus Weisner
Cc: r-help@r-project.org
Subject: Re: [R] how to match exact phrase using gsub (or similar function)
In most regexs the carrot( ^ ) signifies the start of a line and the
dollar sign ( $ ) signifies the end.
gsub('^S S', 'S', a)
gsub('^S S', 'S', '3421 BIGS St')
you can use logical or inside your pattern too:
gsub('^S S|S S$| S S ', 'S', a)
the S S condition is difficult.
gsub('^S S|S S$| S S ', 'S', 'foo S S bar')
gives the wrong output. as does:
gsub('^S S | S S$| S S ', ' S ', 'foo S S bar')
gsub('^S S | S S$| S S ', ' S ', a)
so you might have to catch that with a second gsub.
gsub(' S S ', ' S ', 'foo S S bar')
On Wed, Mar 28, 2012 at 12:32 PM, Markus Weisner r...@themarkus.com wrote:
trying to switch out addresses that have double directions, such as the
following example:
a = S S Main St Interstate 95
a = gsub(pattern=S S , replacement=S , a)
. the problem is that I don't want to affect instances where this might be
a correct address such as the following:
3421 BIGS St
what I want to say is switch out only if this is either of the following
situations
[beginning of char]S S
S S
S S[end of char]
Is there anyway of making gsub or a similar function make the replacements
I want? Thanks in advance for your help.
~Markus
[[alternative HTML version deleted]]
__
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Re: [R] Convert day of year back into a date format.
There may very well be a better solution, but this works.
format(strptime(dayofyear, format=%j), format=%m-%d)
On Tue, Mar 27, 2012 at 11:12 AM, Sam Albers tonightstheni...@gmail.comwrote:
Hello,
I am having trouble figuring out how to convert a Day of Year integer
back into a Date format. For example I have the following:
date -
c('2008-01-01','2008-01-02','2008-01-03','2008-01-04','2008-01-05','2008-01-06','2008-01-07',
'2008-01-08','2008-01-09','2008-01-10','2008-01-11','2008-01-12','2008-01-13','2008-01-14','2008-01-15',
'2008-01-16','2008-01-17','2008-01-18','2008-01-19','2008-01-20','2008-01-21','2008-01-22','2008-01-23')
## this is then converted into a number corresponding to the day of
the year like so:
dayofyear - strptime(date, format=%Y-%m-%d)$yday + 1
## Now my question is how do I get back to a date format (obviously
omitting the year).
## The end result is that I'd like to be able to have axis labels as
something like Month-Day or just Month
## instead of just an integers which isn't always intuitive for people
but I can't seem to figure out how to tell R
## to recognize an integer as a date.
Any suggestions?
Many thanks in advance!
Sam
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[R] Unable to specify order of a factor
Hi all:
I'm attempting to create a faceted plot with ggplot2 and I'm having issues
with a factor's order that is used to define the facet_grid().
The factor (named total.density) has three levels - 8, 16, and 32 - and I
would like them presented in that order. Running
order(levels(total.density)) yields the incorrect order of the facet grid -
2 3 1, corresponding with 16, 32, and 8.
I have attempted correcting the order with the following solutions (of
course, not run at once):
#total.density - relevel(total.density, '8')
#total.density - as.numeric(levels(total.density)[total.density])
#total.density - factor(total.density, levels = c('8','16','32'))
#total.density - factor(total.density, levels =
levels(total.density)[c(3,1,2)])
#library(gregmisc)
#total.density - reorder.factor(total.density, c('8', '16', '32'),
order = T)
The data are as follows:
total.density -
c(8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32)
I'm running R 2.14.2 with all packages up-to-date as of 21.3.2012.
Any help would be greatly appreciated.
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to specify order of a factor
Actually I've try that too, Sarah The test is to run order(levels(total.density)), which I need to be 1 2 3, not 2 3 1, and your solution still gives me 2 3 1. I also don't know how to reply to this thread with the previous message below... - Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to specify order of a factor
I think I understand, but I believe my original interest is in the order of levels(total.density), since ggplot appears to be using that to order the facets. Thus, I'm still getting three graphs, ordered (and displayed as) 16 to 32 to 8, rather than the more intuitive, 8 to 16 to 32. I'm sorry if I wasn't clear and/or I've missed your message. - Justin Montemarano Graduate Student Kent State University - Biological Sciences http://www.montegraphia.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to specify order of a factor
Ista:
Your attached code did work for me; moreover, the facets were presented in
the desired order with facet_wrap() and facet_grid(), which is what I'm
using because I have a second factor used in facet_grid().
Still, my plots with total.density as a facet are coming out in 16, 32, 8,
and I'm not seeing why. Below is my plot code -
ggplot(ag.tab[ag.tab$plant.sp == 'EC',], aes(x = days.out, y = per.remain))
+ facet_grid(total.density ~ prop.ec) +
#add point and error bar data
theme_set(theme_bw()) +
geom_point() + geom_errorbar(aes(ymin = per.remain - se, ymax =
per.remain + se), width = 3) +
#add predicted model data
geom_line(data = se.predict.data[se.predict.data$plant.sp == 'EC',],
aes(x = x.values, y = predicted.values), colour = c('red')) +
geom_line(data = dc.predict.data[dc.predict.data$plant.sp == 'EC',],
aes(x = x.values, y = predicted.values), colour = c('blue'), linetype =
c('dashed')) +
xlab('Day') + ylab('Percent Mass Remaining') + opts(panel.grid.major =
theme_blank(), panel.grid.minor = theme_blank())
Is there anything odd about it that might be producing the odd ordering
problem? FYI, avoiding subsetting ag.tab doesn't do the trick.
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
On Wed, Mar 21, 2012 at 11:42 AM, Ista Zahn istaz...@gmail.com wrote:
Hi Justin,
this gives the correct order (8, 16, 32) on my machine:
total.density -
c(8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32)
total.density - factor(total.density, levels=c(8, 16, 32), ordered=TRUE)
str(total.density)
order(levels(total.density))
dat - data.frame(td = total.density, v1 = rnorm(1:length(total.density)))
ggplot(dat, aes(x = v1)) +
geom_density() +
facet_wrap(~td)
Does it work for you? If yes, then you need to tell us what you're
doing that is different from this example. If no, please give use the
output of sessionInfo().
best,
Ista
On Wed, Mar 21, 2012 at 11:16 AM, Justin Montemarano jmont...@kent.edu
wrote:
I think I understand, but I believe my original interest is in the order
of
levels(total.density), since ggplot appears to be using that to order the
facets. Thus, I'm still getting three graphs, ordered (and displayed as)
16 to 32 to 8, rather than the more intuitive, 8 to 16 to 32. I'm sorry
if
I wasn't clear and/or I've missed your message.
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unable to specify order of a factor
Hi all:
I've got it... it appears that total.density was also defined in two
separate data frames (se.predict.data and dc.predict.data) with levels
order 16, 32, 8. Using relevel(), I moved 8 to the first position and it's
solved the plotting problem.
Ista's 'minimal' reproducible code request prompted me to discover my
error; thanks all.
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
On Wed, Mar 21, 2012 at 12:42 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
You'll also want to use dput() to send us an exact encoding of your
data when making that reproducible example: there might be something
subtle at play here that print methods won't show.
Michael
On Wed, Mar 21, 2012 at 12:28 PM, Ista Zahn istaz...@gmail.com wrote:
On Wed, Mar 21, 2012 at 12:00 PM, Justin Montemarano jmont...@kent.edu
wrote:
Ista:
Your attached code did work for me; moreover, the facets were presented
in
the desired order with facet_wrap() and facet_grid(), which is what I'm
using because I have a second factor used in facet_grid().
Still, my plots with total.density as a facet are coming out in 16, 32,
8,
and I'm not seeing why. Below is my plot code -
ggplot(ag.tab[ag.tab$plant.sp == 'EC',], aes(x = days.out, y =
per.remain)) + facet_grid(total.density ~ prop.ec) +
#add point and error bar data
theme_set(theme_bw()) +
geom_point() + geom_errorbar(aes(ymin = per.remain - se, ymax =
per.remain + se), width = 3) +
#add predicted model data
geom_line(data = se.predict.data[se.predict.data$plant.sp ==
'EC',],
aes(x = x.values, y = predicted.values), colour = c('red')) +
geom_line(data = dc.predict.data[dc.predict.data$plant.sp ==
'EC',],
aes(x = x.values, y = predicted.values), colour = c('blue'), linetype =
c('dashed')) +
xlab('Day') + ylab('Percent Mass Remaining') +
opts(panel.grid.major =
theme_blank(), panel.grid.minor = theme_blank())
Is there anything odd about it that might be producing the odd ordering
problem? FYI, avoiding subsetting ag.tab doesn't do the trick.
I don't know. Please create a minimal example that isolates the
problem. You can start with
levels(ag.tab$total.density)
ggplot(ag.tab[ag.tab$plant.sp == 'EC',], aes(x = days.out, y =
per.remain)) +
facet_grid(total.density ~ prop.ec) +
geom_point()
Best,
Ista
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
On Wed, Mar 21, 2012 at 11:42 AM, Ista Zahn istaz...@gmail.com wrote:
Hi Justin,
this gives the correct order (8, 16, 32) on my machine:
total.density -
c(8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,16,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32)
total.density - factor(total.density, levels=c(8, 16, 32),
ordered=TRUE)
str(total.density)
order(levels(total.density))
dat - data.frame(td = total.density, v1 =
rnorm(1:length(total.density)))
ggplot(dat, aes(x = v1)) +
geom_density() +
facet_wrap(~td)
Does it work for you? If yes, then you need to tell us what you're
doing that is different from this example. If no, please give use the
output of sessionInfo().
best,
Ista
On Wed, Mar 21, 2012 at 11:16 AM, Justin Montemarano
jmont...@kent.edu
wrote:
I think I understand, but I believe my original interest is in the
order
of
levels(total.density), since ggplot appears to be using that to order
the
facets. Thus, I'm still getting three graphs, ordered (and displayed
as)
16 to 32 to 8, rather than the more intuitive, 8 to 16 to 32. I'm
sorry
if
I wasn't clear and/or I've missed your message.
-
Justin Montemarano
Graduate Student
Kent State University - Biological Sciences
http://www.montegraphia.com
[[alternative HTML version deleted]]
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Re: [R] Remove a word from a character vector value XXXX
Hadley's package stringr is wonderful for all things string. library(stringr) ?str_trim and ?str_replace are what you want. (the base R equivalent of these two would be ?gsub and some regular expressions) str_trim(str_replace(d5.Region, 'Average', '')) should do the trick. hope that helps, Justin On Wed, Mar 7, 2012 at 8:03 AM, Dan Abner dan.abne...@gmail.com wrote: Hi everyone, What is the easiest way to remove the word Average and strip leading and trailing blanks from the character vector (d5.Region) below? .nrow.d5. d5.Region 1 1 Central Average 2 2 Coastal Average 3 3 East Average 4 4 Metro East Average 5 5 Metro North Average 6 6 Metro South Average 7 7 Metro West Average 8 8 Northeast Average 9 9 Northwest Average Thanks! Dan __ 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] logical to vector?
?as.numeric as.numeric(c(TRUE, FALSE)) [1] 1 0 On Wed, Mar 7, 2012 at 8:02 AM, Ed Siefker ebs15...@gmail.com wrote: I am trying to use the coXpress function from the coXpress package. This function requires numerical vectors indicating which columns are in which group. The problem is, I can only figure out how to get a logical structure, not a numerical one. In other words, coXpress wants something like: 1:3 I have something like: TRUE TRUE TRUE FALSE FALSE Can I convert one into the other easily? __ 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] GPS handling libraries or (String manipulation)
Take a look at:
http://cran.r-project.org/web/views/Spatial.html
But I've always just parsed the string...
This is from the last time I did this, its not quite the same but you
can see the similarities.
## if data is presented as 43°02'46.60059 N need to split on the °
symbol, ' and .
to.decimal - function(vec){
# convert all symbols to _
vec - gsub('°','_',vec)
vec - gsub('\'','_',vec)
vec - gsub('\','_',vec)
split - str_split(vec,'_')
deg - as.numeric(sapply(split,'[',1))
min - as.numeric(sapply(split,'[',2))
sec - as.numeric(sapply(split,'[',3))
deg - deg + min/60 + sec/3600
return(deg)
}
On Wed, Mar 7, 2012 at 8:28 AM, Alaios ala...@yahoo.com wrote:
Dear all,
I would like to ask you if R has a library that can work with different GPS
formats
For example
I have a string of this format
N50° 47.513 E006° 03.985
and I would like to convert to GPS decimal format.
that means for example converting the part N50° 47.513
to 50 + 47/60 + 513/3600.
Is it possible to do that with R?
What is the name of such a library?
I would like to thank you in advance for your help
B.R
Alex
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Re: [R] GPS handling libraries or (String manipulation)
Wow... that is WAY better! Thanks Gabor! On Wed, Mar 7, 2012 at 8:51 AM, Gabor Grothendieck ggrothendi...@gmail.com wrote: On Wed, Mar 7, 2012 at 11:28 AM, Alaios ala...@yahoo.com wrote: Dear all, I would like to ask you if R has a library that can work with different GPS formats For example I have a string of this format N50° 47.513 E006° 03.985 and I would like to convert to GPS decimal format. that means for example converting the part N50° 47.513 to 50 + 47/60 + 513/3600. Is it possible to do that with R? What is the name of such a library? Use strapply to extract the digits and convert them to numeric followed by matrix multiplication to apply the formula: library(gsubfn) x - N50° 47.513 c(1, 1/60, 1/3600) %*% strapply(x, \\d+, as.numeric, simplify = TRUE) -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] regular expression
gsub('.+; (.+);.+','\\1',x)
or if you just want the value out:
gsub('.+; Surv\\(months\\): ([0-9]+);.+','\\1',x)
You can also look at strsplit:
strsplit(x,';')
[[1]]
[1] 99-625: Cell type: S Surv(months): 21
STATUS(0=alive, 1=dead): 1
lapply(strsplit(x,';'),'[',2)
[[1]]
[1] Surv(months): 21
But i would follow David's second suggestion and just read them in with
sep=';' instead.
Justin
On Wed, Feb 29, 2012 at 11:24 AM, Fred G bayespoker...@gmail.com wrote:
Computer Friends,
with the following example lines:
[107] 98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1
[108] 99-625: Cell type: S; Surv(months): 21; STATUS(0=alive, 1=dead): 1
i want to be able to isolate the number of months of survival for each row.
is there a regular expression that can find the first instance of a ;,
delete everything in front of it-- and find the second instance of an ;
and delete everything behind it? in python there is a function line.find(),
would be grateful to hear the R equiv; or, any other better alternatives to
get the number of months of survival stored as a variable.
Much Thank You!
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Re: [R] Problem building up ggplot graph in a loop.
ggplot is looking for thisData as a column of coffs. the most
'ggplotesque' way of doing this would be:
# melt your data to a long format:
coffs.melt - melt(coffs, id.vars = 'levels')
# plot using colour aes parameter:
ggplot(coffs.melt, aes(x=levels, y=value, colour=variable)) + geom_line() +
ylab('Total Chargeoffs')
this is untested since there is no sample data!
Justin
On Thu, Feb 16, 2012 at 2:50 PM, Keith Weintraub kw1...@gmail.com wrote:
Folks,
I want to automate some graphing using ggplot.
Here is my code
graphChargeOffs2-function(coffs) {
ggplot(coffs, aes(levels))
dataNames-names(coffs)[!names(coffs) == levels]
for(i in dataNames) {
thisData-coffs[[i]]
last_plot() + geom_line(aes(y = thisData, colour = i))
}
last_plot() + ylab(Total Chargeoffs)
}
coffs is a data.frame.
I get the following error:
Error in eval(expr, envir, enclos) : object 'thisData' not found
As little as I know about environments in R I am pretty sure that the
geom_line in the loop is not able to see the thisData variable.
Any help you could provide would be appreciated. I would be surprised if
there wasn't a way to pass the data into the geom_line function without
using environments. Of course I have been wrong once or twice in the past.
:)
Note that geom_line also can't see the input variable coffs.
Thanks for any and all heo
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Re: [R] Change dataframe-structure
There is probably a more ellegant way, but: df - data.frame(p1=c(1,2,1),p2=c(3,3,2),p3=c(2,1,3),p4=c(5,6,4),p5=c(4,4,6),p6=c(6,5,5)) as.data.frame(t(apply(df,1,function(x) names(x)[match(1:6,x)]))) V1 V2 V3 V4 V5 V6 1 p1 p3 p2 p5 p4 p6 2 p3 p1 p2 p5 p6 p4 3 p1 p2 p3 p4 p6 p5 On Mon, Feb 13, 2012 at 2:07 PM, David Studer stude...@gmail.com wrote: Hello everybody, I have the following problem and have no idea how to solve it: In my dataframe I have six columns representing six societal problems (p1, p2, ..., p6). The values are ranks between 1 (worst problem) and 6 (best problem) p1 p2 p3 p4 p5 p6 1 3 2 5 4 6 2 3 1 6 4 5 1 2 3 4 6 5 but I'd like the dataframe the other way round: 123456 p1 p3 p2 p4 p4 p6 p3 p1 p2 p5 p6 p4 p1 p2 p3 p4 p6 p5 Can anyone help? Thanks! [[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] debug in a loop
You can add
if(is.na(tab[i])) browser()
or
if(is.na(tab[i])) break
see inline
On Fri, Feb 10, 2012 at 7:22 AM, ikuzar raz...@hotmail.fr wrote:
Hi,
I'd like to debug in a loop (using debug() and browser() etc but not
print()
). I'am looking for the first occurence of NA.
For instance:
tab = c(1:300)
tab[250] = NA
len = length(tab)
for (i in 1:len){
if(i != len){
if(is.na(tab[i])) browser()
tab[i] = tab[i]+tab[i+1]
}
}
I do not want to do Browse[2] n for each step ... I'd like to declare a
browser() in the loop with a condition. But how to write stop running
when you encounter NA ?
Thanks for your help
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Re: [R] Memory allocation problem (again!)
32 bit windows has a memory limit of 2GB. Upgrading to a computer thats less than 10 years old is the best path. But short of that, if you're just generating random data, why not do it in two or more pieces and combine them later? mat.1 - matrix(rnorm(5*2000),nrow=5) mat.2 - matrix(rnorm(5*2000),nrow=5) mat.3 - matrix(rnorm(5*2000),nrow=5) mat.1.sums - rowSums(mat.1) mat.2.sums - rowSums(mat.2) mat.3.sums - rowSums(mat.3) mat.sums - c(mat.1.sums,mat.2.sums,mat.3.sums) On Wed, Feb 8, 2012 at 8:37 AM, Christofer Bogaso bogaso.christo...@gmail.com wrote: Dear all, I know this problem was discussed many times in forum, however unfortunately I could not find any way out for my own problem. Here I am having Memory allocation problem while generating a lot of random number. Here is my description: rnorm(5*6000) Error: cannot allocate vector of size 2.2 Gb In addition: Warning messages: 1: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 2: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 3: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 4: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) memory.size(TRUE) [1] 15.75 rnorm(5*6000) Error: cannot allocate vector of size 2.2 Gb In addition: Warning messages: 1: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 2: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 3: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) 4: In rnorm(5 * 6000) : Reached total allocation of 1535Mb: see help(memory.size) And the Session info is here: sessionInfo() R version 2.14.0 (2011-10-31) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] graphics grDevices utils datasets grid stats methods base other attached packages: [1] ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6 zoo_1.7-6 loaded via a namespace (and not attached): [1] lattice_0.20-0 I am using Windows 7 (home version) with 4 GB of RAM (2.16GB is usable as my computer reports). So in my case, is it not possible to generate a random vector with such length? Note that generating such vector is my primary job. Later I need to do something on that vector. Those Job includes: 1. Create a matrix with 50,000 rows. 2. Get the row sum 3. then report some metrics on that sum values (min. 50,000 elements must be there). Can somebody help me with some real solution/suggesting? Thanks and regards, __ 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] Force printing of excluded axis annotations
Howdy, This should be simple, but I am finding that I can't find a simple solution. I have a plot to which I am manually adding the annotations to the y-axis with this command: axis(2, c(-4,-3,-2,-1,0,1,2,3,4,5,6,7),labels=c(-4,-3,-2,-1,0,1,2,3,4,5,6,7),cex.axis=8) The issue is that, apparently, R doesn't think that the -1 can fit, even though there is most certainly enough space. Is there a way to force R to print all the annotations I give it, regardless of proximity or to reduce the space it believes it needs? Thank you. - Fincher __ 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 need
Instead of a for loop, why not use the vectorization inherent in R?
sigmasqaured - 1
i - complex(real = 0, imaginary =1)
f - seq(0,0.5,0.1)
spectrum
-
(sigmasqaured)/(abs(1-2.7607*exp(2*pi*i*f)+3.8106*exp(4*pi*i*f)-2.6535*exp(6*pi*i*f)+0.9258*exp(8*pi*i*f))^2)
spectrum
[1] 9.632720e+00 1.411130e+03 2.947753e+00 6.479994e-02 1.295175e-02
8.042731e-03
On Tue, Feb 7, 2012 at 1:08 PM, Jaymin Shah jayminsh...@live.com wrote:
I have mad a for loop to try and output values which i have named
spectrum. However, I cannot seem to get the answers to come out as a
vector which is what i need. They come out as separate values which I am
then unable to join together. Thank you
for(f in seq(0,0.5,0.1)) {
sigmasqaured - 1
i = complex(real = 0, imaginary = 1)
spectrum -
(sigmasqaured)/(abs(1-2.7607*exp(2*pi*i*f)+3.8106*exp(4*pi*i*f)-2.6535*exp(6*pi*i*f)+0.9258*exp(8*pi*i*f))^2)
print(spectrum)
}
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Re: [R] I bet apply has a solution
How bout: apply(Data..,1, function(vec) !all(vec==vec[1])) [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE On Mon, Feb 6, 2012 at 10:34 AM, LCOG1 jr...@lcog.org wrote: Hi all For the data below, I would like to return a logical value indicating differences in the data. #Create data Data..-data.frame(a=rep(1,10),b=c(rep(1,9),2),c=c(rep(1,8),2,2)) a b c 1 1 1 1 2 1 1 1 3 1 1 1 4 1 1 1 5 1 1 1 6 1 1 1 7 1 1 1 8 1 1 1 9 1 1 2 10 1 2 2 So what I want is to return logical value telling me if all the values are the same. So the result would be a b c DidChange 1 1 1 1 FALSE 2 1 1 1 FALSE 3 1 1 1 FALSE 4 1 1 1 FALSE 5 1 1 1 FALSE 6 1 1 1 FALSE 7 1 1 1 FALSE 8 1 1 1 FALSE 9 1 1 2 TRUE 10 1 2 2 TRUE I bet apply could handle this elegantly but that family of functions is still not 100% intuitive to me. Thoughts. Thanks everyone Cheers, Josh -- View this message in context: http://r.789695.n4.nabble.com/I-bet-apply-has-a-solution-tp4362294p4362294.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. [[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] Summary.formula question
Dear all, Before my question, I wish to all of you my very best wishes for 2012. I'm using summary.formula to make table. I have something like this : s1-summary(fdh~cup5+cup6+schef+cpro1+stratify(id2),data=dat,na.include=F) the output give the marginal row named overall, but is it possible to add a marginal column ? Sincerly Justin BEM BP 1917 Yaoundé Tél (237) 76043774 [[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] Calculate the natural log of cdf between 2 intervals
Hello all, I was wondering if there is an R function to do the following: [*] log(pnorm(x)-pnorm(y)), where xy. I don't want all the area under the natural log of the normal pdf less than x, I only want the area between y and x. I am aware of the ability to specify log.p=TRUE, which gives me the log of the probability that X=x. This does not help me, because the following code: pnorm(x, log.p=TRUE)-pnorm(y,log.p=TRUE) is not the same as [*] mathematically. I cannot use [*] because some of my x's are far less than the mean, more than 10 sd. This causes me to take the log(0) which is an error. Thus, I need to stay in the log scale, since, for z less than 10 sd below the mean, log(pnorm(z)) is an error, and pnorm(z,log.p=TRUE) is stable even though theoretically they are equivalent. Thanks for your time Justin [[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] Select elements from text
how bout using read.table(... , sep= ).
That would give you a vector of single words. then
grepl(\\[[9-z]+\\],x)
will return a boolean vector
x-c('test','[bracket]','hi]','[blah','foo','[bar]')
grepl('\\[[9-z]+\\]',x)
[1] FALSE TRUE FALSE FALSE FALSE TRUE
x[grepl('\\[[9-z]+\\]',x)]
[1] [bracket] [bar]
You might need a more complex reg-ex to catch them all incase of
([citation]) instances for example.
Justin
On Tue, Jan 24, 2012 at 6:52 AM, mdvaan mathijsdev...@gmail.com wrote:
Hi,
I have a series of MS word files and each file contains plain text. From
these texts I would like to extract only those elements (read: words) that
are between square brackets. Example of a text:
Most fundamentally, it has led to an effort to clarify the organizational
form concept. According to them [see also Smith, Jones and Carroll 2002],
categories emerge as audience members recognize dissimilarities among
groups
of consumers and label them as members of a common set [Nicol 2000].
Now I would like to get the following selection:
see also Smith, Jones and Carroll 2002
Nicol 2000
Any ideas on how to do this? What would be the best way to import the text
in R? The entire text as an element in a dataframe? Thank you very much!
Best,
Mathijs
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Re: [R] drop columns whose rows are all 0
dataset-data.frame(a=1:10,b=c(0,0,0,1,0,0,0,0,1,0),c=rep(0,10))
apply(dataset,2,function(x) all(x==0))
a b c
FALSE FALSE TRUE
dataset[,!apply(dataset,2,function(x) all(x==0))]
a b
1 1 0
2 2 0
3 3 0
4 4 1
5 5 0
6 6 0
7 7 0
8 8 0
9 9 1
10 10 0
On Tue, Jan 24, 2012 at 8:14 AM, Francisco franciscororol...@google.comwrote:
Hello,
I have a dataset with 40 variables, some of them are always 0 (each row).
I would like to make a subset containing only the columns which values are
not all 0, but I don't know how to do it.
I tried:
for(cut_column in 1:40) {
if(sum(dataset[,cut_column])!=**0) {
columns_useful-c(columns_**
useful,dataset[cut_column])
}
}
sorted_dataset-subset(**dataset, select=columns_useful)
But it doesn't work.
Thank you
Francisco
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Re: [R] How can I access information stored after I run a command in R?
?str tells you about the object. str(MAX3(a,'asy',1)) from that you can see the names of the various parts including p.value. foo - MAX3(a,'asy',1)$p.value On Mon, Jan 23, 2012 at 9:32 AM, Tiago V. Pereira tiago.pere...@mbe.bio.brwrote: Dear all, Supposed I run the following command: ### #install.packages(Rassoc, dependencies=TRUE) library(Rassoc) ca=c(139,249,112) co=c(136,244,120) a=rbind(ca,co) MAX3(a,asy,1) ## I get: The MAX3 test using the asy method data: a statistic = 0.5993, p-value = 0.7933 How can one save the result 0.7933 into a file? say: foo - 0.7933 write.table(foo, file =/home/foo.txt, sep = , row.names=FALSE,col.names=TRUE, quote=FALSE, qmethod = double) However, instead of typing the value above, I would like to replace it by the macro (scalar, local) that has the accurate p-value. thanks in advance for your help. Tiago __ 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] colored outliers
TOC_NI-read.csv2(C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv, sep=;, dec=,, encoding=UTF-8) circ-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,] plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3) ## this line is coloring all points because you're using TOC_NI still points(NI~TOC,data=circ,col='red',pch=1,size=3) ## now we're only plotting the four points in circ. sorry for the confusion. however, in the future please provide a reproducible data set along with your question so we can more easily help. Justin On Fri, Jan 20, 2012 at 5:49 AM, Geophagus falk.hilli...@twain-systems.comwrote: Dear Petr and Justin, my problem ist, that I only want to have the 4 highest values for Ni as a red point or with a red circle. The other points should not be modificated. In your proposals always all points get a red circle or a red point not only the 4 highest Ni values! I hope you could understand me! Thanks for your help! GeO -- View this message in context: http://r.789695.n4.nabble.com/colored-outliers-tp4282207p4313278.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. [[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] Stacked barchart in ggplot (or other library)
to use ggplot:
dat-data.frame(num=1:3,usage=c(4,2,5),cap=c(10,20,10),diff=c(6,18,5))
dat.melt-melt(dat,id.var=c('num','cap'))
ggplot(dat.melt)+geom_bar(aes(x=num,y=value,fill=variable),stat='identity')
On Fri, Jan 20, 2012 at 12:30 PM, Jean V Adams jvad...@usgs.gov wrote:
Bart6114 wrote on 01/20/2012 08:54:39 AM:
Hey,
I want to create a stacked barchart in R for the following dataset
(http://pastebin.com/pyHUNgr2):
# usage capacity diff
1 4 10 6
2 2 20 18
3 5 10 5
The stacked barchart should, in one plot show each line of the dataset
as a
stacked bar using data from 'usage' and 'diff' to create the stacked
bar.
I can't find a good example of how to do this on the ggplot2 site.
Thanks in advance!
See the help on barplot:
?barplot
For example:
df - data.frame(usage=c(4, 2, 5), capacity=c(10, 20, 10), diff=c(6, 18,
5))
barplot(t(as.matrix(df[, 1:2])))
Jean
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Re: [R] Establishing groups using something other than ifelse()
how bout
levels(df$z)[grep('A',levels(df$z))] - 'A'
levels(df$z)[grep('B',levels(df$z))] - 'B'
levels(df$z)[grep('C',levels(df$z))] - 'C'
does that do what you're wanting?
On Thu, Jan 19, 2012 at 3:05 PM, Sam Albers tonightstheni...@gmail.comwrote:
Hello all,
This is one of those Is there a better way to do this questions. Say
I have a dataframe (df) with a grouping variable (z). This is my base
data. Now I know that there is a higher order level of grouping that
exist for my group variable. So what I want to do is create a new
column that express that higher order level of grouping based on
values in the sub-group (z in this case). In the past I have used
ifelse() but this tends to get fairly redundant and messy with a large
amount of sub-groupings (z). I've created a sample dataset below. Can
anyone recommend a better way of achieving what I am currently
achieving with ifelse()? A long series of ifelse statements makes me
think that there is something better for this.
## Dataframe creation
df - data.frame(x=runif(36, 0, 120),
y=runif(36, 0, 120),
z=factor(c(A1,A1,A2,A2,B1,B1,B2,B2,C1,C,C2,C2))
)
## Current method is grouping
df$Big.Group - with(df, ifelse(df$z==A1,A, ifelse(df$z==A2,A,
ifelse(df$z==B1, B, ifelse(df$z==B2, B, C)
So any suggestions? Thanks in advance!
Sam
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[R] png output on a server?
I've got R running on a gentoo server that doesn't have X11 installed. Its a custom build to keep those dependencies at bay! However, some of my scripts use the base png() function and ggplot2. But, png uses X11. A google search suggests using the Cairo package, which works... but changes the fonts (specifically the size of the font). Adjusting the pointsize doesn't seem to have much effect. Aside from tuning the CairoPNG function to make my graphs look right, has anyone found a good way to avoid the X11 dependency but still use the base png function? If anyone has experience with CairoPNG and making it look like the base png function, id love to hear what you've learned! Thanks, Justin capabilities() jpeg png tifftcltk X11 aqua http/ftp sockets libxml fifo clediticonv NLS profmem FALSEFALSEFALSEFALSEFALSEFALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUEFALSE cairo FALSE sessionInfo() R version 2.14.1 (2011-12-22) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8LC_MESSAGES=en_US.UTF-8LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets grid methods base other attached packages: [1] Cairo_1.5-1 ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.7.1 loaded via a namespace (and not attached): [1] tools_2.14.1 [[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] Points inside a polygon
On Wed 11 Jan 2012 08:28:03 PM PST, Hasan Diwan wrote: I have a list of bounds for a series of polygons. I do understand the formula to determine whether point i is within polygon X (X[x1] i[x] X[x2] i[x] X[y1] i[y] X[y2] i[y]), and I can apply this throughout the dataset. However, this naive algorithm doesn't scale very well. The data set contains 10,000 points consisting of (n,e) pairs where I'm interested in which are inside polygons denoted by vertices (V[x1]/V[y1],V[x2],V[y2]). Is there a shortcut to accomplish this goal? Many thanks! -- H Check out the splancs package. particularly the inout function. Justin __ 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] relative frequency plot using ggplot or other function
On Thu 12 Jan 2012 09:02:27 AM PST, Mary Kindall wrote: Hi I have a data frame in the following form. There are two groups and for each 'width' relative frequency for group1 and group2 is given. How to plot this in R using ggplot or other package. Width relativeFrequency1 relativeFrequency2 1 100 0.0006388783 0.02265428 2 200 0.0022677303 0.02948625 3 300 0.0061182673 0.01739936 4 400 0.0152237225 0.02569902 5 500 0.0300215262 0.03639880 6 600 0.0597610250 0.07717765 Thanks not sure exactly what you're looking for but... dat-data.frame(width=1:6*100,rel1=runif(6), rel2=runif(6)) dat.melt-melt(dat,id.var='width') ggplot(dat.melt,aes(x=factor(width),y=value,fill=variable))+geom_bar(stat='identity',position='dodge') __ 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] relative frequency plot using ggplot or other function
ggplot(dat.melt,aes(x=width,y=value,fill=variable,colour=variable))+geom_density(stat='identity',alpha=0.5) the fill and colour variables can be removed if you want. or ggplot(dat.melt,aes(x=width,y=value,fill=variable))+geom_density(stat='identity',alpha=0.5)+facet_wrap(~variable,ncol=1) same with this version. On Thu, Jan 12, 2012 at 9:35 AM, Mary Kindall mary.kind...@gmail.comwrote: Hi this is exactly what i am looking for but I do not like to draw as histogram instead I want two separate plot for this data. Something like the ones shown in the following link. Please disregard the legends of the following fig. http://had.co.nz/ggplot2/graphics/55078149a733dd1a0b42a57faf847036.png http://had.co.nz/ggplot2/graphics/90983232ced45a93d9fbbe40afffd69a.png Thanks On Thu, Jan 12, 2012 at 12:13 PM, Justin Haynes jto...@gmail.com wrote: On Thu 12 Jan 2012 09:02:27 AM PST, Mary Kindall wrote: Hi I have a data frame in the following form. There are two groups and for each 'width' relative frequency for group1 and group2 is given. How to plot this in R using ggplot or other package. Width relativeFrequency1 relativeFrequency2 1 100 0.0006388783 0.02265428 2 200 0.0022677303 0.02948625 3 300 0.0061182673 0.01739936 4 400 0.0152237225 0.02569902 5 500 0.0300215262 0.03639880 6 600 0.0597610250 0.07717765 Thanks not sure exactly what you're looking for but... dat-data.frame(width=1:6*100,**rel1=runif(6), rel2=runif(6)) dat.melt-melt(dat,id.var='**width') ggplot(dat.melt,aes(x=factor(**width),y=value,fill=variable))** +geom_bar(stat='identity',**position='dodge') -- - Mary Kindall Yorktown Heights, NY USA [[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] Add color to Boxplot by value
how bout:
dat-data.frame(val=rnorm(100,12,10),x=letters[1:4])
col.val-ddply(dat,.(x),summarise,mean(val))
col.val$breaks-cut(col.val$..1,c(0,9,15,Inf))
dat.merge-merge(dat,col.val)
ggplot(dat.merge,aes(x=x,y=val,colour=breaks))+geom_boxplot()+scale_color_manual(values=c('green','yellow','red'))
On Thu, Jan 12, 2012 at 7:45 AM, KWyshak kwys...@illumina.com wrote:
I have a boxplot of Production run rates per 10 minute intervals and I
would
like to color code them by the average (i.e. 15ppm = green, 9ppm = red,
everything else yellow).
Is there a way to do this?
http://r.789695.n4.nabble.com/file/n4289381/RunRateBoxWhisker.png
--
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Sent from the R help mailing list archive at Nabble.com.
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Re: [R] colored outliers
# find top 4 points circ - TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,]TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,] # add them to your plot! plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3) Justin On Tue, Jan 10, 2012 at 7:11 AM, Geophagus falk.hilli...@twain-systems.comwrote: Hi @ all, I have question how to mark significant outliers in R. This is my very simple script to plot a regression: TOC_NI-read.csv2(C:/Users/XYZ/Desktop/Master/Daten/Statistik/TOC-NI.csv, sep=;, dec=,, encoding=UTF-8) plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) summary(lm(NI~TOC,data=TOC_NI)) The result is the following pic: http://r.789695.n4.nabble.com/file/n4282207/nickel_TOC_5f.png nickel_TOC_5f.png Now I want to make small red circles around the four highest values of Ni. Does anyone has an idea how to do that? Thanks a lot! Best Regards Geophagus -- View this message in context: http://r.789695.n4.nabble.com/colored-outliers-tp4282207p4282207.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. [[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] colored outliers
woops! see inline. Hope that helps, and enjoy R. Justin On Tue, Jan 10, 2012 at 8:40 AM, Geophagus falk.hilli...@twain-systems.comwrote: Hi Justin, thanks a lot for your quick answer. If I use your code, all points become red. How do you include the sorted and separated four values into the points argument? The variable in your script is called circ but this is not fronted up anymore. Here the script again: TOC_NI-read.csv2(C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv, sep=;, dec=,, encoding=UTF-8) this line just needs trimming. not sure how i missed that on my copy... anyway, order puts the data.frame in order of the given vector, default behavior sorts in ascending order unless you specify decreasing=TRUE. circ-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,] and it should work plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3) Thanks a lot for your help! GeO -- View this message in context: http://r.789695.n4.nabble.com/colored-outliers-tp4282207p4282481.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. [[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] match matrices of different lengths
see ?merge merge(xx,aa,by.x='x',by.y='a') x y b 1 2.00112e+11 1.0 1.2 2 2.00112e+11 1.1 1.9 making the two matricies time series does not mean that R knows that the first column is a datetime. and depending on your desired result, that may not be important. hope that helps, Justin On Thu, Jan 5, 2012 at 5:51 AM, Thijs vanden Bergh bergh.thijsvan...@gmail.com wrote: was trying to match different matrices of different lengths with in the first collumn date and time info (yearmonthdayhourminute). the routine needs to return NA´s where data of either of the matrices is non existent. have been trying the following: x - c(200112030003, 200112030004, 200112030005, 200112030006) y - c(0.1, 1, 1.1, 1.5) a - c(200112030004, 200112030005, 200112030007, 200112030008, 200112030009) b - c(1.2, 1.9, 2.0, 2.5, 2.1) xx - cbind(x, y) aa - cbind(a, b) xxnew - ts(xx) aanew - ts(aa) cc - ts.union(xxnew, aanew) cc this does however not give the wished for result as it simply cbinds the two matrices and filles up empty spots that are created due to the one matrix being shorter then the other at the bottom end of the shortest matrix. i realy want the routine to match matrix xx and aa to time in the first collumn of both matrices. any help towards this end would be much appreciated, th. __ 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] ggplot2 - tricky problem
how bout:
dat-data.frame(id=1:4,city=c('berlin','munich'),likeability=c(5,4,6,5),uniqueness=c(3,4,4,4))
ggplot(ddply(melt(dat,
id.vars=c('id','city')),
.(variable,city),
summarise,
value=mean(value)),
aes(x=factor(city),y=value)) +
geom_point() +
facet_wrap(~variable)
the line drawing is a bit more tricky... Since the x values are factors
rather than continuous, fitting a line to them is kind of nonsense. It
matters which order they are in for example. If instead you want to plot
something like:
ggplot(dat,aes(x=likeability,y=uniqueness,colour=city))+geom_point()+geom_smooth(aes(group=city),method='lm')
You could draw fit lines that make a bit more sense. Forgive me if I'm
over simplifying your problem!
Justin
On Thu, Jan 5, 2012 at 7:46 AM, Mario Giesel rr.gie...@yahoo.de wrote:
Hello, R friends,
I've been struggling quite a bit with ggplot2.
Having worked through Hadleys book twice I still wonder how to solve this
task.
1. Short example Dataframe:
idcityLikeabilityUniqueness
1Berlin53
2Munich44
3Berlin64
4Munich54
2. Task:
a) Facetting plots for each attitude (1 plot for likeability and
uniqueness each, horizontally on one page)
b) Showing Berlin and Munich together on x axis
c) Showing the means of Berlin and Munich on y axis (means of cities in
likeability on first plot, means of cities in uniqueness on second plot)
d) Drawing a line through mean points on each plot
Hope I could explain it understandably. Any help is appreciated!
Thanks a lot,
Mario
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Re: [R] [newbie] stack operations, or functions with side effects (or both)
do s[1] and s[-1] do what you're looking for?
those are just to display... if you want to change s, you need to reassign
it or fiddle with namespacing. however, I'd say it is better to write R
code as though data structures are immutable until you explicitly re-assign
them rather than trying to deal with side effects and state...
pop - function(vec){
+ print(vec[1])
+ print(vec[-1])
+ return(vec[-1])
+}
s - 1:5
s - pop(s)
[1] 1
[1] 2 3 4 5
s
[1] 2 3 4 5
On Wed, Jan 4, 2012 at 1:22 PM, Tom Roche tom_ro...@pobox.com wrote:
summary: Specifically, how does one do stack/FIFO operations in R?
Generally, how does one code functions with side effects in R?
details:
I have been a coder for years, mostly using C-like semantics (e.g.,
Java). I am now trying to become a scientist, and to use R, but I don't
yet have the sense of good R and R idiom (i.e., expressions that are
to R what (e.g.) the Schwartzian transform is to Perl).
I have a data-assimilation problem for which I see a solution that
wants a stack--or, really, just a pop(...) such that
* s - c(1:5)
* print(s)
[1] 1 2 3 4 5
* pop(s)
[1] 1
* print(s)
[1] 2 3 4 5
but in fact I get
pop(s)
Error: could not find function pop
and Rseek'ing finds me nothing. When I try to write pop(...) I get
pop1 - function(vector_arg) {
+ length(vector_arg) - lv
+ vector_arg[1] - ret
+ vector_arg - vector_arg[2:lv]
+ ret
+ }
pop1(s)
[1] 1
print(s)
[1] 1 2 3 4 5
i.e., no side effect on the argument
pop2 - function(vector_arg) {
+ length(vector_arg) - lv
+ vector_arg[1] - ret
+ assign(vector_arg, vector_arg[2:lv])
+ return(ret)
+ }
pop2(s)
[1] 1
print(s)
[1] 1 2 3 4 5
ditto :-( What am I missing?
* Is there already a stack API for R (which I would expect)? If so, where?
* How to cause the desired side effect to the argument in the code above?
TIA, Tom Roche tom_ro...@pobox.com
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Re: [R] Combining characters
apply(expand.grid(x, y, z, stringsAsFactors=F), 1, paste, collapse=' ') On Wed, Jan 4, 2012 at 8:32 AM, jeremy jeremynamer...@gmail.com wrote: Hi all, I'm trying to combine exhaustively several character arrays in R like: x=c(one,two,three) y=c(yellow,blue,green) z=c(apple,cheese) in order to get concatenation of x[1] y[1] z[1] (one yellow apple) x[1] y[1] z[2] (one yellow cheese) x[1] y[2] z[1](one blue apple) ... x[length(x)] y[length(y)] z[length(z)] (three green cheese) Anyone has a solution ? Thank in advance -- View this message in context: http://r.789695.n4.nabble.com/Combining-characters-tp4261888p4261888.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. [[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] a quick question about rbinom
homework or not, ?rbinom should be plenty. On Wed, Jan 4, 2012 at 1:38 PM, lynn.tsai vernal@gmail.com wrote: Hello, I have the following code using rbinom, but I don't understand what *+1* means in the code. Could someone help? Thanks so much, X1-c(A,B)[rbinom(n,1,0.6)+1] X2-c(C,D)[rbinom(n,1,0.1)+1] -- View this message in context: http://r.789695.n4.nabble.com/a-quick-question-about-rbinom-tp4262977p4262977.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. [[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] Applyiing mode() or class() to each column of a data.frame XXXX
there is also colwise in the plyr package.
library(plyr)
colwise(class)(data6)
v13 v14 v15 f4 v16
1 integer numeric character factor logical
Justin
On Thu, Dec 29, 2011 at 4:47 PM, Jean V Adams jvad...@usgs.gov wrote:
Dan Abner wrote on 12/29/2011 06:13:11 PM:
Hi everyone,
I am attempting to use the apply() function to obtain the mode and class
of
each column in a data frame, however, I am encountering unexpected
results.
I have the following example data:
v13-1:6
v14-c(1,2,3,3,NA,1)
v15-c(Good,Bad,NA,Good,Bad,Bad)
f4-factor(rep(c(Blue,Red,Green),2))
v16-c(F,T,F,F,T,F)
data6-data.frame(v13,v14,v15,f4,v16)
data6
Here is my function definition:
contents-function(x){
output-data.frame(Varnum=1:ncol(x),
Name=names(x),
Mode=apply(x,2,mode),
Class=apply(x,2,class))
print(output)
}
Use sapply() instead of apply(). In the help file for apply() it says:
If X is not an array but an object of a class with a non-null dim value
(such as a data frame), apply attempts to coerce it to an array via
as.matrix if it is two-dimensional (e.g., a data frame) or via as.array.
This coercion to a matrix might be causing the unexpected result. sapply()
and lapply() are designed specifically for lists (which a data frame is).
I also simplified the function a bit ...
contents-function(x){
data.frame(Varnum=1:ncol(x), Name=names(x),
Mode=sapply(x,mode), Class=sapply(x,class))
}
Jean
When I call the function, I obtain the following:
contents(data6)
Varnum Name Mode Class
v13 1 v13 character character
v14 2 v14 character character
v15 3 v15 character character
f4 4 f4 character character
v16 5 v16 character character
=
Any help is appreciated.
Thank you,
Dan
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Re: [R] Help with code
the short answer... which is a guess cause you didn't provide a
reproducible example... is:
your column (i think its called t1d_ptype[1:25]) is a factor and using
factors is dangerous at best.
you can check with ?str.
see ?factor for how to convert back to strings and see if your code works.
to answer your second question, yes I'm sure there is a better simple way
to do this, but i can't follow what you're doing... for example, I don't
know what c1 is...
but, the place I would look is at the plyr package. its excellent at
splitting and reordering data.
and one final note, you should avoid naming things with pre-existing R
functions (e.g. data).
Justin
On Tue, Dec 20, 2011 at 11:14 AM, 1Rnwb sbpuro...@gmail.com wrote:
hello gurus,
i have a data frame like this
HTN HTN_FDR Dyslipidemia CAD t1d_ptype[1:25]
1Y YY T1D
2 T1D
3 Ctrl_FDR
4 T1D
5Y Ctrl
6 Ctrl
7 Ctrl_FDR
8 T1D
9YY T1D
10 T1D
11 Ctrl_FDR
12 YY T1D
13 Y YY T1D
14 T1D
15 Ctrl
16 Ctrl
17 Ctrl_FDR
18 T1D
19 T1D
20 Y T1D
21 Ctrl_FDR
22 Ctrl_FDR
23 Ctrl
24 Ctrl
25 T1D
i am converting it to define the groups more uniformly using this code:
for( i in 1:dim(c1)[1])
{
num_comp-0
for (j in 1:dim(c1)[2])
if (c1[i,j]==2) num_comp=num_comp+1 #Y=2
for (j in 1:dim(c1)[2])
if(num_comp0)
{
if (data$t1d_ptype[i] == T1D c1[i ,j] == 2) c2[i,j]-T1D_w
if (data$t1d_ptype[i] == T1D c1[i, j] == 1) c2[i,j]-T1D_oc
if(substr(data$t1d_ptype[i],1,4) == Ctrl c1[i,j] == 2)
c2[i,j]-Ctrl_w
if (substr(data$t1d_ptype[i],1,4) == Ctrl c1[i,j] == 1)
c2[i,j]-Ctrl_oc
}
else
{
if(data$t1d_ptype[i] == T1D) c2[i,j]-T1D_noc
if(substr(data$t1d_ptype[i],1,4) == Ctrl) c2[i,j]-Ctrl_noc
}
}
it is giving me error
In `[-.factor`(`*tmp*`, iseq, value = structure(c(NA, ... :
invalid factor level, NAs generated
Also it there a simple way to do this.
Thanks
Sharad
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Re: [R] Help with code
Fair enough and good point. How about, dangerous when used unknowingly!
On Tue, Dec 20, 2011 at 1:01 PM, William Dunlap wdun...@tibco.com wrote:
Re
your column (i think its called t1d_ptype[1:25]) is a factor and using
factors is dangerous at best.
This depends on how you want to define dangerous. If t1d_ptype ought
take values from a certain set of strings then making it a factor gives
you some safety, since it warns you when you go outside of that set and
try to give it an illegal value. E.g.,
sex - factor(c(M,F,F), levels=c(F, M))
sex[2] - no
Warning message:
In `[-.factor`(`*tmp*`, 2, value = no) :
invalid factor level, NAs generated
It does take more work to set up, since you need to enumerate the set
of good strings. That is tedium, not danger.
If t1d_ptype might take any value, then make it a character vector.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Justin Haynes
Sent: Tuesday, December 20, 2011 11:54 AM
To: 1Rnwb
Cc: r-help@r-project.org
Subject: Re: [R] Help with code
the short answer... which is a guess cause you didn't provide a
reproducible example... is:
your column (i think its called t1d_ptype[1:25]) is a factor and using
factors is dangerous at best.
you can check with ?str.
see ?factor for how to convert back to strings and see if your code
works.
to answer your second question, yes I'm sure there is a better simple way
to do this, but i can't follow what you're doing... for example, I don't
know what c1 is...
but, the place I would look is at the plyr package. its excellent at
splitting and reordering data.
and one final note, you should avoid naming things with pre-existing R
functions (e.g. data).
Justin
On Tue, Dec 20, 2011 at 11:14 AM, 1Rnwb sbpuro...@gmail.com wrote:
hello gurus,
i have a data frame like this
HTN HTN_FDR Dyslipidemia CAD t1d_ptype[1:25]
1Y YY T1D
2 T1D
3 Ctrl_FDR
4 T1D
5Y Ctrl
6 Ctrl
7 Ctrl_FDR
8 T1D
9YY T1D
10 T1D
11 Ctrl_FDR
12 YY T1D
13 Y YY T1D
14 T1D
15 Ctrl
16 Ctrl
17 Ctrl_FDR
18 T1D
19 T1D
20 Y T1D
21 Ctrl_FDR
22 Ctrl_FDR
23 Ctrl
24 Ctrl
25 T1D
i am converting it to define the groups more uniformly using this code:
for( i in 1:dim(c1)[1])
{
num_comp-0
for (j in 1:dim(c1)[2])
if (c1[i,j]==2) num_comp=num_comp+1 #Y=2
for (j in 1:dim(c1)[2])
if(num_comp0)
{
if (data$t1d_ptype[i] == T1D c1[i ,j] == 2)
c2[i,j]-T1D_w
if (data$t1d_ptype[i] == T1D c1[i, j] == 1)
c2[i,j]-T1D_oc
if(substr(data$t1d_ptype[i],1,4) == Ctrl c1[i,j] == 2)
c2[i,j]-Ctrl_w
if (substr(data$t1d_ptype[i],1,4) == Ctrl c1[i,j] == 1)
c2[i,j]-Ctrl_oc
}
else
{
if(data$t1d_ptype[i] == T1D) c2[i,j]-T1D_noc
if(substr(data$t1d_ptype[i],1,4) == Ctrl)
c2[i,j]-Ctrl_noc
}
}
it is giving me error
In `[-.factor`(`*tmp*`, iseq, value = structure(c(NA, ... :
invalid factor level, NAs generated
Also it there a simple way to do this.
Thanks
Sharad
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Re: [R] how to manually enter an double quote as data feed?
\ is how its displayed on the screen. however, if you write your object to a csv it will be correct. r cant display as it is so it is escaping the second double quote for you however, ' (double quote single quote double quote) does display correctly as well as save correctly. If that doesn't answer your question, some more back story on what you're trying to do would help. Justin On Tue, Dec 13, 2011 at 2:03 PM, bonnieyuan bby2...@columbia.edu wrote: I'm doing a text mining project where I have to manually enter a double quote as an element inside a vector. I tried char[10]=''#where i enclosed the double quote in a pair of single quotes. But the result is [1] \. Somehow a back slash is added automatically. I also tried to enclose the double quote in a pair of double quotes. That didn't work either. I'm using Mac and latest release of R. Thank you! Bonnie Yuan -- View this message in context: http://r.789695.n4.nabble.com/how-to-manually-enter-an-double-quote-as-data-feed-tp4192283p4192283.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. [[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] using sample
Emma,
If you haven't spent much time on the r-help forums, please do read the
posting guide.
You need to provide reproducible examples for us to help you.
We don't know anything about your data...
what is event.details, (if you can't provide the data often ?str will do)
since I don't know what event.details is, I can't figure out waht the line:
obs = (1:133429)[event.details[,2] == i]
is supposed to do.
But if I had to guess... ?sample says it expects the first argument as a
vector. I assume obs is not a vector but a larger structure?
Feel free to post more info about your data (see ?str and ?dput) or if you
can generate made up data that replicates your problem that works too.
Justin
On Wed, Dec 7, 2011 at 9:16 AM, bevare emma.ra...@jbaconsulting.co.ukwrote:
Hi,
Can anyone help sort out the problem with the following script - I am a R
newbie and I am self taught.
obs.all = c()
for(i in 1:386){
if (n.sim[i]0){
obs = (1:133429)[event.details[,2] == i]
obs.all = c(obs.all, sample(obs[obs n.sim[i]], size = n.sim[i],
replace=T))
}
Basically, in the sample bit, I only want to get obs.all if the value of
obs
is less than the value of n.sim[i]. I get the error message
Error in sample(obs[obs n.sim[i]], size = n.sim[i], replace = T) :
invalid first argument
length(n.sim) is 386
Thanks in advance for your suggestions
Emma
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[R] adding hyperlinked text to pdf plot
Howdy, I have read that if you put a URL in the text of a plot being saved into pdf, the result is a functional hyperlink. I am interested in having text in a plot that is linked to a URL, but I would like the text to be something other than the URL. Is this possible? Thank you. - Fincher __ 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] adding hyperlinked text to pdf plot
For example, say I am plotting some data that is genomic and therefore maps to a specific locus on the human genome, like a gene. I was hoping to have the title of the plot display the gene name, but have it be a link so that clicking on it would take you to those coordinates on a public browser, like USCS's genome browser. So basically, I was hoping to have text in a plot generated by R function as a normal html-style link. - Fincher On Mon, Dec 5, 2011 at 14:09, Yihui Xie x...@yihui.name wrote: It seems I missed the context of this post -- who is you, and what is something other than the URL? I feel the tikzDevice package should be an option for the task. Regards, Yihui -- Yihui Xie xieyi...@gmail.com Phone: 515-294-2465 Web: http://yihui.name Department of Statistics, Iowa State University 2215 Snedecor Hall, Ames, IA On Mon, Dec 5, 2011 at 12:39 PM, Justin Fincher finc...@cs.fsu.edu wrote: Howdy, I have read that if you put a URL in the text of a plot being saved into pdf, the result is a functional hyperlink. I am interested in having text in a plot that is linked to a URL, but I would like the text to be something other than the URL. Is this possible? Thank you. - Fincher __ 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. -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean. __ 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] hour in x-axis
without knowing much about your data or the base plotting... I'd use the library ggplot2. First, you'll need to format your dates to POSIXct AggData$time - as.POSIXct(AggData$time,format='%H:%M') Then plotting is trivial. ggplot(AggData,aes(x=time,y=value))+geom_points() or +geom_line() if you'd rather. Hope that helps, Justin On Tue, Nov 29, 2011 at 10:07 AM, threshold r.kozar...@gmail.com wrote: Dear R useres, got the following problem. Given the AggData (listed below) I need to plot AggData[,2] vs time (AggData[,1]) for chosen 'rows'. Ive done already: plot(AggData[rows,2], xaxt='n') axis(1,at=seq(1,length(rows),1),sub(,, AggData[rows,1])) which works, but I need to list only chosen data points, say full hours or every 60th point, something like: axis(1,at=seq(1,seq(1,length(rows),60)),sub(, , AggData[day.rows[seq(1,length(rows),60)],2])) but does not work. Could be nice if time on the x-axis is in H:m format (no seconds). In the original data time bout is 1 minute, e.g. 17:19:35, 17:20:35, 17:21:35 . Taken every 100th for brevity yields (AggData[seq(1,length(rows),100),c(2,7)]) time value 117:19:3580.68327 101 18:59:3580.97230 201 20:39:3578.30810 301 22:19:3580.41558 401 23:59:3577.01051 501 01:39:3577.19687 601 03:19:3578.20762 701 04:59:3577.13315 801 06:39:3576.29110 901 08:19:3575.32090 1001 09:59:3585.32890 1101 11:39:3579.86978 1201 13:19:3583.32418 1301 14:59:3578.26018 1401 16:39:3579.06434 Thanks in advance. Best, robert -- View this message in context: http://r.789695.n4.nabble.com/hour-in-x-axis-tp4120142p4120142.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. [[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] aggregate syntax for grouped column means
look at just your data that is in that first id category and I bet you can figure it out! myData[myData$id=='0m11',] var1 var2 id 10 30.79 32.15 0m11 11 30.79 32.39 0m11 12 30.94NA 0m11 aggregate performs the na.rm step on the entire row thus, a mean of 30.79. data.table and plyr perform the na.rm on each column. Justin On Tue, Nov 29, 2011 at 12:21 PM, Juliet Hannah juliet.han...@gmail.comwrote: I am calculating the mean of each column grouped by the variable 'id'. I do this using aggregate, data.table, and plyr. My aggregate results do not match the other two, and I am trying to figure out what is incorrect with my syntax. Any suggestions? Thanks. Here is the data. myData - structure(list(var1 = c(31.59, 32.21, 31.78, 31.34, 31.61, 31.61, 30.59, 30.84, 30.98, 30.79, 30.79, 30.94, 31.08, 31.27, 31.11, 30.42, 30.37, 30.29, 30.06, 30.3, 30.43, 30.61, 30.64, 30.75, 30.39, 30.1, 30.25, 31.55, 31.96, 31.87, 30.29, 30.15, 30.37, 29.59, 29.52, 28.96, 29.69, 29.58, 29.52, 30.21, 30.3, 30.25, 30.23, 30.29, 30.39), var2 = c(33.78, 33.25, NA, 32.05, 32.59, NA, 32.24, NA, NA, 32.15, 32.39, NA, 32.4, 31.6, NA, 30.5, 30.66, NA, 30.6, 29.95, NA, 31.24, 30.73, NA, 30.51, 30.43, 31.17, 31.44, 31.17, 31.18, 31.01, 30.98, 31.25, 30.44, 30.47, NA, 30.47, 30.56, NA, 30.6, 30.57, NA, 31, 30.8, NA), id = c(0m4, 0m4, 0m4, 0m5, 0m5, 0m5, 0m6, 0m6, 0m6, 0m11, 0m11, 0m11, 0m12, 0m12, 0m12, 205m1, 205m1, 205m1, 205m4, 205m4, 205m4, 205m5, 205m5, 205m5, 205m6, 205m6, 205m6, 205m7, 205m7, 205m7, 600m1, 600m1, 600m1, 600m3, 600m3, 600m3, 600m4, 600m4, 600m4, 600m5, 600m5, 600m5, 600m7, 600m7, 600m7)), .Names = c(var1, var2, id), row.names = c(NA, -45L), class = data.frame) head(myData) var1 var2 id 1 31.59 33.78 0m4 2 32.21 33.25 0m4 3 31.78NA 0m4 4 31.34 32.05 0m5 5 31.61 32.59 0m5 6 31.61NA 0m5 results1 - aggregate(. ~ id ,data=myData,FUN=mean,na.rm=T) head(results1,1) #id var1 var2 # 1 0m11 30.79 32.27 library(data.table) mydt - data.table(myData) setkey(mydt,id) results2 - mydt[,lapply(.SD,mean,na.rm=TRUE),by=id] head(results2,1) # id var1 var2 # [1,] 0m11 30.84 32.27 library(plyr) results3 - ddply(myData,.(id),colwise(mean),na.rm=TRUE) head(results3,1) #id var1 var2 # 1 0m11 30.84 32.27 sessionInfo() R version 2.14.0 (2011-10-31) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] plyr_1.6 data.table_1.7.3 __ 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] tip: large plots
Very cool. Sadly, as far as I can tell, it doesn't work with ggplot though :( x-runif(1e6) y-runif(1e6) system.time(plot(x,y,pch='.')) user system elapsed 0.824 0.012 0.845 system.time(plot(x,y)) user system elapsed 33.422 0.016 33.545 system.time(print(qplot(x,y))) user system elapsed 45.142 0.228 45.687 system.time(print(qplot(x,y,pch='.'))) user system elapsed 47.483 1.060 49.040 system.time(print(qplot(x,y,shape='.'))) user system elapsed 44.807 0.689 45.710 On Fri, Nov 18, 2011 at 11:03 AM, Sarah Goslee sarah.gos...@gmail.comwrote: Hi all, I'm working with a bunch of large graphs, and stumbled across something useful. Probably many of you know this, but I didn't and so others might benefit. Using pch=. speeds up plotting considerably over using symbols. x - runif(100) y - runif(100) system.time(plot(x, y, pch=.)) user system elapsed 1.042 0.030 1.077 system.time(plot(x, y)) user system elapsed 37.865 0.033 38.122 If you have enough points, the result is also more legible. Choice of which pch symbol makes a difference too, the default pch=1 being the slowest of what I tried, but . is by far the speediest. system.time(plot(x, y, pch=0)) user system elapsed 11.191 0.011 11.270 system.time(plot(x, y, pch=1)) user system elapsed 38.024 0.008 38.245 system.time(plot(x, y, pch=2)) user system elapsed 14.140 0.027 14.270 system.time(plot(x, y, pch=3)) user system elapsed 15.696 0.011 15.799 system.time(plot(x, y, pch=4)) user system elapsed 18.770 0.007 18.888 This is a vanilla R session, 2.13.1 for x86_64-redhat-linux-gnu. I haven't tried it on any other OS, but it's making my life a lot smoother right now. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ 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] tip: large plots
That is a function I did not know about, thanks Hadley!
I still don't see the speed increase that you do with the base plot
package, but I'm sticking with ggplot anyway!
x-runif(1e6)
y-runif(1e6)
system.time(print(qplot(x,y)))
user system elapsed
42.234 0.520 43.061
system.time(print(qplot(x,y,pch=I('.'
user system elapsed
32.370 0.204 33.868
On Fri, Nov 18, 2011 at 12:39 PM, Hadley Wickham had...@rice.edu wrote:
You need: system.time(print(qplot(x,y,pch=I('.'
Hadley
On Fri, Nov 18, 2011 at 1:30 PM, Justin Haynes jto...@gmail.com wrote:
Very cool. Sadly, as far as I can tell, it doesn't work with ggplot
though
:(
x-runif(1e6)
y-runif(1e6)
system.time(plot(x,y,pch='.'))
user system elapsed
0.824 0.012 0.845
system.time(plot(x,y))
user system elapsed
33.422 0.016 33.545
system.time(print(qplot(x,y)))
user system elapsed
45.142 0.228 45.687
system.time(print(qplot(x,y,pch='.')))
user system elapsed
47.483 1.060 49.040
system.time(print(qplot(x,y,shape='.')))
user system elapsed
44.807 0.689 45.710
On Fri, Nov 18, 2011 at 11:03 AM, Sarah Goslee sarah.gos...@gmail.com
wrote:
Hi all,
I'm working with a bunch of large graphs, and stumbled across
something useful. Probably many of you know this, but I didn't and so
others might benefit.
Using pch=. speeds up plotting considerably over using symbols.
x - runif(100)
y - runif(100)
system.time(plot(x, y, pch=.))
user system elapsed
1.042 0.030 1.077
system.time(plot(x, y))
user system elapsed
37.865 0.033 38.122
If you have enough points, the result is also more legible.
Choice of which pch symbol makes a difference too, the default pch=1
being
the slowest of what I tried, but . is by far the speediest.
system.time(plot(x, y, pch=0))
user system elapsed
11.191 0.011 11.270
system.time(plot(x, y, pch=1))
user system elapsed
38.024 0.008 38.245
system.time(plot(x, y, pch=2))
user system elapsed
14.140 0.027 14.270
system.time(plot(x, y, pch=3))
user system elapsed
15.696 0.011 15.799
system.time(plot(x, y, pch=4))
user system elapsed
18.770 0.007 18.888
This is a vanilla R session, 2.13.1 for x86_64-redhat-linux-gnu. I
haven't tried it on any other OS, but it's making my life a lot
smoother right now.
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
__
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
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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.
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
[[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.
Re: [R] apply on rows and columns?
To expand on what Sarah and Michael said: if you have a 3d array: x-array(1:4,c(2,2,4)) x , , 1 [,1] [,2] [1,]13 [2,]24 , , 2 [,1] [,2] [1,]13 [2,]24 , , 3 [,1] [,2] [1,]13 [2,]24 , , 4 [,1] [,2] [1,]13 [2,]24 apply(x,c(1,2),sum) [,1] [,2] [1,]4 12 [2,]8 16 a margin of c(1,2) makes more sense. Hope that clarifies things. Justin On Wed, Nov 16, 2011 at 12:18 PM, Sarah Goslee sarah.gos...@gmail.com wrote: Hi, On Wed, Nov 16, 2011 at 3:13 PM, rkevinbur...@charter.net wrote: I have the following scenario: m - matrix(1:4, ncol=2) m [,1] [,2] [1,] 1 3 [2,] 2 4 apply(m, 2, sum) [1] 3 7 apply(m, 1, sum) [1] 4 6 So I can apply to rows *or* columns. According to the documentation (?apply) MARGIN a vector giving the subscripts which the function will be applied over. E.g., for a matrix 1 indicates rows, 2 indicates columns, c(1, 2) indicates rows and columns. Where X has named dimnames, it can be a character vector selecting dimension names. But I get the following results: apply(m, c(1,2), sum) [,1] [,2] [1,] 1 3 [2,] 2 4 How am I to interpret this result? I'm pretty sure R is taking the sum of m[1,1] and putting it [1,1], and the sum of m[1,2] and putting it in [1,2] and so on. You instructed apply() to work on rows and columns *simultaneously*, rather than sequentially. apply() on c(1,2) is useful if you have a matrix that's three-dimensional, but not so much if it's two dimensional. What are you trying to accomplish? Sarah -- Sarah Goslee http://www.functionaldiversity.org __ 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] Extract pattern from string
take a look at the structure of what Sys.time returns. str(Sys.time) and now at ?strptime! format(Sys.time(),format='%d-%H-%M-%S') [1] 15-09-55-55 format(Sys.time(),format='%Y') [1] 2011 format(Sys.time(),format='%m') [1] 11 Hope that helps, Justin On Tue, Nov 15, 2011 at 9:48 AM, syrvn ment...@gmx.net wrote: Hello, with Sys.time() you get the following string: 2011-11-15 16:25:55 GMT How can I extract the following substrings: year - 2011 month - 11 day_time - 15_16_25_55 Cheers, Syrvn -- View this message in context: http://r.789695.n4.nabble.com/Extract-pattern-from-string-tp4073432p4073432.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] Create design matrix
?expand.grid expand.grid(c(M,F),c(Y,O)) Var1 Var2 1MY 2FY 3MO 4FO Justin On Thu, Nov 3, 2011 at 10:56 AM, Bond, Stephen stephen.b...@cibc.com wrote: Greetings useRs, What is the easiest way to create a design matrix of several factor variables? Function gendata in Design seems to do that for a fitted model, but how to do that only on several factor vectors?? The result should be a df with one row for each distinct combination of levels of factors eg for (M,F) (Y,O) We get M Y M O F Y F O In reality I will have more than 1000 rows so doing by hand not good. Maybe there is a way with outer, but I couldn't see it. All the best to everybody. Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-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] mysterious warning message regarding bytecode...
While running a long script which source()s other scripts I get the following warning: Warning message: In t(object$S[[1]]) : bytecode version mismatch; using eval I cannot replicate it if I run the sourced files line by line though... What is that error? And do I care about it? It doesn't seem to affect my output as far as I can tell. Thanks! Justin sessionInfo() R version 2.13.2 (2011-09-30) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] grid stats graphics grDevices utils datasets methods base other attached packages: [1] mgcv_1.7-9stringr_0.5 RPostgreSQL_0.2-0 biglm_0.8 DBI_0.2-5 doMC_1.2.3multicore_0.1-7 [8] foreach_1.3.2 codetools_0.2-8 iterators_1.0.5 cairoDevice_2.19 pixmap_0.4-11 gridExtra_0.8.5 splancs_2.01-29 [15] sp_0.9-91 ellipse_0.3-5 ggplot2_0.8.9 proto_0.3-9.2 reshape_0.8.4 plyr_1.6 MASS_7.3-14 loaded via a namespace (and not attached): [1] compiler_2.13.2 digest_0.5.1lattice_0.19-33 Matrix_1.0-1nlme_3.1-102 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] factor level issue after subsetting
first of all, the subsetting line is overly complicated. dat.sub-dat[dat$treat!='cont',] will work just fine. R does exactly what you're describing. It knows the levels of the factor. Once you remove 'cont' from the data, that doesn't mean that the level is removed from the factor: df-data.frame(let=factor(sample(letters[1:5],100,replace=T)),num=rnorm(100)) str(df) 'data.frame': 100 obs. of 2 variables: $ let: Factor w/ 5 levels a,b,c,d,..: 1 5 1 4 3 5 2 2 1 3 ... $ num: num 0.224 -0.523 0.974 -0.268 -0.61 ... df.sub-df[df$let!='a',] str(df.sub) 'data.frame': 82 obs. of 2 variables: $ let: Factor w/ 5 levels a,b,c,d,..: 5 4 3 5 2 2 3 3 5 3 ... $ num: num -0.523 -0.268 -0.61 -1.383 -0.193 ... unique(df.sub$let) [1] e d c b Levels: a b c d e df.sub$let-factor(df.sub$let) unique(df.sub$let) [1] e d c b Levels: e d c b str(df.sub$let) Factor w/ 4 levels e,d,c,b: 1 2 3 1 4 4 3 3 1 3 ... by redefining your factor you can eliminate the problem. the other option, if you don't want factors to begin with is: options(stringsAsFactors=FALSE) # to set the global option or dat-read.csv(~/MyFiles/data.csv,stringsAsFactors=FALSE) # to set the option locally for this single read.csv call. On Tue, Nov 1, 2011 at 2:28 PM, Schreiber, Stefan stefan.schrei...@ales.ualberta.ca wrote: Dear list, I cannot figure out why, after sub-setting my data, that particular item which I don't want to plot is still in the newly created subset (please see example below). R somehow remembers what was in the original data set. A work around is exporting and importing the new subset. Then it's all fine; but I don't like this idea and was wondering what am I missing here? Thanks! Stefan P.S. I am using R 2.13.2 for Mac. dat-read.csv(~/MyFiles/data.csv) class(dat$treat) [1] factor dat treat yield 1 cont 98.7 2 cont 97.2 3 cont 96.1 4 cont 98.1 5 10 103.0 6 10 101.3 7 10 102.1 8 10 101.9 9 30 121.1 10 30 123.1 11 30 119.7 12 30 118.9 13 60 109.9 14 60 110.1 15 60 113.1 16 60 112.3 plot(dat$treat,dat$yield) dat.sub-dat[which(dat$treat!='cont')] class(dat.sub$treat) [1] factor dat.sub treat yield 5 10 103.0 6 10 101.3 7 10 102.1 8 10 101.9 9 30 121.1 10 30 123.1 11 30 119.7 12 30 118.9 13 60 109.9 14 60 110.1 15 60 113.1 16 60 112.3 plot(dat.sub$treat,dat.sub$yield) [[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.
