[R] Feature or bug?
Hi,
if I run
update <- function (newtime) { ginput <<- list(time=newtime)}
server <- function (input) {
print(paste("Before", input$time))
update(1)
print(paste("After:", input$time))
}
ginput <- list(time=0)
server(ginput)
then I get as result
[1] "Before 0"
[1] "After: 0"
If I uncomment the first print
update <- function (newtime) { ginput <<- list(time=newtime) }
server <- function (input) {
#print(paste("Before", input$time))
update(1)
print(paste("After:", input$time))
}
ginput <- list(time=0)
server(ginput)
then I get
[1] "After: 1"
Even when I use a side effect (by assign some new value to a global
variable) I would have expected the same behaviour in both cases.
Sigbert
--
http://u.hu-berlin.de/sk
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Re: [R] Feature or bug?
G'day Sigbert,
long time no see :)
How is Berlin these days?
On Thu, 21 May 2015 11:45:26 +0200
Sigbert Klinke wrote:
It is a feature.
> if I run
>
> update <- function (newtime) { ginput <<- list(time=newtime)}
>
> server <- function (input) {
> print(paste("Before", input$time))
> update(1)
> print(paste("After:", input$time))
> }
>
> ginput <- list(time=0)
> server(ginput)
>
> then I get as result
>
> [1] "Before 0"
> [1] "After: 0"
The first print command evaluates input and after this the function
server has an object named "input" in its local environment. The
second print command reuses this object and extracts the component time
from it (which has not changed). The change of the global variable has
no effect.
> If I uncomment the first print
>
> update <- function (newtime) { ginput <<- list(time=newtime) }
>
> server <- function (input) {
> #print(paste("Before", input$time))
> update(1)
> print(paste("After:", input$time))
> }
>
> ginput <- list(time=0)
> server(ginput)
>
> then I get
>
> [1] "After: 1"
Because the global variable is changed before input is evaluated. R
has lazy argument evaluation, arguments are only evaluated once they
are needed. You are essentially getting bitten by R's lazy evaluation
plus "pass by value" syntax.
> Even when I use a side effect (by assign some new value to a global
> variable) I would have expected the same behaviour in both cases.
To get the behaviour that you expect, you would have to write your code
along the following lines:
R> update <- function (newtime) { ginput <<- list(time=newtime)}
R> server <- function(input){
+ inp <- as.name(deparse(substitute(input)))
+ print(paste("Before", eval(substitute(XXX$time, list(XXX=inp)
+ update(1)
+ print(paste("After:", eval(substitute(XXX$time, list(XXX=inp)
+ }
R> ginput <- list(time=0)
R> server(ginput)
[1] "Before 0"
[1] "After: 1"
A cleaner way is perhaps to use environments, as these are passed by
reference:
R> update <- function(env, newtime) env$time <- newtime
R> server <- function(input){
+ print(paste("Before", input$time))
+ update(input, 1)
+ print(paste("After:", input$time))
+ }
R> ginput <- new.env()
R> ginput$time <- 0
R> server(ginput)
[1] "Before 0"
[1] "After: 1"
HTH.
Cheers,
Berwin
== Full address
A/Prof Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr)
School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
The University of Western Australia FAX : +61 (8) 6488 1028
35 Stirling Highway
Crawley WA 6009 e-mail: berwin.turl...@gmail.com
Australiahttp://www.maths.uwa.edu.au/~berwin
http://www.researcherid.com/rid/A-4995-2008
__
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[R] The R Foundation announces new mailing list 'R-package-devel'
New Mailing list: R-package-devel -- User R Packages Development At last week's monthly meeting, the R foundation has decided to create a new mailing list in order to help R package authors in their package development and testing. The idea is that some experienced R programmers (often those currently helping on R-devel or also R-help) will help package authors and thus unload some of the burden of the CRAN team members. Please read the detailed description of the mailing list here, https://stat.ethz.ch/mailman/listinfo/r-package-devel or also the more extended announcement of the list on R-devel, archived at https://stat.ethz.ch/pipermail/r-devel/2015-May/071208.html For the R foundation, Martin Maechler, Secretary General ___ r-annou...@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ 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] Vincentizing Reaction Time data in R
In line John Kane Kingston ON Canada > -Original Message- > From: yishinlin...@gmail.com > Sent: Thu, 21 May 2015 10:13:54 +0800 > To: gabriel.wein...@gmail.com > Subject: Re: [R] Vincentizing Reaction Time data in R > > On Wed, 20 May 2015 18:13:17 +0800, > Hi Gabriel, > > As far as I could recall, there isn't an R package that has explicitly > implemented "vincentization". You definitively can find some code > segments/functions that have implemented "vincentize" on the web. But you > should verify if they do exactly what you wish to do. If you could look > at the question from percentile/quantle perspective, it would not take > you too much time to realise that they are similar. I would suggest you > to read, as John Kane suggested, Prof. Ratcliff's 1979 paper. Another > paper that may be very helpful is Prof van Zandt's 2000 RT paper. > > However, you should be aware that there are some different implementation > of "vincentization", and it is debatable, if not problematic, to use it, > rather than other more general quantile methods. It would help you to > understand not only how to do vincentization, but also why/why not if you > could read papers from Jeff Rouder's as well as from Heathcote's and > Brown's lab. > > Sorry that I hesitate to give you the code, because this looks like part > of your course works. It would be more rewarding for you, if you could > figure out by yourself. > > Yishin > While I agree the exercise is likely to be a good learning experience I don't see this as the equivalent of course work. If Gabriel (the OP) was tasked with implementing "vincentization" in R then, strictly speaking it is course work but if I understand him the requirement is to do his work in R rather than Minitab. If such a function existed in an existing R package than he could have simply plugged in the numbers et voilà, done. The tenor of the question did not suggest this and it would require the stats instructor to know that there was no "vincentization" function anywhere among the, what, a thousand or so packages? And if the OP was working on his own data as part of the course then the instructor might have little or no idea of exactly what functions are needed The course strikes me more as an effort to get psychologists away from SPSS which often seems to be the only software package anyone knows. > Gabriel WEINDEL wrote: >> >> Dear all, >> >> For my master thesis, I'm currently working in cognitive neuroscience >> on executive control through measurement of reaction time and I need >> to get my data 'vincentized' with an exclusive use of R set by my >> statistic teacher for a test purpose, for this reason I can't use the >> python code the lab team usually uses. >> Despite a dozen hours of research I couldn't find any package or >> R-code which would allow the use of vincentization, that's why I'm >> querying help on the R forum. >> >> So has anyone ever used vincentization in R ? >> >> Best regards, >> >> -- >> Gabriel Weindel >> Master student in Neuropsychology - Aix-Marseille University (France) >> Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords & protects your account. __ 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] Feature or bug?
Well..
"Because the global variable is changed before input is evaluated. R
has lazy argument evaluation, arguments are only evaluated once they
are needed. You are essentially getting bitten by R's lazy evaluation
plus "pass by value" syntax."
While I may be either wrong or just picking on semantics, I don't
think so. It is merely what you stated previously: input was assigned
a value in the local server function environment, and that assignment
was not affected by the subsequent assignment to the global
environment. So it is a matter of R's semantics -- where it looks for
the values bound to symbols -- rather than lazy evaluation.
Obviously then, a simple way to do what the OP seemed to want would be
to simply assign the updated value in the local function environment,
rather than the global. I get nervous whenever I see constructs with
eval(substitute...)) or global assignments from within a function.
Both have their place, of course, but (the latter especially) can be
dangerous, and my experience both on the list and with my own code, is
that if you think you need them, you probably should rethink what you
need to do.
Corrections and/or criticism of these comments are welcome.
Best,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Thu, May 21, 2015 at 3:50 AM, Berwin A Turlach
wrote:
> G'day Sigbert,
>
> long time no see :)
> How is Berlin these days?
>
> On Thu, 21 May 2015 11:45:26 +0200
> Sigbert Klinke wrote:
>
> It is a feature.
>
>> if I run
>>
>> update <- function (newtime) { ginput <<- list(time=newtime)}
>>
>> server <- function (input) {
>> print(paste("Before", input$time))
>> update(1)
>> print(paste("After:", input$time))
>> }
>>
>> ginput <- list(time=0)
>> server(ginput)
>>
>> then I get as result
>>
>> [1] "Before 0"
>> [1] "After: 0"
>
> The first print command evaluates input and after this the function
> server has an object named "input" in its local environment. The
> second print command reuses this object and extracts the component time
> from it (which has not changed). The change of the global variable has
> no effect.
>
>> If I uncomment the first print
>>
>> update <- function (newtime) { ginput <<- list(time=newtime) }
>>
>> server <- function (input) {
>> #print(paste("Before", input$time))
>> update(1)
>> print(paste("After:", input$time))
>> }
>>
>> ginput <- list(time=0)
>> server(ginput)
>>
>> then I get
>>
>> [1] "After: 1"
>
> Because the global variable is changed before input is evaluated. R
> has lazy argument evaluation, arguments are only evaluated once they
> are needed. You are essentially getting bitten by R's lazy evaluation
> plus "pass by value" syntax.
>
>> Even when I use a side effect (by assign some new value to a global
>> variable) I would have expected the same behaviour in both cases.
>
> To get the behaviour that you expect, you would have to write your code
> along the following lines:
>
> R> update <- function (newtime) { ginput <<- list(time=newtime)}
> R> server <- function(input){
> + inp <- as.name(deparse(substitute(input)))
> + print(paste("Before", eval(substitute(XXX$time, list(XXX=inp)
> + update(1)
> + print(paste("After:", eval(substitute(XXX$time, list(XXX=inp)
> + }
> R> ginput <- list(time=0)
> R> server(ginput)
> [1] "Before 0"
> [1] "After: 1"
>
>
> A cleaner way is perhaps to use environments, as these are passed by
> reference:
>
> R> update <- function(env, newtime) env$time <- newtime
> R> server <- function(input){
> + print(paste("Before", input$time))
> + update(input, 1)
> + print(paste("After:", input$time))
> + }
> R> ginput <- new.env()
> R> ginput$time <- 0
> R> server(ginput)
> [1] "Before 0"
> [1] "After: 1"
>
> HTH.
>
> Cheers,
>
> Berwin
>
> == Full address
> A/Prof Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr)
> School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
> The University of Western Australia FAX : +61 (8) 6488 1028
> 35 Stirling Highway
> Crawley WA 6009 e-mail: berwin.turl...@gmail.com
> Australiahttp://www.maths.uwa.edu.au/~berwin
> http://www.researcherid.com/rid/A-4995-2008
>
> __
> 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, sel
Re: [R] Feature or bug?
On 21 May 2015, at 16:26 , Bert Gunter wrote:
> Well..
>
> "Because the global variable is changed before input is evaluated. R
> has lazy argument evaluation, arguments are only evaluated once they
> are needed. You are essentially getting bitten by R's lazy evaluation
> plus "pass by value" syntax."
>
> While I may be either wrong or just picking on semantics, I don't
> think so. It is merely what you stated previously: input was assigned
> a value in the local server function environment, and that assignment
> was not affected by the subsequent assignment to the global
> environment. So it is a matter of R's semantics -- where it looks for
> the values bound to symbols -- rather than lazy evaluation.
>
> Obviously then, a simple way to do what the OP seemed to want would be
> to simply assign the updated value in the local function environment,
> rather than the global. I get nervous whenever I see constructs with
> eval(substitute...)) or global assignments from within a function.
> Both have their place, of course, but (the latter especially) can be
> dangerous, and my experience both on the list and with my own code, is
> that if you think you need them, you probably should rethink what you
> need to do.
>
> Corrections and/or criticism of these comments are welcome.
>
Berwin is right, or rather: There are two issues. The "input" argument to
server() is evaluated when first used. The difference between the two examples
is whether this happens before or after update(), and that is the effect of
lazy evaluation. The fact that it in the first case the value is unchanged by
the update is due to scoping: Once evaluate "input" becomes a local variable
an is unchanged by assignment to the global variable.
-pd
> Best,
> Bert
>
>
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> Clifford Stoll
>
>
>
>
> On Thu, May 21, 2015 at 3:50 AM, Berwin A Turlach
> wrote:
>> G'day Sigbert,
>>
>> long time no see :)
>> How is Berlin these days?
>>
>> On Thu, 21 May 2015 11:45:26 +0200
>> Sigbert Klinke wrote:
>>
>> It is a feature.
>>
>>> if I run
>>>
>>> update <- function (newtime) { ginput <<- list(time=newtime)}
>>>
>>> server <- function (input) {
>>> print(paste("Before", input$time))
>>> update(1)
>>> print(paste("After:", input$time))
>>> }
>>>
>>> ginput <- list(time=0)
>>> server(ginput)
>>>
>>> then I get as result
>>>
>>> [1] "Before 0"
>>> [1] "After: 0"
>>
>> The first print command evaluates input and after this the function
>> server has an object named "input" in its local environment. The
>> second print command reuses this object and extracts the component time
>> from it (which has not changed). The change of the global variable has
>> no effect.
>>
>>> If I uncomment the first print
>>>
>>> update <- function (newtime) { ginput <<- list(time=newtime) }
>>>
>>> server <- function (input) {
>>> #print(paste("Before", input$time))
>>> update(1)
>>> print(paste("After:", input$time))
>>> }
>>>
>>> ginput <- list(time=0)
>>> server(ginput)
>>>
>>> then I get
>>>
>>> [1] "After: 1"
>>
>> Because the global variable is changed before input is evaluated. R
>> has lazy argument evaluation, arguments are only evaluated once they
>> are needed. You are essentially getting bitten by R's lazy evaluation
>> plus "pass by value" syntax.
>>
>>> Even when I use a side effect (by assign some new value to a global
>>> variable) I would have expected the same behaviour in both cases.
>>
>> To get the behaviour that you expect, you would have to write your code
>> along the following lines:
>>
>> R> update <- function (newtime) { ginput <<- list(time=newtime)}
>> R> server <- function(input){
>> + inp <- as.name(deparse(substitute(input)))
>> + print(paste("Before", eval(substitute(XXX$time, list(XXX=inp)
>> + update(1)
>> + print(paste("After:", eval(substitute(XXX$time, list(XXX=inp)
>> + }
>> R> ginput <- list(time=0)
>> R> server(ginput)
>> [1] "Before 0"
>> [1] "After: 1"
>>
>>
>> A cleaner way is perhaps to use environments, as these are passed by
>> reference:
>>
>> R> update <- function(env, newtime) env$time <- newtime
>> R> server <- function(input){
>> + print(paste("Before", input$time))
>> + update(input, 1)
>> + print(paste("After:", input$time))
>> + }
>> R> ginput <- new.env()
>> R> ginput$time <- 0
>> R> server(ginput)
>> [1] "Before 0"
>> [1] "After: 1"
>>
>> HTH.
>>
>> Cheers,
>>
>>Berwin
>>
>> == Full address
>> A/Prof Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr)
>> School of Maths and Stats (M019)+61 (8) 6488 3383 (self)
>> The University of Western Australia FAX : +61 (8) 6488 1028
>> 35 Stirling Highway
>> Crawley WA 6009 e-mail: berwin.turl...@gmail.com
>> Austral
Re: [R] Feature or bug?
If you add a print statement to trace the evaluation of the input argument
you can see how the lazy evaluation works:
> update <- function (newtime) {
cat("# update() is changing global ginput's time from", ginput$time,
"to", newtime, "\n")
ginput <<- list(time = newtime)
}
> server <- function(input, doFirstPrint) {
if (doFirstPrint) {
cat("# Before calling update(1): input$time=", input$time,
"ginput$time=", ginput$time, "\n")
}
update(1)
cat("# After calling update(1): input$time=", input$time,
"ginput$time=", ginput$time, "\n")
}
> ginput <- list(time=0)
> server({cat("# Evaluating server's 'input' argument\n"); ginput},
doFirstPrint=TRUE)
# Evaluating server's 'input' argument
# Before calling update(1): input$time= 0 ginput$time= 0
# update() is changing global ginput's time from 0 to 1
# After calling update(1): input$time= 0 ginput$time= 1
> ginput <- list(time=0)
> server({cat("# Evaluating server's 'input' argument\n"); ginput},
doFirstPrint=FALSE)
# update() is changing global ginput's time from 0 to 1
# Evaluating server's 'input' argument
# After calling update(1): input$time= 1 ginput$time= 1
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 21, 2015 at 7:48 AM, peter dalgaard wrote:
>
> On 21 May 2015, at 16:26 , Bert Gunter wrote:
>
> > Well..
> >
> > "Because the global variable is changed before input is evaluated. R
> > has lazy argument evaluation, arguments are only evaluated once they
> > are needed. You are essentially getting bitten by R's lazy evaluation
> > plus "pass by value" syntax."
> >
> > While I may be either wrong or just picking on semantics, I don't
> > think so. It is merely what you stated previously: input was assigned
> > a value in the local server function environment, and that assignment
> > was not affected by the subsequent assignment to the global
> > environment. So it is a matter of R's semantics -- where it looks for
> > the values bound to symbols -- rather than lazy evaluation.
> >
> > Obviously then, a simple way to do what the OP seemed to want would be
> > to simply assign the updated value in the local function environment,
> > rather than the global. I get nervous whenever I see constructs with
> > eval(substitute...)) or global assignments from within a function.
> > Both have their place, of course, but (the latter especially) can be
> > dangerous, and my experience both on the list and with my own code, is
> > that if you think you need them, you probably should rethink what you
> > need to do.
> >
> > Corrections and/or criticism of these comments are welcome.
> >
>
> Berwin is right, or rather: There are two issues. The "input" argument to
> server() is evaluated when first used. The difference between the two
> examples is whether this happens before or after update(), and that is the
> effect of lazy evaluation. The fact that it in the first case the value is
> unchanged by the update is due to scoping: Once evaluate "input" becomes a
> local variable an is unchanged by assignment to the global variable.
>
> -pd
>
>
> > Best,
> > Bert
> >
> >
> >
> > Bert Gunter
> > Genentech Nonclinical Biostatistics
> > (650) 467-7374
> >
> > "Data is not information. Information is not knowledge. And knowledge
> > is certainly not wisdom."
> > Clifford Stoll
> >
> >
> >
> >
> > On Thu, May 21, 2015 at 3:50 AM, Berwin A Turlach
> > wrote:
> >> G'day Sigbert,
> >>
> >> long time no see :)
> >> How is Berlin these days?
> >>
> >> On Thu, 21 May 2015 11:45:26 +0200
> >> Sigbert Klinke wrote:
> >>
> >> It is a feature.
> >>
> >>> if I run
> >>>
> >>> update <- function (newtime) { ginput <<- list(time=newtime)}
> >>>
> >>> server <- function (input) {
> >>> print(paste("Before", input$time))
> >>> update(1)
> >>> print(paste("After:", input$time))
> >>> }
> >>>
> >>> ginput <- list(time=0)
> >>> server(ginput)
> >>>
> >>> then I get as result
> >>>
> >>> [1] "Before 0"
> >>> [1] "After: 0"
> >>
> >> The first print command evaluates input and after this the function
> >> server has an object named "input" in its local environment. The
> >> second print command reuses this object and extracts the component time
> >> from it (which has not changed). The change of the global variable has
> >> no effect.
> >>
> >>> If I uncomment the first print
> >>>
> >>> update <- function (newtime) { ginput <<- list(time=newtime) }
> >>>
> >>> server <- function (input) {
> >>> #print(paste("Before", input$time))
> >>> update(1)
> >>> print(paste("After:", input$time))
> >>> }
> >>>
> >>> ginput <- list(time=0)
> >>> server(ginput)
> >>>
> >>> then I get
> >>>
> >>> [1] "After: 1"
> >>
> >> Because the global variable is changed before input is evaluated. R
> >> has lazy argument evaluation, arguments are only evaluated once they
> >> are needed. You are essentially getting bitten by R's lazy evaluation
> >> plus "pass by value" syntax.
> >>
> >>> Even when I use a side effect (by assign some new value to a glob
[R] Question regarding different R versions on an enterprise network server
Hi all, I represent R users vs. IT dept. at my workplace (yes, an enviable task :) We've managed to get a workable network-based R application, for people who work remotely, or don't have a machine (i.e., they use a VDI terminal). Everything in this organization is staunchly Windows and Microsoft. We've agreed to upgrade only once yearly, to save IT resources. Now we're upgrading, and I would like users to be able to keep their old R 3.1.0 directory trees like it's available on a single Windows machine. At least for a few months, so that people can evaluate back-compatibility if they need. In fact, we have an even older server-based 3.0.1, which happens to be the only R version for which the Tableaux-R connection works (at least according to my colleagues, I don't use Tableaux). Anyway, long story short. That was just the motivating example. The problem I'm dealing with is whether a network application that has several versions of R (3.1.z, 3.2.z), etc., all available, and each reading and installing libraries to a different folder tree. The libraries right now are installed into each user's "personal" share drive. It's pretty stable. However, obviously the 3.2.z libraries will now overwrite the 3.1.z. My IT contact says it's impossible, because the Windows app name is always just Rgui.exe, and they can only have one set of instructions associated with the same app name (i.e., what folders to go to, etc.) I wonder whether anyone has had experience with this, or I should just give up and alert people that if they want to explore various historical layers of R and the associated packages, they will have to work around and/or install and uninstall lots of packages each time. Thanks! Assaf -- Assaf P. Oron, Ph.D. Senior Statistician, Children's Core for Biomedical Statistics (206)884-1236, assaf.o...@seattlechildrens.org Consulting statistician, Seattle DEEDS Project http://www.duwamishdiesel.org/ Instructor, UW Certificate for Statistical Analysis with R as...@uw.edu [[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] Vincentizing Reaction Time data in R
Bert : Thank you for your advice, it would be a little bit difficult to do it for my master thesis but, if I want to go further with a PhD thesis (and I do want), I would probably follow your advice and get in touch with a statistician. Yishin : Thank you very much for the references, I will definitively read the papers you quote. I'm already a little bit aware of the misuses possible with the vincentization in particular thanks to the paper of Rouder and Speckman (2004) and it seems to fit with my design. No problem if you want to keep the code but I have to tell you that it's our first semester using R and the teacher surely didn't thought that we will run out of available code with our experiment. Like John guessed the purpose of the course was to give a first view of R to get over the temptation of SPSS, my bad if I want to avoid biased statistics like sample mean ANOVA's on RT. Dan : Thank you for your tip, this sure will help but I'm quiet at the beginning of my R skills so I hardly trust myself to do it on my own, but I can sure give it a try. John : I had the same assumption but my research director warned me that I might run out of time for my first presentation by doing so but fairly enough for my master thesis. But again like I said to Dan I'm quiet concerned by my actual R skill. Anyway I have to say that I'm really glad to see how much help you can get by using the r-help mailing-list. Regards, Gabriel Le 21/05/2015 15:52, John Kane a écrit : In line John Kane Kingston ON Canada -Original Message- From: yishinlin...@gmail.com Sent: Thu, 21 May 2015 10:13:54 +0800 To: gabriel.wein...@gmail.com Subject: Re: [R] Vincentizing Reaction Time data in R On Wed, 20 May 2015 18:13:17 +0800, Hi Gabriel, As far as I could recall, there isn't an R package that has explicitly implemented "vincentization". You definitively can find some code segments/functions that have implemented "vincentize" on the web. But you should verify if they do exactly what you wish to do. If you could look at the question from percentile/quantle perspective, it would not take you too much time to realise that they are similar. I would suggest you to read, as John Kane suggested, Prof. Ratcliff's 1979 paper. Another paper that may be very helpful is Prof van Zandt's 2000 RT paper. However, you should be aware that there are some different implementation of "vincentization", and it is debatable, if not problematic, to use it, rather than other more general quantile methods. It would help you to understand not only how to do vincentization, but also why/why not if you could read papers from Jeff Rouder's as well as from Heathcote's and Brown's lab. Sorry that I hesitate to give you the code, because this looks like part of your course works. It would be more rewarding for you, if you could figure out by yourself. Yishin While I agree the exercise is likely to be a good learning experience I don't see this as the equivalent of course work. If Gabriel (the OP) was tasked with implementing "vincentization" in R then, strictly speaking it is course work but if I understand him the requirement is to do his work in R rather than Minitab. If such a function existed in an existing R package than he could have simply plugged in the numbers et voilà, done. The tenor of the question did not suggest this and it would require the stats instructor to know that there was no "vincentization" function anywhere among the, what, a thousand or so packages? And if the OP was working on his own data as part of the course then the instructor might have little or no idea of exactly what functions are needed The course strikes me more as an effort to get psychologists away from SPSS which often seems to be the only software package anyone knows. Gabriel WEINDEL wrote: Dear all, For my master thesis, I'm currently working in cognitive neuroscience on executive control through measurement of reaction time and I need to get my data 'vincentized' with an exclusive use of R set by my statistic teacher for a test purpose, for this reason I can't use the python code the lab team usually uses. Despite a dozen hours of research I couldn't find any package or R-code which would allow the use of vincentization, that's why I'm querying help on the R forum. So has anyone ever used vincentization in R ? Best regards, -- Gabriel Weindel Master student in Neuropsychology - Aix-Marseille University (France) Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords & protects your account. Check it out at http://mysecurelogon.com/manager __ 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 provi
Re: [R] Subset and 0 replace?
Thanks William/Duncan!
Duncan - Yes - I am using the doBy package.
running this line on the sample data below gives weights for V5,V44, & V2.
Ideally I would like 0's for V8 and V10 in the output.
So it would look like:
e<-structure(matrix(c("V5", "0.008714910", "V8", "0", "V10", "0", "V44",
"0.004357455", "V2", "0.008714910"),nrow = 2))
This is far as I've gotten by subsetting and summing:
a<-t(summaryBy(Wgt.sum~as.numeric(.id),data=subset(ldply(c,function(x)
summaryBy(Wgt ~ SPCLORatingValue, data=x,
FUN=c(sum))),SPCLORatingValue>16),FUN=c(sum),order=FALSE))
All help/guidance is much appreciated! Thanks Vince!
Sample data example:
c<-structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
), row.names = 12:22, class = "data.frame"), V8 = structure(list(
WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
12L, 15L, 12L, 13L, 15L, 14L, 15L, 14L)), .Names = c("WgtBand",
"Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
V10 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingValue = c(15L, 13L, 14L, 14L, 13L, 13L, 13L,
15L, 15L, 13L, 14L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
), row.names = 12:22, class = "data.frame"), V44 = structure(list(
WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
c(0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083), SPCLORatingValue = c(13L, 14L,
16L, 15L, 14L, 14L, 18L, 13L, 16L, 15L, 11L)), .Names = c("WgtBand",
"Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
V2 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingValue = c(13L, 14L, 15L, 15L, 15L, 14L, 12L,
16L, 17L, 15L, 19L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
), row.names = 12:22, class = "data.frame")), .Names = c("V5",
"V8", "V10", "V44", "V2"))
structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
), row.names = 12:22, class = "data.frame"), V8 = structure(list(
WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
12L, 15L, 12L, 13L, 15L, 14L, 15L, 14L)), .Names = c("WgtBand",
"Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
V10 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083,
0.0043574552083, 0.0043574552083, 0.0043574552083
), SPCLORatingValue = c(15L, 13L, 14L, 14L, 13L, 13L, 13L,
15L, 15L, 13L, 14L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
From: wdun...@tibco.com
Date: Wed, 20 May 2015 22:12:01 -0700
Subject: Re: [R] Subset and 0 replace?
To: newrnew...@hotmail.com
CC: r-help@r-project.org
Can you show a small self-contained example of you data
Re: [R] Question regarding different R versions on an enterprise network server
On 21/05/2015 12:21 PM, Assaf P. Oron wrote: Hi all, I represent R users vs. IT dept. at my workplace (yes, an enviable task :) We've managed to get a workable network-based R application, for people who work remotely, or don't have a machine (i.e., they use a VDI terminal). Everything in this organization is staunchly Windows and Microsoft. We've agreed to upgrade only once yearly, to save IT resources. Now we're upgrading, and I would like users to be able to keep their old R 3.1.0 directory trees like it's available on a single Windows machine. At least for a few months, so that people can evaluate back-compatibility if they need. In fact, we have an even older server-based 3.0.1, which happens to be the only R version for which the Tableaux-R connection works (at least according to my colleagues, I don't use Tableaux). Anyway, long story short. That was just the motivating example. The problem I'm dealing with is whether a network application that has several versions of R (3.1.z, 3.2.z), etc., all available, and each reading and installing libraries to a different folder tree. The libraries right now are installed into each user's "personal" share drive. It's pretty stable. However, obviously the 3.2.z libraries will now overwrite the 3.1.z. My IT contact says it's impossible, because the Windows app name is always just Rgui.exe, and they can only have one set of instructions associated with the same app name (i.e., what folders to go to, etc.) If you are doing the install, you can rename Rgui.exe to something else, e.g. rename the old one to Rgui31.exe. The default setup already installs user packages into a local directory with a versioned name, so that shouldn't be a problem. See ?R_LIBS for details on that. Duncan Murdoch I wonder whether anyone has had experience with this, or I should just give up and alert people that if they want to explore various historical layers of R and the associated packages, they will have to work around and/or install and uninstall lots of packages each time. Thanks! Assaf __ 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] Question regarding different R versions on an enterprise network server
3.2 library is in a different directory than 3.1 library. You might benefit from reading the discussion about packages in the installation manual for R. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On May 21, 2015 9:21:28 AM PDT, "Assaf P. Oron" wrote: >Hi all, > >I represent R users vs. IT dept. at my workplace (yes, an enviable task >:) > >We've managed to get a workable network-based R application, for people >who >work remotely, or don't have a machine (i.e., they use a VDI terminal). >Everything in this organization is staunchly Windows and Microsoft. > >We've agreed to upgrade only once yearly, to save IT resources. Now >we're >upgrading, and I would like users to be able to keep their old R 3.1.0 >directory trees like it's available on a single Windows machine. At >least >for a few months, so that people can evaluate back-compatibility if >they >need. In fact, we have an even older server-based 3.0.1, which happens >to >be the only R version for which the Tableaux-R connection works (at >least >according to my colleagues, I don't use Tableaux). > >Anyway, long story short. That was just the motivating example. The >problem >I'm dealing with is whether a network application that has several >versions >of R (3.1.z, 3.2.z), etc., all available, and each reading and >installing >libraries to a different folder tree. > >The libraries right now are installed into each user's "personal" share >drive. It's pretty stable. However, obviously the 3.2.z libraries will >now >overwrite the 3.1.z. > >My IT contact says it's impossible, because the Windows app name is >always >just Rgui.exe, and they can only have one set of instructions >associated >with the same app name (i.e., what folders to go to, etc.) > >I wonder whether anyone has had experience with this, or I should just >give >up and alert people that if they want to explore various historical >layers >of R and the associated packages, they will have to work around and/or >install and uninstall lots of packages each time. > >Thanks! > >Assaf __ 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] Question regarding different R versions on an enterprise network server
Thanks for the quick response. @Duncan: The IT person says he cannot rename the binary. Naturally that's the first suggestion I made to him. I found it odd but he's the IT person not me. OTOH if I have authoritative word from the R team that it's perfectly doable, I can get back to him :) @Jeff: Since it's a network app, on their side it doesn't sit in a 'folder' so there's no way for the different Rgui.exe's to reference different folders. If they are called the same, they will all read off of the same initialization files. Assaf On Thu, May 21, 2015 at 9:54 AM, Duncan Murdoch wrote: > On 21/05/2015 12:21 PM, Assaf P. Oron wrote: > >> Hi all, >> >> I represent R users vs. IT dept. at my workplace (yes, an enviable task :) >> >> We've managed to get a workable network-based R application, for people >> who >> work remotely, or don't have a machine (i.e., they use a VDI terminal). >> Everything in this organization is staunchly Windows and Microsoft. >> >> We've agreed to upgrade only once yearly, to save IT resources. Now we're >> upgrading, and I would like users to be able to keep their old R 3.1.0 >> directory trees like it's available on a single Windows machine. At least >> for a few months, so that people can evaluate back-compatibility if they >> need. In fact, we have an even older server-based 3.0.1, which happens to >> be the only R version for which the Tableaux-R connection works (at least >> according to my colleagues, I don't use Tableaux). >> >> Anyway, long story short. That was just the motivating example. The >> problem >> I'm dealing with is whether a network application that has several >> versions >> of R (3.1.z, 3.2.z), etc., all available, and each reading and installing >> libraries to a different folder tree. >> >> The libraries right now are installed into each user's "personal" share >> drive. It's pretty stable. However, obviously the 3.2.z libraries will now >> overwrite the 3.1.z. >> >> My IT contact says it's impossible, because the Windows app name is always >> just Rgui.exe, and they can only have one set of instructions associated >> with the same app name (i.e., what folders to go to, etc.) >> > > If you are doing the install, you can rename Rgui.exe to something else, > e.g. rename the old one to Rgui31.exe. > > The default setup already installs user packages into a local directory > with a versioned name, so that shouldn't be a problem. > See ?R_LIBS for details on that. > > Duncan Murdoch > > >> I wonder whether anyone has had experience with this, or I should just >> give >> up and alert people that if they want to explore various historical layers >> of R and the associated packages, they will have to work around and/or >> install and uninstall lots of packages each time. >> >> Thanks! >> >> Assaf >> >> > -- Assaf P. Oron, Ph.D. Senior Statistician, Children's Core for Biomedical Statistics (206)884-1236, assaf.o...@seattlechildrens.org Consulting statistician, Seattle DEEDS Project http://www.duwamishdiesel.org/ Instructor, UW Certificate for Statistical Analysis with R as...@uw.edu [[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] Question regarding different R versions on an enterprise network server
Assaf: they are named differently when you run different versions. 3.1 and 3.2 are different, but 3.1.1 and 3.1.2 are both in the 3.1 directory. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On May 21, 2015 11:09:25 AM PDT, "Assaf P. Oron" wrote: >Thanks for the quick response. > >@Duncan: The IT person says he cannot rename the binary. Naturally >that's >the first suggestion I made to him. I found it odd but he's the IT >person >not me. >OTOH if I have authoritative word from the R team that it's perfectly >doable, I can get back to him :) > >@Jeff: Since it's a network app, on their side it doesn't sit in a >'folder' >so there's no way for the different Rgui.exe's to reference different >folders. If they are called the same, they will all read off of the >same >initialization files. > >Assaf > > > >On Thu, May 21, 2015 at 9:54 AM, Duncan Murdoch > >wrote: > >> On 21/05/2015 12:21 PM, Assaf P. Oron wrote: >> >>> Hi all, >>> >>> I represent R users vs. IT dept. at my workplace (yes, an enviable >task :) >>> >>> We've managed to get a workable network-based R application, for >people >>> who >>> work remotely, or don't have a machine (i.e., they use a VDI >terminal). >>> Everything in this organization is staunchly Windows and Microsoft. >>> >>> We've agreed to upgrade only once yearly, to save IT resources. Now >we're >>> upgrading, and I would like users to be able to keep their old R >3.1.0 >>> directory trees like it's available on a single Windows machine. At >least >>> for a few months, so that people can evaluate back-compatibility if >they >>> need. In fact, we have an even older server-based 3.0.1, which >happens to >>> be the only R version for which the Tableaux-R connection works (at >least >>> according to my colleagues, I don't use Tableaux). >>> >>> Anyway, long story short. That was just the motivating example. The >>> problem >>> I'm dealing with is whether a network application that has several >>> versions >>> of R (3.1.z, 3.2.z), etc., all available, and each reading and >installing >>> libraries to a different folder tree. >>> >>> The libraries right now are installed into each user's "personal" >share >>> drive. It's pretty stable. However, obviously the 3.2.z libraries >will now >>> overwrite the 3.1.z. >>> >>> My IT contact says it's impossible, because the Windows app name is >always >>> just Rgui.exe, and they can only have one set of instructions >associated >>> with the same app name (i.e., what folders to go to, etc.) >>> >> >> If you are doing the install, you can rename Rgui.exe to something >else, >> e.g. rename the old one to Rgui31.exe. >> >> The default setup already installs user packages into a local >directory >> with a versioned name, so that shouldn't be a problem. >> See ?R_LIBS for details on that. >> >> Duncan Murdoch >> >> >>> I wonder whether anyone has had experience with this, or I should >just >>> give >>> up and alert people that if they want to explore various historical >layers >>> of R and the associated packages, they will have to work around >and/or >>> install and uninstall lots of packages each time. >>> >>> Thanks! >>> >>> Assaf >>> >>> >> __ 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] configure script claims that pkg-config doesn't know about cairo
and yet I checked manually : pkg-config --print-variables cairo exec_prefix prefix libdir includedir something weird is going on with this __ 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] compiling R with tuned BLAS
I am looking at the instructions on http://cran.r-project.org/doc/manuals/r-patched/R-admin.html#ATLAS I have noticed that ATLAS produces two shared libs in addition to the *.a files: http://math-atlas.sourceforge.net/atlas_install/node22.html contents of the ATLAS lib directory: libatlas.a libcblas.a libf77blas.a liblapack.a libptcblas.a libptf77blas.a libsatlas.so libtatlas.so The instructions do not appear to match up with the *.a files & *.so files as described. (it appears to want me to use shared libs, but the names defined are static libs, not shared libs). should I simply be having it link against libtatlas.so (and pthreads) for shared threaded atlas and libsatlas.so for shared sequential atlas? do I need shared versions of the other static libraries? I think the help is a bit out of date, or at least unclear as to what it intends of me. M. Gooch __ 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] Subset and 0 replace?
I renamed your 'c' to be 'toyData' and your 'e' to be 'desiredResult'. Do
you
want the following, which uses only base R code?
> vapply(toyData,
FUN=function(V)with(V, sum(Wgt[SPCLORatingValue>16])),
FUN.VALUE=0)
V5 V8 V10 V44 V2
0.008714910 0.0 0.0 0.004357455 0.008714910
It what is in your desired result but in a more useful format (e.g., numbers
instead of character strings for sum).
> desiredResult
[,1] [,2] [,3] [,4] [,5]
[1,] "V5" "V8" "V10" "V44" "V2"
[2,] "0.008714910" "0" "0" "0.004357455" "0.008714910"
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 21, 2015 at 9:50 AM, Vin Cheng wrote:
> Thanks William/Duncan!
>
> Duncan - Yes - I am using the doBy package.
>
> running this line on the sample data below gives weights for V5,V44, &
> V2. Ideally I would like 0's for V8 and V10 in the output.
>
> So it would look like:
> e<-structure(matrix(c("V5", "0.008714910", "V8", "0", "V10", "0", "V44",
> "0.004357455", "V2", "0.008714910"),nrow = 2))
>
>
> This is far as I've gotten by subsetting and summing:
> a<-t(summaryBy(Wgt.sum~as.numeric(.id),data=subset(ldply(c,function(x)
> summaryBy(Wgt ~ SPCLORatingValue, data=x,
> FUN=c(sum))),SPCLORatingValue>16),FUN=c(sum),order=FALSE))
>
> All help/guidance is much appreciated! Thanks Vince!
>
> Sample data example:
> c<-structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
> 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083
> ), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
> 13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
> ), row.names = 12:22, class = "data.frame"), V8 = structure(list(
> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
> c(0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
> 12L, 15L, 12L, 13L, 15L, 14L, 15L, 14L)), .Names = c("WgtBand",
> "Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
> V10 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2,
> 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083
> ), SPCLORatingValue = c(15L, 13L, 14L, 14L, 13L, 13L, 13L,
> 15L, 15L, 13L, 14L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
> ), row.names = 12:22, class = "data.frame"), V44 = structure(list(
> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
> c(0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083), SPCLORatingValue = c(13L, 14L,
> 16L, 15L, 14L, 14L, 18L, 13L, 16L, 15L, 11L)), .Names =
> c("WgtBand",
> "Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
> V2 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2,
> 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083
> ), SPCLORatingValue = c(13L, 14L, 15L, 15L, 15L, 14L, 12L,
> 16L, 17L, 15L, 19L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
> ), row.names = 12:22, class = "data.frame")), .Names = c("V5",
> "V8", "V10", "V44", "V2"))
> structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
> 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083
> ), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
> 13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
> ), row.names = 12:22, class = "data.frame"), V8 = structure(list(
> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
> c(0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083, 0.0043574552083, 0.0043574552083,
> 0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
>
Re: [R] Subset and 0 replace?
This is perfect! Thanks William!!!
Vince
> On May 21, 2015, at 3:36 PM, William Dunlap wrote:
>
> I renamed your 'c' to be 'toyData' and your 'e' to be 'desiredResult'. Do you
> want the following, which uses only base R code?
>
> > vapply(toyData,
> FUN=function(V)with(V, sum(Wgt[SPCLORatingValue>16])),
> FUN.VALUE=0)
> V5 V8 V10 V44 V2
> 0.008714910 0.0 0.0 0.004357455 0.008714910
>
> It what is in your desired result but in a more useful format (e.g., numbers
> instead of character strings for sum).
>
> > desiredResult
> [,1] [,2] [,3] [,4] [,5]
> [1,] "V5" "V8" "V10" "V44" "V2"
> [2,] "0.008714910" "0" "0" "0.004357455" "0.008714910"
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
>> On Thu, May 21, 2015 at 9:50 AM, Vin Cheng wrote:
>> Thanks William/Duncan!
>>
>> Duncan - Yes - I am using the doBy package.
>>
>> running this line on the sample data below gives weights for V5,V44, & V2.
>> Ideally I would like 0's for V8 and V10 in the output.
>>
>> So it would look like:
>> e<-structure(matrix(c("V5", "0.008714910", "V8", "0", "V10", "0", "V44",
>> "0.004357455", "V2", "0.008714910"),nrow = 2))
>>
>>
>> This is far as I've gotten by subsetting and summing:
>> a<-t(summaryBy(Wgt.sum~as.numeric(.id),data=subset(ldply(c,function(x)
>> summaryBy(Wgt ~ SPCLORatingValue, data=x,
>> FUN=c(sum))),SPCLORatingValue>16),FUN=c(sum),order=FALSE))
>>
>> All help/guidance is much appreciated! Thanks Vince!
>>
>> Sample data example:
>> c<-structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
>> 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083
>> ), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
>> 13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
>> ), row.names = 12:22, class = "data.frame"), V8 = structure(list(
>> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
>> c(0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083), SPCLORatingValue = c(14L, 15L, 15L,
>> 12L, 15L, 12L, 13L, 15L, 14L, 15L, 14L)), .Names = c("WgtBand",
>> "Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
>> V10 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2,
>> 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083
>> ), SPCLORatingValue = c(15L, 13L, 14L, 14L, 13L, 13L, 13L,
>> 15L, 15L, 13L, 14L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
>> ), row.names = 12:22, class = "data.frame"), V44 = structure(list(
>> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
>> c(0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083), SPCLORatingValue = c(13L, 14L,
>> 16L, 15L, 14L, 14L, 18L, 13L, 16L, 15L, 11L)), .Names = c("WgtBand",
>> "Wgt", "SPCLORatingValue"), row.names = 12:22, class = "data.frame"),
>> V2 = structure(list(WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2,
>> 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083
>> ), SPCLORatingValue = c(13L, 14L, 15L, 15L, 15L, 14L, 12L,
>> 16L, 17L, 15L, 19L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
>> ), row.names = 12:22, class = "data.frame")), .Names = c("V5",
>> "V8", "V10", "V44", "V2"))
>> structure(list(V5 = structure(list(WgtBand = c(2, 2, 2, 2, 2,
>> 2, 2, 2, 2, 2, 2), Wgt = c(0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083,
>> 0.0043574552083, 0.0043574552083, 0.0043574552083
>> ), SPCLORatingValue = c(11L, 15L, 14L, 15L, 14L, 14L, 16L, 19L,
>> 13L, 17L, 11L)), .Names = c("WgtBand", "Wgt", "SPCLORatingValue"
>> ), row.names = 12:22, class = "data.frame"), V8 = structure(list(
>> WgtBand = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), Wgt =
>> c(0.0043574552083,
