Re: [R] OT: distribution of a pathological random variate
Thank you Ted, Daniel, and Charles. I thought Cauchy distribution has to do
with resonance. Didn't know this nice extension.
H.
>>> Ted Harding <[EMAIL PROTECTED]> 8/29/2007 11:02 AM >>>
On 29-Aug-07 17:39:17, Horace Tso wrote:
> Folks,
>
> I wonder if anything could be said about the distribution of a random
> variate x, where
>
> x = N(0,1)/N(0,1)
>
> Obviously x is pathological because it could be 0/0. If we exclude this
> point, so the set is {x/(0/0)}, does x have a well defined
> distribution? or does it exist a distribution that approximates x.
>
> (The case could be generalized of course to N(mu1, sigma1)/N(mu2,
> sigma2) and one still couldn't get away from the singularity.)
>
> Any insight or reference to related discussion is appreciated.
>
> Horace Tso
A good question -- but it has a long-established answer. X has the
Cauchy distribution, whose density function is
f(x) = 1/(pi*(1 + x^2))
Have a look at ?dcauchy
It is also the distribution of t with 1 degree of freedom.
See also ?dt
You don;t need to exclude the point (0,0) explicitly, since
it has zero probabilityof occurring. But the chance that the
denominator could be small enough to give a very large value
of X is quite perceptible.
Try
X<-rcauchy(1000)
max(X)
and similar. Play around!
Best wishes,
ted.
E-Mail: (Ted Harding) <[EMAIL PROTECTED]>
Fax-to-email: +44 (0)870 094 0861
Date: 29-Aug-07 Time: 19:02:32
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[R] OT: distribution of a pathological random variate
Folks,
I wonder if anything could be said about the distribution of a random variate
x, where
x = N(0,1)/N(0,1)
Obviously x is pathological because it could be 0/0. If we exclude this point,
so the set is {x/(0/0)}, does x have a well defined distribution? or does it
exist a distribution that approximates x.
(The case could be generalized of course to N(mu1, sigma1)/N(mu2, sigma2) and
one still couldn't get away from the singularity.)
Any insight or reference to related discussion is appreciated.
Horace Tso
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Re: [R] small issue with densityplot
Thank you Prof. Bates and Deepayan. As the title of this thread suggests, it is a minor issue once the user is familar with where to look for hints. But better documentation certainly won't hurt. H. >>> "Douglas Bates" <[EMAIL PROTECTED]> 8/21/2007 1:44:11 PM >>> On 8/21/07, Horace Tso <[EMAIL PROTECTED]> wrote: > Deepayan, you're right. Now I realize anyone could write a densityplot > function to apply on a different class of objects. I guess I should write to > the author of lme4 which from what I could see does not describe a > densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). You're right. It's my bad for not documenting that the coda package is required for plots of the results of mcmcsamp. I was trying to piggy-back on the work that the authors of the coda package had done on diagnostics, etc. It was actually Deepayan and I who added the lattice-based plots to the coda package for exactly this purpose (did you remember that, Deepayan?). In some ways it might be better to remove the dependence on coda and write the classes and methods in the lme4 package. That way I can use S4 classes and irritate all those people who rail against S4 classes and methods (and you know who you are). > >>> "Deepayan Sarkar" <[EMAIL PROTECTED]> 8/21/2007 11:17:39 AM >>> > On 8/21/07, Horace Tso <[EMAIL PROTECTED]> wrote: > > Hi folks, > > > > This is really minor but to someone not familiar with the various tentacles > > of the lmer package it could be really annoying. I was trying to plot the > > posterior density of the fixed effect parameters of a lmer model, > > > > > hr.mcmc = mcmcsamp(hr.lmer, n=5) > > > densityplot(hr.mcmc, plot.points=F) > > > > There is this error, > > > > "Error in densityplot(hr.mcmc, plot.points = F) : > > no applicable method for "densityplot" " > > > > It kind of smells like something I've come across before. So I checked the > > mcmcsamp help page, and alas, the example suggests that the package coda is > > needed. > > > > >From the help page of densityplot alone, there is no way one could figure > > >out this dependency. It says, together with histogram, it is part of > > >lattice. > > > > Could the function author *please* make clarification in future editions of > > lattice. > > There is nothing to clarify. densityplot() is a generic function, and > it is not possible for the author of the generic function to > anticipate and document all possible methods, especially those in > other packages. I would say that since you are using mcmcsamp(), it's > perfectly reasonable to expect you to look at its help page to figure > out what you can do with the results. > > What gave you the idea that densityplot would work on the result of > mcmcsamp in the first place? > > -Deepayan > > __ > R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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@stat.math.ethz.ch 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@stat.math.ethz.ch 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] small issue with densityplot
Deepayan, you're right. Now I realize anyone could write a densityplot function to apply on a different class of objects. I guess I should write to the author of lme4 which from what I could see does not describe a densityplot. By the way, it is used in Baayen, Davidson, and Bates (2006). H. >>> "Deepayan Sarkar" <[EMAIL PROTECTED]> 8/21/2007 11:17:39 AM >>> On 8/21/07, Horace Tso <[EMAIL PROTECTED]> wrote: > Hi folks, > > This is really minor but to someone not familiar with the various tentacles > of the lmer package it could be really annoying. I was trying to plot the > posterior density of the fixed effect parameters of a lmer model, > > > hr.mcmc = mcmcsamp(hr.lmer, n=5) > > densityplot(hr.mcmc, plot.points=F) > > There is this error, > > "Error in densityplot(hr.mcmc, plot.points = F) : > no applicable method for "densityplot" " > > It kind of smells like something I've come across before. So I checked the > mcmcsamp help page, and alas, the example suggests that the package coda is > needed. > > >From the help page of densityplot alone, there is no way one could figure > >out this dependency. It says, together with histogram, it is part of lattice. > > Could the function author *please* make clarification in future editions of > lattice. There is nothing to clarify. densityplot() is a generic function, and it is not possible for the author of the generic function to anticipate and document all possible methods, especially those in other packages. I would say that since you are using mcmcsamp(), it's perfectly reasonable to expect you to look at its help page to figure out what you can do with the results. What gave you the idea that densityplot would work on the result of mcmcsamp in the first place? -Deepayan __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] small issue with densityplot
Hi folks, This is really minor but to someone not familiar with the various tentacles of the lmer package it could be really annoying. I was trying to plot the posterior density of the fixed effect parameters of a lmer model, > hr.mcmc = mcmcsamp(hr.lmer, n=5) > densityplot(hr.mcmc, plot.points=F) There is this error, "Error in densityplot(hr.mcmc, plot.points = F) : no applicable method for "densityplot" " It kind of smells like something I've come across before. So I checked the mcmcsamp help page, and alas, the example suggests that the package coda is needed. >From the help page of densityplot alone, there is no way one could figure out >this dependency. It says, together with histogram, it is part of lattice. Could the function author *please* make clarification in future editions of lattice. Thanks. Horace W. Tso __ R-help@stat.math.ethz.ch 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] pvals.fnc unhappy about lmer objects
Dear folks (or Dear Professor Bates), I'm quite confused as to the current status of some of the available functions applicable to lmer objects. Following the examples in Baayen, Davidson, Bates (2006), my plan is to run mcmcsamp on a random effect model created by lmer in package lme4, then use the (perhaps outdated) pvals to estimate p-value. But then I couldn't find pvals anywhere. So question number one is : Has pvals been replaced? by pvals.fnc in the package languageR perhaps? >From the help page, it's stated that pvals.fnc takes a model fitted with lmer, in contrast with the pvals I've read about in Baayen et al, which takes an mcmc object. So I then tried, hr.lmer = lmer(hr ~ tg + spl + (1+tg|M), data=dat) pvals.fnc(hr.lmer) An error is reported, "Error in get(x, envir, mode, inherits) : variable "Gen.2001" of mode "function" was not found" First I suspected it must be my model, so I run through the example in pvals.fnc help page, as follow, > data(primingHeid) > primingHeid = primingHeid[primingHeid$RT < 7.1,] > primingHeid = primingHeid[primingHeid$RT < 7.1,] > primingHeid.lmer = lmer(RT ~ RTtoPrime * ResponseToPrime + + Condition + (1|Subject) + (1|Word), data = primingHeid) > primingHeid.pvals = pvals.fnc(primingHeid.lmer) The same error is encountered, "Error in get(x, envir, mode, inherits) : variable "Gen.2001" of mode "function" was not found" Thanks in advance. Horace W. Tso *- PS: I'm on Windows XP, with R2.5.1 (2007-06-27) > sessionInfo() R version 2.5.1 (2007-06-27) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] "splines" "graphics" "grDevices" "datasets" "utils" "stats" "methods" "base" other attached packages: languageRrpart MASS Design survival Hmisce1071class cluster "0.2" "3.1-37" "7.2-34" "2.1-1" "2.32" "3.4-2" "1.5-16" "7.2-34" "1.11.7" zipfR coda xlsReadWrite lme4 Matrix lattice R.oo zoo "0.6-0" "0.12-1" "1.3.2" "0.99875-7" "0.999375-1" "0.15-11" "1.2.7" "1.2-2" > __ R-help@stat.math.ethz.ch 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] Constructing correlation matrices (follow up)
Greg, in light of Doug Bates' question, what i have suggested a little early in response to your question is known as a Wishart matrix with n degree of freedom, which is guarenteed to be positive definite. If this is not what you want, you have to be more specific about the property of this correlation matrix you want to simulate. H. = Greg, I take it that you're trying to generate a random correlation matrix, so first create a covariance matrix, p = 6 v = matrix(rnorm(p*p), ncol=p) cov = t(v) %*% v Then convert it to a correlation matrix, cov2cor(cov) HTH. Horace >>> Gregory Gentlemen <[EMAIL PROTECTED]> 7/29/2007 7:31:36 PM >>> Greetings, I have a seemingly simple task which I have not been able to solve today and I checked all of the help archives on this and have been unable to find anything useful. I want to construct a symmetric matrix of arbtriray size w/o using loops. The following I thought would do it: p <- 6 Rmat <- diag(p) dat.cor <- rnorm(p*(p-1)/2) Rmat[outer(1:p, 1:p, "<")] <- Rmat[outer(1:p, 1:p, ">")] <- dat.cor However, the problem is that the matrix is filled by column and so the resulting matrix is not symmetric. I'd be grateful for any adive and/or solutions. Gregory - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] getting the name of variables passed to a function
Thanks to Prof. Brian Ripley, Marc, and Andy. match.call() is what i need.
After spending a weekend away from email, I stumbled on a thread on exactly the
same topic this morning,
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/101445.html
Horace
>>> "Liaw, Andy" <[EMAIL PROTECTED]> 7/30/2007 9:57:47 AM >>>
Here's one possibility:
R> f <- function(...) { call <- match.call(); sapply(as.list(call[-1]),
deparse) }
R> f(x, y)
[1] "x" "y"
R> f(x=x, y=y)
x y
"x" "y"
You basically need to know how to manipulate call objects. The relevant
section in the R Language Definition should help.
Andy
From: Horace Tso
>
> Folks,
>
> I've entered into an R programming territory I'm not very
> familiar with, thus this probably very elementary question
> concerning the mechanic of a function call.
>
> I want to know from within a function the name of the
> variables I pass down. The function makes use of the "..." to
> allow for multiple unknown arguments,
>
> myfun = function(...) { do something }
>
> In the body I put,
>
> {
> nm <- names(list(...))
> nm
> }
>
> When the function is called with two vectors x, and y
>
> myfun(x, y)
>
> It returns NULL. However, when the call made is,
>
> >myfun(x=x, y=y)
>
> The result is
> [1] "x" "y"
>
> Question : how do i get the names of the unknown variables
> without explicitly saying x=x...
>
> Thanks in advance.
>
> Horace
>
> __
> R-help@stat.math.ethz.ch 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.
>
>
>
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Re: [R] Constructing correlation matrices
Greg, I take it that you're trying to generate a random correlation matrix, so first create a covariance matrix, p = 6 v = matrix(rnorm(p*p), ncol=p) cov = t(v) %*% v Then convert it to a correlation matrix, cov2cor(cov) HTH. Horace >>> Gregory Gentlemen <[EMAIL PROTECTED]> 7/29/2007 7:31:36 PM >>> Greetings, I have a seemingly simple task which I have not been able to solve today and I checked all of the help archives on this and have been unable to find anything useful. I want to construct a symmetric matrix of arbtriray size w/o using loops. The following I thought would do it: p <- 6 Rmat <- diag(p) dat.cor <- rnorm(p*(p-1)/2) Rmat[outer(1:p, 1:p, "<")] <- Rmat[outer(1:p, 1:p, ">")] <- dat.cor However, the problem is that the matrix is filled by column and so the resulting matrix is not symmetric. I'd be grateful for any adive and/or solutions. Gregory - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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 the name of variables passed to a function
Folks,
I've entered into an R programming territory I'm not very familiar with, thus
this probably very elementary question concerning the mechanic of a function
call.
I want to know from within a function the name of the variables I pass down.
The function makes use of the "..." to allow for multiple unknown arguments,
myfun = function(...) { do something }
In the body I put,
{
nm <- names(list(...))
nm
}
When the function is called with two vectors x, and y
myfun(x, y)
It returns NULL. However, when the call made is,
>myfun(x=x, y=y)
The result is
[1] "x" "y"
Question : how do i get the names of the unknown variables without explicitly
saying x=x...
Thanks in advance.
Horace
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Re: [R] create matrix from comparing two vectors
In case you really want to use the apply variety, here is another one, sapply(fac,function(x)ifelse(test/x<1, 1, 0)) H. >>> "Van Campenhout Bjorn" <[EMAIL PROTECTED]> 6/26/2007 11:13:14 AM >>> hi all, sorry for this basic question, I think I know I should use ?apply, but it is really confusing me... I want to create a matrix by comparing two vectors. Eg: test<-seq(1:10) fac<-c(3,6,9) and i want to end up with a 10*3 matrix with a boolean that tests if testhttps://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@stat.math.ethz.ch 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] Imputing missing values in time series
Thanks to Mark and Erik for different versions of locf, also Erik's pointer to
archive where I found another function due to Simon Fear. I haven't tested the
zoo locf function. The following shows their performance. Interestingly, Erik's
use of a while loop is the fastest.
HT.
x = 1:1e5
x[sample(1:1e5, 1)] = NA
>system.time(z2<-locf.iverson2(x))
user system elapsed
0.070.000.06
> system.time(z1<-locf.iverson(x))
user system elapsed
0.110.000.11
> system.time(z3<-locf.sfear(x))
user system elapsed
1.130.001.12
==
# Due to Erik Iverson
locf.iverson2 = function(x) {
while(any(is.na(x))) {
x[is.na(x)] <- x[which(is.na(x))-1]
}
x
}
# Due to Simon Fear (Fri Nov 14 17:28:57 2003)
locf.sfear = function(x) {
assign("stored.value", x[1], envir=.GlobalEnv)
sapply(x, function(x) {
if(is.na(x))
stored.value
else {
assign("stored.value", x, envir=.GlobalEnv)
x
}})
}
# Due to Erik Iverson
locf.iverson = function(x, unkn=-1) {
x[is.na(x)] = unkn #something that is not a possible price
run = rle(x)
run$values[run$values==unkn] = run$values[which(run$values==unkn)-1]
inverse.rle(run)
}
>>> "Horace Tso" <[EMAIL PROTECTED]> 6/22/2007 12:21 PM >>>
Mark, thanks for the tips. I thought you financial folks must have run into
things like these before. Just wonder why this problem wasn't asked more often
on this list.
H.
>>> "Leeds, Mark (IED)" <[EMAIL PROTECTED]> 6/22/2007 12:16 PM >>>
I have a function that does this type of thing but it works off a pure
vector so it wouldn have to be modified.
If you make your object a zoo object, the that object has many functions
associated with it and na.locf would
Do what you need, I think.
-Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erik Iverson
Sent: Friday, June 22, 2007 3:02 PM
To: Horace Tso
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Imputing missing values in time series
I think my example should work for you, but I couldn't think of a way to
do this without an interative while loop.
test <- c(1,2,3,NA,4,NA,NA,5,NA,6,7,NA)
while(any(is.na(test)))
test[is.na(test)] <- test[which(is.na(test))-1]
test
[1] 1 2 3 3 4 4 4 5 5 6 7 7
Horace Tso wrote:
> Folks,
>
> This must be a rather common problem with real life time series data
> but I don't see anything in the archive about how to deal with it. I
> have a time series of natural gas prices by flow date. Since gas is
> not traded on weekends and holidays, I have a lot of missing values,
>
> FDate Price
> 11/1/2006 6.28
> 11/2/2006 6.58
> 11/3/2006 6.586
> 11/4/2006 6.716
> 11/5/2006 NA
> 11/6/2006 NA
> 11/7/2006 6.262
> 11/8/2006 6.27
> 11/9/2006 6.696
> 11/10/20066.729
> 11/11/20066.487
> 11/12/2006NA
> 11/13/2006NA
> 11/14/20066.725
> 11/15/20066.844
> 11/16/20066.907
>
> What I would like to do is to fill the NAs with the price from the
> previous date * gas used during holidays is purchased from the week
> before. Though real simple, I wonder if there is a function to perform
> this task. Some of the imputation functions I'm aware of (eg. impute,
> transcan in Hmisc) seem to deal with completely different problems.
>
> 2.5.0/Windows XP
>
> Thanks in advance.
>
> HT
>
> __
> R-help@stat.math.ethz.ch 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.
__
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This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
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Re: [R] Imputing missing values in time series
Mark, thanks for the tips. I thought you financial folks must have run into
things like these before. Just wonder why this problem wasn't asked more often
on this list.
H.
>>> "Leeds, Mark (IED)" <[EMAIL PROTECTED]> 6/22/2007 12:16 PM >>>
I have a function that does this type of thing but it works off a pure
vector so it wouldn have to be modified.
If you make your object a zoo object, the that object has many functions
associated with it and na.locf would
Do what you need, I think.
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Erik Iverson
Sent: Friday, June 22, 2007 3:02 PM
To: Horace Tso
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Imputing missing values in time series
I think my example should work for you, but I couldn't think of a way to
do this without an interative while loop.
test <- c(1,2,3,NA,4,NA,NA,5,NA,6,7,NA)
while(any(is.na(test)))
test[is.na(test)] <- test[which(is.na(test))-1]
test
[1] 1 2 3 3 4 4 4 5 5 6 7 7
Horace Tso wrote:
> Folks,
>
> This must be a rather common problem with real life time series data
> but I don't see anything in the archive about how to deal with it. I
> have a time series of natural gas prices by flow date. Since gas is
> not traded on weekends and holidays, I have a lot of missing values,
>
> FDate Price
> 11/1/2006 6.28
> 11/2/2006 6.58
> 11/3/2006 6.586
> 11/4/2006 6.716
> 11/5/2006 NA
> 11/6/2006 NA
> 11/7/2006 6.262
> 11/8/2006 6.27
> 11/9/2006 6.696
> 11/10/20066.729
> 11/11/20066.487
> 11/12/2006NA
> 11/13/2006NA
> 11/14/20066.725
> 11/15/20066.844
> 11/16/20066.907
>
> What I would like to do is to fill the NAs with the price from the
> previous date * gas used during holidays is purchased from the week
> before. Though real simple, I wonder if there is a function to perform
> this task. Some of the imputation functions I'm aware of (eg. impute,
> transcan in Hmisc) seem to deal with completely different problems.
>
> 2.5.0/Windows XP
>
> Thanks in advance.
>
> HT
>
> __
> R-help@stat.math.ethz.ch 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.
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Re: [R] Imputing missing values in time series
Erik, indeed it gets the work done. I was hoping to avoid the dreaded looping, though. Thanks. Horace >>> Erik Iverson <[EMAIL PROTECTED]> 6/22/2007 12:01 PM >>> I think my example should work for you, but I couldn't think of a way to do this without an interative while loop. test <- c(1,2,3,NA,4,NA,NA,5,NA,6,7,NA) while(any(is.na(test))) test[is.na(test)] <- test[which(is.na(test))-1] test [1] 1 2 3 3 4 4 4 5 5 6 7 7 Horace Tso wrote: > Folks, > > This must be a rather common problem with real life time series data > but I don't see anything in the archive about how to deal with it. I > have a time series of natural gas prices by flow date. Since gas is not > traded on weekends and holidays, I have a lot of missing values, > > FDate Price > 11/1/2006 6.28 > 11/2/2006 6.58 > 11/3/2006 6.586 > 11/4/2006 6.716 > 11/5/2006 NA > 11/6/2006 NA > 11/7/2006 6.262 > 11/8/2006 6.27 > 11/9/2006 6.696 > 11/10/20066.729 > 11/11/20066.487 > 11/12/2006NA > 11/13/2006NA > 11/14/20066.725 > 11/15/20066.844 > 11/16/20066.907 > > What I would like to do is to fill the NAs with the price from the > previous date * gas used during holidays is purchased from the week > before. Though real simple, I wonder if there is a function to perform > this task. Some of the imputation functions I'm aware of (eg. impute, > transcan in Hmisc) seem to deal with completely different problems. > > 2.5.0/Windows XP > > Thanks in advance. > > HT > > __ > R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Imputing missing values in time series
Folks, This must be a rather common problem with real life time series data but I don't see anything in the archive about how to deal with it. I have a time series of natural gas prices by flow date. Since gas is not traded on weekends and holidays, I have a lot of missing values, FDate Price 11/1/2006 6.28 11/2/2006 6.58 11/3/2006 6.586 11/4/2006 6.716 11/5/2006 NA 11/6/2006 NA 11/7/2006 6.262 11/8/2006 6.27 11/9/2006 6.696 11/10/2006 6.729 11/11/2006 6.487 11/12/2006 NA 11/13/2006 NA 11/14/2006 6.725 11/15/2006 6.844 11/16/2006 6.907 What I would like to do is to fill the NAs with the price from the previous date * gas used during holidays is purchased from the week before. Though real simple, I wonder if there is a function to perform this task. Some of the imputation functions I'm aware of (eg. impute, transcan in Hmisc) seem to deal with completely different problems. 2.5.0/Windows XP Thanks in advance. HT __ R-help@stat.math.ethz.ch 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] Viewing a data object
Stephen and Christophe,
I'm aware of fix and edit and the few issues with fix. Thus my reluctance to
use them. Emacs may be the way to go, but from what I heard here it has a steep
learning curve. The autocompletion feature in 2.5.1 is great. Andy Liaw points
me to JGR which I'm just about to jump in.
I was hoping something like the head/tail function with a little more
flexibility may also be useful to a lot of folks here.
vw = function( , location, nlines=10 ) {
#'...' gives the string fragments that identify an object
# location : 0.5 = middle, 0.25 = the first quartile, etc
# nlines : the number of lines to show
}
vw(Auro, 0.5)
returns the middle part of the first data frame it finds with name string
starting with "Auro". This function should be easy to write with all thse
regular expression functions.
Well, I'll save it for my Christmas wish list.
H.
>>> Stephen Tucker <[EMAIL PROTECTED]> 6/12/2007 11:25 PM >>>
Hi Horace,
I have also thought that it may be useful but I don't know of any Object
Explorer available for R.
However, (you may alread know this but)
(1) you can view your list of objects in R with objects(),
(2) view objects in a spreadsheet-like table (if they are matrices or data
frames) with invisible(edit(objectName)) [which isn't easy on the fingers].
fix(objectName) is also a shorter option but it has the side effect of
possibly changing your object when you close the viewing data. For instance,
this can happen if you mistakenly type something into a cell; it can also
change your column classes when you don't - for example:
> options(stringsAsFactors=TRUE)
> x <- data.frame(letters[1:5],1:5)
> sapply(x,class)
letters.1.5. X1.5
"factor""integer"
> fix(x) # no user-changes made
> sapply(x,class)
letters.1.5. X1.5
"factor""numeric"
(3) I believe Deepayan Sarkar contributed the tab-completion capability at
the command line. So unless you have a lot of objects beginning with
'AuroraStoch...' you should be able to type a few letters and let the
auto-completion handle the rest.
Best regards,
ST
--- Horace Tso <[EMAIL PROTECTED]> wrote:
> Dear list,
>
> First apologize that this is trivial and just betrays my slothfulness at
> the keyboard. I'm sick of having to type a long name just to get a glimpse
> of something. For example, if my data frame is named
> 'AuroraStochasticRunsJune1.df" and I want to see what the middle looks
> like, I have to type
>
> AuroraStochasticRunsJune1.df[ 400:500, ]
>
> And often I'm not even sure rows 400 to 500 are what I want to see. I
> might have to type the same line many times.
>
> Is there sort of a R-equivalence of the Object Explorer, like in Splus,
> where I could mouse-click an object in a list and a window pops up? Short
> of that, is there any trick of saving a couple of keystrokes here and
> there?
>
> Thanks for tolerating this kind of annoying questions.
>
> H.
>
> __
> R-help@stat.math.ethz.ch 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.
>
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[R] Viewing a data object
Dear list, First apologize that this is trivial and just betrays my slothfulness at the keyboard. I'm sick of having to type a long name just to get a glimpse of something. For example, if my data frame is named 'AuroraStochasticRunsJune1.df" and I want to see what the middle looks like, I have to type AuroraStochasticRunsJune1.df[ 400:500, ] And often I'm not even sure rows 400 to 500 are what I want to see. I might have to type the same line many times. Is there sort of a R-equivalence of the Object Explorer, like in Splus, where I could mouse-click an object in a list and a window pops up? Short of that, is there any trick of saving a couple of keystrokes here and there? Thanks for tolerating this kind of annoying questions. H. __ R-help@stat.math.ethz.ch 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] Excel calling R functions
Thanks to Marc, Duncan, Tobias, Alberto. I'm aware of rcom and have in fact downloaded it somewhere. But unfortunately it won't work for me as I'm in a corporate environment and don't have admin rights to set up the R(D)COM (background) server. If there is a way to work around it, pls let me know. Horace >>> Duncan Temple Lang <[EMAIL PROTECTED]> 6/1/2007 11:27:35 AM >>> And just for completeness for the thread, there is the RDCOMServer package which has a more general mechanism for making R functionality available to other applications that does not require writing R commands but uses the language independent spirit of DCOM. And it also allows event handlers for things like Excel actions to be written as R functions Tobias Verbeke wrote: > Marc Schwartz wrote: >> On Fri, 2007-06-01 at 08:50 -0700, Horace Tso wrote: >>> Hi folks, >>> >>> Is it possible to have Excel call a R function. If not, how about >>> making Excel send off a command to call a R script and then read the >>> result back into Excel. >>> >>> I know, I know, this should belong to some Excel forum, but i just try >>> my luck here. >>> >>> Thanks in advance. >>> >>> Horace W. Tso >> >> See the R-Excel add-in linked from here: >> >> http://www.sciviews.org/_rgui/projects/RDcom.html >> > > Some of the files listed in the link on that page appear > to be quite outdated. The following link brings you to > the current portal of the R(D)COM server and the > rcom package. > > http://sunsite.univie.ac.at/rcom/ > > See the Excel heading for your question. You might be > interested as well by the recent OpenOffice.org plugin > (heading OOo). > > HTH, > Tobias > -- Duncan Temple Lang[EMAIL PROTECTED] Department of Statistics work: (530) 752-4782 4210 Mathematical Sciences Bldg. fax: (530) 752-7099 One Shields Ave. University of California at Davis Davis, CA 95616, USA __ R-help@stat.math.ethz.ch 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] Excel calling R functions
Charles, thanks for the thought. In my case that's quite impractical/impossible because I'm working from a monstrous spreadsheet having dozens of sheets and numerous links. In some way that's the irreplaceable aspect of a spreadsheet program like Excel. H. >>> "Charles Annis, P.E." <[EMAIL PROTECTED]> 6/1/2007 9:03:20 AM >>> You might consider having R do everything: R can read the Excel sheet, do what needs to be done, and write the results to an Excel sheet. Charles Annis, P.E. [EMAIL PROTECTED] phone: 561-352-9699 eFax: 614-455-3265 http://www.StatisticalEngineering.com -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Horace Tso Sent: Friday, June 01, 2007 11:51 AM To: r-help@stat.math.ethz.ch Subject: [R] Excel calling R functions Hi folks, Is it possible to have Excel call a R function. If not, how about making Excel send off a command to call a R script and then read the result back into Excel. I know, I know, this should belong to some Excel forum, but i just try my luck here. Thanks in advance. Horace W. Tso __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Excel calling R functions
Hi folks, Is it possible to have Excel call a R function. If not, how about making Excel send off a command to call a R script and then read the result back into Excel. I know, I know, this should belong to some Excel forum, but i just try my luck here. Thanks in advance. Horace W. Tso __ R-help@stat.math.ethz.ch 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] converting a row of a data.frame to a vector
probably something like, unlist(x[1,]) HTH. H. >>> "Andrew Yee" <[EMAIL PROTECTED]> 5/15/2007 4:37 PM >>> I've searched for the answer to this in the help list archive, but wasn't able to get the answer to work. I'm interested in converting a row of a data.frame into a vector. However, when I use as.vector(x,[1,]) I get another data.frame, instead of a vector. (On the other hand, when I use as.vector(x,[,1]), I get a vector.) Thanks, Andrew [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] statistics/correlation question NOT R question
Mark, I suppose you make the usual assumptions, ie. E[x]=0, E[x*epsilon]=0, the
correlation is just simply,
corr(x,y) = beta * ( var(x) / var(y) )
And you could get var(y) from var(x) and var(epsilon).
HTH.
Horace
>>> "Leeds, Mark (IED)" <[EMAIL PROTECTED]> 5/8/2007 10:25:11 AM >>>
This is not an R question but if anyone can help me, it's much
appreciated.
Suppose I have a series ( stationary ) y_t and a series x_t ( stationary
)and x_t has variance sigma^2_x and epsilon is normal
(0, sigma^2_epsilon )
and the two series have the relation
y_t = Beta*x_t + epsilon
My question is if there are particular values that sigma^2_x and
sigma^2_epsilon have to take in order for corr(x_t,y_t) to equal Beta ?
I attempted to figure this out using two different methods and in one
case I end up involving sigma^2_epsilon and in the other I don't
and I'm not sure if either method is correct. I think I need to use
results form the conditional bivariate normal but i'm really not sure.
Also, it's not a homework problem because I am too old to have homework.
Thanks for any insights/solutions.
This is not an offer (or solicitation of an offer) to buy/se...{{dropped}}
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Re: [R] How to identify and exclude the outliers with R?
It depends on the nature of your data set. There is a package simply called 'outliers', which has the Grubbs/Dixon/Cochran tests. There is also the Bonferroni outlier test in 'car' package. I'm sure there are more in the hundreds of packages on CRAN. HTH Horace >>> Shao <[EMAIL PROTECTED]> 4/25/2007 6:27:37 AM >>> Hello, everyone, I want to ask a simple question. If I have a set of data,and I want to identify how many outliers there are in the data.Which packages and functions can I use? Thanks. Shao chunxuan. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Log-Returns
diff(log(x)) And this should be made into a function. H. >>> "Soare Marcian-Alin" <[EMAIL PROTECTED]> 4/24/2007 2:17 PM >>> Hello, I have a Problem to make Log-Returns of the dataset EuStockMarkets. Is there any function which could calculate it for me? data(EuStockMarkets) Thanks! Alin Soare [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Size of an object in workspace
Aaaah, wonderful. Thanks Ben. lls appears below. Should these functions be a
permanent part of R? Just a thought.
H.
> lls
function (pos = 1, pat = "") {
dimx <- function(dd) if (is.null(dim(dd)))
length(dd)
else dim(dd)
lll <- ls(pos = pos, pat = pat)
cat(formatC("mode", 1, 15), formatC("class", 1, 18), formatC("name",
1, max(nchar(lll)) + 1),
"size\n-\n")
if (length(lll) > 0) {
for (i in 1:length(lll)) {
cat(formatC(eval(parse(t = paste("mode(", lll[i],
")"))), 1, 15), formatC(paste(eval(parse(t = paste("class(",
lll[i], ")"))), collapse = " "), 1, 18), formatC(lll[i],
1, max(nchar(lll)) + 1), " ", eval(parse(t = paste("dimx(",
lll[i], ")"))), "\n")
}
}
}
>>> "Ben Fairbank" <[EMAIL PROTECTED]> 4/24/2007 9:10:20 AM >>>
Here is a copy of a correspondence I had with Jim Holtman -- his
function, which I now use almost daily, does exactly what I think you
need. Could you send me a copy of Bendix Carstensen's function>
Thanks,
Ben Fairbank
From: jim holtman <[EMAIL PROTECTED]>
To: Ben Fairbank <[EMAIL PROTECTED]>
Date: Sep 27 2006 - 4:42pm
Thanks for the suggestion. I did.
On 9/27/06, Ben Fairbank <[EMAIL PROTECTED]> wrote:
> Nicely done. May I suggest you post it to the list? Others probably
> would benefit also. Or maybe I am the only user dense enough not to
see
> how to write it.
>
> Ben
>
> -Original Message-
> From: jim holtman [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, September 27, 2006 4:36 PM
> To: Ben Fairbank
> Subject: Re: [R] Space required by object?
>
> That was a function I wrote when I needed a similar thing.
>
> On 9/27/06, Ben Fairbank <[EMAIL PROTECTED]> wrote:
> > Thank you! Exactly what I needed!
> >
> > Is that something you wrote?
> >
> > Ben Fairbank
> >
> > -Original Message-
> > From: jim holtman [mailto:[EMAIL PROTECTED]
> > Sent: Wednesday, September 27, 2006 4:23 PM
> > To: Ben Fairbank
> > Subject: Re: [R] Space required by object?
> >
> > try this one:
> >
Note -- Horace, the function starts here --
my.ls <- function(pos=1, sorted=F){
.result <- sapply(ls(pos=pos, all.names=TRUE),
function(..x)object.size(eval(as.symbol(..x
if (sorted){
.result <- rev(sort(.result))
}
.ls <-
as.data.frame(rbind(as.matrix(.result),"**Total"=sum(.result)))
names(.ls) <- "Size"
.ls$Size <- formatC(.ls$Size, big.mark=',', digits=0, format='f')
.ls$Mode <- c(unlist(lapply(rownames(.ls)[-nrow(.ls)],
function(x)mode(eval(as.symbol(x), '---')
.ls
}
And ends there
> >
> >
> > On 9/27/06, Ben Fairbank <[EMAIL PROTECTED]> wrote:
> > > Does R provide a function analogous to LS() or str() that reports
> the
> > > storage space, on disk or in memory, required by objects?
> > >
> > > Ben Fairbank
> > >
> > > __________
> > > R-help@stat.math.ethz.ch 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.
> > >
> >
> >
> > --
> > Jim Holtman
> > Cincinnati, OH
> > +1 513 646 9390
> >
> > What is the problem you are trying to solve?
> >
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Horace Tso
Sent: Tuesday, April 24, 2007 10:59 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Size of an object in workspace
Hi folks,
Is there a function to show the size of an R object? eg. in Kbytes?
Couple months ago Bendix Carstensen posted this marvelous little
function lls(), which shows all objects in the current workspace by
mode, class and 'size'. This is a wonderful enhancement to the build-in
ls() already and I now have it sourced in my Rprofile.site at startup.
The only drawback is, 'size' is just the length/dim of an object. For
matrices and data frames this is good enough. But for a list, knowing
how many elements in there doesn't help much. I need to know the
totality of the content in a common unit, eg. byte.
Thanks in advance.
Horace
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[R] Size of an object in workspace
Hi folks, Is there a function to show the size of an R object? eg. in Kbytes? Couple months ago Bendix Carstensen posted this marvelous little function lls(), which shows all objects in the current workspace by mode, class and 'size'. This is a wonderful enhancement to the build-in ls() already and I now have it sourced in my Rprofile.site at startup. The only drawback is, 'size' is just the length/dim of an object. For matrices and data frames this is good enough. But for a list, knowing how many elements in there doesn't help much. I need to know the totality of the content in a common unit, eg. byte. Thanks in advance. Horace __ R-help@stat.math.ethz.ch 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] Bonferroni p-value greater than 1
Thank you John and Peter. Peter, yes I'm guilty of tacking onto a random mail. I thought you couldn't tell since I got ride of the text from the last mail. Apologize. H. >>> "John Fox" <[EMAIL PROTECTED]> 3/28/2007 5:37 PM >>> Dear Horace, The Bonferonni p-value is obtained from the "unadjusted" p-value by multiplying the latter by the number of observations, and provides a conservative (although usually quite accurate) outlier test. When the adjusted p-value exceeds 1 you can take that as an indication that there are no unusually large studentized residuals (and indeed that the largest studentized residual is smaller than one would expect under the standard linear-model assumptions). I hope this helps, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Horace Tso > Sent: Wednesday, March 28, 2007 6:36 PM > To: 'R R-help' > Subject: [R] Bonferroni p-value greater than 1 > > Hi folks, > > I use the outlier.test in package car to test a lm model and > the bonferroni p value returned is shown as NA. When the > object is typed it indicates the p value is greater than 1. > I'm not sure how to interpret it. > > Thanks in advance. > > Horace W. Tso > > > > outlier.test(mod)$test > max|rstudent|df unadjusted p Bonferroni p >2.04106376 18.0.05618628NA > > > outlier.test(mod) > > max|rstudent| = 2.041064, degrees of freedom = 18, > unadjusted p = 0.05618628, Bonferroni p > 1 > > Observation: 1 > > The lm model looks fine to me, > > > summary(mod) > > Call: > lm(formula = x ~ ind, na.action = na.fail) > > Residuals: > Min 1Q Median 3Q Max > -1.2082 -0.5200 0.1309 0.5725 0.9593 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > (Intercept) 59.845860.31900 187.6 < 2e-16 *** > ind -0.167680.02541-6.6 2.57e-06 *** > --- > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.705 on 19 degrees of freedom > Multiple R-Squared: 0.6963, Adjusted R-squared: 0.6803 > F-statistic: 43.56 on 1 and 19 DF, p-value: 2.57 > > __ > R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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] Bonferroni p-value greater than 1
Hi folks, I use the outlier.test in package car to test a lm model and the bonferroni p value returned is shown as NA. When the object is typed it indicates the p value is greater than 1. I'm not sure how to interpret it. Thanks in advance. Horace W. Tso > outlier.test(mod)$test max|rstudent|df unadjusted p Bonferroni p 2.04106376 18.0.05618628NA > outlier.test(mod) max|rstudent| = 2.041064, degrees of freedom = 18, unadjusted p = 0.05618628, Bonferroni p > 1 Observation: 1 The lm model looks fine to me, > summary(mod) Call: lm(formula = x ~ ind, na.action = na.fail) Residuals: Min 1Q Median 3Q Max -1.2082 -0.5200 0.1309 0.5725 0.9593 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 59.845860.31900 187.6 < 2e-16 *** ind -0.167680.02541-6.6 2.57e-06 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.705 on 19 degrees of freedom Multiple R-Squared: 0.6963, Adjusted R-squared: 0.6803 F-statistic: 43.56 on 1 and 19 DF, p-value: 2.57 __ R-help@stat.math.ethz.ch 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] qr decomposition issue inside lm (solved)
Folks, apologize for such an obvious oversight on my part. The reason qr
fails is, one of the data points has value of -Inf (response is actually
the log of something, and I have a zero in the original set). That
explains the error message in call to dqrls. I should have taken the
mean of the response before proceeding and that would tell me right away
what's wrong.
Thanks.
H.
>>> "Horace Tso" <[EMAIL PROTECTED]> 3/6/2007 1:45:28 PM >>>
Dear list,
It's never happened to me before in such a simple exercise but is not
going away and I've checked my data are good. I want a simple lm model
with one response and one predictor, where N is about 4,200 * data set
not exactly small. Both x and y are nice, continuous variables having NA
filtered out with a call to na.omit. So I did
mod = lm( y ~ x, data=x1)
Then the error,
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...)
:
NA/NaN/Inf in foreign function call (arg 4)
I did a trace back and it turned out it's an error thrown by the
Fortran subroutine that seems to be trying a QR decomposition,
traceback()
3: .Fortran("dqrls", qr = x, n = n, p = p, y = y, ny = ny, tol =
as.double(tol),
coefficients = mat.or.vec(p, ny), residuals = y, effects = y,
rank = integer(1), pivot = 1:p, qraux = double(p), work =
double(2 *
p), PACKAGE = "base")
2: lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...)
1: lm(log.p.sales ~ log.mktcap, data = x1)
My question is why would QR fail since the default in lm.fit is
'singular.ok' ? Furthermore, is there a way to get around presumably a
singularity in my design matrix?
Thanks in advance.
Horace W. Tso
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[R] qr decomposition issue inside lm
Dear list,
It's never happened to me before in such a simple exercise but is not going
away and I've checked my data are good. I want a simple lm model with one
response and one predictor, where N is about 4,200 * data set not exactly
small. Both x and y are nice, continuous variables having NA filtered out with
a call to na.omit. So I did
mod = lm( y ~ x, data=x1)
Then the error,
Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) :
NA/NaN/Inf in foreign function call (arg 4)
I did a trace back and it turned out it's an error thrown by the Fortran
subroutine that seems to be trying a QR decomposition,
traceback()
3: .Fortran("dqrls", qr = x, n = n, p = p, y = y, ny = ny, tol =
as.double(tol),
coefficients = mat.or.vec(p, ny), residuals = y, effects = y,
rank = integer(1), pivot = 1:p, qraux = double(p), work = double(2 *
p), PACKAGE = "base")
2: lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...)
1: lm(log.p.sales ~ log.mktcap, data = x1)
My question is why would QR fail since the default in lm.fit is 'singular.ok' ?
Furthermore, is there a way to get around presumably a singularity in my design
matrix?
Thanks in advance.
Horace W. Tso
__
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Re: [R] putting a column name on a zoo object
Mark, it's not the most elegant solution but here it is,
I have a zoo with a vector of numeric values,
> head(spc.z)
2000-01-03 2000-01-04 2000-01-05 2000-01-06 2000-01-07 2000-01-10
1455.221399.421402.111403.451441.471457.60
> class(spc.z)
[1] "zoo"
Now I want to give it a name,
z = zoo(data.frame(mydata=coredata(spc.z)), index(spc.z))
> head(z)
mydata
2000-01-03 1455.22
2000-01-04 1399.42
2000-01-05 1402.11
2000-01-06 1403.45
2000-01-07 1441.47
2000-01-10 1457.60
HTH
H.
>>> "Leeds, Mark (IED)" <[EMAIL PROTECTED]> 11/14/2006
1:49:24 PM >>>
I only have one column so when I try your suggestion, I get dsays,
"attempt to set colnames on object with less than two dimensions.
It is of class zoo but I guess something changes during that addition
that I am too much of a novice to understand.
This is what a dput gives. There is no dim in the dput which is
probably
what the problem is but I don't know how to fix it. Thanks.
structure(c(4.77126598671015, 4.7716359319335, 4.77176152623894,
4.77196709472722, 4.77218067995761, 4.77213077515643, 4.77200790109121,
4.77203963822312, 4.7719973218234, 4.771903278968, 4.77183227075241,
4.77182803831544, 4.77181216651724, 4.77182803831544, 4.77197193112401,
4.77198938729909, 4.77201612933324, 4.77202905929107, 4.7720184802471,
4.77188623275365, 4.77181745714463, 4.77187036187909, 4.77187036187909,
4.77187036187909, 4.77182803831544, 4.77182803831544, 4.77174338581391,
4.77179982161115, 4.77174338581391, 4.77182199194583, 4.77179417817475,
4.77174338581391, 4.77174338581391, 4.77179982161115, 4.77182803831544,
4.77182803831544, 4.77187036187909, 4.77191268365153, 4.7719338438661,
4.77191268365153, 4.77176690111649, 4.77188094249012, 4.77189857659306,
4.77212752040897, 4.77224143026734, 4.77240630536398), index =
structure(c(1144022520,
1144022580, 1144022640, 1144022700, 1144022760, 1144022820, 1144022880,
1144022940, 1144023000, 1144023060, 1144023120, 1144023180, 1144023240,
1144023300, 1144023360, 1144023420, 1144023480, 1144023540, 1144023600,
1144023660, 1144023720, 1144023780, 1144023840, 1144023900, 1144023960,
1144024020, 1144024080, 1144024140, 1144024200, 1144024260, 1144024320,
1144024380, 1144024440, 1144024500, 1144024560, 1144024620, 1144024680,
1144024740, 1144024800, 1144024860, 1144024920, 1144024980, 1144025040,
1144025100, 1144025160, 1144025220), class = c("POSIXt", "POSIXct"
)), class = "zoo", .Names = c("1144022520", "1144022580", "1144022640",
"1144022700", "1144022760", "1144022820", "1144022880", "1144022940",
"1144023000", "1144023060", "1144023120", "1144023180", "1144023240",
"1144023300", "1144023360", "1144023420", "1144023480", "1144023540",
"1144023600", "1144023660", "1144023720", "1144023780", "1144023840",
"1144023900", "1144023960", "1144024020", "1144024080", "1144024140",
"1144024200", "1144024260", "1144024320", "1144024380", "1144024440",
"1144024500", "1144024560", "1144024620", "1144024680", "1144024740",
"1144024800", "1144024860", "1144024920", "1144024980", "1144025040",
"1144025100", "1144025160", "1144025220"))
-Original Message-
From: Leeds, Mark (IED)
Sent: Tuesday, November 14, 2006 4:39 PM
To: 'Horace Tso'; r-help@stat.math.ethz.ch
Subject: RE: [R] putting a column name on a zoo object
let me try colnames. I thought I tried that earlier and got an error
but I've been wrong before.
Thanks.
-Original Message-
From: Horace Tso [mailto:[EMAIL PROTECTED]
Sent: Tuesday, November 14, 2006 4:29 PM
To: Leeds, Mark (IED); r-help@stat.math.ethz.ch
Subject: Re: [R] putting a column name on a zoo object
Mark,
What's wrong with the following,
> head(sp.z)
OpenHigh Low Close Volume Adj..Close.
2000-01-03 1469.25 1478.00 1438.36 1455.22 93180 1455.22
2000-01-04 1455.22 1455.22 1397.43 1399.42 100900 1399.42
2000-01-05 1399.42 1413.27 1377.68 1402.11 1085500032 1402.11
2000-01-06 1402.11 1411.90 1392.10 1403.45 1092300032 1403.45
2000-01-07 1403.45 1441.47 1400.73 1441.47 122520 1441.47
2000-01-10 1441.47 1464.36 1441.47 1457.60 106480 1457.60
colnames(sp.z)
[1] "Open""High""Low" "Close" "Volume"
[6] "Adj..Close."
> colnames(sp.z)[1] = "OPEN"
> colnames(sp.z)
[1] "OPEN"
Re: [R] putting a column name on a zoo object
Mark,
What's wrong with the following,
> head(sp.z)
OpenHigh Low Close Volume Adj..Close.
2000-01-03 1469.25 1478.00 1438.36 1455.22 93180 1455.22
2000-01-04 1455.22 1455.22 1397.43 1399.42 100900 1399.42
2000-01-05 1399.42 1413.27 1377.68 1402.11 1085500032 1402.11
2000-01-06 1402.11 1411.90 1392.10 1403.45 1092300032 1403.45
2000-01-07 1403.45 1441.47 1400.73 1441.47 122520 1441.47
2000-01-10 1441.47 1464.36 1441.47 1457.60 106480 1457.60
colnames(sp.z)
[1] "Open""High""Low" "Close" "Volume"
[6] "Adj..Close."
> colnames(sp.z)[1] = "OPEN"
> colnames(sp.z)
[1] "OPEN""High""Low" "Close" "Volume"
[6] "Adj..Close."
HTH.
Horace W. Tso
>>> "Leeds, Mark (IED)" <[EMAIL PROTECTED]> 11/14/2006
1:21:33 PM >>>
does anyone know how to put a column name on a zoo object. I think
achim
and gabor are off line or they have gotten totally tired of me
an decided to ignore me ( which is totalyy understandable ).
logbidask<-log((aggfxdata[,"bid"] + aggfxdata[,"ask"])/2.0)
logbidask doesn't have a name and I can't figure out how to get one on
it ?
aggfxdata is a zoo object.
This is not an offer (or solicitation of an offer) to
buy/se...{{dropped}}
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[R] Including time in a zoo
dear list, I have these hourly price data over a 20 year period. Among other things, I want to plot them with time on the x-axis. The data is in a data frame, > head(OWI.SE.all) Record_Number Name Condition Time_PeriodPrice 1312 WECC-OWI SouthEast All 1/1/2012 Hour: 1 41.21383 2827 WECC-OWI SouthEast All 1/1/2012 Hour: 2 38.38091 4342 WECC-OWI SouthEast All 1/1/2012 Hour: 3 39.97879 5857 WECC-OWI SouthEast All 1/1/2012 Hour: 4 40.14156 7372 WECC-OWI SouthEast All 1/1/2012 Hour: 5 38.21092 8887 WECC-OWI SouthEast All 1/1/2012 Hour: 6 40.09152 And the date-times are in a POSIX object dte1, > class(dte1) [1] "POSIXt" "POSIXlt" When I tried to create a zoo with dte1 as the index, I got an error, > zoo(OWI.SE.all, dte1) Error in order(x, ..., na.last = na.last, decreasing = decreasing) : unimplemented type 'list' in 'orderVector1' I guess that's saying the index of a zoo can't have a time component. Am I correct? If so, is there a way to include an hour component in a zoo object. Second question: If I just attach this dte1 to my data frame as a column named Date and try plotting it as, > plot(OWI.SE.all$Date, OWI.SE.all$Price, type="l") I got non-sensical result. How do I plot hourly data in general? TIA. Horace W. Tso __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rolling around rollapply in a zoo
Hi list,
I'm a little confused about rollapply in zoo. There is written that
rollapply has replaced rapply in version 1.2-0 which is what i have. But
when I tried,
> rollapply(1:100, 10, mean)
I got,
Error in rollapply(1:100, 10, mean) : no applicable method for
"rollapply"
So I went back to the good old rapply
> rapply(1:100, 10, mean)
Error in rapply(1:100, 10, mean) : no applicable method for
"rollapply"
In addition: Warning message:
'rapply' is deprecated.
Use 'rollapply' instead.
See help("Deprecated")
Is there a more updated version of zoo?
Thanks.
Horace W. Tso
===
>sessionInfo()
Version 2.3.1 (2006-06-01)
i386-pc-mingw32
attached base packages:
[1] "splines" "methods" "datasets" "stats" "tcltk" "utils"
[7] "graphics" "grDevices" "base"
other attached packages:
lmtest Hmisctseries quadprogzoo MASS
Rpad
"0.9-18" "3.0-12" "0.10-1""1.4-8""1.2-0" "7.2-27.1"
"1.1.1"
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[R] Issue with Data Editor and Date objects
Dear list, It's a minor issue but for someone who uses the Data Editor once in a while this could be very annoying. I have a dataframe with a Date column. When I bring it up with the (gui) Data Editor (Edit -> Data editor...), do nothing and then close it, the Date column turns into "numeric". Before using Data Editor, >class(gas.gd$Flow.Date) [1] "Date" After Data Editor, >class(gas.gd$Flow.Date) [1] "numeric" Any way to fix this? TIA. Horace W. Tso Portland General Electric Portland, Oregon __ R-help@stat.math.ethz.ch 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] Quickie : unload library
Aah, that works. The missing "package:..."
H.
>>> Sachin J <[EMAIL PROTECTED]> 8/25/2006 1:16 PM >>>
try detach("package:zoo")
Sachin
Horace Tso <[EMAIL PROTECTED]> wrote:
Sachin,
I did try that, ex
detach(zoo)
Error in detach(zoo) : invalid name
detach("zoo")
Error in detach("zoo") : invalid name
But zoo has been loaded,
sessionInfo()
Version 2.3.1 (2006-06-01)
i386-pc-mingw32
attached base packages:
[1] "methods" "datasets" "stats" "tcltk" "utils"
"graphics"
[7] "grDevices" "base"
other attached packages:
tseries quadprog zoo MASS Rpad
"0.10-1" "1.4-8" "1.2-0" "7.2-27.1" "1.1.1"
Thks,
H.
>>> Sachin J 8/25/2006 12:56 PM >>>
see ?detach
Horace Tso wrote:
Dear list,
I know it must be obvious and I did my homework. (In fact I've
RSiteSearched with keyword "remove AND library" but got timed
out.(why?))
How do I unload a library? I don't mean getting ride of it permanently
but just to unload it for the time being.
A related problem : I have some libraries loaded at startup in
.First()
which I have in .Rprofile. Now, I exited R and commented out the lines
in .First(). Next time I launch R the same libraries are loaded again.
I.e. there seems to be a memory of the old .First() somewhere which
refuses to die.
Thanks in adv.
Horace
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-
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-
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Re: [R] Quickie : unload library
Sachin,
I did try that, ex
detach(zoo)
Error in detach(zoo) : invalid name
detach("zoo")
Error in detach("zoo") : invalid name
But zoo has been loaded,
sessionInfo()
Version 2.3.1 (2006-06-01)
i386-pc-mingw32
attached base packages:
[1] "methods" "datasets" "stats" "tcltk" "utils"
"graphics"
[7] "grDevices" "base"
other attached packages:
tseries quadprogzoo MASS Rpad
"0.10-1""1.4-8" "1.2-0" "7.2-27.1""1.1.1"
Thks,
H.
>>> Sachin J <[EMAIL PROTECTED]> 8/25/2006 12:56 PM >>>
see ?detach
Horace Tso <[EMAIL PROTECTED]> wrote:
Dear list,
I know it must be obvious and I did my homework. (In fact I've
RSiteSearched with keyword "remove AND library" but got timed
out.(why?))
How do I unload a library? I don't mean getting ride of it permanently
but just to unload it for the time being.
A related problem : I have some libraries loaded at startup in
.First()
which I have in .Rprofile. Now, I exited R and commented out the lines
in .First(). Next time I launch R the same libraries are loaded again.
I.e. there seems to be a memory of the old .First() somewhere which
refuses to die.
Thanks in adv.
Horace
__
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-
[[alternative HTML version deleted]]
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[R] Quickie : unload library
Dear list, I know it must be obvious and I did my homework. (In fact I've RSiteSearched with keyword "remove AND library" but got timed out.(why?)) How do I unload a library? I don't mean getting ride of it permanently but just to unload it for the time being. A related problem : I have some libraries loaded at startup in .First() which I have in .Rprofile. Now, I exited R and commented out the lines in .First(). Next time I launch R the same libraries are loaded again. I.e. there seems to be a memory of the old .First() somewhere which refuses to die. Thanks in adv. Horace __ R-help@stat.math.ethz.ch 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] An apply and rep question
Thanks Gabor, Andy, and Phil. I learn new trick, particularly the use of gl(). H. >>> "Gabor Grothendieck" <[EMAIL PROTECTED]> 8/11/2006 12:01 PM >>> The approach here is to perform the repetition on the indices (or rownames) rather than on the data frame directly. Using the builtin data frame BOD any of the following would work: BOD[gl(nrow(BOD), 2),] BOD[rep(1:nrow(BOD), each = 2),] BOD[rep(rownames(BOD), each = 2),] On 8/11/06, Horace Tso <[EMAIL PROTECTED]> wrote: > Hi list, > > I'm sure the explanation must be laughably simple to the experts out > there, but I just could figure it out. I have a simple data frame that > looks like, > > >head(da.off) > DDate OffP > 1 2005-01-01 41.23 > 2 2005-01-02 44.86 > 3 2005-01-03 44.86 > 4 2005-01-04 43.01 > 5 2005-01-05 45.47 > 6 2005-01-06 48.62 > > where the first column DDate currently is character, and OffP is > numeric. > > I want to duplicate every row 2 times, so I thought I use apply(), > > x <- apply(da.off, 2, rep, each=2) > > The result is a matrix of all character, > > head(x) > DDateOffP > 1 "2005-01-01" " 41.23" > 1 "2005-01-01" " 41.23" > 2 "2005-01-02" " 44.86" > 2 "2005-01-02" " 44.86" > 3 "2005-01-03" " 44.86" > 3 "2005-01-03" " 44.86" > > To convert it back to numeric, I did > > x <- as.data.frame(x) > x$OffP <- as.numeric(x$OffP) > > However, the OffP column didn't convert correctly, a mystery since they > "look" quite alright above. (I know, I know, there seems to be a space > there. But why?) > > head(x) > DDate OffP > 1 2005-01-01 150 > 1.1 2005-01-01 150 > 2 2005-01-02 202 > 2.1 2005-01-02 202 > 3 2005-01-03 202 > 3.1 2005-01-03 202 > > Is this the wrong way to use apply or rep? > > Horace > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] An apply and rep question
Hi list, I'm sure the explanation must be laughably simple to the experts out there, but I just could figure it out. I have a simple data frame that looks like, >head(da.off) DDate OffP 1 2005-01-01 41.23 2 2005-01-02 44.86 3 2005-01-03 44.86 4 2005-01-04 43.01 5 2005-01-05 45.47 6 2005-01-06 48.62 where the first column DDate currently is character, and OffP is numeric. I want to duplicate every row 2 times, so I thought I use apply(), x <- apply(da.off, 2, rep, each=2) The result is a matrix of all character, head(x) DDateOffP 1 "2005-01-01" " 41.23" 1 "2005-01-01" " 41.23" 2 "2005-01-02" " 44.86" 2 "2005-01-02" " 44.86" 3 "2005-01-03" " 44.86" 3 "2005-01-03" " 44.86" To convert it back to numeric, I did x <- as.data.frame(x) x$OffP <- as.numeric(x$OffP) However, the OffP column didn't convert correctly, a mystery since they "look" quite alright above. (I know, I know, there seems to be a space there. But why?) head(x) DDate OffP 1 2005-01-01 150 1.1 2005-01-01 150 2 2005-01-02 202 2.1 2005-01-02 202 3 2005-01-03 202 3.1 2005-01-03 202 Is this the wrong way to use apply or rep? Horace __ R-help@stat.math.ethz.ch 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] day, month, year functions
Hi list,
I'm trying to turn a date into something productive. (Not what you may be
thinking)
I want three functions so I could take a "date" object and get the day of week,
month, and year from it.
xx <- as.Date("2006-01-05")
month(xx) equal 1
day(xx) equal 5
year(xx) equal 2006
I'm aware of the weekdays() and months() functions in the base package. But
they return a character object which requires some coding to convert into a
numeric value.
I've also tried the sday.of.week() in fCalendar but it doesn't like my date,
> sday.of.week(xx)
Error in Ops.Date(sdates, 1) : %/% not defined for Date objects
Do these functions exist in some package I'm not aware of?
Thanks in adv.
Horace Tso
__
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Re: [R] Extracting from a matrix w/o for-loop
Unless there is another level of complexity that i didn't see here,
wouldn't it be a simply application of sapply as follow,
sapply( 1:dim(M2)[[1]], function(x) M1[M2[x,1], M2[x,2]] )
Hope this helps.
Horace
>>> "Camarda, Carlo Giovanni" <[EMAIL PROTECTED]> 7/28/2006 9:40 AM
>>>
Dear R-users,
likely there is a simple solution for this problem, but I currently
cannot see it.
I basically would like to get from a matrix values in particular
positions which are the rows of another matrix, without using a
for-loop.
In other words: is there any way to avoid the for-loop in the
following
simple example:
M1 <- matrix(1:20, ncol=2)
M2 <- rbind(c(1,1), c(2,1), c(3,2), c(5,2), c(8,1))
v <- numeric(nrow(M2))
for(i in 1:length(v)){
v[i] <- M1[M2[i,1], M2[i,2]]
}
Any suggestion would be welcome,
Ciao,
Carlo Giovanni Camarda
===
Camarda Carlo Giovanni
PhD-Student
Max Planck Institute for Demographic Research
Konrad-Zuse-Strasse 1
18057 Rostock, Germany
Tel: +49 (0)381 2081 172
Fax: +49 (0)381 2081 472
[EMAIL PROTECTED]
===
--
This mail has been sent through the MPI for Demographic
Rese...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Converting data frame to zoo
Thanks Gabor. I figured out what went wrong. The culprit turns out to be
the headers in my data. read.zoo doesn't recognize column headers and
complains
Error in read.zoo("C:\\...\\table.csv", :
index contains NAs
Or is there an option as in read.table(x, header=...) ?
After the header line is removed it works fine.
H.
>>> "Gabor Grothendieck" <[EMAIL PROTECTED]> 7/7/2006 10:22 AM
>>>
Check out read.zoo in the zoo package.
On 7/7/06, Horace Tso <[EMAIL PROTECTED]> wrote:
> Dear list,
>
> I know this is really basic question but I just couldn't get
anything
> to work. (I did a R site search with keywords "zoo" and "data frame"
but
> the server timed out on me.)
>
> I have a time series which has the following (typical) format,
>
> DATE Open High Low Close
> Volume
> 01-JAN-2006 5.25 5.25 5.25 5.25
> 256
>
>
> I read the data in from a csv file with read.csv() and it defaulted
to
> a data frame which I thought is fine. Now I want to convert it to zoo
so
> I did
>
> x <- zoo(my.df)
>
> which works just fine. But the Date column has been turned into a
> factor. Is there a way to make it into a Date. I've tried,
>
> x$Date <- as.Date(x$Date)
>
> but R complains that
>
> Error in fromchar(x) : character string is not in a standard
> unambiguous format
>
> Thanks in advance.
>
> Horace W. Tso
>
>
>
>
>
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Converting data frame to zoo
Dear list, I know this is really basic question but I just couldn't get anything to work. (I did a R site search with keywords "zoo" and "data frame" but the server timed out on me.) I have a time series which has the following (typical) format, DATE Open High Low Close Volume 01-JAN-2006 5.25 5.25 5.25 5.25 256 I read the data in from a csv file with read.csv() and it defaulted to a data frame which I thought is fine. Now I want to convert it to zoo so I did x <- zoo(my.df) which works just fine. But the Date column has been turned into a factor. Is there a way to make it into a Date. I've tried, x$Date <- as.Date(x$Date) but R complains that Error in fromchar(x) : character string is not in a standard unambiguous format Thanks in advance. Horace W. Tso __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Biobass, SAGx, and Jonckheere-Terpstra test
Hi list, I tried to load the package SAGx and failed because it complains it's looking for the Biobass which is not there. Then I looked up the package list and Biobass is not found. I'm trying to run the Jonckheere-Terpstra test and from what I see in the R archive, SAGx is the only place it's been implemented. > library(SAGx) Loading required package: multtest Loading required package: survival Loading required package: splines Loading required package: sma Error in loadNamespace(i, c(lib.loc, .libPaths())) : there is no package called 'Biobase' Error: package/namespace load failed for 'SAGx' Horace W. Tso __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Modeling inverse relationship with copula
Dear r list, I posted this on the S list last week since i'm using some of the FinMetrics functions on copula. Knowing there is a copula package in R, I figure this would be an appropriate forum to ask this question. I want to model inverse relationship between two (non-normal, non-symmetric) marginals with the gumbel copula, or with any copula. Say, x is lognormal and y is norm. Since gumbel's delta must be greater than one, how do I specify the equivalence of a negative correlation? If both are symmetric, I think I could get away by using a positive delta, simulate the bivariate realizations and then flipping the sign on one of them. Or am I completely off. I did search through the archive but found no related posting. Thanks in advance. Horace W. Tso __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
