If you multiply the data for a certain group by the scaling matrix for that
group, the variance matrix will be the identity. E.g.,
> z <- qda(iris[-5], grouping=iris$Species)
> zapsmall(var( as.matrix(subset(iris, Species=="virginica", 1:4)) %*%
z$scaling[,,"virginica"] ))
1 2 3 4
1 1 0 0 0
2 0
You found that not all deviates are larger than m.
Comparing the formulas quoted in Wikipedia and in your email it is obvious that
m is not the same as x_m.
If you don't find the rmutil formulation useful you can try another such as the
one in PtProcess, or write your own.
--
Sent from my pho
This was a change in tidyr 0.7.0 that is causing a lot of confusion,
so we are preparing tidyr 0.7.1 which will back this change out.
If you want a work around in the meantime, you can express your
operation a bit more elegantly as:
library(tidyr)
df <- data.frame(v1 = 1:5, somestring = 6:10, v3
I guess the question that is being asked here is what is the scaling matrix
that is being returned in the qda object. The help file on qda() says:
...
scaling: for each group ‘i’, ‘scaling[,,i]’ is an array which transforms
observations so that within-groups covariance matrix is spherical.
...
T
In https://en.wikipedia.org/wiki/Pareto_distribution, it is clear what the
parameters are for the pareto distribution: *xmin *the scale parameter and
*a* the shape parameter.
I am using rmutil to generate random deviates from a pareto distribution.
It says in the documentation that the probabilty
> On Aug 24, 2017, at 10:20 AM, David Winsemius wrote:
>
>
>> On Aug 23, 2017, at 2:29 AM, Rolf Turner wrote:
>>
>>
>> On 23/08/17 18:33, Stefan Evert wrote:
>>
On 23 Aug 2017, at 07:45, Rolf Turner wrote:
My reading of ?regex led me to believe that
gsub("[:alp
Or perhaps two exclamation points would be better ('unquote' in the
tidyverse lexicon, 3 bangs is 'unquote-splice').
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Aug 24, 2017 at 10:20 AM, William Dunlap wrote:
> Try putting !!! (three exclamation symbols) in front of which(...)==
>
Looks like a bug to me. I think you need to correspond with the package
(tidyr?) maintainer, perhaps by putting a bug report on GitHub.
Next time please make your example reproducible by including the necessary
"library" function calls.
--
Sent from my phone. Please excuse my brevity.
On Aug
Try putting !!! (three exclamation symbols) in front of which(...)==
The non-standard evaluation in the tidyverse can cause confusion.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Aug 24, 2017 at 4:32 AM, Eugeny Melamud <
eugeny.mela...@lanit-tercom.com> wrote:
> Hi all!
>
> The fol
> On Aug 23, 2017, at 2:29 AM, Rolf Turner wrote:
>
>
> On 23/08/17 18:33, Stefan Evert wrote:
>
>>> On 23 Aug 2017, at 07:45, Rolf Turner wrote:
>>>
>>> My reading of ?regex led me to believe that
>>>
>>>gsub("[:alpha:]","",x)
>>>
>>> should give the result that I want.
>> That's look
Hi,
Inline below.
> On Aug 24, 2017, at 5:22 AM, John Kane via R-help
> wrote:
>
>
> IIt might help to read the material at one or both of these links
> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
>
In this case, had Jagan spent about 30 seconds
> On Aug 23, 2017, at 10:59 AM, Elahe chalabi via R-help
> wrote:
>
> Any responds?!
When I look at the original post a I see a question about a function named
`rfcv` but do not see a `library` call to load such a function. I also see a
reference to a help page or vignette, perhaps?, from th
Inline.
-- Bert
On Thu, Aug 24, 2017 at 4:32 AM, Eugeny Melamud
wrote:
> Hi all!
>
> The following code (executed in console)...
> somevar <- data.frame(v1 = 1:5, somestring = 6:10, v3 = 11:15, v4 =
> 16:20);
> somevar %>% gather(key = var, value = val, which(names(somevar) ==
> "som
Hi all!
The following code (executed in console)...
somevar <- data.frame(v1 = 1:5, somestring = 6:10, v3 = 11:15, v4 = 16:20);
somevar %>% gather(key = var, value = val, which(names(somevar) ==
"somestring"):length(somevar)) %>% head(2);
throws...
Error in which(names(somevar) == "so
IIt might help to read the material at one or both of these links
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Reproducibility · Advanced R.
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Reproducibility · Advanced R.
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On Thursday, August 24, 2017, 6:19:25 AM EDT, John Kan
On Thursday, August 24, 2017, 1:50:13 AM EDT, David Winsemius
wrote:
> On Aug 22, 2017, at 11:31 PM, jagan krishnan via R-help
> wrote:
>
> Hi all,
> This is Jagan.i have been provided a task of analyzing sales data of a
> company in R programming...Just wanted to know,how can I pull Tal
> On Aug 23, 2017, at 8:08 AM, PIKAL Petr wrote:
>
> Hi
>
> Well, yes I tried it about two weeks ago but my post did not get through as
> it still awaits moderator approval.
It got through just fine. It appeared on Aug 15. It just didn't get any replies.
As I read your original question in t
> On 24 Aug 2017, at 11:58 , peter dalgaard wrote:
>
> M <- as.matrix(do.call(expand.grid, rep(list(0:1),5)))
> M
> # This is just 0:31 encoded as (little-endian) binary
> apply(M, 1, function(i) sum((2^(0:4))[i]))
>
> # now, use rows of M for logical indexing into "A"-"E"
> mode(M) <- "logical
> On 24 Aug 2017, at 11:58 , peter dalgaard wrote:
>
> apply(M, 1, function(i) sum((2^(0:4))[i]))
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd@cbs.dk Priv: pda...
> On 24 Aug 2017, at 01:25 , Duncan Murdoch wrote:
>
> On 23/08/2017 6:25 PM, Bert Gunter wrote:
>> Doesn't sort by size of subgroup. I interpret the phrase I asterisked as:
>
> You were fooled by Peter's tricky single negative.
>
...
Let's do this more carefully, then:
M <- as.matrix(do.c
OK, I will try that.
I notified the maintainer of boot.stepAIC, so he might fix this in due course.
Thanks,
SGO.
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Heinz Tuechler
Sent: 24 August 2017 00:07
To: r-help@r-project.org
Subject: Re: [R] boot.st
Hello,
I am facing a problem with optimization in R from 2-3 weeks.
I have some Gaussian mixtures parameters and I want to find the maximum in
that
*Parameters are in the form *
mean1mean2mean3 sigma1 sigma2 sigma3 c1c2c3
506.8644 672.8448 829.902 61.02859
Hm.
I used the code from Mark's response and it works correctly. So you either
messed the data or you do not tell us the whole story.
data
> dput(data)
structure(list(COL1 = structure(c(16222, 16252), class = "Date"),
COL2 = structure(c(16556, 16556), class = "Date"), Date_Flag = c(0,
0)
On Thu, Aug 24, 2017 at 7:07 AM, David Winsemius
wrote:
>
> > On Aug 23, 2017, at 10:06 PM, Dhivya Narayanasamy
> wrote:
> >
> > I have a vibration signal coming accelerometer. I converted this signal
> from*
> > m/s^2* to *mm/s*. Now I am supposed to predict this vibration signal in R
> > using
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