Thank you for the explanation, Peter.
Angel
-Mensaje original-
De: peter dalgaard [mailto:pda...@gmail.com]
Enviado el: lun 01/09/2014 20:10
Para: Angel Rodriguez
CC: r-help
Asunto: Re: [R] Unexpected behavior when giving a value to a new variable based
on the value of another variable
I suspect you are seeing old code in rgeos. Please upgrade. If you cannot
do that, try a few other things like
## since you obviously have raster loaded
projection(clip.extent) <- projection(WorldMap)
LSMap <- gIntersection(WorldMap, clip.extent, byid = TRUE)
Does that work? If not possibly
p
Please keep your reply inside the thread. Anyway, it doesn't anwer the question.
Regards,
Pascal
On Tue, Sep 2, 2014 at 7:18 PM, Girija Kalyani wrote:
> Sir,
> The worldmap i here refer to is , the shape file of the complete world map
>
>
>
> On Tue, Sep 2, 2014 at 12:09 PM, Pascal Oettli wrote
On Mon, Sep 1, 2014 at 10:00 PM, Simon Blomberg wrote:
> Is MKL open source software? If not, that could be the sticking point.
>
> Simon.
IANAL, but from reading here:
https://software.intel.com/en-us/en-us/intel-mkl
It is neither Libre nor Gratis. But it is royalty free. IOW, you
cannot give th
Hi,
I'd like to match the beginning and the end of a string. E.g. I want to
extract all strings from a vector that beginn with "12" and end with
"Apples":
a <- "2 green Apples"
b <- "12 green Apples"
c <- "12 Apples and 2 green Bananas"
d <- "12 yellow Bananas"
fruitlist <- c(a,b,c,d)
# This is
Is there a test for an expired RSQLiteConnection? For example, if I run
library(RSQLite)
f <- tempfile()
con <- dbConnect(SQLite(), f)
dbDisconnect(con)
con
then I get
> con
and most operations using it give errors. (In my case I have a
persistent connection object, but if I save the workspac
On Tue, Sep 2, 2014 at 7:12 AM, Johannes Radinger
wrote:
> Hi,
>
> I'd like to match the beginning and the end of a string. E.g. I want to
> extract all strings from a vector that beginn with "12" and end with
> "Apples":
>
> a <- "2 green Apples"
> b <- "12 green Apples"
> c <- "12 Apples and 2 g
Dear all,
I'm looking for a package that allows me to run a Bayesian multivariate
linear regression and extract predicted values. In essence I'm looking for
the equivalent of lm and lm.predict in a Bayesian framework.
I have found several libraries that allow to run Bayesian multivariate
linear r
You should probably pose this question directly to the package author as the
function you are using cites an unpublished manuscript as a reference. It
appears that the function uses random draws to estimate R2 so each result is
approximately correct and no result is exactly correct. You can prob
Look at your correlation matrices. Your variable names make it more difficult
to so I’m abbreviating them:
> dat6 <- dat5
> colnames(dat6) <- abbreviate(colnames(dat5))
> round(cor(dat6[,1:6]), 4)
c8_SM_7 c7_SM_7 c6_SM_7 c5_SM_7 c4_SM_7 c3_SM_7
c8_SM_7 1. -1. -0.1214 0.1214 0.6
Dear R help mailing list,
I am looking for an ID3 implementation in R. I know that there are many
other decision tree algorithms already implemented (via rpart, tree, caret,
C50, etc., etc.), but for research purposes I would like to reproduce the
result of running ID3.
I was not able to find suc
from a list of GO term enrichment results in xml format, what is the best way
to extract term and p-value columns? Which function is the best to use? I used
xmlToDataFrame but got
Error in `[<-.data.frame`(`*tmp*`, i, names(nodes[[i]]), value = c("5", :
duplicate subscripts for columns
Lo
Rweka
On Sep 2, 2014 11:04 AM, "Tal Galili" wrote:
> Dear R help mailing list,
>
> I am looking for an ID3 implementation in R. I know that there are many
> other decision tree algorithms already implemented (via rpart, tree, caret,
> C50, etc., etc.), but for research purposes I would like to re
Hi Wensui,
When I looked at their docs:
http://cran.r-project.org/web/packages/RWeka/RWeka.pdf
It appeared they only have a connection to:
J48
LMT
M5P
DecisionStump
Is it possible to connect it to:
http://www.cs.tufts.edu/~ablumer/weka/doc/weka.classifiers.Id3.html
If so, how?
Thanks.
-
Yes with: make_Weka_classifier(name, class = NULL, handlers = list(),
init = NULL)
HTH, Christian
Hi Wensui,
When I looked at their docs:
http://cran.r-project.org/web/packages/RWeka/RWeka.pdf
It appeared they only have a connection to:
J48
LMT
M5P
DecisionStump
Is it possible to connect it
Even reducing the command to less variables than observations I still get:
set.cor(y = (66:76), x = c(1:6), z = 65, data = dat5)
Error in solve.default(x.matrix, xy.matrix) :
Lapack routine dgesv: system is exactly singular: U[2,2] = 0
On Mon, Sep 1, 2014 at 12:25 PM, David L Carlson wrote:
Thanx for the reply.
It is solved.
It is just a sample file whose dimensions are clipped.
Thanx
On Tue, Sep 2, 2014 at 4:19 PM, Pascal Oettli wrote:
> Please keep your reply inside the thread. Anyway, it doesn't anwer the
> question.
>
> Regards,
> Pascal
>
> On Tue, Sep 2, 2014 at 7:18 PM, Gir
This is explained in the RWeka vignette, but briefly:
## load RWeka
library(RWeka)
## look for a package providing id3
WPM("refresh-cache")
WPM("list-packages", "available") ## look for id3
## install package providing id3
WPM("install-package", "simpleEducationalLearningSchemes")
## load the pack
Dear Ista and Christian - this works wonderfully, thank you!
Contact
Details:---
Contact me: tal.gal...@gmail.com |
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
Greetings,
I am currently exploring some capabilities of the 'Shiny' package. I am
currently working with the most recent version of 'shiny' from the rstudio
github repository (version - 0.10.1.9006) in order to use the most up to
date datatables plugin. Using the ggplot2 diamonds dataset, I can
The data vector, art, given below using dput(), gives a set of discrete
numeric values for 915 observations,
in the range of 0:19. I want to make some plots of the frequency
distribution, but the standard
tools (hist, barplot, table) don't give me what I want to make a custom
plot due to 0 fre
On Sep 2, 2014, at 12:29 PM, Michael Friendly wrote:
> The data vector, art, given below using dput(), gives a set of discrete
> numeric values for 915 observations,
> in the range of 0:19. I want to make some plots of the frequency
> distribution, but the standard
> tools (hist, barplot, ta
I like Marc's answer, and I occasionaly have need for a different idiom.
old <-
structure(list(`0` = 275L, `1` = 246L, `2` = 178L, `3` = 84L,
`4` = 67L, `5` = 27L, `6` = 17L, `7` = 12L, `8` = 1L, `9` = 2L,
`10` = 1L, `11` = 1L, `12` = 2L, `16` = 1L, `19` = 1L), .Names = c("0",
"1", "2", "3
> The best that I can see for R would be if someone were to post a "how
> to use MKL for compiling R" type document.
I build R with MKL and ICC on my Archlinux box[1][2].
If I can help in anything, I will do it.
[1]https://wiki.archlinux.org/index.php/R
[2]https://aur.archlinux.org/packages/r-mk
On 02/09/2014, 1:29 PM, Michael Friendly wrote:
> The data vector, art, given below using dput(), gives a set of discrete
> numeric values for 915 observations,
> in the range of 0:19. I want to make some plots of the frequency
> distribution, but the standard
> tools (hist, barplot, table) don
Hello,
As for table, the help page says that "It is best to supply factors
rather than rely on coercion.", So if you want to include elements in
the range 0:19 with a count of zero, try
table(factor(art, levels = 0:19))
As for hist, use option right = FALSE.
art.hist <- hist(art, breaks=0:
On 8/30/2014 2:11 PM, Felipe Carrillo wrote:
library(plyr)
b <- structure(list(SampleDate = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = "5/8/1996", class = "factor"), TotalCount = c(1L,
2L, 1L, 1L, 4L, 3L, 1L, 10L, 3L), ForkLength = c(61L, 22L, NA,
NA, 72L, 34L, 100L, 23L, 25L), T
Another approach using barplot:
barplot(table(cut(art, breaks= -1:19, labels=0:19)))
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-b
The bottom of the expression is set by the lowest character (which can even
change for subscripted letters with descenders. The solution is to get axis()
to align the tops of the axis labels and move the line up to reduce the space,
e.g.
plot(1:5, xaxt = "n")
axis(1, at = 1:5, labels = c(expres
Dear John,
Thank you for your insights. I think you do understand what I've been
trying to do. Because I am doing a multigroup comparison – specifically,
examining the moderating role of test type on outcome – I needed two
covariance matrices to pass through the model. I wasn't sure how to do this
Hi Michael,
I think that histograms are intrinsically misleading for discrete data, and
that while bar graphs are an improvement, they also invite
misinterpretation. I generally do something like this:
f <- table(factor(art, levels=0:19))
plot(as.numeric(names(f)), as.numeric(f), type="h",
xl
Is there a way to over-write a procedure (subroutine)?
I include a default procedure fixx in a list of procedures which are
compiled into a package. By default, the procedure deliver the data matrix x.
fixx <- function(x){
result <- list(x=x)
return(result)
}
In some applications, I have tran
Dear Chad,
> -Original Message-
> From: Chad Danyluck [mailto:c.danyl...@gmail.com]
> Sent: Tuesday, September 02, 2014 3:29 PM
> To: John Fox
> Cc: r-help@r-project.org
> Subject: Re: [R] Problems bootstrapping multigroup SEM
>
> Dear John,
>
> Thank you for your insights. I think you d
The built-in table method for plot() makes a decent looking plot as
well. Look at
plot(table(art), ylab="Count")
plot(table(factor(art, levels=0:19)), ylab="Count")
plot(table(LETTERS[art+1]), ylab="Count")
plot(table(factor(LETTERS[art+1], levels=LETTERS[1:20])), ylab="Count")
Bill Dunlap
Hi,
Trying to change the way functions inside packages work is almost
certainly not what you want to do. It's possible but usually there is
an easier way.
Your example is so simplified that I suspect it doesn't capture your
actual needs very well, but in the case of your example you can simply
do
Dear Bill,
Yes, that's better -- essentially similar to what I suggested but much less
work. I wasn't aware of it. You could even add the points at the tops of the
spikes via a follow-up points() command.
Thanks,
John
> -Original Message-
> From: William Dunlap [mailto:wdun...@tibco.c
A perhaps better approach would be to have the functions that
currently call fixx accept an argument of a function to use. It could
default to fixx, but if the caller passed in a new function it would
use that function instead.
If you really want to overwrite a function inside of a package
namesp
On Sep 2, 2014, at 5:32 AM, Duncan Murdoch wrote:
> Is there a test for an expired RSQLiteConnection? For example, if I run
>
> library(RSQLite)
> f <- tempfile()
> con <- dbConnect(SQLite(), f)
> dbDisconnect(con)
> con
>
> then I get
>
>> con
>
>
> and most operations using it give errors
DBI 0.3 (just released to CRAN) includes a new generic, dbIsValid(),
for exactly this purpose. Unfortunately no packages implement a method
for it yet, but eventually it will be the right way to detect this
problem.
(I'm now the maintainer for RSQLite, so I added this to my to do list:
https://git
On 02/09/2014, 5:28 PM, David Winsemius wrote:
>
> On Sep 2, 2014, at 5:32 AM, Duncan Murdoch wrote:
>
>> Is there a test for an expired RSQLiteConnection? For example, if I run
>>
>> library(RSQLite)
>> f <- tempfile()
>> con <- dbConnect(SQLite(), f)
>> dbDisconnect(con)
>> con
>>
>> then I ge
Hi,
I wish to simulate some data from a Weibull distribution. The rweibull function
in R uses the parameterisation
'with shape parameter a and scale parameter b has density given by f(x) = (a/b)
(x/b)^(a-1) exp(- (x/b)^a)'.
However, it would be much more useful for me to simulate data using a d
Hi Lucy,
Try the gamlss.dist package, specifically the rWEI2() function.
Cheers,
Josh
On Wed, Sep 3, 2014 at 11:52 AM, Lucy Leigh
wrote:
> Hi,
> I wish to simulate some data from a Weibull distribution. The rweibull
> function in R uses the parameterisation
>
> 'with shape parameter a and sc
On 02/09/2014, 9:52 PM, Lucy Leigh wrote:
> Hi,
> I wish to simulate some data from a Weibull distribution. The rweibull
> function in R uses the parameterisation
>
> 'with shape parameter a and scale parameter b has density given by f(x) =
> (a/b) (x/b)^(a-1) exp(- (x/b)^a)'.
>
> However, it w
Awesome, thankyou!
From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
Sent: Wednesday, 3 September 2014 12:04 PM
To: Lucy Leigh
Cc: r-help@R-project.org
Subject: Re: [R] Simulating from a Weibull distribution
Hi Lucy,
Try the gamlss.dist package, specifically the rWEI2() function.
Cheers,
Josh
I just tested it and this plugin does not seem to work with the new
.DataTable() API in DataTables 1.10.x, so I guess it is unlikely to
make it work in (the current development version of) shiny. It is not
in the official list of plugins, either:
http://www.datatables.net/extensions/index
Regards,
Hi,
I am trying to test a mediation hypothesis using coefficients from logistic
regression analyses (x, m, and y are all dichotomous). I am running a test
of significance using MacKinnon and Dwyer's adaptation of Sobel's test
(i.e., correcting for different scales of coefficients in cases of a
dic
On Sep 2, 2014, at 6:52 PM, Lucy Leigh wrote:
> Hi,
> I wish to simulate some data from a Weibull distribution. The rweibull
> function in R uses the parameterisation
>
> 'with shape parameter a and scale parameter b has density given by f(x) =
> (a/b) (x/b)^(a-1) exp(- (x/b)^a)'.
>
> However
Hi.
I'm taking the long-overdue step of moving from using online calculators to
compute results for Mann-Whitney U tests to a more streamlined system
involving R.
However, I'm finding that R computes a different result than the 3 online
calculators that I've used before (all of which approximatel
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