Hi,
Can anyone tell me how to get source code of a package of R?
I want to integrate elastic search with R if there is any way please let
me know or give me a direction to do that.
Thanks.
[[alternative HTML version deleted]]
__
On 05.09.2014 09:01, madhvi.gupta wrote:
Hi,
Can anyone tell me how to get source code of a package of R?
An R package is distributed in form of source code:
Say you want to see the sources of abc, then go to
http://cran.r-project.org/web/packages/abc/index.html
and download the osurces or
On 04.09.2014 20:02, Karim Mezhoud wrote:
Dear All,
How can I add folder content examples of needed files to run example?
The simple add of folder did not built and reload with the package.
If you need additional data, put the data in the package as described in
Writing R Extensions.
If
On 04.09.2014 15:03, Angel Marley wrote:
Dear R list,
I'm working with recursive tress using packages mvpart and rpart in R in linux
xubuntu (64).
The package performed with no problem under my previous R version (2.14)
I had recently updated my R version to 3.1.1 and when I try to run a
On Fri, Sep 5, 2014 at 12:28 AM, David Winsemius dwinsem...@comcast.net wrote:
If you are accepting feature requests
The R issue tracker has a wishlist section:
On Fri, Sep 5, 2014 at 8:01 AM, madhvi.gupta madhvi.gu...@orkash.com wrote:
Hi,
Can anyone tell me how to get source code of a package of R?
I want to integrate elastic search with R if there is any way please let
me know or give me a direction to do that.
Which R packages do you want source
Read the Writing R Extensions document that comes with R? Use that knowledge to
write a package? Read the Advanced R website (http://adv-r.had.co.nz)? Use a
web search engine to look for others working on a similar package? Read the
Posting Guide mentioned at the bottom of this or any other
Please add it if you think it fits, and expand it as discussed, I am not
creating a package for one single utility function.
T.
On Fri, Sep 5, 2014 at 1:28 AM, David Winsemius dwinsem...@comcast.net
wrote:
On Sep 4, 2014, at 12:54 PM, Greg Snow wrote:
The TeachingDemos package has %% and
I have several lme objects like the ones shown below and I wish to
combine the coefficients and confidence intervals of fixed effects of
several models. Is there a function that could do that job?
m1 - lme(mark1 ~ pm10 + temp + + age + gender + bmi + statin
+ smoke + dow + season
On Thu, Sep 4, 2014 at 10:41 AM, Torbjørn Lindahl
torbjorn.lind...@gmail.com wrote:
Not sure if this is the proper list to propose changes like this, if it
passes constructive criticism, it would like to have a %between% operator
in the R language.
There is a between function in the
I don't mind maintaining this, however I'm not creating a new util package
just for one function, there are already several nice util libraries out
there, adding one more adds to fragmentation more than this single function
provides usefulness.
If anyone wants to adopt this low-maintenance-cost
On 05/09/2014, 6:47 AM, Torbjørn Lindahl wrote:
I don't mind maintaining this, however I'm not creating a new util package
just for one function, there are already several nice util libraries out
there, adding one more adds to fragmentation more than this single function
provides usefulness.
Please add it if you think it fits, and expand it as discussed, I am not
creating a package for one single utility function.
Why not? There's nothing wrong with a package that only provides one function.
Hadley
--
http://had.co.nz/
__
Dear Uwe,
thanks for your prompt reply.
I have run the update.packages, but the error persisted, then I did it again
and it works
Best regards and thank for your time
Angel
On Friday, September 5, 2014 4:15 AM, Uwe Ligges
lig...@statistik.tu-dortmund.de wrote:
On 04.09.2014
Subscribers,
Could someone please indicate correct terminology and relevant manual
sections to achieve the following conceptual workflow, to create a
script to select sequentially parts of a dataframe:
Welcome to this R script program
Select level number below of variable 'X'
[1]
[2]
[3]
You can easily run into precedence problems with the %fun% syntax. E.g., if
1 %% 5 %% 10
returns TRUE then
1 %% 5 %% 10*2
will return 2 because %% has higher precedence than *.
as.list(quote(1 %% 5 %% 10*2))
[[1]]
`*`
[[2]]
1 %% 5 %% 10
[[3]]
[1] 2
There may be a specialized package for this in bioconductor, but it seems that
you could just use aggregate() to calculate the means for each population and
then use the results of that in dist().
?aggregate
-
David L Carlson
Department of Anthropology
Texas
HI, of course.
The a mini-version of my data-set is below, stored in d2. Then the code I'm
working follows.
library(reshape2)
#Create d2
structure(list(row = 1:50, rank1 = structure(c(3L, 3L, 3L, 4L,
3L, 3L, NA, NA, 3L, NA, 3L, 3L, 1L, NA, 2L, NA, 3L, NA, 2L, 1L,
1L, 3L, NA, 6L, NA, 1L, NA,
The big difference between the data sets is that many of your rows (16) have
all missing values. None of mine do. If you run my data and yours, you will see
that decast throws a warning Aggregation function missing: defaulting to
length with your data but not with mine. As a result, instead of
Thanks again John. I think the problem is now resolved.
I changed the information that I passed through the sem() as follows:
MAP.mg.sem - sem(MAP.mg.mod, data=list(stereotype=stereotype.MAP.data,
evaluative=evaluative.MAP.data), group=IAT.factor)
Then, from ?bootSem
The default is the data
How can I parameterize the function call to aggregate() below with colx (a
character string) in the following code? That is, replace both if
statements with aaa - aggregate(colx_something, data=tidy[grpyrs,],sum).
I've tried a couple of ideas for colx_something but get an R-error each
time.
I compile R3.1.0 to install it in centOS without root. The disk partition
format is NFS (network file system). After configure it
($/home/XX/download/R3.1.0/configure --prefix=/home/zj/local/R --enable-R-shlib
--with-x) successfully, I make it ($make). But there always is an error when
make
On 2014/9/4 12:24, David Winsemius wrote:
On Sep 3, 2014, at 10:05 PM, Jinsong Zhao wrote:
On 2014/9/3 21:33, Jinsong Zhao wrote:
On 2014/9/2 11:50, David L Carlson wrote:
The bottom of the expression is set by the lowest character (which can
even change for subscripted letters with
On 2014/9/4 14:58, David L Carlson wrote:
The problem with this approach is that the horizontal positioning of the labels
is based on the width of the label including the phantom part so that the E's
are pushed to the left of the tick mark (at least on my Windows machine). But
it does provide
You could try using the non-formula interface to aggregate.
Note that the following two calls to aggregate are equivalent but
the second (using the non-formula interface) makes the response column
a variable:
df - data.frame(Y1=1:10, Y2=101:110, Group=rep(letters[1:3], c(3,3,4)))
On Sep 5, 2014, at 4:15 PM, Jinsong Zhao wrote:
On 2014/9/4 14:58, David L Carlson wrote:
The problem with this approach is that the horizontal positioning of the
labels is based on the width of the label including the phantom part so that
the E's are pushed to the left of the tick mark
On Sep 5, 2014, at 4:53 PM, David Winsemius wrote:
On Sep 5, 2014, at 4:15 PM, Jinsong Zhao wrote:
On 2014/9/4 14:58, David L Carlson wrote:
The problem with this approach is that the horizontal positioning of the
labels is based on the width of the label including the phantom part so
Hi.
Say I have a model like
y = a + B1*x1 + B2*x2 + B3*x3 + B4*x4 + e
and I want to test
H0: B2/B1 = 0
or
H0: B2/B1=B4/B3
(whatever H1). How can I proceed?
I now about car::linearHypothesis, but I can't figure out a way to do the
tests above.
Any hint?
Thanks.
C
AFAICS you are not testing a linear hypothesis (which is of the form Lb=b0
where L is a matrix and b=(a,B1,B2,B3,B3) is the parameter vector).
If, for simplicity, your model is E(y) = a + bx then -a/b is the x-value for
which y is zero.
When you turn to estimates then u = -a/b is the ratio of
Well:
1) 8th grade algebra tells me B2/B1 == 0 == B2 =0;
2) I suspect you would need to provide more context for the other, as
you may be going about this entirely incorrectly (have you consulted a
local statistician?): your nonlinear hypothesis probably can be made
linear under the right
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