Thanks William,
I've used Dolphin for years and never encountered that phenomenon.
Even so, that description doesn't fit what's going on here. 1.7
kilobytes is hardly a 'large directory'.
The problem seems to be with the way VirtualBox mounts directories
which isn't an R issue, nor is the fact t
On 13/11/2018 12:31, Elahe chalabi wrote:
Hi Robert,
Thanks for your reply but your code returns the number of verbs in each
massage. What I want is a string showing verbs in each massage.
The output of my code (below) is:
# A tibble: 4 x 2
DocumentID verbs
1 478920 has|been|
Hi Sasha,
Your attached table did not come through, please see the posting guidelines:
"No binary attachments except for PS, PDF, and some image and archive
formats (others are automatically stripped off because they can contain
malicious software). Files in other formats and larger ones should
Thanks, Sarah's answer helps the question. Now how to change the gridded
data back to DF1 format? I don't know how to name the format, thanks.
On Tue, Nov 13, 2018 at 10:56 PM David L Carlson wrote:
> Sarah's answer is probably better depending on what you want to do with
> the resulting data, b
Perhaps you got bitten by Dolphin's non-modal dialogs, as described in
https://userbase.kde.org/Dolphin/File_Management:
Non Modal Dialogs
When Moving, Copying or Deleting files/directories the dialog disappears
even when the operation has not yet completed. A progress bar then appears
in the bot
This is getting more strange.
I normally copy from the shared folder to the appropriate directory
using Dolphin, the KDE file manager. If instead I use the standard bash
cp command, no corruption happens -- at least with the limited testing I
have done. There also seems to be no problem copy
There are a couple of important concepts in play here:
1.
You are drawing the map with the 'graphics' system, but 'vwline' works
in the 'grid' system.
2.
In the 'grid' system, you can work with lots of different coordinate
systems. The default is usually "npc" (which is 0 to 1), but you c
On 13/11/2018 12:58 PM, William Dunlap wrote:
You also asked about doing this for the rows of a matrix. unique() give
the unique rows but match operates on a per element, not per row,
basis. You can use split, which operates on rows of a matrix, to help.
> m <- cbind( A=c(i=5,ii=5,iii=5,i
Hi all,
Not sure if the following could be considered a bug, or just a user error
but here goes:
We're teaching our students to use the tidyverse for most of their R stuff
and the following gives problems (code adapted/shortened to pinpoint
problem):
```
iris_long = tidyr::gather(iris, key ="var
Dear All,
I am pretty new to R and would appreciate a help how to calculate
centroids from the latitude and longitude of existing cells (e.g. to get
centroid for a new cell I would need to combine latitude and
9161,9162,9163,9164 to 9160 or 10152, 10154 to 10150 etc.)
Please see attached ta
You also asked about doing this for the rows of a matrix. unique() give
the unique rows but match operates on a per element, not per row,
basis. You can use split, which operates on rows of a matrix, to help.
> m <- cbind( A=c(i=5,ii=5,iii=5,iv=4,v=4,vi=4), B=c(2,3,2,2,2,2) )
> unique(m)
A B
There was a typo in my example. Here is the fixed version:
# initialize matrix
values = c(1,0.725,0,0,0.725,1,0.692,0,0,0.692,1,0.664,0,0,0.664,1)
B = matrix(values, 4,4)
# show that singular values are positive
svd(B)$d
# show that matrix is symmetric
isSymmetric(B)
# B is symmetric positive
Try supplying the order argument to arima. It looks like the default is to
estimate only the mean.
> arima(b, order=c(1,0,0))
Call:
arima(x = b, order = c(1, 0, 0))
Coefficients:
ar1 intercept
0.8871 0.2369
s.e. 0.0145 0.2783
sigma^2 estimated as 1.002: log likelihood
Aren't singular values always positive or zero? Look at eigen(B)$values to
check for positive definiteness.
Fix your example - your B is not symmetric.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Nov 13, 2018 at 7:30 AM, Hoffman, Gabriel
wrote:
> My understanding is that a Cholesky d
It seems like copying the files corrupted them. How did you copy them (with
R
or cp or copy or ftp, etc.)? I don't see how this has anything to do with
R.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Nov 12, 2018 at 7:10 PM, p_connolly
wrote:
> On 2018-11-13 12:55, William Dunlap wrote
Dear All,
There is something I don't understand fundamentally about handling geocoords
with vwlines, as the axample below may illustrate.
vwlines seems to require values somewhere between 0 and 1 as shares of the plot
length and width, but obviously that's not the complete story: Point (1,1) is
> > xtabs(~V1+V2+V3, df) # There are 9 cells with 0 entries. That is the
> > problem.
>
> , , V3 = 9
>
> V2
> V1 8 16 24
> 200 5 0 5
> 350 5 5 0
> 500 0 5 5
thanks for the insight
yah
signature.asc
Description: OpenPGP digital signature
__
R-h
Your understanding is wrong. The eigenvalues, not singular values, must be
positive, and they are not.
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue
Eigen shows that the matrix is not positive definite (it has a negative
eigenvalue).
And isSymmetric() also shows it is not symmetric - compare (3,4) and (4,3)
On Tue, Nov 13, 2018 at 5:39 PM Hoffman, Gabriel
wrote:
> My understanding is that a Cholesky decomposition should work on any
> square,
My understanding is that a Cholesky decomposition should work on any square,
positive definite matrix. I am encountering an issue where chol() fails and
give the error: "the leading minor of order 3 is not positive definite"
This occurs on multiple machines and version of R.
Here is a minimal
Long ago, when R's t.test had var.equal=TRUE by default, I wrote some
class notes showing that the result was equivalent to a one predictor
regression model. Because t.test does not default to var.equal=TRUE
these days, I'm curious to know if there is a way to specify weights
in an lm to obtain th
Sarah's answer is probably better depending on what you want to do with the
resulting data, but here's a way to go from your original DF1 to DF2:
> DF1 <- structure(list(latitude = c(45.5, 45.5, 45.5, 45.5, 46, 46, 46,
+ 46), longitude = c(110.5, 111, 111.5, 112, 110.5, 111, 111.5,
+
If you want an actual spatial dataset, the best place to ask is R-sig-geo
R has substantial capabilities for dealing with gridded spatial data,
including in the sp, raster, and sf packages.
Here's one approach, creating a SpatialGridDataFrame, which can be
exported in any standard raster format u
You might take a look at the reshape package, which switches from 'long' to
'wide' formats and vice versa in a fairly flexible way.
S Ellison
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of lily li
> Sent: 13 November 2018 07:22
> To: R mailing list
Try google'ing for 'variance of an AR(1) process'.
With the same seed, if you set n=100, you will get something that will
compare well with what you discover from your search.
On Tue, Nov 13, 2018 at 2:04 PM Ashim Kapoor wrote:
> Dear All,
>
> Here is a reprex:
>
> set.seed(123)
> b <- arima
Dear All,
Here is a reprex:
set.seed(123)
b <- arima.sim(list(order = c(1,0,0),ar= .9),n=1000,sd=1)
arima(b)
Call:
arima(x = b)
Coefficients:
intercept
0.2250
s.e. 0.0688
sigma^2 estimated as 4.735: log likelihood = -2196.4, aic = 4396.81
>
Should sigma^2 not be equal to
Perfect! This might be a good example to add to the documentation of mgcv
somewhere
Thanks.
Mark
On Thu, 8 Nov 2018 at 22:08, Simon Wood wrote:
> This first derivative penalty spline will do it, but the price paid is
> that the curves are often quite wiggly.
>
>
> library(mgcv); set.seed(
On 13/11/2018 12:35 AM, Pages, Herve wrote:
Hi,
On 11/12/18 17:08, Duncan Murdoch wrote:
The duplicated() function gives TRUE if an item in a vector (or row in
a matrix, etc.) is a duplicate of an earlier item. But what I would
like to know is which item does it duplicate?
For example,
v <-
> PIKAL Petr
> on Tue, 13 Nov 2018 08:42:22 + writes:
> Hi
> similar result (with different numerical values) could
> be achieved by making v a factor.
> > v <- letters[c(2,2,1,2,1,1)]
> > vf<-factor(v)
> > as.numeric(vf)
> [1] 2 2 1 2 1 1
>
> Cheers
> Petr
Yes, as
Hi
similar result (with different numerical values) could be achieved by making v
a factor.
> v <- letters[c(2,2,1,2,1,1)]
> vf<-factor(v)
> as.numeric(vf)
[1] 2 2 1 2 1 1
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Bert Gunter
> Sent: Tuesday, November 13, 2018 6:44
Hi Rebecca,
I think it was me how suggested you the missRanger package, so this is
actually a follow-up of you previous question about censored imputation
of missing values (as far as I can remember).
The missRanger package uses predictive mean matching, so take a look at
?missRanger::pmm an
Hi lily,
Something like this should work:
DF1<-read.table(text=
"latitude longitude Precip
45.5 110.5 3.2
45.5 1115.0
45.5 111.5 1.8
45.5 1122.0
46 110.5 6.1
46 1114.5
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