Displaying 20 results from an estimated 1684 matches for "roye".
Did you mean:
role
2003 May 16
1
kphone fails to register with asterisk (sip)
hi all
when starting kphone, it tries to register with asterisk but fails after a
while. The SIP entry in * for this user is below. This is identical to the
other SIP entries. The other SIP clients are MSN messenger plus one snom.
these work fine. See SIP debug output attached as 'screen-exchange'
thanks
roy
[roy]
type=friend
;insecure=yes
username=roy
;secret=password
host=dynamic
2017 Oct 17
2
ggridges help
yes, thanks, and I was getting close to that. One thing I found is the manual says the height is the distance above the y-line, which should be, but doesn't have to be positive. In fact, the time series are estimates of a cycle, and has negative values, which unfortunately are not included in my sub-sample. And the negative values are not handled properly (the series disappears for
2017 Oct 17
0
ggridges help
The min_height = -0.25 is there to make it show cycle values down to -1/4.
You may want to change it to -1 so it shows more of the cycle values.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Oct 17, 2017 at 1:26 PM, Roy Mendelssohn - NOAA Federal <
roy.mendelssohn at noaa.gov> wrote:
> yes, thanks, and I was getting close to that. One thing I found is the
> manual says the
2017 Jun 01
3
Reversing one dimension of an array, in a generalized case
> On 1 Jun 2017, at 22:42, Roy Mendelssohn - NOAA Federal <roy.mendelssohn at noaa.gov> wrote:
>
> Thanks to all for responses/. There was a question of exactly what was wanted. It is the generalization of the obvious example I gave,
>
>>>> junk1 <- junk[, rev(seq_len(10), ]
>
>
> so that
>
> junk[1,1,1 ] = junk1[1,10,1]
> junk[1,2,1] =
2018 Jan 02
4
httr::content without message
Hi All:
I am using httr to download files form a service, in this case a .csv file. When I use httr::content on the result, I get a message. Since this will be in a package. I want to suppress the message, but haven't figured out how to do so.
The following should reproduce the result:
myURL <-
2017 Aug 29
3
RMarkdown question
Hi All:
In creating a R Notebook I know that in the text I can link to a (sub) section by using the command:
[Header 1](#anchor)
and putting the appropriate anchor name at the appropriate header. But can the same be done for code chunks, if the code chunk is named? What I want to do is say that such and such code chunk is an example of how to do something, and have that link to the
2017 Jun 01
1
Reversing one dimension of an array, in a generalized case
Thanks again. I am going to try the different versions. But I probably won't be able to get to it till next week.
This is probably at the point where anything further should be sent to me privately.
-Roy
> On Jun 1, 2017, at 1:56 PM, David L Carlson <dcarlson at tamu.edu> wrote:
>
> On the off chance that anyone is still interested, here is the corrected function using
2017 Oct 17
2
ggridges help
I have tried:
ggplot(plotFrame, aes(x = time, y = cycle, height = cycle, group = depth)) + geom_ridgeline()
ggplot(plotFrame, aes(x = time, y = depth, height = cycle, group = depth)) + geom_ridgeline()
ggplot(plotFrame, aes(x = time, y = depth, group = depth)) + geom_density_ridges()
none are producing a plot that was a ridgeline for each depth showing the time series at that depth. The plot
2017 Jun 01
5
Reversing one dimension of an array, in a generalized case
Hi All:
I have been looking for an elegant way to do the following, but haven't found it, I have never had a good understanding of any of the "apply" functions.
A simplified idea is I have an array, say:
junk(5, 10, 3)
where (5, 10, 3) give the dimension sizes, and I want to reverse the second dimension, so I could do:
junk1 <- junk[, rev(seq_len(10), ]
but what I am
2017 Jun 01
0
Reversing one dimension of an array, in a generalized case
On the off chance that anyone is still interested, here is the corrected function using aperm():
z <- array(1:120,dim=2:5)
f2 <- function(a, wh) {
idx <- seq_len(length(dim(a)))
dims <- setdiff(idx, wh)
idx <- append(idx[-1], idx[1], wh-1)
aperm(apply(a, dims, rev), idx)
}
all.equal(f(z, 1), f2(z, 1))
# [1] TRUE
all.equal(f(z, 2), f2(z, 2))
# [1] TRUE
2017 Oct 17
0
ggridges help
Does the following work for you?
ggplot2::ggplot(plotFrame, aes(x = time, y = depth, height = cycle,
group = depth)) + ggridges::geom_ridgeline(fill="red", min_height=-0.25)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Oct 17, 2017 at 12:43 PM, Roy Mendelssohn - NOAA Federal <
roy.mendelssohn at noaa.gov> wrote:
> I have tried:
>
> ggplot(plotFrame, aes(x =
2018 Jul 21
2
Failed to establish your Kerberos Ticket cache due time differences with the domain controller
>
> Strange, you say the time is okay, but the error says it isn't.
>
> Try this, open a terminal on both DC's, run 'date' and 'samba-tool
> time' on both. The results should be virtually the same.
>
> e.g.
> root at dc4:~# samba-tool time
> Sat Jul 21 16:47:43 2018 BST
> root at dc4:~# date
> Sat 21 Jul 16:47:46 BST 2018
>
>
2018 Jul 21
4
Failed to establish your Kerberos Ticket cache due time differences with the domain controller
On Sat, 21 Jul 2018 18:59:08 +0100
Rowland Penny via samba <samba at lists.samba.org> wrote:
> On Sat, 21 Jul 2018 18:30:48 +0100
> Roy Eastwood via samba <samba at lists.samba.org> wrote:
>
> > Thanks Rowland.
> >
> > > -----Original Message-----
> > > From: samba [mailto:samba-bounces at lists.samba.org] On Behalf Of
> > > Rowland
2018 Jul 24
0
Failed to establish your Kerberos Ticket cache due time differences with the domain controller
> -----Original Message-----
> From: samba [mailto:samba-bounces at lists.samba.org] On Behalf Of L.P.H. van
> Belle via samba
> Sent: 24 July 2018 09:41
> To: samba at lists.samba.org
> Subject: Re: [Samba] Failed to establish your Kerberos Ticket cache due time
> differences with the domain controller
>
> I did re-read the whole thread again.
>
> Im running out
2017 Jun 01
2
Reversing one dimension of an array, in a generalized case
My error. Clearly I did not do enough testing.
z <- array(1:24,dim=2:4)
> all.equal(f(z,1),f2(z,1))
[1] TRUE
> all.equal(f(z,2),f2(z,2))
[1] TRUE
> all.equal(f(z,3),f2(z,3))
[1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >"
[2] "Mean relative difference: 0.6109091"
# Your earlier example
> z <- array(1:120, dim=2:5)
>
2006 Apr 18
12
Update on SWIG
SWIG 1.3.29 isn''t going to cut it for us. We''ll have to wait until
1.3.30, unless I can hack together a nasty workaround for a problem with
the object tracking. It seem that each SWIG module gets its own hash
for tracking objects. This means that, for example, you create a find
dialog (in FindReplaceDialog.cpp) events you get from it (from
FindDialogEvent.cpp)
2011 May 24
2
[LLVMdev] predicates and conditional execution
2011/5/24 Justin Holewinski <justin.holewinski at gmail.com>:
> On Tue, May 24, 2011 at 3:02 AM, roy rosen <roy.1rosen at gmail.com> wrote:
>>
>> Hi,
>>
>> I was wondering if LLVM supports predicates and conditional execution.
>> Something like we have in IA64.
>> There is a register class of predicates and then every instruction may
>> be
2018 Jan 02
1
httr::content without message
Thanks to all that replied. I had just looked through the httr code and sure enough for a .csv mime time it calls readr::read_csv(). The httr::content docs suggest not using automatic parsing in a package, rather to determine mime type and parse yourself and Ben's suggestion also works if I do:
junk <- readr::read_csv(r1$content, col_types = cols())
Perfect. Using httr rather than
2007 Oct 15
11
What web GUI are people happy with?
Just wondering what web GUI people like for asterisk. I installed
asterisk from source and I was looking at possibly installing web GUI
for system management. So far freepbx.org looks promising anybody else
have any suggestions.
Thanks
Roy Anciso
Director of Technology
Manistee Intermediate School District
1710 Merkey Road
Manistee, MI 49660
Ph: 231-723-4264
Fx: 231-723-1690
roy
2017 Jun 01
0
Reversing one dimension of an array, in a generalized case
How about this:
f <- function(a,wh){ ## a is the array; wh is the index to be reversed
l<- lapply(dim(a),seq_len)
l[[wh]]<- rev(l[[wh]])
do.call(`[`,c(list(a),l))
}
## test
z <- array(1:120,dim=2:5)
## I omit the printouts
f(z,2)
f(z,3)
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into