Displaying 20 results from an estimated 10000 matches similar to: "Catenating strings involving plotmath symbols."
2009 May 01
1
integrate with large parameters
Dear R-users,
i have to integrate the following function
`fun1` <- function (a, l1, l2)
{
exp(log(l1) * (a - 1) - l2 * lgamma(a))
}
but if l1 is large, i get the "non-finite function value" error, so my
idea is to rescale with exp(-l1)
`fun2` <- function (a, l1, l2)
{
exp(log(l1) * (a - 1) - l2 * lgamma(a) - l1)
}
but it seems this doesn't solve the problem, when
2013 Apr 09
1
sorting the VAR model output according to variable names??
I was wondering if one can have the coefficients of VAR model sorted
according to variable names rather than lags. If you notice below, the
output is sorted according to lags.
>VAR(cbind(fossil,labour),p=2,type="const")
VAR Estimation Results:
=======================
Estimated coefficients for equation fossil:
===========================================
Call:
fossil = fossil.l1
2011 Jun 27
4
How many L1/L2 my cpu have ?
Hi
Could anybody explain me how to check how many L1/L2 cache my cpu have.
I'm using CentOS 5.6
*cat /proc/cpuinfo |grep CPU *
model name : Intel(R) Core(TM)2 Duo CPU T9300 @ 2.50GHz
model name : Intel(R) Core(TM)2 Duo CPU T9300 @ 2.50GHz
Diagram of a generic dual-core processor, with CPU-local level 1 caches, and
a shared, on-die level 2 cache.
2010 Aug 12
1
Need help to understand integrate function
Hi,
I'm running into a wall when trying to use the integrate function. I
have the following setting:
powerLaw2 <- function(x,l1,l2,c0,t0) {
idx <- which(x <= 0);
if (length(idx) > 0) {
x[idx] <- 0;
}
xl <- (-l1+l2)*log(x/t0);
L <- log(c0)-l1*log(x)-log(1+exp(xl));
L <- exp(L);
return(L);
}
plCDF2 <- function(x,l1,l2,c0,t0) {
2007 Apr 16
1
Names in vector occurring in another vector
I have a vector of character strings such as
mainnames<-c("CAD","AUD") and another vector say
checknames<-c("CAD.l1","AUD.l1","JPY.l1","EUR.l1","CAD.l2","AUD.l2","JPY
.l2","EUR.l2")
I want a new vector of character strings that is just
2012 Aug 04
1
lme4 / HLM question
I'm hoping that this is a relatively easy question for someone familiar with
the lme4 package.
I'm accustomed to using HLM software and writing a simple 2 level [null]
equation like this:
L1 - Yij = b0 + e
L2 - b0 = B00 + u0
The following command in R provides results that are identical to the HLM
program.
results <- lmer( Y ~ 1 |id , PanelData4)
I can't
2014 Jun 26
2
[LLVMdev] cross-section differences in MC generation
I think that's incorrect. It should to:
.section .foo
.L1:
.L2 = .L1
.section .bar
.long .L3-.L2
.L3:
Because .L3 and .L2 are in different sections.
- Justin
On Thu, Jun 26, 2014 at 2:46 PM, Rafael EspĂndola
<rafael.espindola at gmail.com> wrote:
> This reduces to
>
> .section .foo
> .L1:
> .L2 = .L1
> .section .bar
> .long .L1-.L2
>
>
> Which is fairly
2005 Feb 21
5
Compare rows of two matrices
Hello,
#I have two matrices, eg.:
y <- matrix( c(20, NA, NA, 45, 50, 19, 32, 101, 10, 22, NA, NA, 80, 49, 61, 190), ncol=4 )
x <- matrix( c(20, NA, NA, NA, 50, 19, 32, 101, 10, 22, NA, NA, 80, 49, 61, 190), ncol=4 )
#Whereas x contains all NA?s from y plus some additional NA?s.
#I want to find the index of these additional NA?s. I think, there must be a very
2007 Feb 12
1
Page allocation failure
Hi list,
I have a very strange problem with my network. I have 2 internet
connections: A - 1 Gbit, B - 100Mbps.
Network layout:
A, B
| |
[Brd1]
/ \
[L1] [L2]
\ /
[ GW1]
...................
Clients
.....................
Brd1 runs bgpd, and balances the traffic through L1 and L2.
L1 and L2 do traffic shaping.
GW1 does some packet
2017 Jun 01
3
Post for R
Hello,?
I want to split the dataframe into 1000 groups based on two column values(max value and second max value). First, I made two lists L1 and L2. ?L1 is the list divided into 100 groups based on the range of max value and L2 is divided into 10 groups based on the second max values. Now I want to do the combinations based on L1 and L2. I want to do a for loop for L1 and for each element in L1,
2010 Jan 14
1
lattice dotplot with missing levels in factor variable
Hi,
I am trying to create a dotplot where each panel shows levels vs.
responses; the levels are sorted by responses but levels vary from one
panel to another. However, I run into problems with controlling the
y-limits and y-labels.
In particular, suppose I have a data frame
rsp <- c(10,2,4,0,2,3)
lvl <-
2004 Feb 05
2
correction to the previously asked question (about merging factors)
I have two factors l1, l2, and I'd like to merge them.
(Remark: The factors can not be converted to charaters)
Function c() does not give me the result I want:
> l1 = factor(c('aaaa', 'bbbb'))
> l2 = factor(c('ccc', 'dd'))
> lMerge = factor(c(l1, l2))
> lMerge
[1] 1 2 1 2
Levels: 1 2
>
I'd like to merge l1 and l2 and to get lMerge
2009 Feb 17
4
joining "one-to-many"
Hello list,
I am wondering if a joining "one-to-many" can be done a little bit easier. I tried merge function but I was not able to do it, so I end up using for and if.
Suppose you have a table with locations, each location repeated several times, and some attributes at that location. The second table has the same locations, but only once with a different set of attributes. I would
2015 Mar 11
2
[LLVMdev] How to run two loop passes non-interleaved if they are registered one by one?
Hi LLVM developers,
I want to add LICM pass after loop unrolling pass in current optimization
pipeline. Because both of them are loop passes, so if I registered them one
by one, they will interleaved go through all loops in bottom up way within
same loop pass manager. Loop unroling pass may create new inner loops from
partial unrolling, and those newly created loops can be visited only if the
2018 Feb 08
1
Re: Nested KVM: L0 guest produces kernel BUG on wakeup from managed save (while a nested VM is running)
On 08.02.2018 11:46, Kashyap Chamarthy wrote:
> On Wed, Feb 07, 2018 at 11:26:14PM +0100, David Hildenbrand wrote:
>> On 07.02.2018 16:31, Kashyap Chamarthy wrote:
>
> [...]
>
>> Sounds like a similar problem as in
>> https://bugzilla.kernel.org/show_bug.cgi?id=198621
>>
>> In short: there is no (live) migration support for nested VMX yet. So as
>>
2012 May 23
3
applying cbind (or any function) across all components in a list
#If I have two lists as follows
a1<- array(1:6, dim=c(2,3))
a2<- array(7:12, dim=c(2,3))
l1<- list(a1,a2)
a3<- array(1:4, dim=c(2,2))
a4<- array(5:8, dim=c(2,2))
l2<- list(a3,a4)
#how can I create a new list with the mean across all arrays within the
list, so all components are included? As an example for [[1]];
cbind((l1[[1]][,1]+l2[[1]][,1])/2,
2017 Jun 01
0
Post for R
Hi Carrie,
You may have a problem with this if some subsets are empty:
L3<-lapply(split(df,cut(df$max,seq(0,1,by=0.01))),
split,cut(df$submax,seq(0,0.2,by=0.02)))
Jim
On Thu, Jun 1, 2017 at 12:48 PM, carrie wang via R-help
<r-help at r-project.org> wrote:
>
> Hello,
> I want to split the dataframe into 1000 groups based on two column values(max value and second max value).
2018 Feb 08
5
Re: Nested KVM: L0 guest produces kernel BUG on wakeup from managed save (while a nested VM is running)
>> In short: there is no (live) migration support for nested VMX yet. So as
>> soon as your guest is using VMX itself ("nVMX"), this is not expected to
>> work.
>
> Hi David, thanks for getting back to us on this.
Hi Florian,
(sombeody please correct me if I'm wrong)
>
> I see your point, except the issue Kashyap and I are describing does
> not
2012 Jan 12
1
[LLVMdev] A question of Sparc assembly generated by llc
Hi,
There are some generated Sparc assembly code like this:
main: ! @main
! BB#0:
save %sp, -112, %sp
sethi 0, %l0
or %g0, 5, %l1
st %l0, [%fp+-4]
st %l1, [%fp+-8]
st %l1, [%fp+-12]
sethi %hi(.L.str), %l1
ld [%fp+-8], %o1
add %l1, %lo(.L.str), %l1
or %g0, %l1, %o0
call printf
nop
ld [%fp+-12], %o2
ld [%fp+-8], %l2
sethi %hi(.L.strQ521), %l3
add
2010 Mar 26
3
Using lapply with two lists
Hello guys,
I have a list L1 of matrix. I have another list L2 with the same number of
elements representing the row of the L matrix that I want to delete
(L1[[i]][-L2[[i]],]) but I can't do this with lapply as it iterates through
L1 (first argument) and not L2. Any idea?
-----
Anna Lippel
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