I think you are looking for match or %in% (which is a based on match)
> a <- sample(12)
> b <- c(1, 3, 5, 11, 17)
> a
[1] 10 8 1 4 7 3 6 11 2 12 5 9
> b
[1] 1 3 5 11 17
> [1] 1
> match(a, b)
[1] NA NA 1 NA NA 2 NA 4 NA NA 3 NA
> match(a, b, 0)
[1] 0 0 1 0 0 2 0 4 0 0 3 0
> match
I have two structures. I think they are lists, but I am not sure. Both
structures contain integers. I am trying to find those members of list b that
are found in list a. I have tried to perform the search using grep, but I get
an error. Please see code below. I would appreciate knowing how to se
Dénes:
A fair point! The only reason I have is ignorance -- I have not used
data.table. I am not surprised that it and perhaps other packages
(dplyr maybe?) can do things in a reasonable way very efficiently. The
only problem is that it requires us to learn yet another
package/paradigm. There may
On 17/09/15 09:50, Khedhaouiria Dikra wrote:
Hi everyone,
I'm trying to fit several models with the VGAM packages using the
gamma2 function with an identity function link. ..
For questions concerning a particular package it is usually best to
contact the package maintainer (see maintai
On 09/16/2015 04:41 PM, Bert Gunter wrote:
Yes! Chuck's use of mapply is exactly the split/combine strategy I was
looking for. In retrospect, exactly how one should think about it.
Many thanks to all for a constructive discussion .
-- Bert
Bert Gunter
Use mapply like this on large problem
I assume it's to return the vectors in their original order.
-- Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Wed, Sep 16, 2015 at 2:10 PM, David Winsemius wrote:
>
> On Sep 15, 2015, at 7:20 PM, Charl
On Sep 15, 2015, at 12:09 PM, Huot, Matthieu wrote:
> Hi Tom
>
> I know the post is over 7-8 years old but I am having the same question. How
> to do a post-hoc test like TukeyHSD on coxph type output.
Create a new variable using the `interaction`-function, apply you contrasts to
that object,
On Sep 15, 2015, at 7:20 PM, Charles C. Berry wrote:
> On Tue, 15 Sep 2015, Bert Gunter wrote:
>
>> Thanks to both Davids.
>>
>> I realize that these things are often a matter of aesthetics -- and
>> hence have little rational justification -- but I agree with The Other
>> David: eval(parse) se
> On Sep 16, 2015, at 3:40 PM, Bert Gunter wrote:
>
> Nope. Take it back. I stand uncorrected.
>
>> system.time(z <-sample(1:10,1e6, rep=TRUE))
> user system elapsed
> 0.045 0.001 0.047
>
>> system.time(z <-sample.int(10,1e6,rep=TRUE))
> user system elapsed
> 0.012 0.000 0.013
Dear all,
I’m trying to do a graph,
3 rows, 5 columns, with the design:
# 3 4 5 6
#2
# 7 8 9 10
I had a code for 3 rows, 3 columns, with the design::
# 3 4
#2
# 7 8
and I tried to modify it, but I had no success :(
I suppose the
Dear All,
when using the following code
x <- iris[,1:2] # Sepal.Length Sepal.Width
y <- iris[,3:4] # Petal.Length Petal.Width
vsel <-
varSelection(x=x,y=y,nboot=5,yaiMethod="randomForest",useParallel=FALSE)
I get the following error code
Error in yai(x = xa, y = y, method = yaiMethod, bootst
Hello all,
I've been busy figuring out how to get realisations for a non-standard
distribution in R. Define \theta=o (not 0). Consider f_x(x)=o*x for o in
\sqrt{2/o}.
Yet, I can't seem to find a way to get realisations for custom
distributions. Can anyone help me out here?
--
View this messa
Hi everybody,
>From a questionnaire, I have a dataset like this one with some 40 items:
df1 <- data.frame(subject=c('user1','user2', 'user3', 'user4'),
item1=c(0,1,2,5), item2=c(1,2,1,2), item3=c(2,3,4,0), item4=c(0,3,3,2),
item5=c(5,5,5,5))
Users can choose an answer from 0 to 5 for each item.
Nope. Take it back. I stand uncorrected.
> system.time(z <-sample(1:10,1e6, rep=TRUE))
user system elapsed
0.045 0.001 0.047
> system.time(z <-sample.int(10,1e6,rep=TRUE))
user system elapsed
0.012 0.000 0.013
sample() has to do subscripting in the general case; sample.int d
Yes. Thanks Marc. I stand corrected.
-- Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Wed, Sep 16, 2015 at 1:28 PM, Marc Schwartz wrote:
>
>> On Sep 16, 2015, at 1:06 PM, Bert Gunter wrote:
>>
>> Yikes!
> On Sep 16, 2015, at 1:06 PM, Bert Gunter wrote:
>
> Yikes! The uniform distribution is a **continuous** distribution over
> an interval. You seem to want to sample over a discrete distribution.
> See ?sample for that, as in:
>
> sample(1:4,100,rep=TRUE)
>
> ## or for this special case and fa
You can use the coordinates of the plot region as fractions of the figure
region, par("plt"), to define the adj= argument of mtext(). And you can
use the number of lines of the plot margin to define the line= argument of
mtext(). For example:
plot.figure <- function() {
par(mfrow=c(3, 1), mar=
Yikes! The uniform distribution is a **continuous** distribution over
an interval. You seem to want to sample over a discrete distribution.
See ?sample for that, as in:
sample(1:4,100,rep=TRUE)
## or for this special case and faster
sample.int(4,size=100,rep=TRUE)
Cheers,
Bert
Bert Gunter
"Da
> On Sep 16, 2015, at 12:11 PM, thanoon younis
> wrote:
>
> Dear R- users
>
> I want to generate ordered categorical variable vector with 200x1 dimension
> and from 1 to 4 categories and i tried with this code
>
> Q1=runif(200,1,4) the results are not just 1 ,2 3,4, but the results with
> dec
There is a version of sample especially for integers:
> Q1 <- matrix(sample.int(4, 200, replace=TRUE), 200)
> str(Q1)
int [1:200, 1] 1 4 4 2 3 3 4 4 2 3 ...
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
Hello,
Try ?sample.
Hope this helps,
Rui Barradas
Em 16-09-2015 18:11, thanoon younis escreveu:
Dear R- users
I want to generate ordered categorical variable vector with 200x1 dimension
and from 1 to 4 categories and i tried with this code
Q1=runif(200,1,4) the results are not just 1 ,2 3,4
Actually x is the variable name since your function returned a vector of three
values:
> tbl <- aggregate(warpbreaks[, 1], warpbreaks[, 2:3], function(breaks) c(Min =
> min(breaks),
+ Med = median(breaks), Max = max(breaks)))
> str(tbl)
'data.frame': 6 obs. of 3 variables:
$ wool : Facto
If I understand correctly
?sample
On 16/09/2015 18:11, thanoon younis wrote:
Dear R- users
I want to generate ordered categorical variable vector with 200x1 dimension
and from 1 to 4 categories and i tried with this code
Q1=runif(200,1,4) the results are not just 1 ,2 3,4, but the results wi
Dear R- users
I want to generate ordered categorical variable vector with 200x1 dimension
and from 1 to 4 categories and i tried with this code
Q1=runif(200,1,4) the results are not just 1 ,2 3,4, but the results with
decimals like 1.244, 2.342,4,321 and so on ... My question how can i
generate a
Hello,
for a multiple figures plot I am looking for the syntax to put text in the
top left of the margin (of the plot). I want my testfunction plot.figure to
place mtext in the top left of the red margin (created by box("figure",
col="red")).
Can anybody help?
Thanks
Hermann
plot.figure <- func
Good day,
Yes, exactly. I found that aggregate is another alternative which doesn't
require a package dependency, although the column formatting is less suitable,
always prepending x.
aggregate(warpbreaks[, 1], warpbreaks[, 2:3], function(breaks) c(Min =
min(breaks), Med = median(breaks), Max
I have frequently noticed a strange effect when using the progress bar
in a loop, that the console will sometimes be cleared. Whatever was
printed before will then be inaccessible, which can be annoying when I
want to check the progress and possible issues in a long script. It
seems to be an in
Yes! Chuck's use of mapply is exactly the split/combine strategy I was
looking for. In retrospect, exactly how one should think about it.
Many thanks to all for a constructive discussion .
-- Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not
The data structure returned by optimx (that you're storing as gmmiv) is
like a data frame, and stores the conversion code in the field convcode.
So do something like this:
gmmiv =Optimx()
if (gmmiv$convcode ==0) {
store[j,] = coef(gmmiv)
}
On Tue, Sep 15, 2015 at 3:12 PM, Olu Ola wrote:
> Th
On 15.09.2015 11:54, Pau Marc Muñoz Torres wrote:
Hello everybody,
I want to use Rapidr package, it is an old package that uses the
package requires GenomicRanges version 1.14.4. The current version of the
package is GenomicRanges 1.20.6. There is some way of having both the
actual and the p
I think a more common idiom for the simpler case would be to use indexing
vals <- c(0.1, 0.15, 0.2)
mult <- vals[ASBclass]
(However, some people are on the move to enforce
mult <- vals[as.numeric(ASBclass)]
because people who confuse factors and character variables get even more
confused abo
31 matches
Mail list logo
