Thanks a lot for your answers.
Peter: sorry, here is the missing information:
* I use the function gam() of the package �mgcv�
* Yes, the output changes when I use offset(log_trap_eff) instead of
offset=log_trap_eff. By using offset(log_trap_eff), the output is more coherent
with the obs
> On 22 Nov 2016, at 23:07 , Bert Gunter wrote:
>
> Define "very different." Sounds like a subjective opinion to me, for
> which I have no response. Apparently others are similarly flummoxed.
> Of course they would not in general be identical.
Er? I don't see much reason to disagree that a ran
Well part of the issue is that the negative binomial estimates are for
means and they can differ a fair bit from the raw counts, but I'm also
guessing that part of the issue is that the offset may not be accounted for
with the predict.gam() function.
Brian
Brian S. Cade, PhD
U. S. Geological Sur
> On Nov 22, 2016, at 1:29 PM, Marine Regis wrote:
>
> Hello,
>
>> From capture data, I would like to assess the effect of longitudinal changes
>> in proportion of forests on abundance of skunks. To test this, I built this
>> GAM where the dependent variable is the number of unique skunks and
Define "very different." Sounds like a subjective opinion to me, for
which I have no response. Apparently others are similarly flummoxed.
Of course they would not in general be identical.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticki
Hello,
>From capture data, I would like to assess the effect of longitudinal changes
>in proportion of forests on abundance of skunks. To test this, I built this
>GAM where the dependent variable is the number of unique skunks and the
>independent variables are the X coordinates of the centroid
On Windows 32-bit I think (it's been a while) you can push it to 3 GB but
to go beyond you need to run R on 64-bit Windows (same rule for all
software not just R). I'm pretty sure this is already documented in the R
documentation.
Henrik
On Nov 22, 2016 19:49, "Ista Zahn" wrote:
Not convenient
Ah, you also need to use a 64-bit operating system. Depending on the age of
your hardware this may also mean you need a new computer.
There are ways to process data on disk for certain algorithms, but you will be
glad to leave them behind once the opportunity arises, so you might as well do
so
Not conveniently. Memory is cheap, you should buy more.
Best,
Ista
On Nov 22, 2016 12:19 PM, "Partha Sinha" wrote:
> I am using R 3.3.2 on win 7, 32 bit with 2gb Ram. Is it possible to use
> more than 2 Gb data set ?
>
> Regards
> Partha
>
> [[alternative HTML version deleted]]
>
> ___
Thanks for helping Jim.
I'm actually using the pbapply function together with the print function within
a loop. In earlier versions, the progress bar and the output of the print
function used to appear after each iteration of the loop. But with the 3.3.1.
Version nothing appears, instead the c
Yes.
If you cannot read the dataset with the usual means, using functions like
read.table or read.csv, try the ff package: https://cran.r-
project.org/web/packages/ff/index.html.
Best,
On Tue, Nov 22, 2016 at 2:16 PM, Partha Sinha wrote:
> I am using R 3.3.2 on win 7, 32 bit with 2gb Ram. Is
Depends how you use it. e.g. it can be stored on disk and worked with
in pieces. Or some packages work with virtual memory, I believe.
However, it is certainly not possible to read it into R. In fact, you
probably won't be able to handle more (and maybe much less) than about
500 mb in R.
Cheers,
I am using R 3.3.2 on win 7, 32 bit with 2gb Ram. Is it possible to use
more than 2 Gb data set ?
Regards
Partha
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mai
Thanks Sarah (and all the others who replied) for your precious
suggestions! Matteo
On 22 November 2016 at 14:18, Sarah Goslee wrote:
> Here's how to get one:
>
> x <- c(9,9,1,3,2,7,6,10,5,6)
>> which.min(abs(x - quantile(x, .25)))
> [1] 4
>
> And here's one of the various ways to get the entire
You might try something like
x <- c(9,9,1,3,2,7,6,10,5,6)
p <- (0:4)/4
order(x) [ quantile(seq_along(x),p, type=1) ]
# [1] 3 4 7 1 8
Selecting which value of 'type' works makes my head hurt.
You could also use 1+(p*(length(x)-1) as the index into order(x).
Bill Dunlap
TIBCO Software
w
> On Nov 22, 2016, at 4:21 AM, Matteo Richiardi
> wrote:
>
> Dear R-users,
> a very easy one for you, I guess. I need to extract the indexes of the
> elements corresponding to different quantiles of a vector. When a
> quantile is an interpolation between two adjacent values, I need the
> index
Dear all,
This is my test code:
library(wmtsa)
sunspotsLong <- rep(sunspots, times=3000) ## try "times=30" (or 300)
sunspots.cwt <- wavCWT(sunspotsLong)
plot(sunspots.cwt, series=TRUE)
If you adapt times in the second line with "30", the code works. But 300 or
3000 not so much.
Are there more
Here's how to get one:
x <- c(9,9,1,3,2,7,6,10,5,6)
> which.min(abs(x - quantile(x, .25)))
[1] 4
And here's one of the various ways to get the entire set:
> xq <- quantile(x)
> sapply(xq, function(y)which.min(abs(x - y)))
0% 25% 50% 75% 100%
34718
Sarah
On Tue, Nov 22,
Hi Ferri,
It sounds like the function 'separate' from the tidyr package is what you
look for,
HTH
Ulrik
On Tue, 22 Nov 2016 at 14:49 Ferri Leberl wrote:
Dear All,
I asked for support to deal with a hirarchy within a character separated
list.
I solved the problem crudely but effectively by
-
Dear All,
I asked for support to deal with a hirarchy within a character separated list.
I solved the problem crudely but effectively by
- Choosing for a TSV as input, where in columns that may contain several (or as
well no) items the items are separated via semicolon
- adding semicolons to th
Dear R-users,
a very easy one for you, I guess. I need to extract the indexes of the
elements corresponding to different quantiles of a vector. When a
quantile is an interpolation between two adjacent values, I need the
index of the value which is closer (the lower value - or the higher
value for w
Hi
Heatmap does not work with data frames (at least the version I have)
> b=as.data.frame(matrix(c(3,4,5,8,9,10,13,14,15,27,19,20),3,4))
> heatmap(b)
Error in heatmap(b) : 'x' must be a numeric matrix
With matrix, heatmap works as expected.
> b=matrix(c(3,4,5,8,9,10,13,14,15,27,19,20),3,4)
> he
Hi
Well, thean you could try as I sugested.
.mean <- apply(temp[,-(1:3)],1, mean, na.rm=T)
.max <- apply(temp[,-(1:3)],1, max, na.rm=T)
.min <- apply(temp[,-(1:3)],1, min, na.rm=T)
temp2 <- data.frame(temp[,1:3], maxim = .max, minim = .min, aver = .mean)
You should construct a cycle, read the ye
Dear Carl,
this came through fine, as text only
... but then I did not see any question anymore.
Best regards,
Martin Maechler
(R core and mailing list "operator")
> Carl Sutton via R-help
> on Tue, 22 Nov 2016 05:38:49 + writes:
> Hopefully the attached is text and not htm
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