Okay. Got some lunch, I can think about this with both halves of the brain.
drop_token1<-function(x) {
return(paste(x[2:length(x)],sep="",collapse="."))
}
affnames<-c("X0.Classical.10.11.1_.HuEx.1_0.st.v2..CEL",
"X1.Classical.10.11.1_.HuEx.1_0.st.v2..CEL")
affnames.split<-strsplit(affnames,"[.]"
Transcription. I forgot the "collapse" argument when I wrote the email:
drop_token1<-function(x) {
return(paste(x[2:length(x)],sep="",collapse="."))
}
Jim
On Wed, Mar 23, 2016 at 10:14 AM, Christian T Stackhouse (Campus)
wrote:
> Very close! The header now looks like this: c("10", "11", "1_",
I think it's the "unlist". I can only test this with one set of made
up names at a time.
names(out[[affdf]])<-
lapply(strsplit(names(out[[affdf]]),"[.]"),drop_token1)
Jim
On Wed, Mar 23, 2016 at 9:57 AM, Christian T Stackhouse (Campus)
wrote:
> This is what I ran:
>
>> drop_token1<-function(x
Sorry, should be:
names(out[[affdf]])<-
lapply(unlist(strsplit(names(out[[affdf]]),"[.]")),drop_token1)
Jim
On Wed, Mar 23, 2016 at 9:43 AM, Christian T Stackhouse (Campus)
wrote:
> Thank you, Jim. I got this error returned:
>
> Error in strsplit(names(out[[affdf]])) :
> argument "split" is
Okay, I just snipped off the first token in the header labels assuming
that there would be no more periods. Try this:
drop_token1<-function(x) {
return(paste(x[2:length(x)],sep="."))
}
for(affdf in 1:length(out)) {
names(out[[affdf]])<-lapply(unlist(strsplit(names(out[[affdf]]))),drop_token1)
wri
Hi Christian,
This untested script might get you going (assuming you want a CSV format):
for(affdf in 1:length(out)) {
names(out[[affdf]])<-lapply(strsplit(names(out[[affdf]]),"[.]"),"[",2)
write.csv(out[[affdf]],file=paste("affymetrix",affdf,".txt",sep=""))
}
Jim
On Wed, Mar 23, 2016 at 6:32
Hi Eliza,
I think you only need to change the margins and the placement of the
right axis label:
colours <- c("black", "grey")
par(mar=c(5,4,4,4))
barplot(prop,ylab = "Numbers", cex.lab = 1.5, cex.main = 1.4,
beside=TRUE, col=colours,ylim=c(0,250))
axis(side=3,xlim=c(0,45), at=c(6,12,18,24,30,36,
Hello I have encountered a bug(?) with the parallel package. When run
from within a function, the parLapply function appears to be copying
the entire parent environment (environment of interior of function)
into all child nodes in the cluster, one node at a time - which is
very very slow
Dear users:
Has anyone tried using splm to estimate a fixed effects model with one or
more time-invariant variables?
I may have missed something in the manual, but there isn't a clear way to
switch to estimators (e.g. first difference) suited for data like that in
splm.
My colleagues and I are v
Dear useRs,
I have defined two matrices "prop" and "ELE" in the following manner
> dput(prop)
structure(c(122.4667, 87.1500875, 94.3647755102041, 84.8471625,
95.2767755102041, 84.15558125, 121.8467, 90.75970625, 98.2028979591837,
87.1500875, 88.2953043478261, 72.81219375, 88.234,
Hello!
The overall goal I have is taking a large data frame and splitting it into
several smaller data frames (preserving column headers) which I can save as txt
files to feed into my APACHE ANY23 server for conversion into RDF.
This is what I call to split up the original file:
out <- spli
> On Mar 22, 2016, at 10:00 AM, Martin Maechler
> wrote:
>
>> Roy Mendelssohn <- NOAA Federal >
>>on Tue, 22 Mar 2016 07:42:10 -0700 writes:
>
>> Hi All:
>> I am running prcomp on a very large array, roughly [50, 3650]. The
>> array itself is 16GB. I am running on a Unix mac
On 22/03/2016 1:08 PM, Jan Kacaba wrote:
Hello, is it possible to run kiniter by script instead by clicking on
button compile PDF?
Say I have "texfile.rnw" and "myscript.R". I would like to knit texfile.rnw
by runnig script "myscript.R".
In "myscript.R" I would write something like this:
knit("t
Hello, is it possible to run kiniter by script instead by clicking on
button compile PDF?
Say I have "texfile.rnw" and "myscript.R". I would like to knit texfile.rnw
by runnig script "myscript.R".
In "myscript.R" I would write something like this:
knit("texfile.rnw")
[[alternative HTML ve
> Roy Mendelssohn <- NOAA Federal >
> on Tue, 22 Mar 2016 07:42:10 -0700 writes:
> Hi All:
> I am running prcomp on a very large array, roughly [50, 3650]. The
array itself is 16GB. I am running on a Unix machine and am running “top” at
the same time and am quite surpri
Hi All:
I am running prcomp on a very large array, roughly [50, 3650]. The array
itself is 16GB. I am running on a Unix machine and am running “top” at the
same time and am quite surprised to see that the application memory usage is
76GB. I have the “tol” set very high (.8) so that it s
Thanks all. This is interesting, and for what I am doing worthwhile and
helpful. I have to be careful in each operation whether a copy is made or not,
and knowing this allows me to test on small examples what any command will do
before I use,
Thanks again, I appreciate all the help. I will
dpois(0, lambda) == e^(-lambda)
The wikipedia formula is
ifelse(x == 0, zero + dpois(0, lambda) * (1-zero), dpois(x, lambda) *
(1-zero))
or
ifelse(x == 0, zero + dpois(x, lambda) * (1-zero), dpois(x, lambda) *
(1-zero))
so we can move the dpois() out of the ifelse()
ifelse(x == 0, zero, 0) +
And why is the first term of ifelse(x == 0, zero, 0) + dpois(x, lambda) / (1 -
zero)
ifelse(x == 0, zero, 0)
rather than something corresponding to
zero+(1-zero)e^{-lambda}
https://en.wikipedia.org/wiki/Zero-inflated_model#Zero-inflated_Poisson
> On 22 Mar 2016, at 14:25, Matti Viljamaa wrot
zero = proportion of zero inflation part
lamba = expected value of poisson part
There was a typo in the distribution. It should multiple by (1 - zero)
instead of divide by it.
dzeroinflpois <- function(x, lambda, zero){
ifelse(x == 0, zero, 0) + dpois(x, lambda) * (1 - zero)
}
ir. Thierry Onke
Could you clarify what are the parameters and why it’s formulated that way?
-Matti
> On 22 Mar 2016, at 14:17, Thierry Onkelinx wrote:
>
> Dear Matti,
>
> What about this?
>
> dzeroinflpois <- function(x, lambda, zero){
> ifelse(x == 0, zero, 0) + dpois(x, lambda) / (1 - zero)
> }
> plot(x,
Dear Matti,
What about this?
dzeroinflpois <- function(x, lambda, zero){
ifelse(x == 0, zero, 0) + dpois(x, lambda) / (1 - zero)
}
plot(x, dzeroinflpois(x, lambda = 10, zero = 0.2), type = "l")
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Fo
I’m doing some optimisation that I first did with normal Poisson (only
parameter theta was estimated), but now I’m doing the same with a zero-inflated
Poisson model which
gives me two estimated parameters theta and p (p is also pi in some notation).
My question is, is there something equivalent
On 03/22/2016 12:44 AM, Omar André Gonzáles Díaz wrote:
Hi,I have a DF with a column with "html", like this:
https://ad.doubleclick.net/ddm/trackimp/N344006.1960500FACEBOOKAD/B9589414.130145906;dc_trk_aid=303019819;dc_trk_cid=69763238;ord=[timestamp];dc_lat=;dc_rdid=;tag_for_child_directed_tre
Hi Martin,
On 03/22/2016 10:20 AM, Martin Maechler wrote:
>Dénes Tóth
> on Fri, 18 Mar 2016 22:56:23 +0100 writes:
> Hi Roy,
> R (usually) makes a copy if the dimensionality of an array is modified,
> even if you use this syntax:
> x <- array(1:24, c(2, 3, 4))
> Dénes Tóth
> on Fri, 18 Mar 2016 22:56:23 +0100 writes:
> Hi Roy,
> R (usually) makes a copy if the dimensionality of an array is modified,
> even if you use this syntax:
> x <- array(1:24, c(2, 3, 4))
> dim(x) <- c(6, 4)
> See also ?tracemem, ?data.table:
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