Hi,
I'm trying to interpret the following results, with respect to normality test
using the over identifying moment conditions
where returns have normal distribution
with parameter mu,sd
and i have 4 moment conditions
E[r-mu/sd]=0
E[(r-mu)^2/sd-1]=0
E[(r-mu)^3/sd^3]=0
E[(r-mu)^4/sd^4-3]=0
Probably, it's an obvious info, but I have not found anything in R FAQ
concerning
this feature/bug.
The results of ks.test and wilcoxon.test (in the Mann-Whitney version,
paired = 'FALSE') don't coincide with the results from the other statistical
packages, e.g. Statistica, Medcalc, and (as for
I have a problem with results from difftime being 1 hour different than
expected. 2 examples are given below:
datetime - matrix(data=rbind(c(2012-03-31 21:00:00, 2012-04-01 00:00:00,
2012-04-01 03:00:00, 2012-04-01 06:00:00),
c(2012-10-06 21:00:00, 2012-10-07 00:00:00,
Hello together,
i have a data.frame, like this one:
No. Change Date
A 123 final2013-01-15
B 123 error 2013-01-16
C 123 bug fixed 2013-01-17
D
Hello,
Here is the result I get using your script:
sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: x86_64-suse-linux-gnu (64-bit)
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00
[4] 2012-04-01 06:00:00
Time differences in hours
[1] 3 3 3
attr(,tzone)
[1]
[1] 2012-10-06
Hi
Maybe others can help you but here is my comment. I already use R for many
years and never used such construction. Objects in global environment shall not
be modified by functions it is a bad practice. Imagine you have some data frame
you prepared and controlled in many steps and use some
Dear Gundala,
Try
as.data.frame.table(foo)
HTH,
Jorge.-
On Fri, Feb 1, 2013 at 5:00 PM, Gundala Viswanath wrote:
I have the following data frame:
foo
w x y z
n 1.51550092 1.4337572 1.2791624 1.1771230
q 0.09977303 0.8173761 1.6123402 0.1510737
r
Hi:
Try
DF - structure(list(Gene = structure(1:5, .Label = c(NM_001001130,
NM_001001144, NM_001001152, NM_001001160, NM_001001176
), class = factor), T0h = c(68L, 0L, 79L, 1L, 0L), T0.25h = c(95L,
1L, 129L, 1L, 0L), T0.5h = c(56L, 4L, 52L, 2L, 0L), T1h = c(43L,
0L, 50L, 0L, 0L), T2h = c(66L, 1L,
Hi
see inline
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of staysafe23
Sent: Friday, February 01, 2013 1:01 AM
To: r-help@r-project.org
Subject: [R] Nested loop and output help
Hello Everyone,
My name is Thomas and I
Hi
The same with me on Windows
Most probably an issue of daylight savings time setting.
Sys.timezone()
[1] CET
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Pascal Oettli
Sent: Friday, February 01, 2013 9:18
Hi,
I want to write own R functions for cumulative incidence function and also
for the pseudovalue of the cumulative incidence function.
Can you help me?
Tas.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
In example for function 'stop' in R, there is
options(error = expression(NULL))
with comment
# don't stop on stop(.) Use with CARE!
I was interested, wanted to know how don't stop on stop(.) was. So, I tried
it.
Typing
example(stop)
at the R prompt and pressing ENTER give this.
Thanks a lot, I'll test this package very soon.
As it seems general purpose (ie not specifically fitted to the square
symmetric context), I hope the advantage over the standard routine for
symmetric matrices remains significant.
Pierrick Bruneau
CRP Gabriel Lippmann
On Thursday, January 31,
Dear List,
I have a data structure like:
aa
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5 6
bb
[[1]]
[1] 3
[[2]]
[1] 5
I would like to set differences between aa and bb and get as
result another list of lists like:
res
[[1]]
[1] 1 2
[[2]]
[1] 4 6
I am trying:
lapply(aa, setdiff, bb)
but I simply get
Hi
you have something wrong in your R session, works for me.
aa-list(c(1:3), c(4:6))
aa
[[1]]
[1] 1 2 3
[[2]]
[1] 4 5 6
bb-list(3,5)
bb
[[1]]
[1] 3
[[2]]
[1] 5
lapply(aa, setdiff, bb)
[[1]]
[1] 1 2
[[2]]
[1] 4 6
Regards
Petr
-Original Message-
From:
Impossible to say w/o a reproducible example, but to start with let me suggest
looking at the exact= (both functions) and correct= (wilcox.test) arguments.
Experience shows that some change of the default settings allows you to
reproduce results from other software (and the help pages will
Dear List,
I have a list of lists and on each list I want to apply a function
from the igraph package, graph.knn. I need to calculate graph.knn for
each list on a graph g, then retrieve one of the result called $knn
and calculate its mean. The non-R style code looks like this:
for(i in
Hi Srini,
This is a statistics question, not a question about R, so this may not
be the best place to ask. Try posting at
http://stats.stackexchange.com/ or another statistics help list.
Best,
Ista
On Thu, Jan 31, 2013 at 11:11 PM, Srinivas Iyyer
srini_iyyer_...@yahoo.com wrote:
Hi:
Apologies
There are many plotCI functions in many different packages... which
one are you referring to? Also please construct a reproducible example
illustrating your problem.
Best,
Ista
On Thu, Jan 31, 2013 at 11:58 PM, li li hannah@gmail.com wrote:
Hi all,
In my plotCI function, the argument x
Hi
I do not see any problem with your code. *apply functions are also hidden
cycles and there shall not be substantial improvement in speed.
Why you do not want to use for cycle?
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On
try this:
x - read.table(text = No. Change Date
+ A 123 final2013-01-15
+ B 123 error 2013-01-16
+ C 123 'bug fixed' 2013-01-17
+ D 111
You can also try 'mapply'
aa-list(c(1:3), c(4:6))
bb-list(3,5)
mapply(setdiff, aa, bb)
[,1] [,2]
[1,]14
[2,]26
On Fri, Feb 1, 2013 at 5:16 AM, Simone Gabbriellini
simone.gabbriell...@gmail.com wrote:
Dear List,
I have a data structure like:
aa
[[1]]
[1] 1 2 3
Hi,
Yes I know it works, but I'd like to assign the results like this:
V(g)$meanknn - ONELINER
Where V(g) elencates all the nodes in my graph...
Thanks,
Simone
Sent from my iPhone. Please excuse brevity and odd typos
On 01/feb/2013, at 13:31, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
Dear All,
I am struggling with a simple problem. I did some online research, but
I am still banging my head against the floor.
I read a csv file into a data frame and I have a columns with entries like
2012-04-13 00:23:45
2012-04-22 07:53:09
etc..
Each line posted above is the content of a
newdf-data.frame(k1=rep(row.names(foo),ncol(foo)),stack(foo))
--- On Fri, 1/2/13, Gundala Viswanath gunda...@gmail.com wrote:
From: Gundala Viswanath gunda...@gmail.com
Subject: [R] Transforming 4x3 data frame into 2 column df in R
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Date:
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below is a copy
of the function (renamed to n.for.2means.js). I have inserted a single line of
code towards the bottom of the function which uses the cat function to print
the value of n1. You will
Hello!
I have a list of variable length. One example is:
X=vector(list,3)
X[[1]]=1:2
X[[2]]=1:2
X[[3]]=1:2
How could I run expand.grid on the elements of X so that the results would
be the same as expand.grid(1:2,1:2,1:2)?
Thank you!
Dimitri
--
Dimitri Liakhovitski
gfk.com
Mr Isella,
On 1 February 2013 05:37, Lorenzo Isella lorenzo.ise...@gmail.com wrote:
How can I convert that into Unix time?
format.POSIXct(dateCol, '%s'); -- H
--
Sent from my mobile device
Envoyait de mon portable
__
R-help@r-project.org mailing
On 13-02-01 8:47 AM, John Sorkin wrote:
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below is a copy
of the function (renamed to n.for.2means.js). I have inserted a single line of
code towards the bottom of the function which uses the cat
Hi
So why you do not assign it inside a cycle? As I said lapply does the same as
for cycle does (maybe in some more compact manner).
Regards
Petr
-Original Message-
From: Simone Gabbriellini [mailto:simone.gabbriell...@gmail.com]
Sent: Friday, February 01, 2013 2:29 PM
To: PIKAL
Oops, it was easier than I thought:
expand.grid(X)
Dimitri
On Fri, Feb 1, 2013 at 8:48 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Hello!
I have a list of variable length. One example is:
X=vector(list,3)
X[[1]]=1:2
X[[2]]=1:2
X[[3]]=1:2
How could I run expand.grid on
On Feb 1, 2013, at 7:47 AM, John Sorkin wrote:
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below
is a copy of the function (renamed to n.for.2means.js). I have
inserted a single line of code towards the bottom of the function
which uses
Hello Srini,
It sounds as if you are attempting to establish a prior probability and
compare it to the posterior probability -- a perfect candidate for bayesian
analysis. I would simply do a search for 'bayesian analysis of gene
expression data' -- there are a number of statistical packages that
Hi,
expand.grid(X)
# Var1 Var2 Var3
#1111
#2211
#3121
#4221
#5112
#6212
#7122
#8222 expand.grid(1:2,1:2,1:2)
# Var1 Var2 Var3
#1111
#2211
#3121
#4
Stephen,
I can almost but not quite get my arms around what you are asking. A bit more detail
would help. Like
cph.approve = what kind of object, generated by function __
But, let me make a guess:
cph is the result of coxph, and the model has both covariates and a strata
You
Thanks so much for the reply, Ista. I used plotrix library.
Here is my example:
xx - seq(0.05, 0.95, by=0.05)
lower - c(-2.896865, -2.728416, -2.642574, -2.587724, -2.548672, -2.518994,
-2.495409, -2.476031, -2.459662, -2.445513, -2.433014, -2.421739, -2.411344,
-2.401536, -2.392040, -2.382571,
Hello,
As for the KS test, the op might also want to look at
Simard, L'Ecuyer (2011)
http://www.jstatsoft.org/v39/i11
for an account on the different algorithms available, and their accuracy.
Apparently, R uses different algorithms according to the test in
question. See the details section of
Dear colleagues, I have 2 points: One opinion and one question.
1)
In one paper in a peer-reviewed journal, I read about the idea of using a
logit regression as a surrogate for the log-binomial, just adding the
numerator to the denominator ...
It’s tempting to immediately get the RR instead of
You haven't given any y values to plotCI.
Try, for example,
plotCI(xx, (lower+upper)/2, ui=upper, (etc)
What you got was in effect
plot( seq(along=xx), xx )
which is standard behavior for plot() when no y values are supplied.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
Here is another option using plyr:
library(plyr)
creek - read.csv(creek.csv)
library(ggplot2)
creek[1:10,]
colnames(creek) - c(date,flow)
creek$date - as.Date(creek$date, %m/%d/%Y)
ddply(creek,year,summarise,MED=median(flow),MEAN=mean(flow),SD=sd(flow),MIN=min(flow))
Felipe D. Carrillo
Hello all,
When I tried to plot the following two arrays in one figure with the following:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x),max(x)))
par(new=T)
plot(y, type='l', ylim=c(min(y),max(y)))
Found that the first points and last points from those two
Dear useRs,
Version 1.2 of the R package 'pa' is now available
on CRAN. This package provides tools for conducting
equity portfolio attribution. Two methods included
in the package are the Brinson Method and a
regression-based analysis. The new version allows
users to view attribution results for
library(sqldf)
k1-data.frame(ID=LETTERS[1:4],
No=c(rep(123,3),111),
Change=c(final,error,bug fixed,final),
Date=c(2013-01-15,2013-01-16,2013-01-17,2013-01-12))
k1$Date=as.Date(as.character(k1$Date),tz=UK)
sqldf(select *
from k1
group by No
having max(Date))
--- On Fri, 1/2/13, Mat
Hi!
I get the following error messages when trying to run the package 'arm'.
library(arm)
Error : Functions found when exporting methods from the namespace arm
which are not S4 generic: fixef, ranef
Error: package/namespace load failed for arm
My R version is:
version
_
Hi Folks!
I have a question regarding an issue on which I´ve read a few threads but
can´t resolve -- Labelling every other tick mark on the x-axis using
factors. Here´s an example:
db-data.frame(date=as.factor(c(1:50)),var=rnorm(50))
Has anyone implemented a reader and writer for data in the vowpal wabbit
format? It's a format for representing sparse data in ASCII files, see
https://github.com/JohnLangford/vowpal_wabbit
- Roy
[[alternative HTML version deleted]]
__
Hi,
Perhaps, (#Untested)
do.call(rbind,lapply(split(dat1,dat1$No),function(x) tail(x,1)))
#or
library(plyr)
ddply(dat1,.(No), function(x) x[nrow(x),])
A.K.
- Original Message -
From: Mat matthias.we...@fnt.de
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 3:04 AM
Subject:
Dear R Help Mailing List Members,
I have some baseline data (attached) on which I plan to run a DFA in R to find
out the best predictor variables that best allow discrimination between the
different stocks. I then need to apply this DFA function to the mixed data set
(attached), which
It is strongly discouraged in R to have functions that change data values
in the global workspace (or any location other than their local
environment).
The usual procedure in R is to have your function return a modified version
of the object and the user then decides what to do with it. They can
I am sure, that this is not a pure Poisson! Huge overdispersion!
You get inflated confidence intervals!
(although, the point estimates of the regression coefficients stay the same)
Try to look for the causes of overdispersion! It may be geteroscedastisity?
What is the nature of the response, is
Ivan,
In reference to your part 2), in 1989 Li and Duan published a paper where
they examined the effect of using the wrong link function (
http://projecteuclid.org/DPubS?service=UIversion=1.0verb=Displayhandle=euclid.aos/1176347254).
The short version is that they found that in common models
On Thu, Jan 31, 2013 at 2:13 PM, Wim Kreinen wkrei...@gmail.com wrote:
Hello,
I have a question about modelling via glm.
I think you are way off track. Either the data, glm, or both, are not what
you think they are.
I have a dataset
skn300.tab - structure(list(n = 1:97, freq = c(0L, 0L,
Running on Slackware here with R-2.15.2. There is a data frame I need to
edit to correct mis-spellings and to add a row. I used the edit() command
with emacs but could not find the mis-spelled strings nor figure out how to
add another row.
The data frame is stored in
One method is to follow correct usage of base graphics and only use the plot
function once for each plot. To overlay, use a function like points or lines.
Another approach is to use lattice graphics or ggplot2 and give the data to
higher-level plot routines that know how to plot multiple groups
Hello David,
When I used the same ylim for both plots, I got what I need:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x,y),max(x,y)))
par(new=T)
plot(y, type='l', ylim=c(min(x,y),max(x,y)))
Thanks,
Rebecca
-Original Message-
From: David Winsemius
Hi,
library(reshape)
foo$id-row.names(foo)
melt(foo,id.var=id)
id variable value
#1 n w 1.51550092
#2 q w 0.09977303
#3 r w 1.17083866
#4 n x 1.43375720
#5 q x 0.81737610
#6 r x 1.24693470
#7 n y 1.27916240
#8 q y
Not sure if you are aware of the OpenMx SEM package
(http://openmx.psyc.virginia.edu/). It's a very full-featured structural
equation modeling package.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Hi,
You could use:
creek - read.csv(creek.csv,sep=\t)
colnames(creek) - c(date,flow)
creek$date - as.Date(creek$date, %m/%d/%Y)
creek1 - within(creek, year - format(date, '%Y'))
library(data.table)
creek2- data.table(creek1)
Dr. Therneau,
You are correct about the fit:
(cph.approve -
coxph(Surv(fundterm,resp)~I(CommitAmt/1e5)+res+CommitedRate+dayflag+mth+strata(termfac),
data=dfa,
subset=(HedgeDate2012-11-15 p1!=FixedOpen), method=efron))
However the output from survfit has individuals in each column and the
Hello List
I was wondering if it is possible to make the individual 'tiles' in a heatmap
larger. Often when I plot heatmaps and want to label the rows with eg gene
names I either have to shrink the text or leave it out altogether as it becomes
so small as to be unreadable. I wondered if there
I've compared the solutions.
*Solution 1:*
myf - function( df1, df2 ){
cond - df2$a min(df1$a)
if( cond )
{
idx - which( df1$a == min(df1$a) )
df1[idx, ] - df2[1, ]
}
df1
}
# On a larger example,
set.seed(4530)
tst - data.frame(item = 1:1000,a = rnorm(1000),b = rnorm(1000)) #
Dear R users,
I'd like to change the default order of colors line types.
Especially I am using ggplot2 and using color Set1.
In Set1, the default color order is red, blue, green, violet,.. ect.
However, I want to put red in fourth (not first).
Likewise, I want to change the order of default
R-helpers:
Say I have a list:
myvariables - list(a=1:10,b=20)
Is there a way to load the list components into the environment as
variables based on the component names? i.e. by applying this theoretical
function to myvariables I would have the variables a and b loaded into the
environment
Is there anyway with some experience in using armadillo in R C++ extensions?
My problem is the following:
I programmed a function in a header looking like
#include armadillo
inline arma::vec foo(input) {
... do something
return an arma::vec object
}
compiling this
Dear R users,
I'd like to change the default order of colors line types.
Especially I am using ggplot2 and using color Set1.
In Set1, the default color order is red, blue, green, violet,.. ect.
However, I want to put red in fourth (not first).
Likewise, I want to change the order of default
Hi I'm having a simple issue with the order function. When I use the following
code my data is not ordered correctly output[order(output[,3]),]
Namebeta pval
881 9.09303277751237 0.000100253350483199
74026.40553461638365 0.00010228641631914
4879
Hi I'm having a simple issue with the order function. When I use the following
code my data is not ordered correctly
output[order(output[,3]),]
Namebeta pval
881 9.09303277751237 0.000100253350483199
74026.40553461638365 0.00010228641631914
4879
I am new to this mailing list. Is everything I posted seen for everyone in
the mailing list? or just you can see the post?
On Fri, Feb 1, 2013 at 11:33 AM, arun kirshna [via R]
ml-node+s789695n465731...@n4.nabble.com wrote:
HI,
Could you show the modified code and also str(dataset)?
A.K.
Thank you very much Petr,
I believe I have fixed my inquiry to not use floating points in my cycle as
you pointed out and to use the list structure to keep my results. I am
still at a loss as to how to run the multiple loops. I have tried quite a
few different strategies but my failure seems to
On 13-02-01 12:15 PM, Kripa R wrote:
Hi I'm having a simple issue with the order function. When I use the following
code my data is not ordered correctly
output[order(output[,3]),]
Namebeta pval
881 9.09303277751237 0.000100253350483199
7402
On Fri, Feb 1, 2013 at 5:24 PM, Jonathan Greenberg j...@illinois.edu wrote:
R-helpers:
Say I have a list:
myvariables - list(a=1:10,b=20)
Is there a way to load the list components into the environment as
variables based on the component names? i.e. by applying this theoretical
function
Look up the terribly wonderful RcppArmadillo package.
MW
On Feb 1, 2013, at 8:38 PM, Simon Zehnder szehn...@uni-bonn.de wrote:
Is there anyway with some experience in using armadillo in R C++ extensions?
My problem is the following:
I programmed a function in a header looking like
On 1 February 2013 at 21:38, Simon Zehnder wrote:
| Is there anyway with some experience in using armadillo in R C++ extensions?
Yes, sure -- we wrote (and use) a package RcppArmadillo that provides very
easy access to Armadillo via Rcpp.
| My problem is the following:
|
| I programmed a
Hello,
Something like this?
myfun - function(x, envir = .GlobalEnv){
nm - names(x)
for(i in seq_along(nm))
assign(nm[i], x[[i]], envir)
}
myvariables - list(a=1:10,b=20)
myfun(myvariables)
a
b
Hope this helps,
Rui Barradas
Em 01-02-2013 22:24, Jonathan
Sure. Start by reading [1] and take its advice to heart.
Then study the many ways to create factors with values ordered in sync with the
color ordering. The color ordering can be set using the default, using any of
the various color picker functions, or manually.
[1]
Hi, I have been trying to install R on my mac which is running mountain lion.
It is partially working, but one error message I am getting is:
Error in function () : object '.activeModel' not found.
I cannot find anything on this by googling. It appears, for example, when I try
to 'add
I am trying to figure out how to calculate p-values for the difference in
prevalence of a risk factor between men and women. For example, I find that
277 out of 710 male patients and 125 out of 305 female patients have
obesity, what is the p-value for their difference?
If there is a package that
This is very basic stuff, so I think your main problem is statistical,
not R related. Try posting on stats.stackexchange.com to get
statistical advice or do some reading about differences in
proportions, contingency tables, and the like.
Incidentally, off the top of my head, I'd say there's no
Perhaps you would benefit from joining R-sig-mac, and/or perusing their
archives.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#.
Dear Claire,
This error is due to a bug in the previous version of the Rcmdr package; it
is fixed in the current version on CRAN, version 1.9-4. Simply reinstall the
Rcmdr package from CRAN and verify that you have the correct version.
I hope this helps,
John
-Original Message-
From:
Dear Jeff,
As I just explained to Claire, this error is caused by a bug in an earlier
version of the Rcmdr package; the bug affects the Rcmdr on all platforms,
not just Mac OS X, and is fixed in the current version 1.9-4 of the Rcmdr on
CRAN.
Best,
John
-Original Message-
From:
On Feb 1, 2013, at 9:30 AM, Alfredo Tello wrote:
Hi Folks!
I have a question regarding an issue on which I´ve read a few
threads but
can´t resolve -- Labelling every other tick mark on the x-axis using
factors. Here´s an example:
Why not:
Hi there,
Just wondering why my post was rejected?
cheersRachel
Subject: repeating autocovariate functions
From: r-help-ow...@r-project.org
To: moy...@hotmail.com
Date: Sat, 2 Feb 2013 02:56:27 +0100
Message rejected by filter rule match
--Forwarded Message Attachment--
Date: Fri, 1 Feb
This smacks of homework to me.
cheers,
Rolf Turner
On 02/02/2013 12:45 PM, Jin Choi wrote:
I am trying to figure out how to calculate p-values for the difference in
prevalence of a risk factor between men and women. For example, I find that
277 out of 710 male patients and 125
The Subject line mostly says it. I'm designing it as a semester-long, 3
hours per week, course
that takes in students who got the basics of R in stats classes, but don't
really know how to
program in it. Translation: if their own examples don't look enough like
examples from
previous work,
Hi,
Saw your reply on Nabble:
#Your code:
library(zoo)
res1- do.call(rbind,lapply(lapply(split(d,list(d$m1,d$n1)),function(x)
{x$cterm1_p0L[x$Qm=c11]- cumsum(x$term1_p0[x$Qm=c11]);
x$cterm1_p0H[x$Qn=c12]- cumsum(x$term1_p0[x$Qn=c12]);
x$cterm1_p1L[x$Qm=c11]- cumsum(x$term1_p1[x$Qm=c11]);
HI,
I think this should be more correct:
maxN-9
c11-0.2
c12-0.2
p0L-0.05
p0H-0.05
p1L-0.20
p1H-0.20
d - structure(list(m1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
n1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
The short answer is ... because it was from Nabble.
The longer answer is that the filters needed to be tightened on Nabble
postings because of a an autobot spammer and the moderation queue was
being swamped with cases.
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