hi: no. it's not the same. if you read the paper that I referenced last
night, that explains how to do the following test :
Ho: R2 = R1
H1: R2 != R1
that's a different test from what you did but i think it's what you want.
On Jul 8, 2010, chen jia chen_1...@fisher.osu.edu
wow chuck. you really know how to dig up the archives. I don't know if it's
exactly relevant for what the OP is asking but i did use the
( or atleast a )Â paper by hotelling and it was titled the selection of
variates for use in prediction with some comments on the general problem of
Hi: I'm not familar with prcomp but with the principal components function
in bill revelle's psych package , one can specify the number of components
one wants to use to build the closest covariance matrix I don't know
what tol is doing in your example but it's not doing that.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hi: John Fox's CAR book has some very nice examples of how the multinomial
likelihood is estimated computationally speaking. But you mentioned
multilevel earlier which sounds more complex ?
On Aug 2, 2009, nikolay12 nikola...@gmail.com wrote:
Thanks a lot. The info about
Hi All: I get the following error when trying to install the rjags package.
I've installed the
jags software and I'm using Fedora 10.0 and my sessionInfo is at the bottom
of this email.
I'm also sorry if this email ends up having control A's all over it. I still
haven't figured
agresti's book ( forget which one. there are two ) has an appendix about the
history of categorical data that i remember being quite interesting. that's
the only one i know of.
On Jun 18, 2009, Michael comtech@gmail.com wrote:
Hi all,
I apologize for this off-topic
Hi Bill: I was trying to do below myself but was having problems. So I took
your solution and made another one. yours was working
a little weirdly because I don't think the person wants to keep rows where
there are 2 dnv's in a row and he/she also wanted to keep
the row if the
Hi:Â you have to return the dataframe inside the lapply. I also changed =
to - but I doubt that matters.
d1 - data.frame(x1=1,x3=4)
d2 - data.frame(x1=2,x3=5)
d3 - data.frame(x1=3,x3=6)
d4 - data.frame(x1=4,x3=7)
a = list(d1,d2,d3,d4)
print(a)
lapply(a,function(.df) {
Hi Kynn: this oddity is discussed in Patrick Burn's document called The R
Inferno. I don't recall the fix so I'm not sure if below is the same as
what his book says to do but it seems to do what you want.
x - list()
x[[1]] - 2
x
length(x)
print(str(x))
x[2] -
I hate to start a whole war about this but isn't there some percent chance (
not much but
non zero ) that she's willing to pay the 300.00Â so that she can get a nice
solution that she can then
learn from ? I'm definitely guilty of this behavior as a non-student and i
forget to
to whomever that was. i deleted your email but I think below does what you
want. as always, there's probably some
improvement that could be done. whether it's  minor or major, i'm not sure
?
also, 3 things:Â Â
A)there are some Inf's in the output because of division by zero but
Hi: you've got to create a setdiff in both directions in order to get the
lone ones in each column because setdiff is
not commutative meaning that setdiff(a,b) does not equal setdiff(b,a). once
you do that, then
( setdiff1 + setdiff2 - intersect ) should equal the union.
if
Hi kevin: one use ( there are probably many others ) is for use inside
a vector autoregression model where the RHS is
lags of the independent variable. So, x_t is the 3rd column, x_t-1 is the
second, x_t-2 is the third etc.
On Apr 3, 2009, rkevinbur...@charter.net wrote:
I
Below works but it has two backslashes in the word. maybe someone can
explain why the 4 and 2 works but 2 1 doesn't ? thanks.
gsub(,,Hello\\World,perl=TRUE)
On Apr 3, 2009, Andrew Conway agc...@nyu.edu wrote:
I am trying to check for backslashes in data, then remove them
Hi: I have had a similar issue so below are ways that I deal with
that.  i don't think there are manuals/ documentation for
becoming more of a developer ( someone can correct me if I'm wrong and I'd
be happy to be wrong ) but there are other ways:
1) Staying on this
Hi Stephen: If you want to take tem out of quotes, you can use deparse
substitute as in below. Maybe there are other ways
also but that's the one I often see used. I hope this email ends
up looking okay because my mailer has been acting strangely lately.
         Â
Hi Everyone: I've been going through Alan Lenarcic's package tutorial but
when I did
R CMD SHLIB Xdemo.cc Xdemo_main.cc -o Xdemo.dll
I got the following error:
XDemo_res.rc:15:38: warning: missing terminating character
XDemo_res.rc:23:34: warning: missing terminating '
apologies for my screwed up send. it looks really funked when I read it
so my mail settings must be messed up. I'll resend it later tonight from
a different computer.
On Mar 31, 2009, markle...@verizon.net wrote:
Hi Everyone: I've been going through Alan Lenarcic's package
Hi: Here's my revised attempt at explaining my problem. Hopefully it won't
get sent out weirdly like it did earlier. Â The link to what I'm trying to
follow is
http://www.stat.columbia.edu/~gelman/stuff_for_blog/AlanRPackageTutorial.pdf
 I've been going through the tutorial but
by definition, the one tailed p-value has to be = 0.5 so there is
still something wrong with your OpenEpi calc. Most likely it's
calculating the
2 tailed p-value and then mistakenly multiplying by 2. For example:
A) Suppose you are testing
Ho: u = u_0
at 10:42 PM, Ben Bolker wrote:
markleeds at verizon.net writes:
by definition, the one tailed p-value has to be = 0.5 so there is
still something wrong with your OpenEpi calc. Most likely it's
calculating the
2 tailed p-value and then mistakenly multiplying by 2. For example:
For what
Thanks berwin. See what I sent to Ben. I better start looking up my
stat textbooks or going back to class before I start using stat terms.
You guys are on top of things .
On Sat, Mar 21, 2009 at 10:50 PM, Berwin A Turlach wrote:
G'day Mark,
On Sat, 21 Mar 2009 22:08:21 -0500 (CDT)
Hi:
For A, you can use head and tail but you have to add a zero the front.
For B, you can use the same function, but put it inside an sapply and
run over the columns and then cbind it back with the original dataframe.
A)
nm2 - c(rep(SPZ8, 10), rep(SPX9, 10))
-1.0*c(0,as.numeric((head(nm2,-1)
I think that Gabor sent something earlier today ( or atleast something
related ) so check the archives for his solution.
On Thu, Mar 12, 2009 at 4:29 PM, Ferry wrote:
Hi,
Is there a package I can use to write to multiple sheets on xls file,
other than using the paid version of
Jorge was kind enough to point out that I made a mistake in referencing
gabor's solution.
See below. Thanks Jorge.
Begin forwarded message
Subject: Re: [R] Writing xls - multiple sheets
Date: 3/12/2009 6:03:13 PM
From: Jorge Ivan Velez
To: markle...@verizon.net
Hi Mark,
Hi: Below works but it's extremely ugly and overly complicated. i'm sure
someone else will send you something better and I'll be waiting also.
Also, the way I named the rows and columns works for below but it won't
hold in the general case if you don't have nice ordered names like
you do
i think there's confusion here between a time series that reverts to its
long term mean
and an ornstein uhlenbeck type of mean reversion. they're not the same
thing and
I don't want to go into the difference because I would probably just add
to the confusion.
you might be better off sending
below is one link for an explanation of the equivalence between AR(1)
and OU so I was wrong about needing an AR(2). at a quick glance, i think
OU's that can always be written as AR(1)'s
( with a correct mapping of parameters ) but not every AR(1) can be
written as an OU ?
Hi: what you want to do is not recommended if you're doing a statistical
computation ( many threads
on this. search the archives if you want more details ) but if you
really want to do that below should work:
mat[is.na(mat)]-0
On Thu, Mar 5, 2009 at 11:22 PM, Manli Yan wrote:
I have a
can someone show me how to use a regular expression to break the string
at the bottom up into its three components :
(-0.791,-0.263]
(-38,-1.24]
(0.96,2.43]
I tried to use strplit because of my regexpitis ( it's not curable. i've
been to many doctors all over NYC. they tell me there's no cure
David, Wacek: Just so everyone knows, I just looked and this is
explained quite clearly in the R Language Reference manual, very
similarly to what Wacek did below.
On Wed, Feb 25, 2009 at 3:42 AM, Wacek Kusnierczyk wrote:
David Winsemius wrote:
On Feb 24, 2009, at 11:36 PM, Fuchs Ira
Hi Wacek: Somewhere I remember reading that environments have
functionality like lists EXCEPT for the names part. IIRC, I think that I
read this in the R Language Reference manual also.
On Wed, Feb 25, 2009 at 4:32 AM, Wacek Kusnierczyk wrote:
a quick follow-up:
e = new.env()
Thanks Berwin. You're correct in that I meant the R Language
Definition. Well, it may be a draft but I read it for the first time a
few months ago and
it was very enlightening so, whether it's a draft or not, I highly
recommend it. ( but not for total beginners. The R-intro is better for
a
Hi: John Fox's recode function in his car package provides a convenient
way for doing what
you need. I don't know what your factor is specifically but below is
mostly taken out
of the help for recode and shows how to take a factor and recode it to
make it a new
factor. you can apply that for
oops, then i guess i should not have sent the recode suggestion.
choonhong: I only
sent it as an example of how to recode your factor. I didn't mean to
imply ( nor
did i even give it much thought ) that what you're doing is
statistically/philosophically
correct.
I'm a friend but I think what
There's probably something built in to R but you can change the values
of the percentiles (p) below to get the value that corresponds to it.
rounding might be problematic also.
temp - c(1,4,8,3,5)
p - 0.8
temp[order(temp)][round(length(temp)*p)]
On Tue, Feb 24, 2009 at 2:36 PM, Peterko
One way is to use %in% like below:
df-data.frame(var.a=rnorm(10), var.b=rnorm(10),var.c=rnorm(10))
print(df)
df - df[,!(names(df) %in% c('var.a','var.b')),drop=FALSE]
print(df)
On Tue, Feb 24, 2009 at 3:10 PM, Sean Zhang wrote:
Dear R-helpers:
I am an R novice and would appreciate answer
Hi: Below is a TOTAL HACK and I don't recommend it but it does seem to
do what you want. I think that I remember Gabor saying that you can
merge multiple data frames using zoo but I don't know the specifics. I'm
sure he'll respond with the correct way. Below uses a global variable
to access
Hi Simon: In below , test1 spelled out is count ~ siteall + yrs +
district + yrs:district so this is fine.
but in test2 , you have years interacting with district but not the main
effect for years. this is against the rules of marginality so I still
think there's a problem. I would wait for
Hi Gundala: Below works but I was trying to figure out a way to not have
to add the last line and I gave up.
dat-c(0.00377467,0.00377467,0.00377467,0.00380083,0.00380083,0.00380083,0.00380959,
0.00380959,0.00380959,0.00380083,0.00380083,0.00380083)
# MAKE A TEMPORARY MATRIX
temp -
Doug Martin et al ( he may not be the first author ) has a book called
Robust Regression
and I vaguely recall seeing a chapter in there related to outlier
detection. in time series.
Given that there's an S+ package associated with the book, maybe that
package has been or can be translated to
Hi Jason: below seems to work. you have to take the transpose because
the apply
returns the rows transposed. i'm also not sure how to make the NAs be
the last
ones but maybe someone can show us how to do that.
mat - matrix(c(2,7,2,7,9,10,10,6,8,6,1,9,7,2,0),byrow=TRUE,nrow=3)
print(mat)
Thanks Rolf. very nice but pretty easy is ALWAYS a relative
statement.
On Thu, Feb 12, 2009 at 3:59 PM, Rolf Turner wrote:
On 13/02/2009, at 9:06 AM, markle...@verizon.net wrote:
Hi Jason: below seems to work. you have to take the transpose because
the apply
returns the rows transposed.
it does and you get exactly what monica wanted if you take out the sp
and just return the whole thing. thanks.
On Thu, Feb 12, 2009 at 5:52 PM, David Winsemius wrote:
aggregate and by are convenience functions of tapply.
Consider this alternate solution:
xveg[which(xveg$tot %in%
hi: maybe it's minimizing the negative of the likeihood rather than
maximizing the likelihood ?
other than that, i don't see that being possible because of what you
said. see what
happens if you estimate an arima(1,1,3) ?
On Tue, Feb 10, 2009 at 12:51 PM, Stephen Collins wrote:
All -
I
Hi: Bert: can you do that because the null is that they are equal
before and after,
not that the proportion is zero ? Thank for any clarification to my
lack of understanding.
On Tue, Feb 10, 2009 at 5:43 PM, Bert Gunter wrote:
Ah, experimental units,again ... a subject little taught by
Hi Bert: I think what you said about a prior guess for the NULL is
also similar to what Chuck said about people looking with a blank stare.
Thanks for the clarification.
On Tue, Feb 10, 2009 at 7:06 PM, Bert Gunter wrote:
The only question at issue (i.e. capable of being addressed)
this is a bad question but I can't figure it out and i've tried. if i
sort the 2 column
matrix , temp1, by the first column, then things work as expected. But,
if I sort the 1 column matrix, temp2, then it gets turned coerced to a
vector. I realize that I
need to use drop=FALSE but i've put it
i'm sorry. i had an error in my previous code because i left out a
letter in the rownames.
while fixing that, i also found a solution. so i'm sorry for the
confusion.
below is my fix.
temp2 - matrix(rnorm(10),nc=1,nrow=10)
rownames(temp2) - c(a,b,c,d,e,f,g,h,i,j)
print(temp2)
temp2 -
hi: it's not clear to me what you're trying to do but maybe outer is
what you want ?
outer(x,y)
it takes every value in x and pairs it with every value in y and the
default operation is multiply. see details by doing ?outer.
On Wed, Feb 4, 2009 at 10:36 PM, cruz wrote:
Hi,
I compute
I'm sure below is fine but john fox's CAR book has some nice examples
of how to compute the logit parameters and variances from scratch using
iteratively weighted least squares.
On Thu, Jan 29, 2009 at 1:54 AM, justin bem wrote:
Run
outfit-nlm(..., hessian=T) and then standards error
I'm still going over old emails and trying to get my head around
evaluation so I'm persistent if nothing else.
A while back , an expert sent me below as an exercise in understanding
and I only got around to it tonight. I understand some of the output but
not all of it and I put Why not Zero
Hi: if i understand, i think
newx-x[ x %in% key]
should give you what you want.
On Sun, Jan 25, 2009 at 10:27 PM, Akshaya Jha wrote:
Hi,
I have the following datasets:
x=data I am looking through
key=a set of data with the codes I want
I have the following issue:
I want the subset of x
Hi: I think I saw a link where the author clarified the original article
and explained more clearly that the design of R had it roots in S/S+. I
don't
remember where I saw it but it's somewhere. Also, I think it's jumping
the gun to claim that anyone lied to anyone before doing the research
I've been going back to old difficult R-list evaluation emails that I
save in order to understand evaluation better and below still confuses
me. Could someone explain why A) works and B) doesn't. A variant of
below is in the Pat's Inferno book also but I'm still not clear on what
is
Hi Harold: Below works on your data set but check it a lot because I am
a little worried that
I could have missed something. Hopefully someone can send a a little
clearer way.
dat - data.frame(id = c(1,1,2,2,2), var1 = c(10,10,20,20,25), var2 =
c('foo', 'foo', 'foo', 'foobar', 'foo'))
I am trying to learn the basics of Sweave so I read some things and a
friend gave me his Rnw file
to play with. I am able to do R CMD Sweave mark-example.Rnw. That works.
But,
then when I do latex mark-example.tex, I get the message below. I looked
in the archives
and Martin Morgan mentioned
In my original message, I forot to include my Session Info so it is
below. I apologize for that.
sessionInfo()
R version 2.8.0 (2008-10-20)
i386-redhat-linux-gnu
locale:
Martin Morgan kindly explained to me off-list that I need to use
R CMD pdflatex mark-example.tex.
That fixed my problem.
On Mon, Jan 12, 2009 at 11:25 PM, markle...@verizon.net wrote:
In my original message, I forot to include my Session Info so it is
below. I apologize for that.
Charlotte: I ran your code because I wasn't clear on it and your way
would cause more matrices than the person requested. So
I think the code below it, although not too short, does what the person
asked. Thanks though because I understand outer better now.
temp - matrix(c(1,2,3,4,5,6),ncol=2)
Thanks Kingsford. I thought the column power was supposed to be just for
that column but you're probably correct. English has its oddities
because if one reads the actual sentence the person wrote it's still not
clear, atleast to me.
Actually I want to have a matrix with p columns such that
hi: i tried regular expressions and , as usual, failed. but below does
do the job uglily using strsplit. i'd still be curious and appreciate if
someone could do the regular expression method. thanks.
dts - c(1990m12, 1992m8) #March 1990 and Aug 1992
#SPLIT IT USING m
temp - strsplit(dts,m)
#
Hi: Veslot: I'm too tired to even try to figure out why but I think
that there is something wrong with your sl function. see below for an
empirical
proof of that statement. OR maybe you're definition of sliding window
is different than rollapply's definition but rollapply's answer makes
more
i guess it depends on what you have installed on your linux OS but
xpdf works for me. i use fedora.
On Sun, Dec 14, 2008 at 7:18 PM, Dennis Fisher wrote:
Colleagues,
I am interesting in opening a PDF document via the command line from
both Windows, OS X, and Linux ( R version 2.8.0).
hi: change your dots to NAs and then use na.locf in the zoo package. i
didn't test it but i think that should work.
DF$ID[DF$ID == .]-NA
DF$ID-na.locf(DF$ID)
On Sun, Dec 14, 2008 at 8:56 PM, Zhixin Liu wrote:
Hi R helpers,
If I have a dataset looks like:
ID record 120
.
andrew has a point which makes my solution wrong. you'd have to change
the factors to numerics and I'm not sure what
would happen when you did that. if you want to send a sample file of
your data, that would be best but andrew's suggestion may
work right off the bat.
On Sun, Dec 14, 2008 at
could someone explain what is happening below ? I was trying to solve a
related question on the list and then , as I was solving it,
I was getting strange answers and then I noticed below. It's obviously
not a bug but I don't get it. Thanks.
m - data.frame(class = c(birds, planes), feather =
disregard my last question. i forgot that d1 was a dataframe and was
thinkiing it was a vector. i think i can fix what i was doing now.
On Fri, Dec 12, 2008 at 8:52 PM, markle...@verizon.net wrote:
could someone explain what is happening below ? I was trying to solve
a related question
to the person who just asked about the feather and jet problem: i lost
your email but below works. there should be a shorter way by
avoiding the last statement and doing everything in the lapply but I've
had enough of feathers and jets tonight.
m - data.frame(class = c(birds, planes), feather
could someone explain why the name of FPVAL gets .value concatenated
onto it when the code below is run and temp is returned.
I've been trying to figure this out for too long. It doesn't matter when
I put the FPVAL in the return statement. It happens regardless of
whether it's first or last.
a friend of mine sent me the following and I'm thinking that there are
people on this list that might be interested and could also hit the 10%
improvement mark possibly ? i don't know any more about it than the
article says so, if someone is interested, I guess you would have to
contact
hi chris: i'm betting that there is a better/shorter more R'ish way to
do it ( if someone could provide that, it's appreciated ) but below will
get the output in the format you need. the all function tests whether
all the elements are equal so you don't need setequal although i guess
it's a
Hi Oliver: if I understood Chris's email correctly , he wanted to
compare all possible row ( row1 with the other 4 rows of the other
matrix,
row 2 with the 4 rows of the other matrix, etc, etc, ) combinations of
the two matrices, not the individual elements of the matrix. That's just
my
hi: below almost gets you there except for the endpoints. see filter for
more information.
d=0.5
L=20
x=seq(20, by=1, length.out=20)
temp - as.numeric(filter(x,filter=c(d,(1-2*d),d),sides=2))
print(temp)
On Sun, Nov 9, 2008 at 10:57 PM, stephen sefick wrote:
#Is there a way to vectorize
my problem is more complex than below but I think below can suffice. i
have a list and the name of it at the top level is GGG. so, if i do an
lapply and operate on lower components in the sublist, then I can do as
shown in EXAMPLE 1 and what will come back will be named GGG at the top
level.
Hi Rolf: it's not what's inside the lapplys that i was interested in
which probably made my question more confusing. all i was trying to show
was that,
in EXAMPLE 1, the name comes back at the top level of the result.
in EXAMPLE 2, i can use the name GGG inside the lapply functon to do
Hi: Gabor's solution does do it in a single line. he just used paste to
make the line. see below. John's is sort of a single line also but he
called sub twice.
I doubt that it's possible to make it shorter than those solutions.
# Gabor's solution spelled out.
patReg1 - (^[ *]+)
patReg2 - ([
hi ricardo: you're english is fine. i'll cc the list because i
haven't done that and i don't know what a geodata object is but my
guess is that you can use write.table. you can see the details of that
function by doing ?write.table.
On Sun, Nov 2, 2008 at 9:02 PM, Ricardo Bandin wrote:
Hi Phil: That's EXACTLY what it is. Thanks so much. It's nice to know
that the R Gods don't hate me. I hope it's okay that I'm going to cc
r-help
in case this thread comes up in the future and also so that other people
who might want to help know that it's solved. Thanks again.
On Wed,
thanks Rolf. Yes, I meant temp.R. I was going to use test.R but then
I realized that I already had a program named that. I think the R gods
are
really hating me !!! it's a very odd thing. I'll grep the file
because maybe the output is in there somewhere and i'm missing it ?
On Wed,
the str function shows that x is an int and y is a num so it's probably
not a bug. or maybe the conversion to num is but probably not the
identical.
x = 1:10
y = 1:10
all.equal(x,y)
identical(x,y)
y[11] = 11
y = y[1:10]
all.equal(x,y)
identical(x,y)
print(str(y))
print(str(x))
On
just to add to ted's explanation in case it helps to fix the email
server problem: below are the two Recipients and one of the two is
always contained in the repeated emails that I'm receiving. Also, of
course i don't mean to claim that the names contained in below are
doing anything to
if you do below, then d[[1]] is the first, d[[2]] is the second etc.
d - lapply(1:50,function(.index) {
data.frame(x=factor(),y=numeric())
})
print(d)
print(str(d))
On Fri, Oct 24, 2008 at 4:36 AM, tedzzx wrote:
Dear R users,
I want to creat a group of data frames, such as:
below i think semi does what you want but it's never going to repeat a
value because it's sampling without replacement ? in that sense,
it's not as general as what you asked for. if you need the repeating
thing, then i'm not sure how to do that. hopefully someone else does.
lapply(1:30,
megh: assuming that i understand what you want, i think below does it
but check it carefully because i didn't.
i = 1:9
a = matrix(1:9, 3)
tempa - lapply(i,function(.index) {
a^.index
})
tempb - lapply(seq(1,dim(a)[1]*dim(a)[2],by=dim(a)[1]),function(.index)
{
that's extremely rude , especially to all the people who made and make
R what it is. If you don't like R, noone is forcing you to use it.
On Thu, Oct 16, 2008 at 5:50 PM, repkakala Gazeta.pl wrote:
On 10/16/2008 10:50 AM, culpritNr1 wrote:
Now, modern high level languages like the
this seems l ike it shouldn't be that hard but i give up.
if i have a string say, temp-01, I want to increase it by 1 so that
it becomes 02. but the following code obviously won't work when the
input string is say 10 because then it gives 011 when I just want
11. uuugh.
does someone know
If I have the string below. does someone know a regular expression to
just get the BLC.NYSE. I bought the O'Reilley
book and read it when I can and I study the solutions on the list but
I'm still not self sufficient with these things. Thanks.
hi: you should probably send below to R-Sig-Finance because there are
some econometrics people over there who could also possibly give you
a good answer and may not see this email ? Also, there's package called
mar ( I think that's the name ) that may do what you want ?
Finally, I don't
I am estimating a multinomial model with two quantitative predictors, X1
and X2, and 3 responses. The responses are called neutral, positive and
negative with neutral being the baseline. There are actually many models
being estimated because I estimate the model over time and also for
various
DF-cbind(c(a,b,a),c(4,3,6))
DF[(DF[,1] %in% names(which(table(DF[,1]) = 2))),]
On Sun, Aug 31, 2008 at 2:19 AM, Yuan Jian wrote:
Hi,
�
I have a matrix.
a-cbind(c(a,b,a),c(4,3,6))
[,1] [,2]
[1,] a� 4 [2,] b� 3 [3,] a� 6
I want to remove rows in matrix a whose first column has
this is for the person who asked me about prediction confidence
intervals in a GLM because I lost your email. Below follows a simple
example in CAR and the variance covariance of the beta coefficients is
in the summary. So, I think, given that output, it should be pretty
straightforward to do
Duncan: Just one more thing which Heinz alerted me to. Suppose that we
changed the game so that instead of being double or half of X,
we said that one envelope will contain X + 5 and the other contains X-5.
So someone opens it and sees 10 dollars. Now, their remaining choices
are 5 and 15 so
beautiful. now, i think i got it. the fact that the calculation works
in the additive case doesn't make the calculation correct. the expected
value calculation is kind of assuming that the person putting the
things in the envelopes chooses what's in the second envelope AFTER
knowing
i haven't looked at your code and I'll try when I have time but, as
you stated, that's an EXTREMELY famous problem that has tried to be
posed
in a bayesian way and all sorts of other things have been done to try
solve it. Note that if you change the utility function so that its
log(X) rather
Hi: I think you want regexpr so below does what you want but it doesn't
handle the case when L isn't in the second column. I'm still trying to
figure that out but don't count on it. Hopefully someone else will reply
with that piece.
DF - data.frame(col1=L,col2=MAIL,col3=PLOY)
print(DF)
index
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