This is for ACroske but I can't find his email so I'll just send it to
the list. Hi ACroske: The code below takes a zeros and ones matrix and
puts ones
in the places you wanted. It can be made shorter ( maybe ?. i haven't
thought about that ) but first let me know if that's what you wanted ?
The
Just to let everyone know: John Chambers R book is no longer at
pre-order status on Amazon which is a good thing. The only problem
is that now it is temporarily out of stock. There must have been many
pre-orders so now I'm kicking myself for not having done that.
Mark
below is another way ( maybe the same ? )but with an extra line to make
roworder. i'm also not clear on why I have to take the transpose of
the result that
comes back form the apply call. in ?apply, it says that the function
tries to convert the result back to a matrix. that's fine but why doe
Is anyone aware of a way to check whether multinom has converged by
checking a component of the output ? I'm
not familar with nnet but, since multinom calls nnet , maybe there is an
extra argument once can send to multinom to capture this
information. Thanks.
___
if you mean matrix multiplication, it's %*% but i'm not sure if that's
what you meant by below ?
On Mon, Aug 11, 2008 at 3:44 PM, dott wrote:
Matlab has .*, so that
t(x1,x2).*t(x3,x4).* t(x5,x6)
but what is .* in R?
Thanks.
Dot.
--
View this message in context:
http://www.nabble.com/how-c
Hi: you can do
result<-lapply(agg, function(.df) {
mean(.df$Discharge)
})
this will give you the mean of the Discharge column in the various
dataframes. aggregate s used more when you want to do calculations
on various subset by another variable. if you want to do that, you can.
like say ther
I have a string such as
fileName<-"Agg.20.20.20-all-01".
All I want to do is pull the "20.20.20" and the "all" as strings.
Obviously, they aren't always those values.
The "20.20.20" can be "30.30.30" but it's always after the . which is
next to the second g in Agg and it's always the same len
i don't think i understood what you were trying to do, atleast based
on Henrique's solution which I haven't cut and pasted yet in order
to understand. Did Henrique's solution do what you wanted ?
On Wed, Aug 13, 2008 at 2:45 PM, Ralph S. wrote:
I tried this - I get an empty set:
<0 rows> (o
sorry ralph. i meant the OR instead of the AND so that was my bad
mistake. the subset function should also work with the OR.
i think i understand better what you want now also. the approach below
for doing what you want assumes that , if there are 2 rows associated
with the
values in the fi
Ralph: I looked at Henrique's solution and he does 2 things which make
it better than mine.
1) He splits based off the first two columns where I just split based on
the second. So, my split assumes that the "same rows" are next to each
other
which is an unnecessary assumption.
2) He actually
Hi: If I remember correctly, I think I gave you something like mat[mat
== 0]<-NA. I think what you're doing below is pretty different from
that but I may not be understanding what you want ? Let me know if I can
clarify more because my intention was not to guide you
into doing below. Looping is
I was trying to do what is on page 70 of John Chambers' new book namely
using trace to invoke the browser by doing
trace(zapsmall, edit = TRUE)
but , typing above at an R prompt, I get
trace(zapsmall, edit=TRUE)
sh: EMACS: command not found
Error in edit(name, file, title, editor) :
problem
My previous problem with the emacs editor was that I had
options(editor="EMACS", when it should have been small letters as in
options(editor="emacs",papersize="letter", width=180, htmlhelp=FALSE,
browser="/usr/bin/firefox")
Thanks for the help that I'm confident would have been there if I di
I don't know if it's the best way but you can do
d<-d[complete.cases(d),]
plot(d, type="b")
plot doesn't connect the points when there are NAs between them.
On Sun, Aug 17, 2008 at 9:55 PM, stephen sefick wrote:
d <- structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 1,
2, 3, 4, 5
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)
Hi John: I didn't realize that that was your problem. You can make it
work for any number of rows by putting it in lapply as below.
I'm sorry for the misunderstanding. I'll send to the list also since I
guess my last solution was kind of bad now that I understand what you
want.
DF <-
data.fr
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 t
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 the
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 what's
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
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 whos
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
__
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 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 h
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 wrote:
Thanks a lot. The info about computing the gradient
__
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: 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
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
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] <- list
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 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 sec
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 wrote:
Hi all,
I apologize for this off-topic question but I need your
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 ' charact
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 tu
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 wh
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 list
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 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 h
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 wrote:
I am trying to check for backslashes in data, then remove them when I
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 it
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
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: 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 wrote:
Thanks, Chuck.
for learning purposes and also to help someone, i used roger peng's
document to get the mle's of the gamma where the gamma is defined as
f(y_i) = (1/gammafunction(shape)) * (scale^shape) * (y_i^(shape-1)) *
exp(-scale*y_i)
( i'm defining the scale as lambda rather than 1/lambda. various books
maybe this should also be a FAQ because i've seen it a lot since I've
been on this list ?
assign each data.frame to a component of a list, rather than to a
component of a dataframe. you also need paste to get the number.
listDFs<-list()
for ( y in 3:10 ) {
listDFs[[y]]<-read.table(paste("te
hi: you simulate the distribution of the test statistic by simulating
the hypothesis test over and over and calculating the test statistic
each time.
the details of how you simulate the hypothesis test depends on the
particular problem you are dealing with but you didn't describe that.
if you d
i think you should resend your question to the list (but this is
sounding more and more like a homework problem so i don't know if you
will get much response. all i'll say is that P(T > t ) = 1 - P( T < t
). But P( T < t ) is the cumulative distribution of T ( and any random
variable ) is
u
Thanks Duncan. The person didn't mention the exponential in their
private email to me so I was pretty confused. Now, I'm unconfused.
On Sat, May 24, 2008 at 10:02 PM, Duncan Murdoch wrote:
[EMAIL PROTECTED] wrote:
The context is "goodness-of-fit techniques"
Suppose I want to know whether
can someone show me how to convert a table to a data.frame or a matrix ?
I tried below and as.data.frame rearranges the columns
similarly to a melt from reshape and as.matrix didn't change it. I
actually would prefer to change it to a dataframe but if someone
can show me how to convert it to a
I have the data structure below and I'm attempting to send it into
barchart using the R code below it. I don't get an error but I don't get
any output either. Deepyan's new Lattice book is amazing and there are
some examples sort of similar to what i'm doing but I couldn't see a
way of using t
hi: you can do below but i don't know if it's worth it ?
newx <- data.matrix(data.frame(x))
print(newx)
On Mon, Jun 16, 2008 at 4:22 PM, Alberto Monteiro wrote:
Why does as.numeric convert matrices and arrays to vectors?
as.numeric(matrix(c("1", "2", "3", "4"), 2, 2))
[1] 1 2 3 4
I could
thanks for your correction Gavin. i read ?data.matrix and neglected to
pay attention to the last line of the description:
"Factors and ordered factors are replaced by their internal codes".
On Mon, Jun 16, 2008 at 5:30 PM, Gavin Simpson wrote:
On Mon, 2008-06-16 at 15:35 -0500, [EMAIL PR
If one estimates a model using multinom, is it possible to perform the
omnibus LR test ( the analogue to omnibus F in linear models ) using
the output
from multinom ? The residual deviance is there but I was hoping I could
somehow pull out the deviance based on just using an intercept ?
Sample
This is not exactly an R question but the R code below may make my
question more understandable.
If one plots sin(x) where x runs from -pi to pi , then the curve hovers
around zero obviously. so , in a"stationary in the mean" sense,
the series is stationary. But, clearly if one plots the acf,
I have a simple table below called temptable and i want to obtain the
same structure that prop.table creates except get the counts
rather than the proportions. margin.table seems to create one table with
columns and rows whereas I am looking for the three table
type structure that prop.table gi
Please disregard my previous question. I realize that I don't need such
a function because I already have what I need ( duuuh ).
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide htt
I'm creating a lattice barchart based off a pretty complicated data
structure. The barchart comes out quite nice ( thanks
to lattice ) but the problem is that the horizontal axis comes out all
scrunched because the barchart doesn't know that the intervals
of Var.1 are really "associated" with th
below is milton's example with the text command changed to be the values
of y itself ? is that what you wanted ?
x<-runif(20)
y<-rnorm(20)
df<-data.frame(cbind(x,y))
plot(y~x, data=df, type="n")
text(df$x, df$y, df$y)
On Sun, Jul 6, 2008 at 6:59 PM, hippie dream wrote:
Thanks Milton,
I
milton: that was totally your solution. i just changed it for what
hippie dream wanted. hippie dream: you can do ?text to get a better
explanation but
the basic idea is that the first two parameters in text specify the
coordinates at which you put the text and the third param is the text.
so
someone else can blast me if this is not correct but i think that 2
step procedure only gives the same answer as the regular regression if
X1 and
X2 perfectly uncorrelated. If they are at all correlated, then what john
pointed out messes the procedure up. i was asked that question on
an intervi
In Julian Faraway's text on pgs 117-119, he gives a very nice, pretty
simple description of how a glm can be thought of as linear model
with non constant variance. I just didn't understand one of his
statements on the top of 118. To quote :
"We can use a similar idea to fit a GLM. Roughly spea
A friend of mine can't send emails to the
R-list from his work and he had a question that
he asked me to send because I don't know the answer
to his question. I did suggest the cran packages
list and the search function but I think he wasn't
successful ?
He needs to do a quick prototype in R with
>From: "Wang, Xiaojing" <[EMAIL PROTECTED]>
>Date: 2008/03/26 Wed AM 10:08:01 CDT
>To: "'r-help@r-project.org'"
>Subject: [R] pseudo R square and/or C statistic in R logistic regression
The analogue to the regression R rsquared in a GLM
is the deviance. I don't know what C is but hopefully someo
>From: "Ivaha C (AT)" <[EMAIL PROTECTED]>
>Date: 2008/04/10 Thu AM 08:51:14 CDT
>To: R Help
>Subject: [R] Structural Modelling in R-project
if you have a univariate time series and you
want to break it into its various components,
then you get the scalars based on a decomposition.
if you have
This is probably basic but I want to condition based
on the name of the component inside an lapply.
So, in the simple example below, if .elem was x
i want to do one thing but if it's y a
different thing etc. Can someone tell me
how to do that without using names(temp)
as the thing one sends into t
>From: [EMAIL PROTECTED]
>Date: 2008/04/11 Fri PM 04:16:56 CDT
>To: [EMAIL PROTECTED]
>Subject: [R] conditioning inside an lapply
in my previous example, the last name in temp should
have been z not x. i didn't intend to have
duplicate names in temp. I apologize for any
confusion that caused.
>Th
>From: Jorge Velez <[EMAIL PROTECTED]>
>Date: 2008/04/11 Fri PM 04:36:58 CDT
>To: R
>Subject: [R] How to fill out some columns?
below works but my guess is that there's a better
way.
set.seed(4)
dx=matrix(rnorm(6*5),ncol=6)
colnames(dx)=LETTERS[1:6]
dy=matrix(rnorm(3*5),ncol=3)
colnames(dy)=c('
I have strings of the form
TICKER.GGG.XX.dat
but GGG is not always three characters so I can't use substr to pull it
out of the string.
Could someone tell me how to use sub to pull out the GGG but more
generally the string between the dot after the R in TICKER and the next
dot. I still
Since we're on the topic of book reviews, I just received Phil Spector's
new R book called "Data Manipulation with R" and it is also quite a
nice book. I haven't gone through it all and I won't give a detailed
review but I have gotten a lot out of the first 100 pages that I have
read.
Note t
you can do below but there's no way of getting arounf having NAs in your
final matrix.
d1<-as.matrix(rnorm(5))
d2<-as.matrix(rnorm(3))
d3<-as.matrix(rnorm(6))
templist <- list(d1,d2,d3)
maxnum <- max(sapply(templist,length))
print(maxnum)
temp <- lapply(templist,function(.mat) {
if (nrow(.ma
Spencer Graves has discussed this recently ( in the last couple of
months ) in a thread and I think his FinTS package has a variant of the
Box.test
function that does what you need.
On Fri, May 16, 2008 at 11:45 AM, Nuno Prista wrote:
Dear colleagues,
I am new to R and statistics so ple
I was just curious if anyone knew of an alternative model to logistic
regression where the probabilities seems pretty linear to the predictor rather
than having that S shape that probit and logit assume.
Maybe there is there some kind of other GLM that could accomplish that. Any
textbook refere
>From: Prof Brian Ripley <[EMAIL PROTECTED]>
>Date: 2007/11/16 Fri AM 09:44:59 CST
>To: [EMAIL PROTECTED]
>Cc: Terry Therneau <[EMAIL PROTECTED]>, r-help@r-project.org
>Subject: Re: Re: [R] alternative to logistic regression
Thanks Brian: I'll look at the MASS book example
for sure but I don't thi
>From: Prof Brian Ripley <[EMAIL PROTECTED]>
>Date: 2007/11/16 Fri AM 09:28:27 CST
>To: Terry Therneau <[EMAIL PROTECTED]>
>Cc: [EMAIL PROTECTED], r-help@r-project.org
>Subject: Re: [R] alternative to logistic regression
Thanks to both of you, Terry and Brian for your comments. I'm not sure what I
>From: Matthias Bindernagel <[EMAIL PROTECTED]>
>Date: 2007/11/19 Mon PM 02:11:45 CST
>To: r-help@r-project.org
>Subject: [R] Search for a usable pan manual
Joe Schafer wrote a text on missing data, the title
of which escapes me and I can't say if it will be useful for your particular
problem. My
>From: Matthias Bindernagel <[EMAIL PROTECTED]>
>Date: 2007/11/19 Mon PM 02:11:45 CST
>To: r-help@r-project.org
>Subject: [R] Search for a usable pan manual
Joe Schafer wrote a text on missing data, the title
of which escapes me and I can't say if it will be useful for your particular
problem. My
>From: Serguei Kaniovski <[EMAIL PROTECTED]>
>Date: 2007/11/28 Wed AM 08:53:34 CST
>To: [EMAIL PROTECTED]
>Subject: [R] Replacing values job
newY<-sapply(1:length(Y), function(.element)
match(Y[.element],X))
i hope it helps you.
>
>Hallo,
>
>I have two vectors of different lengths which conta
>From: Bernd Jagla <[EMAIL PROTECTED]>
>Date: 2007/11/30 Fri PM 09:24:34 CST
>To: r-help@r-project.org
>Subject: [R] compare strings
below assumes that there are only 2 columns in
the dataframe or that the string columns are the
first two columns ?
which(df[,1] == df[,2])
>Sorry for the questi
>From: "Doran, Harold" <[EMAIL PROTECTED]>
>Date: 2007/11/30 Fri PM 02:12:36 CST
>To: R Help <[EMAIL PROTECTED]>
>Subject: [R] Rating R Helpers
Of course, it's just MHO, but I think your suggestion is more work than
neccesary because it becomes pretty obvious who the R-experts are once you've
be
>From: Nuno Prista <[EMAIL PROTECTED]>
>Date: 2007/12/07 Fri PM 03:10:01 CST
>To: [EMAIL PROTECTED]
>Subject: [R] Cannot insert label on top axis
see mtext by doing ?mtext.
>Hi,
>
>
>
>I am not being able to find code to insert a label on the top graphic, can
>anyone help?
>
>
>
>Thanks in adv
>From: Duncan Murdoch <[EMAIL PROTECTED]>
>Date: 2007/12/09 Sun AM 11:40:45 CST
>To: Christophe Genolini <[EMAIL PROTECTED]>
>Cc: r-help@r-project.org
>Subject: Re: [R] Oriented object programming
Roger Peng has some objected oriented documentation
at his site as part of a powerpoint presentation.
>From: Neil Stewart <[EMAIL PROTECTED]>
>Date: 2007/12/10 Mon PM 05:41:34 CST
>To: [EMAIL PROTECTED]
>Subject: [R] Neat conditional assignment
d <- ifelse(a == "A",b,c)
>I would like to make a new vector with values taken from two existing
>vectors conditional upon a third. At present I am doing
>From: Johannes Graumann <[EMAIL PROTECTED]>
>Date: 2007/12/18 Tue PM 04:40:37 CST
>To: [EMAIL PROTECTED]
>Subject: [R] All anchored series from a vector?
lapply(1:length(myvector) function(.length) {
c(myvector[1}:myvector[.length])
})
but test it because i didn't.
>Hi all,
>
>What may be a s
>From: [EMAIL PROTECTED]
>Date: 2007/12/18 Tue PM 02:50:52 CST
>To: Johannes Graumann <[EMAIL PROTECTED]>
>Cc: r-help@r-project.org
>Subject: Re: [R] All anchored series from a vector?
i'm sorry. i tested it afterwards and of course
it had some problems. below is the working version.
myvector<-c
>From: Waverley <[EMAIL PROTECTED]>
>Date: 2007/12/20 Thu PM 01:29:04 CST
>To: [EMAIL PROTECTED]
>Subject: [R] Question how to get up triangle of a matrix
?upper.tri
>Is there a simple way to get up triangle of a matrix and return as a vector?
>
>Thanks much.
>
>--
>Waverley @ Palo Alto
>
>_
I was playing around with a simple example using solve.qp ( function is in the
quadprog package ) and the code is below. ( I'm not even sure there if there is
a reasonable solution because I made the problem up ).
But, when I try to use solve.QP to solve it, I get the error that D in the
quad
>From: "Kondamani, Arjun (GMI - NY Corporate Bonds)" <[EMAIL PROTECTED]>
>Date: 2007/12/26 Wed PM 03:23:51 CST
>To: r-help@r-project.org
>Subject: [R] Rbind-ing a list into one item
try do.call(rbind,res)
>Hi,
>
>I am doing the following:
>
>1. I have a list of files.. Files1=list.files("some
>d
>From: Max <[EMAIL PROTECTED]>
>Date: 2008/01/03 Thu AM 09:31:40 CST
>To: [EMAIL PROTECTED]
>Subject: Re: [R] Multivariate response methods question
Hi Max: multinom allows ( and probably polr also ) allows a categorical
response that can take on any number of values so I am unsure what you mean
>From: eugen pircalabelu <[EMAIL PROTECTED]>
>Date: 2008/01/10 Thu PM 02:48:32 CST
>To: R-help <[EMAIL PROTECTED]>
>Subject: [R] question regarding kpss tests from urca, uroot and tseries
>packages
Schwert has an algorithm for deciding on the
number of lags to use in an adf test but i
can't say
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 k
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.
stock<-"/opt/limsrv/mark/research/equity/project
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 so
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 cont
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)
{
do.call(rbind,tempa[.index
Hi Rolf: yup, i'm getting copies over and over of a few different
posts and yes one is yours. they have been coming over and over since
about 5pm and it's now about midnight ( US Eastern time )
On Thu, Oct 23, 2008 at 8:01 PM, Rolf Turner wrote:
I keep getting repeated copies of ``R-hel
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, functio
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 cau
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:
d1
1 - 100 of 171 matches
Mail list logo