Dear users
This is a message I was directing to Harold Baize but because I pressed the
wrong button the message got lost g!!!
So I’m doing it all over again:
Lets suppose I have three batches of data:
a - rnorm(50,2500,300)
b - rnorm(50,3500,250)
c - rnorm(50,4000,200)
# Now I want to
Hi Ben,
Sorry (still a little out-of-tune), perhaps what you really need to know
about is ?[
HTH, Mark.
Mark Difford wrote:
Hi Ben,
If you wouldn't mind, how do I access the individual components inside
coefficients matrix?
What you want to know about is ?attributes
##
Hi Ben,
If you wouldn't mind, how do I access the individual components inside
coefficients matrix?
What you want to know about is ?attributes
##
attributes(model)
model$coefficients
model$coefficients[1]
model$coefficients[2:4]
model$coefficients[c(1,5)]
HTH, Mark.
ascentnet wrote:
Hi,
In order to use the mle2-function, one has to define the likelihood function
itself. As we know, the likelihood function is a sum of the logarithm of
probability density functions (pdf). I have implemented myself the pdfs
that I am using. My problem is, that the pdfs values are negative and
Thanks for the answers,
but I'm looking for an editor like Tinn-R.
What about VM Fusion and continue to use Tinn-R then?
--
Regards,
Hans-Peter
PS: I use TextMate (and like it)
__
R-help@r-project.org mailing list
Hi All,
It really comes down to a question of attitude: you either want to learn
something fundamental or core and so bootstrap yourself to a better place
(at least away from where you are), or you don't. As Marc said, Michal seems
to have erected a wall around his thinking.
I don't think it's
this is an off-topic, of course!
by definition, pdf MUST BE POSITIVE and log function is defined ONLY
for positive values.
check your pdf code and find out the errors.
regards
Patrizio Frederic
+-
| Patrizio Frederic
| Research associate in
CG == Christophe Genolini [EMAIL PROTECTED]
on Tue, 22 Jul 2008 19:04:37 +0200 writes:
CG Prof Brian Ripley [EMAIL PROTECTED] a écrit :
On Tue, 22 Jul 2008, [EMAIL PROTECTED] wrote:
Hi the list (well, half of the list, only the one who
are not on holidays...)
Alessandro,
Have a look at the gstat package.
library(gstat)
demo(examples)
demo(krige)
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel
Sascha,
Have a look at ?krige in the gstat package.
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg
Dear R users,
I use the glm() function to fit a generalized linear model with gamma
distribution function and log link.
I have read in the help page that the default method used by R is glm.fit
(iteratively reweighted least squares, IWLS).
Is it possible to use maximum likelihood method?
Hi RichardP
I guess you'll increase your chances of a response if you include some
affiliation info. This could also avoid getting you confused with other people -
I thought for a minute I'd been sleep-emailing :-)
Cheers
A N Other RichardP
RichardP wrote:
hi,
I am using rpart with a
Now I also want to generate two correlated variables where the error
variance vary over the variable-correlation.
And I want to plot this for showing heteroscedasticity.
Like shown here:
http://upload.wikimedia.org/wikipedia/de/1/1b/Heteroske2.png
Is that possible with R?
of course it
Dear R-nautes,
I installed the 2.7.1 version of R and found it is not possible to modify
graphical parameters with par().
for example;
par(mfrow=c(2,2))
Error in c(2, 2) : unused argument(s) (2)
par$mfrow
NULL
Does any one know what is the cause of this problem?
Regards,
Fabio
--
Fabio
Hi Fabio
Works OK for me.
par(mfrow=c(2,2))
par(mfrow)
[1] 2 2
But then I am using Windows ...
sessionInfo()
R version 2.7.1 (2008-06-23)
i386-pc-mingw32
locale:
LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United
Kingdom.1252;LC_MONETARY=English_United
On Wed, 23 Jul 2008, silvia narduzzi wrote:
Dear R users,
I use the glm() function to fit a generalized linear model with gamma
distribution function and log link. I have read in the help page that
the default method used by R is glm.fit (iteratively reweighted least
squares, IWLS). Is it
Hi Fabio
Have you used the name c for something else? Try typing in c at the
command line, you should see something like:
c
function (..., recursive = FALSE) .Primitive(c)
Regards
John
---
for example;
par(mfrow=c(2,2))
Error in c(2, 2) : unused argument(s) (2)
par$mfrow
NULL
dear Silvia,
quoting Venables WN and Ripley DB (1994) Modern Applied Statistics
with S-plus, sringer, pag 185:
Since explicit expressions for the maximum likelihood estimators are
not usually available estimates MUST be calculate iteratively
means that glm.fit performs MLE indirectly and
Perhaps you have a function called 'par' in your workspace.
On Wed, Jul 23, 2008 at 6:56 AM, Fabio Sanchez [EMAIL PROTECTED] wrote:
Dear R-nautes,
I installed the 2.7.1 version of R and found it is not possible to modify
graphical parameters with par().
for example;
par(mfrow=c(2,2))
Senthil,
sending a 12Mb file to the list is not a good idea.
I've run the code in my previous email without any problem, so you need
to be a bit more specific about what went wrong for you.
This is what I get:
library(igraph)
tab - read.csv(/tmp/Test.csv)
dim(tab)
[1] 304711 2
?bxp
This is the underlying routine called by boxplot and you can supply
your own values to the 5 that define a boxplot.
On Wed, Jul 23, 2008 at 2:24 AM, Fernando Marmolejo-Ramos
[EMAIL PROTECTED] wrote:
Dear users
This is a message I was directing to Harold Baize but because I pressed the
Maybe:
sapply(lapply(dir(pattern=data), read.table), '[[', 4)
On Wed, Jul 23, 2008 at 5:21 AM, Daren Tan [EMAIL PROTECTED] wrote:
Better approach than this brute force ?
mm - NULL
for (i in dir(pattern=data)) { m - readTable(i); mm - cbind(mm, m[,4]) }
Fabio Sanchez wrote:
Dear R-nautes,
I installed the 2.7.1 version of R and found it is not possible to modify
graphical parameters with par().
for example;
par(mfrow=c(2,2))
Error in c(2, 2) : unused argument(s) (2)
That's not an error from par(), it's an error from c(). I
On Tue, 2008-07-22 at 16:31 -0400, Jia Ying Mei wrote:
Hi everyone,
I am not new to R, but its been over a year since I've used it, so I
don't remember how to do some things.
I have a data set (in excel currently, but will be converted to text),
that has date, country and price, with a
On Tue, 8 Jul 2008, Duncan Murdoch wrote:
On 7/8/2008 10:53 AM, Sergi M.Garrido wrote:
Hello,
I have tried to create a package, and I have got it. I checked the files
and
I built the package. Nevertheless, I want a .pdf file with the package's
documentation. Anyone know what I have to
simultron
On Wed, Jul 23, 2008 at 4:04 AM, Hans-Peter Suter [EMAIL PROTECTED] wrote:
Thanks for the answers,
but I'm looking for an editor like Tinn-R.
What about VM Fusion and continue to use Tinn-R then?
--
Regards,
Hans-Peter
PS: I use TextMate (and like it)
Thank you all,
I accidentally loaded a variable named c.
All the best,
Fabio
On Wed, Jul 23, 2008 at 1:08 PM, Prof Brian Ripley [EMAIL PROTECTED]
wrote:
Did you start R with --vanilla to check that this is not something you have
done? It is good practice, and I suspect will show you that
Fabio Sanchez wrote:
Dear R-nautes,
I installed the 2.7.1 version of R and found it is not possible to modify
graphical parameters with par().
for example;
par(mfrow=c(2,2))
Error in c(2, 2) : unused argument(s) (2)
par$mfrow
NULL
Does any one know what is the cause of
It would be helpful if you included a sample of the data so that we
could understand what you would like to do with it (before/after
pictures).
?aggregate
On Tue, Jul 22, 2008 at 9:57 PM, Kaposi-Novak, Pal
[EMAIL PROTECTED] wrote:
Hi,
Could somebody tell me how I can average expression values
On Tue, 2008-07-22 at 16:18 -0400, [EMAIL PROTECTED] wrote:
Hi,
I have a CSV file with various biological reactions. Subscripts, superscripts,
and italics are encoded in carats, and I was wondering if R can actually
recognize those and print actual superscripts, etc. Here's an example:
Hello,
I am new to the RODBC package, but I have looked over the PDF as well as a few
websites that go over the SQL language. I can connect to my database fine using
channel-odbcConnect(Oracle ODBC)
# then am prompted to enter my user id and password
After that I'd like to use the odbcQuery
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
Thanks, Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
__
I think the argument supporting the use of bootstrap to determine
coefficients, as opposed to just running linear regression on the whole
dataset, is the comparison of Rsq and prediction errors between these
two approaches - page 1502. There's a substantial difference in favor of
the bootstrap
If I understand:
sort(c(x,y))
or if you want combine the elements:
as.vector(mapply(c, x, y))
On Wed, Jul 23, 2008 at 9:54 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do
Shubba
I'm confused. Your first post said the result should be c(1,2,3,4,5,6)
when x and y are combined. The code I sent does that. But here you say
your result should be c(4,1,2,5,2,3).
What do you want your result to actually be?
-Original Message-
From: Shubha Vishwanath Karanth
Thank you Gustaf,
I apologize for not including an example data in my first email.
Nevertheless, your code worked for me excellently - I only added 55 as
the size of sample.
I must admit this code looks so much simpler, compared to SAS. I am
beginning to love R, despite some disrespectful
Guess this should be fine
c(rbind(x,y))
Please let me know if there are some better ways...
Thanks, Shubha
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Shubha Vishwanath Karanth
Sent: Wednesday, July 23, 2008 6:25 PM
To: [EMAIL PROTECTED]
x - c(1,3,5)
y - c(2,4,6)
sort(c(x,y))
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Shubha
Vishwanath Karanth
Sent: Wednesday, July 23, 2008 8:55 AM
To: [EMAIL PROTECTED]
Subject: [R] Simple... but...
Hi R,
If
x=c(1,3,5)
Shubha Vishwanath Karanth wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
the absolutely best way is:
?c
vQ
__
R-help@r-project.org
This thread may help
http://www.nabble.com/adding-the-mean-and-standard-deviation-to-boxplots-td15271398.html
--- On Wed, 7/23/08, Fernando Marmolejo-Ramos [EMAIL PROTECTED] wrote:
From: Fernando Marmolejo-Ramos [EMAIL PROTECTED]
Subject: Re: [R] adding the mean and standard deviation to
Z=NULL;for (i in 1:3) Z=c(Z,c(x[i],y[i]))
On Wed, Jul 23, 2008 at 8:54 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
Thanks, Shubha
This
OK,
Let x=c(4,2,2)
y=c(1,5,3)
My result should be c(4,1,2,5,2,3)
Thanks, Shubha
-Original Message-
From: Doran, Harold [mailto:[EMAIL PROTECTED]
Sent: Wednesday, July 23, 2008 6:47 PM
To: Shubha Vishwanath Karanth; [EMAIL PROTECTED]
Subject: RE: [R] Simple... but...
x -
This is definitely the best way:
c(lapply(1:length(x), function(i, x, y) c(x[i], y[i]), x, y),
recursive=TRUE)
JS
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Shubha Vishwanath Karanth
Sent: 23 July 2008 13:55
To: [EMAIL PROTECTED]
For a function that takes an argument as a list of lists of parameters,
I'd like to be able to convert that
to a data.frame and vice versa, but can't quite figure out how.
pats - list(structure(list(shape = 0, shape.col = black, shape.lty = 1,
cell.fill = white, back.fill = white, label = 1,
How about this
as.vector(t(cbind(x,y)))
-Original Message-
From: Shubha Vishwanath Karanth [mailto:[EMAIL PROTECTED]
Sent: Wednesday, July 23, 2008 9:17 AM
To: Doran, Harold; [EMAIL PROTECTED]
Subject: RE: [R] Simple... but...
OK,
Let x=c(4,2,2)
y=c(1,5,3)
My
It seems you have accidentally hit a surgeons' mailing list where all
you wanted was some advice how to use this scalpel on your body.
Sorry if we can't be of any help without intimidating you with
unrelated and pompous terms -- like coagulation.
Michal J. Figurski [EMAIL PROTECTED] [Wed, Jul 23,
check the following:
x - c(1,3,5)
y - c(2,4,6)
c(t(cbind(x, y)))
c(matrix(c(x, y), 2, by = TRUE))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
You should at least start with higher-level functions like sqlFetch(). Oh,
and reading the posting guide (you show us no examples let alone error
message) and the tests.R examples file which comes with RODBC.
SQL syntax is DBMS-specific, so we can't comment on abstract examples (and
it is a
Hi Shubha,
Assuming you are after ordering by position in x and y (and not
values), how about
as.vector(t(cbind(x, y)))
HTH,
Jim Porzak
Responsys, Inc.
San Francisco, CA
http://www.linkedin.com/in/jimporzak
On Wed, Jul 23, 2008 at 5:54 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
x - c(1,3,5)
y - c(2,4,6)
xy - sort(c(x,y))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Shubha Vishwanath Karanth
Sent: Wednesday, July 23, 2008 8:55 AM
To: [EMAIL PROTECTED]
Subject: [R] Simple... but...
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
On 23-Jul-08 12:54:49, Shubha Vishwanath Karanth wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
Thanks, Shubha
Your query is ambiguous, in that it is not clear whether you want
a) The elements of the
Try c(rbind(x, y))
On Wed, Jul 23, 2008 at 8:54 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
Thanks, Shubha
This e-mail may contain
On Wed, Jul 23, 2008 at 3:23 PM, Doran, Harold [EMAIL PROTECTED] wrote:
Shubba
I'm confused. Your first post said the result should be c(1,2,3,4,5,6)
when x and y are combined. The code I sent does that. But here you say
your result should be c(4,1,2,5,2,3).
What do you want your result to
Shubha Vishwanath Karanth wrote:
Hi R,
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
How do I do it? I mean the best way
... but seriously, rbind(x,y)[1:6] is one concise way to do it (x and y
are bound into rows in a matrix which
Megan J Bellamy schrieb:
After that I'd like to use the odbcQuery function and the SQL statement:
SELECT * FROM table_name WHERE SAMPLE_YEA LIKE 1965%
According to the RODBC PDF, the odbcQuery command should accept any
valid SQL statement. Is there something wrong with my syntax? I have
Firas Swidan frsswdn at gmail.com writes:
Hi,
I am having difficulties in finding ways to analyse scatter plots and
quantitatively differentiate between them.
Try quantile regression, implemented in Roger Koenker's quantreg
package. There's a very thorough vignette
On Wed, Jul 23, 2008 at 3:14 PM, Michal Figurski
[EMAIL PROTECTED] wrote:
I think the argument supporting the use of bootstrap to determine
coefficients, as opposed to just running linear regression on the whole
dataset, is the comparison of Rsq and prediction errors between these
two
On Wed, 23 Jul 2008, Shubha Vishwanath Karanth wrote:
If
x=c(1,3,5)
y=c(2,4,6)
I need a vector which is c(1,2,3,4,5,6) from x and y.
I am assuming that you want to interleave the vectors, but that is not the
only algorithm giving that result.
How do I do it? I mean the best way
Thank you very much. As for the second part, for every price change, the
percentage of every price change can be calculated (i.e. amount of
change divided by previous price). I was looking for a way to merely
find the mean and median of all price change percentages for each
country. I don't
it's clear, he wants to combine x and y s that the result is (x[1],
y[1], x[2], y[2], ...), unless i am confused as well ;)
now, both rbind(x,y)[1:length(x)] and as.vector(mapply(c,x,y)) do the
job, but the former is *much* faster:
t1 = function(n) as.vector(mapply(c, 1:n, 1:n))
t2 = function(n)
I thinks this is the solution that you want!
as.vector(matrix(c(x,y),byrow=T,ncol=3))
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de Wacek Kusnierczyk
Enviada em: quarta-feira, 23 de julho de 2008 10:25
Para: R help
Assunto: Re: [R] Simple... but...
Shubha
On Wed, Jul 23, 2008 at 4:08 PM, Michal Figurski
[EMAIL PROTECTED] wrote:
Gustaf,
I am sorry, but I don't get the point. Let's just focus on predictive
performance from the cited passage, that is the number of values predicted
within 15% of the original value.
So, the predictive performance
Try this:
pats.df - do.call(rbind, pats)
On Wed, Jul 23, 2008 at 10:23 AM, Michael Friendly [EMAIL PROTECTED] wrote:
For a function that takes an argument as a list of lists of parameters, I'd
like to be able to convert that
to a data.frame and vice versa, but can't quite figure out how.
try this
colSums(matrix(x,8))
regards,
PF
+-
| Patrizio Frederic
| Research associate in Statistics,
| Department of Economics,
| University of Modena and Reggio Emilia,
| Via Berengario 51,
| 41100 Modena, Italy
|
| tel: +39 059 205 6727
|
try
?unlist
it may help
regards
+-
| Patrizio Frederic
| Research associate in Statistics,
| Department of Economics,
| University of Modena and Reggio Emilia,
| Via Berengario 51,
| 41100 Modena, Italy
|
| tel: +39 059 205 6727
| fax: +39 059
Many Thanks Jorge... That was one more way...Is it possible if I can do
this without using rep(1:10,each=8) or the groupingbecause I feel
the number 8 here is fixed... If there is some technique of tracking the
position of first 8, then next 8... don't know whether I am clear in
conveying...
on 07/23/2008 09:03 AM Shubha Vishwanath Karanth wrote:
Hi R,
Let,
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8 elements. So, my new vector should look like:
c(36,100,164,228,292,356,420,484,548,612)
I used:
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8 elements. So, my new vector should look like:
c(36,100,164,228,292,356,420,484,548,612)
I used:
aggregate(x,list(rep(1:10,each=8)),sum)[-1]
or
Dear Shubha,
Try this:
x=1:80
tapply(x,rep(1:10,each=8),sum)
1 2 3 4 5 6 7 8 9 10
36 100 164 228 292 356 420 484 548 612
HTH,
Jorge
On Wed, Jul 23, 2008 at 10:03 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Hi R,
Let,
x=1:80
I want to sum up first
Hi all,
Again I have searched the net and so on, without finding an answer to this
surely simple problem. A short bit of code would be appreciated.
I have a object named `data' with the following column headings.
Date, maxitemp, minitemp, admissions, d.o.w.
Where d.o.w. is day of the week,
At 7:33 PM +0530 7/23/08, Shubha Vishwanath Karanth wrote:
Hi R,
Let,
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8 elements. So, my new vector should look like:
c(36,100,164,228,292,356,420,484,548,612)
I used:
Maybe:
tapply(x, gl(10, 8), sum)
or
unlist(lapply(split(x, gl(10, 8)), sum))
On Wed, Jul 23, 2008 at 11:03 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED] wrote:
Hi R,
Let,
x=1:80
I want to sum up first 8 elements of x, then again next 8 elements of x,
then again another 8
Have u checked skewness of data?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Firas Swidan
Sent: 23 July 2008 18:38
To: r-help@r-project.org
Subject: [R] Quantitative analysis of non-standard scatter plots.
Hi,
I am having difficulties in finding
Dear Michael,
Perhaps,
data.frame(do.call(rbind,lapply(pats,function(x) t(as.matrix(x,ncol=10)
HTH,
Jorge
On Wed, Jul 23, 2008 at 9:23 AM, Michael Friendly [EMAIL PROTECTED] wrote:
For a function that takes an argument as a list of lists of parameters, I'd
like to be able to convert
Hi, I have a huge data set to deal with. Sometimes I got warning message
about limitation of memory, sometimes R reads in the data but it is very
slow.
My question is, is there anything I can do to lift the memory limitation (I
did memory.limit(4095), anything else I can do?). I know I can also
Oh, you are rigth Marc, thanks.
Another option is:
pats.df - do.call(rbind.data.frame, pats)
On Wed, Jul 23, 2008 at 1:10 PM, Marc Schwartz
[EMAIL PROTECTED] wrote:
on 07/23/2008 09:42 AM Henrique Dallazuanna wrote:
Try this:
pats.df - do.call(rbind, pats)
Henrique,
Take note of the
I have written some programs in Common Lisp and I have been using SAS to pipe
those programs to my lisp compiler in batch mode by using the %xlog and
%xlst SAS commands. I wonder if there is in R a similar way to pipe commands
to LISP so that all my work would be concentrated in R even when I
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal
a.string. Is this right?
I am not sure cat() correctly prints strings which are generated by
triple-shQuote():
shQuote(shQuote(a))
[1] a
cat(shQuote(shQuote(shQuote(a))), '\n')
Hi,
I am using an if else statement inside a function …. If I use that function I
have no problems …. If I use the function with the if else statement inside a
second function I get the following waring:
Warning message:
In if (pval == 0) p_value - 2.2e-16 else p_value - pval :
the
Thank you all for your words of wisdom.
I start getting into what you mean by bootstrap. Not surprisingly, it
seems to be something else than I do. The bootstrap is a tool, and I
would rather compare it to a hammer than to a gun. People say that
hammer is for driving nails. This situation is
Dear Patrizio and all,
Do you know a similar approach to this, so that *any* function is sequentially
applied on a matrix?
Particularly, instead of wanting to sum columns every 8 rows, how could we
apply a linear regression to the columns of a matrix every 8 rows?
ie, if we have a matrix say of
Hello,
I have a problem to flip a 200x200 matrix, which is imported by a .asc file.
I want to flip the matrix like in a created example below:
b - matrix(1:9,3,3,byrow=T)
b
[,1] [,2] [,3]
[1,]123
[2,]456
[3,]789
b1 - apply(t(b),1,rev)
b1
[,1]
You would have something that looks like this:
if (data$d.o.w == Sat) data$admission - round(data$admission * 1.21)
if (data$d.o.w == Sun) ...
On Wed, Jul 23, 2008 at 11:44 AM, Robin Williams [EMAIL PROTECTED] wrote:
Hi all,
Again I have searched the net and so on, without finding an
My fingers slipped on the keyboard. Here what they intended to write.
This sets up a list of the data and the matches on a subset for
processing
days - list(list(Sat, 1.21), list(Sun, 1.22), list(Mon, 0.91))
for (i in days){
.subset - data$d.o.w == i[[1]] # subset of data that matches
One more try and I quit: This is what happens if someone does not
sent a sample of data you have to create things on the fly without
testing.
days - list(list(Sat, 1.21), list(Sun, 1.22), list(Mon, 0.91))
for (i in days){
.subset - data$d.o.w == i[[1]] # subset of data that matches
jjdow - sample(c('Sun', 'Mon', 'Tues'), 10, replace=TRUE)
jjdata - data.frame(admission=1:10, d.o.w.=jjdow)
# jjdata
# admission d.o.w.
#1 1 Tues
#2 2 Tues
#3 3 Tues
#4 4Sun
#5 5Mon
#6 6Mon
#7 7Sun
#8
Monica -
Monica Pisica wrote:
Hi,
I am using an if else statement inside a function …. If I use that
function I have no problems …. If I use the function with the if else
statement inside a second function I get the following waring:
Warning message: In if (pval == 0) p_value - 2.2e-16 else
Just to clarify,
if you have two data.frame, one with your data, other with data-admissions,
just use
data.merge-merge(my.df, data.weigth, by.x=data, by.y=data, all=T)
miltinho
On 7/23/08, milton ruser [EMAIL PROTECTED] wrote:
Hi Rob Williams
I think it is one way of you do the job.
How do I check that, and what quantification does it provide?
Thanks,
Firas.
On Wed, 2008-07-23 at 21:20 +0530, Bogaso Cristofer wrote:
Have u checked skewness of data?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Firas Swidan
Sent: 23 July 2008
On 7/23/2008 12:20 PM, Vadim Organovich wrote:
Dear R-users,
It is my understanding that cat(shQuote(a.string)) should print the origintal
a.string. Is this right?
No, cat(a.string) should print the original a.string. shQuote(a.string)
adds quotes so that a.string can be used in the shell.
Hi Erik,
Thanks for your answer. I did print the p-value just before the statement and
it is only one value – or at least I see only one value ….. that is strange.
I am doing a test and I want to see what is it's p-value ….for example my
function myf is like that:
myf - function(m, se,
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Michal Figurski
Sent: Wednesday, July 23, 2008 10:22 AM
To: r-help@r-project.org
Subject: Re: [R] Coefficients of Logistic Regression from
bootstrap - how to get them?
Thank you all for your words of
Hi.
On Wed, Jul 23, 2008 at 9:23 AM, Chris82 [EMAIL PROTECTED] wrote:
Hello,
I have a problem to flip a 200x200 matrix, which is imported by a .asc file.
I want to flip the matrix like in a created example below:
b - matrix(1:9,3,3,byrow=T)
b
[,1] [,2] [,3]
[1,]123
I second that - quantile regression seems to be what you want.
on 07/23/2008 10:10 AM Ben Bolker said the following:
Firas Swidan frsswdn at gmail.com writes:
Hi,
I am having difficulties in finding ways to analyse scatter plots and
quantitatively differentiate between them.
Try
A few things that will help you, if not now then in the future:
1) Preallocate the result object. This allow you to avoid using
cbind()/rbind(), which constantly creates a new large copy in each
iteration. That will eventually bite you if you have a lot of data.
In your case you know the number
Hi,
how can I treat data organised in classes and frequencies?
Ex.
class frequency
20-23 9
23-25 7
26-28 5
29-31 5
32-34 3
Thanks
Angelo Scozzarella
Hi again,
It seems one line in my function came chopped for whatever reason
so the if else statement below should be:
if (pval==0) p_value - 2.2e-16 else p_value - pval
Thanks,
Monica
Date: Wed, 23 Jul 2008 13:24:45 -0500 From: [EMAIL PROTECTED] Subject: Re:
[R] Warning
Ok, thanks!
Gabor Grothendieck [EMAIL PROTECTED] wrote:
See ?title where its mentioned that calls are ok and a formula
is a call. Actually the first ~ is not needed since its already
a formula.
On Tue, Jul 22, 2008 at 1:12 PM, Mike Prager [EMAIL PROTECTED] wrote:
Gabor Grothendieck
Hello,
I have the following data frame (DF):
V5V5.1V5.2 V5.3 V5.4 V5.5
2 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
3 -5890.18905 -6019.84665 -6211.06545 -6198.9353 -6616.8677 -6498.7183
4 -5890.18905 -6019.84665
Greetings. This is very basic but we can't figure it out. The following
simple code counts how many values land in the tail (below a quantile) of
a standardized normal distribution of random numbers. First two commands
are inputs:
c -0.05
i -2500
Second two commands are formulas I want to
1 - 100 of 157 matches
Mail list logo
