HI,
If I understand your question correctly, this should give the result:
dat1-read.table(text=
A B C
a b a
x y z
,sep=,header=TRUE,stringsAsFactors=FALSE)
within(dat1,{new_column-ifelse(A==C|B==C,y,n)})
# A B C new_column
#1 a b a y
#2 x y z n
A.K.
- Original Message
Hi
I am trying to export dataframes to Stata with stata's labels, I have looked
though the documentation for the foreign library and code for write.dta, (but
not the complied code) and I can't find anything. Ideally what I want is
something like write.dta(myDataFrame, labels = listOfLabels,
Hi,
Try this:
dat1-read.table(text=
id price distance
1 2 4
1 3 5
2 4 8
2 5 9
3 6 3
3 4 8
,sep=,header=TRUE)
dat2-split(dat1,dat1$id)
lapply(dat2,function(x) cor(x[2],x[3],method=spearman))
A.K.
- Original Message -
From: Yi liuyi.fe...@gmail.com
To:
Have a look at the documentation for ifelse()
http://stat.ethz.ch/R-manual/R-devel/library/base/html/ifelse.html that should
do what you need.
Rhydwyn
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sapana Lohani
Sent:
On Aug 28, 2012, at 9:54 PM, arun wrote:
HI,
If I understand your question correctly, this should give the result:
dat1-read.table(text=
A B C
a b a
x y z
,sep=,header=TRUE,stringsAsFactors=FALSE)
within(dat1,{new_column-ifelse(A==C|B==C,y,n)})
# A B C new_column
#1 a b a y
#2 x y z
On Tue, Aug 28, 2012 at 9:05 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 28/08/2012 2:16 PM, Liviu Andronic wrote:
I came up with a modified version of the above:
print_noattr - function(x, keep.some=T, ...){
if(keep.some) xa - attributes(x)[c('names', 'row.names', 'class')]
Dear all
Are LaTeX \subfloat{} commands incompatible with Sweave code? I cannot
get the following code to compile properly:
\begin{table}
\subfloat[asdfa]{=
2+2
@
}
\caption{asdf}
\end{table}
If I replace the Sweave chunk with a random string or a table, the
compilation works fine. Any ideas
On 08/29/2012 01:05 AM, Geophagus wrote:
Hi @ all,
I have a problem concerning the barplot (barchart lattice) of a dataframe. I
work with the attached dataframe.
When I try to plot this dataframe I only get two rows plottet. When I try
to plot the whole dataframe, there is message, that it is
Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit :
Hi Rainer,
You could try:
subs - expression(dead==FALSE recTreat==FALSE)
lme(formula, subset = eval(subs))
Not tested, but something along those lines should work.
Out of curiosity, why isn't subset (and weights, which is
On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr wrote:
Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit :
Hi Rainer,
You could try:
subs - expression(dead==FALSE recTreat==FALSE)
lme(formula, subset = eval(subs))
Not tested, but something along those
Hello,
Function Biostrings::matchPattern can be called with an algorithm =
boyer-moore argument.
I've never used it, this is the return value of
library(sos)
r1 - findFn('boyer')
r2 - findFn('moore')
r1 r2
I have implemented the Boyer-Moore algorithm a couple of times, the
first(!) of all
Hi,
On Wed, Aug 29, 2012 at 6:56 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
Are LaTeX \subfloat{} commands incompatible with Sweave code? I cannot
get the following code to compile properly:
\begin{table}
\subfloat[asdfa]{=
2+2
@
}
\caption{asdf}
\end{table}
If I
On Wed, Aug 29, 2012 at 1:34 PM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
This isn't exactly what you want, but I'm using kintr and building and
saving my figures in the their own chunks then just inlining the
path to the generated figure in the \subloat{..}. Things are working
Hello,
You seem to be misreading the help pages for lm and predict.lm, argument
'terms'.
A much simpler way of solving your problem should be to invert the
fitted model using lm():
model - lm(area ~ concn, data) # Your original model
inv.model - lm(concn ~ area, data = data) # Your
Le mercredi 29 août 2012 à 04:01 -0700, Joshua Wiley a écrit :
On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr
wrote:
Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit :
Hi Rainer,
You could try:
subs - expression(dead==FALSE recTreat==FALSE)
Thanks for your suggestions,
The solution was as proposed by Mr. Nordlund, I had to convert the timestamps
again using as.POSIXct() I will have to remember that anytime I do any kind of
filtering/subselection of dates to reconvert them. Lesson learned.
Also thank you for the dput()
Folks,
I just upgraded my Mac to Mountain Lion and on running R.app I get the
following message:
mach_override: some instructions unknown! Need to update mach_override.c
err = f801
The type=1 importance measure in RF compares the prediction error of each
tree on the OOB data with the prediction error of the same tree on the OOB data
with the values of one variable randomly shuffled. If the variable has no
predictive power, then the two should be very close, and there's
On 24.08.2012 21:53, Ivan Alves wrote:
Hi all,
I am encountering an RODBC problem in R 2.15.1 in windows 64 bit which I do not
encountered in the same set up in windows 32 bit (the latest binary version of
RODBC in both cases from the same depository gotten by
install.packages(‘RODBC’),
Dear all,
Please suggest me how can I do it.
I have a matrix which look like following:
x1 x2 x3 t1 .01 0.3 0 t2 0 0.1 0.01 t3 0 .01 .01 t4 0 0 t5 5 0 0 t6
0 0 0 t7 0 0 0 t8 0 0 0 t9 0.6 0 0 t10 0 0 0.66 t11 0 0.6 0.11 t12 0
0.4 0
I want to sort decreasing order in each column
Perhaps you could supply the matrix using dput() ? It is a handy way to supply
sample data. Just do dput(mydata), copy the results and paste into email.
At the moment your matrix is almost unreadable.
John Kane
Kingston ON Canada
-Original Message-
From: nicome...@gmail.com
Sent:
Thank you - I'll try it today and give you a reply!!
Thanks a lot
Geo
--
View this message in context:
http://r.789695.n4.nabble.com/barchart-with-3-rows-tp4641572p4641697.html
Sent from the R help mailing list archive at Nabble.com.
__
Dear All
I have the following code set up:
Code #1
a -matrix(seq(0,8, by = sign(8-0)*0.25))
b -matrix(seq(8,16, by = sign(16-8)*0.25))
c -runif(1000,50,60)
d -exp(-c*a)+exp(-c*b)
This will give me the obvious error message of lengths not matching. What I am
trying to do here is to have 33
Yes, it seems that Prof Meyer hasn't pushed the changes to R forge
yet.
I now did. (There are no other repositories).
David
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Here's the info Michael Weylandt requested:
installed.packages()[c(lme4,nlme,Matrix),c(2,3,12)]
LibPath Version Built
lme4 /usr/lib/R/site-library 0.999375-40 2.13.1
nlme /usr/lib/R/library 3.1-104 2.15.0
Matrix /usr/lib/R/library 1.0-6
Le mercredi 29 août 2012 à 14:26 +0200, Milan Bouchet-Valat a écrit :
Le mercredi 29 août 2012 à 04:01 -0700, Joshua Wiley a écrit :
On Wed, Aug 29, 2012 at 3:56 AM, Milan Bouchet-Valat nalimi...@club.fr
wrote:
Le mardi 21 août 2012 à 07:51 -0700, Joshua Wiley a écrit :
Hi Rainer,
Hello john,
thanks for the suggestion. Please find an example:
X-matrix(rnorm(5*10),nrow=5)
dim(X)
[1] 5 10
X
[,1] [,2][,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] 1.2774431 -1.2427735 0.81933548 -2.1098586 -1.6726799
On 29-08-2012, at 16:08, Nico Met wrote:
Hello john,
thanks for the suggestion. Please find an example:
X-matrix(rnorm(5*10),nrow=5)
dim(X)
[1] 5 10
X
[,1] [,2][,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] 1.2774431
On Wed, Aug 29, 2012 at 8:11 AM, Scott Raynaud scott.rayn...@yahoo.com wrote:
Here's the info Michael Weylandt requested:
installed.packages()[c(lme4,nlme,Matrix),c(2,3,12)]
LibPath Version Built
lme4 /usr/lib/R/site-library 0.999375-40 2.13.1
nlme
On Wed, Aug 29, 2012 at 7:45 AM, Keith Weintraub kw1...@gmail.com wrote:
Folks,
I just upgraded my Mac to Mountain Lion and on running R.app I get the
following message:
mach_override: some instructions unknown! Need to update mach_override.c
err = f801
Please find the require info:
set.seed(12345)
X-matrix(rnorm(5*10),nrow=5)
dim(X)
X
[,1] [,2][,3] [,4] [,5] [,6]
[,7] [,8] [,9] [,10]
[1,] 1.2774431 -1.2427735 0.81933548 -2.1098586 -1.6726799 -2.2994684
-0.28823228
On Wed, Aug 29, 2012 at 4:23 AM, Liviu Andronic landronim...@gmail.com wrote:
On Tue, Aug 28, 2012 at 9:05 PM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 28/08/2012 2:16 PM, Liviu Andronic wrote:
I came up with a modified version of the above:
print_noattr - function(x, keep.some=T,
On 29/08/2012 12:50 AM, Sam Steingold wrote:
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-08-28 21:06:33 -0400]:
On 12-08-28 5:55 PM, Sam Steingold wrote:
* R. Michael Weylandt zvpunry.jrlyn...@tznvy.pbz [2012-08-28 13:45:35
-0500]:
always you shouldn't need manual garbage collection.
Salma,
I don't know much about survival modelling, but many smart folks do,
so I'm forwarding this back to R-help so it gets a wider audience. In
fact, we generally like to keep replies on list 1) so that they are
archived for future googlers; 2) so that you have access to a wider
pool of
Hi
I am not sure if I understand it correctly but maybe
a -seq(0,8, by = sign(8-0)*0.25)
b -seq(8,16, by = sign(16-8)*0.25)
const -runif(1000,50,60)
const - rep(const, 33)
d -exp(-const*a)+exp(-const*b)
dim(d) - c(33,1000)
gives you matrix 33x1000
Maybe you can simplyfy the example to 10
Hi
But in your example there are no row names? How would you structure the result?
Each column has different order and therefore each column shall have different
sequence of row names?
Send also your responses to r-help as somebody could have better answer for you.
Regards
Petr
From: Nico
It depends very much on what you consider a small amount of error. Unless
you specify starting centroids, K-means does not necessarily produce a
unique partition for a particular data set unless you specify the starting
seeds. In other words, you can get different results using Matlab's kmeans
id price distance
1 2 4
1 35
...
2 4 8
2 5 9
I would like to calculate the rank-order correlation between
price and distance for each id.
cor(price,distance,method = spearman) calculate a
correlation for all.
Try by()
#Example
d - data.frame(g=gl(5, 10),
On 8/24/12 9:59 PM, Yihui Xie x...@yihui.name wrote:
On Fri, Aug 24, 2012 at 3:17 PM, Roebuck,Paul L proeb...@mdanderson.org
wrote:
On 8/24/12 2:59 PM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
On Fri, Aug 24, 2012 at 2:48 PM, Roebuck,Paul proeb...@mdanderson.org
wrote:
Yes that is one possible solution, but the filename is hard-coded
somehow. The key to this problem is a missing new line before =,
which was addressed in
https://github.com/downloads/yihui/knitr/knitr-subfloats.pdf
The LyX and Rnw source files can be checked out with GIT under
Christian,
my late answer on this one:
On 21.08.2012 22:28, Christian Hennig wrote:
Dear list,
I want to update my prabclus package which I haven't done for quite a
while.
In the previous version, I had .dat files in my data subdirectory, which
I read using .R files. Now R CMD check gives
Well, then I guess there is no such function in base R, but it should
be something like escapeshellcmd = function(x)
gsub('[metacharacters]', '\\1', x)
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
Hello,
predict(..., type = terms) returns the predicted regressors multiplied
by the respective beta centered. The centering constant is also returned:
model - lm(area ~ concn, data) # Already run
predict(model, type = terms)
concn
1 -11542.321
2 -8244.515
3 1648.903
4 18137.934
Hello,
Inline.
Em 29-08-2012 16:06, John Thaden escreveu:
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted? If so,
what numbers would I fill it with? Or should my newdata object include
the original data, too? Is
Hello,
I'm trying to transform reaction times which are not normally distributed
to an ex gaussian or an
inverse gaussian distribution, but I don't really know how to use the
exGAUS() function.
Can someone show me a script in which data has been transformed?
Thanks in advance
k
Hi
From: Nico Met [mailto:nicome...@gmail.com]
Sent: Wednesday, August 29, 2012 5:27 PM
To: PIKAL Petr
Cc: r-help
Subject: Re: [R] Sorting of columns of a matrix
Yes, each column shall have different sequence of row names
And what? I do not understand? You still did not describe how do you
Hi,
Can someone help me on the following problem?
I have nearly 103,000 records with 22 variables (some are factors). When I ran
a stepwise glm on the data set, the R program stops with an error message
Error: cannot allocate vector of size 171.3Mb.
I have an R 2.12.0 on my PC (Windows XP)
I think I may be misreading the help pages, too, but misreading how?
I agree that inverting the fitted model is simpler, but I worry that I'm
misusing ordinary least squares regression by treating my response, with
its error distribution, as a predictor with no such error. In practice,
with my
Hi
I found an example in R to create a lagged panel data set which works
fine. The only problem is that it adds the lagged variable as follows
wage2.dat
year personwagelag(wage, -1)
1.1 1980 1 -0.75843997NA
1.2 1981 1 0.27233048
I think that what the OP is looking for comes under the heading of
inverse regression or the calibration problem. One reference
with a simple explanation including confidence intervals is Applied
regression analysis by Draper and Smith. (It's in section 3.2 in
my 3rd edition).
Peter Ehlers
On
Hello
I have a huge data frame with three columns 'Roof' 'Month' and 'Temp'
i want to run analyses on the numerical Temp data by the factors Roof and
Month, separately and together.
For using more than one factor i understand i should use aggregate, but i am
struggling with the tapply for single
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Nico Met
Sent: Wednesday, August 29, 2012 4:25 PM
To: Berend Hasselman
Cc: r-help@r-project.org
Subject: Re: [R] Sorting of columns of a matrix
Please find the require
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted? If so,
what numbers would I fill it with? Or should my newdata object include
the original data, too? Is there another mailing list I can ask?
Thanks,
-John
On Wed,
Yes, each column shall have different sequence of row names
N
On Wed, Aug 29, 2012 at 5:21 PM, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
** **
But in your example there are no row names? How would you structure the
result? Each column has different order and therefore each
HI,
Not sure I understand the last part.
Try this:
dat1-structure(c(0.585528817843856, 0.709466017509524, 0, -0.453497173462763,
0.605887455840393, -1.81795596770373, 0.630098551068391,
-0.276184105225216,
-0.284159743943371, -0.919322002474128, -0.116247806352002,
1.81731204370422,
HI,
I guess this might be what you want.
dat1-structure(c(0.585528817843856, 0.709466017509524, 0, -0.453497173462763,
0.605887455840393, -1.81795596770373, 0.630098551068391,
-0.276184105225216,
-0.284159743943371, -0.919322002474128, -0.116247806352002,
1.81731204370422,
0.370627864257954,
Petr,
thank you very much for the help, I think this will work... It seems like this
will generate 33 columns and 1000 rows with calculated values, which is fine. I
also simply used the function x=d + dnext, which generated another 33 columns
and 1000 rows object with the cell contents added
y -matrix(1:50,ncol=5)
x - rowSums(y 13) # sum of logicals counts number of TRUEs
x # your first 'x':
[1] 3 3 3 4 4 4 4 4 4 4
sum(0.25 * x 0.8) # sum of logicals counts number of TRUEs
[1] 7
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From:
Le mercredi 29 août 2012 à 07:37 -0700, andyspeak a écrit :
Hello
I have a huge data frame with three columns 'Roof' 'Month' and 'Temp'
i want to run analyses on the numerical Temp data by the factors Roof and
Month, separately and together.
For using more than one factor i understand i
Of course:
colnames(dats)[4] - new name
M
On Aug 29, 2012, at 9:34 AM, Alok K Bohara, PhD boh...@unm.edu wrote:
Hi
I found an example in R to create a lagged panel data set which works fine.
The only problem is that it adds the lagged variable as follows
wage2.dat
year
Dear Uwe,
Many thanks for the reply.
On 1, the problem is that RODBC on 32 bit ' interprets' factors correctly,
whereas on 64 bit it gives the error below. On both systems forcing characters
(via colClasses = character in read.csv), results in no problems. I still
see this as a problem of
On Aug 29, 2012, at 8:06 AM, John Thaden wrote:
Could it be that my newdata object needs to include a column for the
concn term even though I'm asking for concn to be predicted?
The new data argument MUST contain a column with the name area. If
it does not hten the original data is used.
On Wed, Aug 29, 2012 at 1:32 PM, Ivan Alves papu...@me.com wrote:
Dear Uwe,
Many thanks for the reply.
On 1, the problem is that RODBC on 32 bit ' interprets' factors correctly,
whereas on 64 bit it gives the error below. On both systems forcing
characters (via colClasses = character in
Hi guys,
I've spent a few days trying to install Rmpi with no luck. I originally
tried using mpich, moved on to mpich2, and then to openmpi. I've gotten
the furthest with openmpi, though am still running into this problem and
can't figure it out. Can someone help!? Thanks so much in advanced.
On Aug 29, 2012, at 7:37 AM, andyspeak wrote:
Hello
I have a huge data frame with three columns 'Roof' 'Month' and 'Temp'
i want to run analyses on the numerical Temp data by the factors
Roof and
Month, separately and together.
For using more than one factor i understand i should use
Hello,
Ok, got it, thanks.
Apparently I was right, the estimator is still x.pred = (y.new - b0)/b1
which can be obtained from predict.lm with the inverse regression model
formula. According to the litterature, the main differences are in the
confidence intervals for the true value of x.pred,
On Aug 29, 2012, at 10:23 AM, Siddeek, Shareef (DFG) wrote:
Hi,
Can someone help me on the following problem?
I have nearly 103,000 records with 22 variables (some are factors).
When I ran a stepwise glm on the data set, the R program stops with
an error message
Stepwise methods are
Dear all
I was wondering what old hacks on this list were thinking about the
shiny new wind map, which The Economist describes as breathtaking
for its elegance and rich data presentation. [1]
What do you think of it? And of authors' other graphs? Can something
similar be done in R?
Regards
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-08-29 10:30:10 -0400]:
On 29/08/2012 12:50 AM, Sam Steingold wrote:
* Duncan Murdoch zheqbpu.qha...@tznvy.pbz [2012-08-28 21:06:33 -0400]:
On 12-08-28 5:55 PM, Sam Steingold wrote:
my observation is that gc in R sucks.
(it cannot
On 2012-08-29 08:45, S Ellison wrote:
id price distance
1 2 4
1 35
...
2 4 8
2 5 9
I would like to calculate the rank-order correlation between
price and distance for each id.
cor(price,distance,method = spearman) calculate a
correlation for all.
Try by()
#Example
d -
HI,
Try this:
wage2.dat-read.table(text=
year person wage lag(wage,-1)
1980 1 -0.75843997 NA
1981 1 0.27233048 -0.75843997
1982 1 -1.58335767 0.27233048
1983 1 0.36805926 -1.58335767
1984 1 -0.52312153 0.36805926
1980 2
I have just installed the latest version of R on a openSUSE 12.1 system
running on an ORacle VM VirtualBox and have encountered a problem with
installing ChemometricsWithR. Here is the output:
library(compiler)
install.packages(ChemometricsWithR)
Installing package(s) into
Hi,
I'm trying to introduce a (spatial) exponential correlation
structure (with range=200 and nugget.effet of 0.3) in a lme model of
this form: lme(ARBUS~YEAR, random=~1|IDSOUS).
The structure of the data is IDSOUS XMIN YMAX YEAR ARBUS
with 2 years of data and 5600 points for each
As the posting guide asked, please discuss this with the maintainer. He
knows about this.
In short: Rmpi does not work if OpenMPI was compiled to dynamically load
extensions, which is nowadays the default installation. If your OpenMPI
is = 1.5 (and for some installs of 1.4.x) you will need
Le mercredi 29 août 2012 à 16:00 -0400, Sabrina Plante a écrit :
Hi,
I'm trying to introduce a (spatial) exponential correlation
structure (with range=200 and nugget.effet of 0.3) in a lme model of
this form: lme(ARBUS~YEAR, random=~1|IDSOUS).
The structure of the data is IDSOUS
Le mercredi 29 août 2012 à 14:12 -0400, Stephen P. Molnar a écrit :
I have just installed the latest version of R on a openSUSE 12.1 system
running on an ORacle VM VirtualBox and have encountered a problem with
installing ChemometricsWithR. Here is the output:
library(compiler)
Hey Yihui
On Wed, Aug 29, 2012 at 6:17 PM, Yihui Xie x...@yihui.name wrote:
Yes that is one possible solution, but the filename is hard-coded
somehow. The key to this problem is a missing new line before =,
which was addressed in
https://github.com/downloads/yihui/knitr/knitr-subfloats.pdf
On 29 August 2012 at 11:37, Linh Tran wrote:
| I've spent a few days trying to install Rmpi with no luck. I originally
| tried using mpich, moved on to mpich2, and then to openmpi. I've gotten
| the furthest with openmpi, though am still running into this problem and
| can't figure it out. Can
I have a dataset w/ 184K obs 16 variables. In SAS I proc sort nodupkey it
in seconds by 11 variables.
I tried to do the same thing in R using both the unique then the
!duplicated functions but it just hangs there I get no output. Does
anyone know how to solve this?
This is how I tried to do
Hi,
I am conducting a survey on how people learn and teach R. I think the results
of the survey could lead to interesting insights that could benefit the R
community, and the educational practices in specific. The data and the results
of the survey will be shared with everyone.
Please help
Hi all,
is there a way to extract the name of a function, i.e. do the reverse
of match.fun applied to a character string? I would like to print out
the name of a function supplied to another function as an argument.
For example:
myFunc = function(x) { x+1 }
applyFunc = function(fnc, x)
{
fnc
Do you know what environments are allowed inside \subfloat{}? The
graphics example works because it is nothing but a simple
\includegraphics{} command. The table example you gave is much more
complicated than that.
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web:
On Wed, Aug 29, 2012 at 3:36 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
Hi all,
is there a way to extract the name of a function, i.e. do the reverse
of match.fun applied to a character string? I would like to print out
the name of a function supplied to another function as an
You can find out which rows of a data.frame called dataFrame
are duplicates of previous rows with
dups - duplicated(dataFrame)
To make a new data.frame without them do
duplessDataFrame - dataFrame[!dups, ]
You could use unique(dataFrame), but, as in your examples, I
think one often wants to
On 12-08-29 04:00 PM, Peter Langfelder wrote:
On Wed, Aug 29, 2012 at 3:36 PM, Peter Langfelder
peter.langfel...@gmail.com wrote:
Hi all,
is there a way to extract the name of a function, i.e. do the reverse
of match.fun applied to a character string? I would like to print out
the name of a
deparse(substitute(fnc)) will get you part way there.
myFunc - function(x) x + 1
applyFunc - function(fnc, x) {
cat(fnc is, deparse(substitute(fnc)), \n)
fnc - match.fun(fnc)
fnc(x)
}
applyFunc(myFunc, 1:3)
fnc is myFunc
[1] 2 3 4
applyFunc(function(x)x*10, 1:3)
fnc is function(x)
On Wed, Aug 29, 2012 at 4:14 PM, William Dunlap wdun...@tibco.com wrote:
deparse(substitute(fnc)) will get you part way there.
...
I like to let the user override what substitute might say by
making a new argument out of it:
applyFunc(function(x)x*10, 1:3)
fnc is function(x) x * 10
[1] 10 20
On 30/08/12 10:36, Peter Langfelder wrote:
Hi all,
is there a way to extract the name of a function, i.e. do the reverse
of match.fun applied to a character string? I would like to print out
the name of a function supplied to another function as an argument.
For example:
myFunc = function(x)
Draper Smith sections (3.2, 9.6) address prediction interval issues, but
I'm also concerned with the linear fit itself. The Model II regression
literature makes it abundantly clear that OLS regression of x on y
frequently yields a different line than of y on x. The example below is not
so
Please can someone advise me how I can adjust correlations using bonferroni's
correction? I am doing manny correlation tests as part of an investigation of
the validity/reliability of a psychometric measure.
Help would be so appreciated!
Cheers,
Louise
Thanks for the input. I wanted to avoid counting the column number.
In any case, in the script -- wage.lag1 = lag(wage, -1) seems to do the
trick.
Alok
On 8/29/2012 12:29 PM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
Of course:
colnames(dats)[4] - new name
M
On Aug 29,
On Wed, Aug 29, 2012 at 8:20 PM, Alok Bohara, PhD boh...@unm.edu wrote:
Thanks for the input. I wanted to avoid counting the column number. In
any case, in the script -- wage.lag1 = lag(wage, -1) seems to do the trick.
The important thing is that 4 in my example below can be any
computable
On Wed, Aug 29, 2012 at 6:23 PM, Louise Cowpertwait
louisecowpertw...@gmail.com wrote:
Please can someone advise me how I can adjust correlations using bonferroni's
correction? I am doing manny correlation tests as part of an investigation of
the validity/reliability of a psychometric
Thank you for the advice. I tried using the command, and it still
wouldn't load.
I tried uninstalling all of the MPI interfaces, reinstalled openmpi
using the --enabled-shared --disable-dlopen command, and Rmpi was able
to install sucessfully inside R.
Thank you to Prof Ripley for pointing
On Aug 29, 2012, at 4:08 PM, John Thaden wrote:
Draper Smith sections (3.2, 9.6) address prediction interval
issues, but
I'm also concerned with the linear fit itself. The Model II regression
literature
Citations?
makes it abundantly clear that OLS regression of x on y
frequently yields
Hi Yihui
For figures in latex my bible is
Keith Reckdahl Using Imported Graphics in LATEX and pdfLATEX Version
3.0.1 January 12, 2006
I think there is a 2009 version but I cannot find it
Below is an example of a 3 X 1 figure
\begin{figure}[h]
\centering
% Figure A
\subfloat[A]{
On Aug 29, 2012, at 4:23 PM, Louise Cowpertwait wrote:
Please can someone advise me how I can adjust correlations using
bonferroni's correction? I am doing manny correlation tests as part
of an investigation of the validity/reliability of a psychometric
measure.
Help would be so
Indeed...
But it seems clear that,
1. This is off topic for this list.
2. Ms. Cowpertwait would be wise to seek the help of a local statistician.
does it not?
Cheers,
Bert
On Wed, Aug 29, 2012 at 6:48 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
On Wed, Aug 29, 2012 at 6:23 PM,
On Wed, Aug 29, 2012 at 6:48 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
On Wed, Aug 29, 2012 at 6:23 PM, Louise Cowpertwait
louisecowpertw...@gmail.com wrote:
Please can someone advise me how I can adjust correlations using
bonferroni's correction? I am doing manny correlation
Hi: Bonferroni can be used for any hypothesis test or confidence interval
where a statistic
is calculated. The idea behind it is that, if a statistic is being
calculated many times ( as in the case of say anova where multiple
differences between groups can be calculated ), then the critical value
1 - 100 of 101 matches
Mail list logo
