On Mon, Jul 2, 2012 at 6:16 PM, Spencer Maynes smayne...@gmail.com wrote:
I have a vector d of unknown length, and a list b of unknown length. I
would like to replace every element of b with d. Simply writing b-d does
not work as R tries to fit every element of d to a different element of d,
Dear list members,
I am trying to create a subset of a data frame based on conditions in two
columns, and after spending much time trying (and search R-help) have not had
any luck. Essentially, I have a data frame that is something like this:
Hi
Dear list members,
I am trying to create a subset of a data frame based on conditions in
two
columns, and after spending much time trying (and search R-help) have
not
had any luck. Essentially, I have a data frame that is something like
this:
date-as.POSIXct(as.character(c
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to achieve the
Sounds like operating system troubles, not R troubles.
If you want help with invoking R, please take a deep breath, read the FAQs and
the posting guide, and tell us exactly what you did at the command line if you
still think R is the problem.
Note that any problems you may be having with the
HI,
I have the price and volume data from own product and competitor's product:
Year_Month Volume own product's price Volume
competitor's price
1 201011 17583469.03 NA
NA
2 201012 33899
Hello,
Try:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Regards
Le 12/07/03 16:18, jin...@ga.gov.au a écrit :
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else
On 07/03/2012 06:21 AM, Joseph Clark wrote:
These carpet plots are also called heat maps and there's a current thread with the
subject line Heat Maps in which I've given a couple of examples of code for them. The R function
image() is very easy to use:
image( x=(x values), y=(y values),
Hello,
Sorry, but it was you that misread some of the suggestions. I have
written raw=TRUE not raw=raw. Just see
m - matrix(1:6, ncol=2) # your example
p2 - poly(m, degree=2, raw=TRUE) # it's raw=TRUE, not raw=raw !!!
deg2 - attr(p2, 'degree') == 2
p2[, deg2]
p6 - poly(m, degree=6,
b - rep(list(d), length(b))
On 02/07/2012 23:16, Spencer Maynes wrote:
I have a vector d of unknown length, and a list b of unknown length. I
would like to replace every element of b with d. Simply writing b-d does
not work as R tries to fit every element of d to a different element of d,
and
Hello,
Inline.
Em 03-07-2012 01:15, jim holtman escreveu:
You will have to change the 'i1' expression as follows:
i1 - grepl(^([0D]|[0d])*$, dd$ch)
i1 # matches strings with d D in them
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
# second string had 'd' 'D' in it
Hello,
This is here for some days now, and I've decided to give it a try.
I've rewritten your fitfunction(), making it simpler. And include the
gamma distribution in the list.
require(MASS)
fitfunction - function(Type, x) list(Type=Type, Fit=fitdistr(x, Type))
fun - function(x, data){
Hi
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to remove the loop from above code to
Hello,
In order to avoid messing up the data, use dput. See below, in the end.
As for your question, try this:
set.seed(1234)
xyz - data.frame(x=sample(20, 10), y=sample(20, 10), z=sample(0:1, 10,
TRUE))
# pch=16 -- solid circle; cex=4 -- 4 fold expansion
with(xyz, plot(x, y, col=z+1,
Do you know how can I run a script on R from Excel without rExcel but with
VBA and batch?
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__
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1 data
converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else a1$h2[i]-0
}
My question: is it possible to
Thanks Joshua for the clear explanation.
I've previously posted in r sig mixed, but I got no response... :(
Cheers,
- Camila
On Mon, Jul 2, 2012 at 9:31 PM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi Camila,
In mixed equation form instead of multilevel, it would be:
Y_it = gamma_00 +
Homework?? We don't do homework here. If not, ?optim or look at the CRAN
Optimize task view for optimizers. There is even a maxLik package that
might be useful.
-- Bert
On Mon, Jul 2, 2012 at 8:58 PM, Ali Tamaddoni alicivilizati...@gmail.comwrote:
Hi All
I have a data frame called nbd
On Mon, Mar 5, 2012 at 2:47 PM, vincent.deluard
vincentdelua...@gmail.com wrote:
Hi Jim,
Please disregard my earlier post -- I have done some research and realized
the space between after program was the the issue. I can now open
RScript.exe from the command prompt using the abbreviated
On Jul 2, 2012, at 10:09 PM, Tjun Kiat Teo wrote:
I am trying to to write a wrapper function for the ode solver (under
the package desolve) to enable it to take multivariate arrays. I know
how to do it for 1 dimension arrays but my code breaks down when I try
to do it for 2 dimensional arrays.
On Jul 3, 2012, at 5:08 AM, Jim Lemon wrote:
On 07/03/2012 05:18 PM, jin...@ga.gov.au wrote:
Hi all,
I would like create a new column in a data.frame (a1) to store 0, 1
data converted from a factor as below.
a1$h2-NULL
for (i in 1:dim(a1)[1]) {
if (a1$h1[i]==H) a1$h2[i]-1 else
Your data is effectively unreadable. Please use dput() to supply sample data.
Here is one way to do what you want for data frames using the ggplot2 package.
library(ggplot2)
mydata - data.frame(x = 1:10, y = rnorm(10), z = c(rep(1,4), rep(2, 6)))
ggplot(mydata, aes(x, y, colour= as.factor(
Any sample data for us to work with? See ?dput for a good method of supplying
sample data.
John Kane
Kingston ON Canada
-Original Message-
From: denissearchun...@yahoo.com.mx
Sent: Mon, 2 Jul 2012 14:04:53 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] how to do a graph with
Anything here that might help
http://learnr.wordpress.com/2010/01/26/ggplot2-quick-heatmap-plotting/
John Kane
Kingston ON Canada
-Original Message-
From: joeclar...@hotmail.com
Sent: Mon, 2 Jul 2012 13:21:25 -0700
To: mueller.eisb...@googlemail.com, r-help@r-project.org
Subject:
{ I think this message got rejected at the 1st attempt - trying again}
R 2.15.1 , windows XP
I have a very non-stationary bivariate time-series - say {xt,yt} t=1 ...
lots.
I want to do a bivariate density contour-plot of the whole series and then step
through the series 1 second at a
Dear all,
I am trying to write figures directly to a file using the jpeg() function.
this_ylab=expression(paste(P,sep=)~lambda)
this_xlab=expression(rho)
jpeg(file_name.jpeg,width=100,height=100,units=mm,res=300)
And one more alternative:
a1$h2 - apply(a1,1, function(x) if (x[h1]==H) 1 else 0 )
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hi
I have a data like thursday, November 20,2012. I'm not able to convert
into data format in R
Can anyone please help
-
Thanks in Advance
Arun
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Sent from the R help mailing list
I have already fitted several models
using R code; arima(rates,c(p,d,q))
As I heard, best model produce the
smallest AIC value, but maximum likelihood estimation procedure optimizer
should converge.
How to check whether maximum likelihood estimation procedure optimizer has
converged or not?
Hi,
If you want to assign a vector to every element of a list,
vec1-11:20
list1-split(LETTERS[1:10],1:length(LETTERS[1:10]))
list2-lapply(1:10,function(x) vec1)
or,
list3-lapply(list1,function(x) list1=vec1)
or
list4-list()
vec2-1:5
list4[1:length(list1)]-list(vec2)
# if you want to assign
Hello
I have dataframes.
mydata1 -data.frame(value=c(15,20,25,30,45,50),dates=c(2005-05-25 07:00:00
,2005-05-25 19:00:00,2005-06-25 07:00:00,2005-06-25 19:00:00
,2005-07-25 07:00:00,2005-8-25 19:00:00))
or
mydata2 -data.frame(value=c(15,20,25,30,45,50),dates=c(2005-05-25 00:00:00
,2005-05-25
In R you should slashes instead of backslashes:
C:\PROGRA~1\R\R-2.11.1\bin\RScript.exe
C:/Users/Vincent/Documents/temp/test.r
Bart
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R Graph Gallery contains lots of plot examples with source code:
http://addictedtor.free.fr/graphiques/
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That I correct. I have only 9.
Thanks for the explanation :)
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__
hi list,
used versions: 2.12.1 and 2.14.0 under ubuntu and macosx.
I recently stumbled over a problem with `nls', which occurs if the model
is not specified explicitly but via an evaluation of a 'call' object.
simple example:
Try using polygon() or grid.polygon() from the grid package.
You can find code example on the R project title page.
Go to http://r-project.org and click on the plot (the example plot showing
PCA, clustering and factors, lower left are two graphs, similar to what you
want).
This will bring you
Thank you for your help
So i have tried many ways on different computers, but i believe i have an
operating system, as I open RScript up in my local directory it just
flashes at me and them disappears
I tried to task schedule, using the following code
C:\Program
Dear,
I would like to use the siar program, when I lodging package the message
appear, so I couldn't use the related calculations.
The following object(s) are masked from package:spatstat:convexhull
Could you please write me what shoud I do?
Sukran Yalcin Ozdilek
[[alternative HTML
On Jul 3, 2012, at 4:01 AM, georgeshirreff wrote:
Dear all,
I am trying to write figures directly to a file using the jpeg()
function.
this_ylab=expression(paste(P,sep=)~lambda)
this_xlab=expression(rho)
jpeg(file_name.jpeg,width=100,height=100,units=mm,res=300)
Try instead something
On 03/07/2012 10:36 AM, Robert Douglas Kinley wrote:
{ I think this message got rejected at the 1st attempt - trying again}
R 2.15.1 , windows XP
I have a very non-stationary bivariate time-series - say {xt,yt} t=1 ...
lots.
I want to do a bivariate density contour-plot of the whole
Hello,
Try, with 'x' your date(s),
as.Date(x, format=%A, %B %d,%Y)
strptime(x, format=%A, %B %d,%Y)
Note: this is in the help page for ?strptime
Hope this helps,
Rui Barradas
Em 03-07-2012 07:04, arunkumar escreveu:
hi
I have a data like thursday, November 20,2012. I'm not able to
Hello,
Inline.
Em 03-07-2012 09:22, Sajeeka Nanayakkara escreveu:
I have already fitted several models
using R code; arima(rates,c(p,d,q))
As I heard, best model produce the
smallest AIC value, but maximum likelihood estimation procedure optimizer
should converge.
How to check whether maximum
Many thanks for these ideas ... I'll try them, and report back
Cheers Bob Kinley
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 03 July 2012 15:54
To: Robert Douglas Kinley
Cc: r-help@r-project.org
Subject: Re: [R] saving contour() plot info
Hello,
The trick is to use seq() for date classes.
First of all, when creating data.frames use stringsAsFactors=FALSE, in
order not to convert them to factors. I've added this option to your
data.frame() instructions. And then,
mydata1$dates - as.POSIXct(mydata1$dates)
mydata1b$dates -
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly because there
are no instances of 'Dd' or 'dD' in the data set (that I would/not want to
include/exclude)... but I understand that 'i1new' targets precisely what I
want.
Why isn't a leader of zero's required for either 'i1' or 'i1new',
Hello,
I'm glad it helped. See answer inline.
Em 03-07-2012 17:09, Claudia Penaloza escreveu:
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
because there are no instances of 'Dd' or 'dD' in the data set (that I
would/not want to include/exclude)... but I understand that 'i1new'
On Jul 3, 2012, at 4:43 AM, cindy.dol wrote:
Do you know how can I run a script on R from Excel without rExcel
but with
VBA and batch?
It would seem that this question should be directed to a VBA mailing
list.
--
David Winsemius, MD
West Hartford, CT
I did not suggest that you don't HAVE an operating system... simply that you
don't know how to use yours.
However you accomplish starting programs automatically, you WILL need to
specify both the R interpreter (which you do seem to be accomplishing) and the
name of the script you wish to run
Thanks guys for the help, I'm going to go with Patrick Burns answer because
it seems to work the best for my situation, but these all seem like they
should work.
On Tue, Jul 3, 2012 at 2:51 AM, Patrick Burns pbu...@pburns.seanet.comwrote:
b - rep(list(d), length(b))
On 02/07/2012 23:16,
Dear David,
In fact just updating R seems to have fixed the problem. There's a lesson.
Thanks a lot,
George
On 3 July 2012 17:02, David Winsemius [via R]
ml-node+s789695n4635292...@n4.nabble.com wrote:
On Jul 3, 2012, at 4:01 AM, georgeshirreff wrote:
Dear all,
I am trying to write
Dear all,
I produced the following graph with ggplot which is almost fine, yet I don't
like that the legend for Means and Observations includes a line, though no
line is used in the plot for those two (the line for Overall Mean on the
other hand is wanted):
library(ggplot2)
ddf - data.frame(x
Hi,
Hope this helps.
date1- c(thursday November 20, 2012, friday November 21, 2012, saturday
November 22, 2012)
date2- as.Date(date1, format= %A %B %d, %Y)
date2
[1] 2012-11-20 2012-11-21 2012-11-22
A.K.
- Original Message -
From: arunkumar akpbond...@gmail.com
To:
Hi everyone.
I have these data :
myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type
= c('x','x','y','z'))
which gives me:
Name length type
1a 1x
2a 2x
3b 3y
4b 4 z
I would group (mean) this DF using 'Name' as
#I have a dataframe called tests that contain character expressions. These
characters are rules that use data from within another dataframe. Is there
any way within R I can access the rules in the dataframe called tests, and
then evaluate these rules?
#An example may better explain what I am
I have water chemistry data with censored values (i.e., those less than
reporting levels) in a data frame with a narrow (i.e., database table)
format. The structure is:
$ site: Factor w/ 64 levels D-1,D-2,D-3,..: 1 1 1 1 1 1 1 1 ...
$ sampdate: Date, format: 2007-12-12 2007-12-12 ...
$
Hi guys,
I'm trying to use the the integral function to estimate the area under a
PDF and a crossing curve. first I stated the function with several vectors
in it:
fn=function(a,b,F,mu,alpha,xi)
{
x-vector()
fs-function(x)
{
c - (mu+(alpha*(1-(1-F)^xi)/xi))
tmp - (1 + (xi * (x - mu))/alpha)
On 03/07/2012 06:56, Erin Hodgess wrote:
Dear R People:
I'm back to installing R from source, this time on a 64 bit machine.
What OS?
I'm using the R-Patched.tar.gz as my source.
When I have the openmp option set to -fopenmp, I get the error that
libgomp.spec is not found. Ok, so I
Al Ehan wrote
Hi guys,
I'm trying to use the the integral function to estimate the area under a
PDF and a crossing curve. first I stated the function with several vectors
in it:
fn=function(a,b,F,mu,alpha,xi)
{
x-vector()
fs-function(x)
{
c - (mu+(alpha*(1-(1-F)^xi)/xi))
tmp -
Hi,
Glad all of them worked. In my reply to you, my first solution was:
list2-lapply(1:10,function(x) vec1)
The more generic form should be:
list2-lapply(1:length(list1),function(x) vec1)
A.K.
- Original Message -
From: Spencer Maynes smayne...@gmail.com
To: r-help@r-project.org
Cc:
I have a general question about coefficients estimation of the mixed model.
I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b follows
N(0,\psi) #i.e. bivariate normal
where b is the latent variable, Z and X are
Dear All,
have a general question about coefficients estimation of the mixed model.
I simulated a very basic model: Y|b=X*\beta+Z*b +\sigma^2* diag(ni);
b follows
N(0,\psi) #i.e. bivariate normal
where b is the latent variable, Z
On 02.07.2012 22:48, Mathias Worni wrote:
Sorry for the misunderstanding. What I meant to say is that I cannot get
the code for the specific graphic that I am running, so that I can keep
my code reproducible.
Which code for the graphics?
To be reproducible, you need the R vesion, the package
Dear all,
I have an excel file that contains 6 sheets
1,2,3,4,5,6
The analysis is repeated every 3 sheets
Sheets 1, 2, 3:
I want to add (horizontally) the data contained in the matrix : sheet2
(5:end,3:end )
of *Sheet2 * to the sheet3 such that the first element of the matrix
*sheet2
On 2012-07-03 09:47, Spencer Maynes wrote:
Thanks guys for the help, I'm going to go with Patrick Burns answer because
it seems to work the best for my situation, but these all seem like they
should work.
Patrick's solution is similar to Gabor's, but, personally,
I favour Gabor's. Seems
Got it! Thank you Rui!
cp
On Tue, Jul 3, 2012 at 10:14 AM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
I'm glad it helped. See answer inline.
Em 03-07-2012 17:09, Claudia Penaloza escreveu:
Thank you Rui and Jim, both 'i1' and 'i1new' worked perfectly
because there are no instances
I would like to remove a loop to speed up my code.
I want to remove a loop which references the last row.
In general I want to a remove a loop which looks something like this:
for 2 to number of rows in a matrix do{
if indextrow-1 is currentIndexRow then do something.
}
My R code:
Hello,
I've changed the way you create data.frame 'tests', like it was the
conditions were factors, which are coded as integers. We need them of
class character to be parsed. Note that the same could be done with the
data.frame 'info' but it's not absolutely needed.
tests -
I've taken another look, and I actually think you're right. I'm going with
Gabor's. Changing d to a list is such a simple solution that I can't
believe I didn't try it earlier.
On Tue, Jul 3, 2012 at 1:39 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2012-07-03 09:47, Spencer Maynes wrote:
Vectorize vectorize vectorize!
if(x[-length(x)] x[-1]) {...}
(where x is the whole vector of entries)
Bill Dunlap has posted some elegant code within the last month or 2 aimed
at this sort of thing, so search on his posts in the archive.
-- Bert
On Tue, Jul 3, 2012 at 12:10 PM, jcrosbie
Of course it _should_ be:
ifelse(x[-length(x)] x[-1], ...,...)
Sorry...
-- Bert
On Tue, Jul 3, 2012 at 1:00 PM, Bert Gunter bgun...@gene.com wrote:
Vectorize vectorize vectorize!
if(x[-length(x)] x[-1]) {...}
(where x is the whole vector of entries)
Bill Dunlap has posted some
Hello,
I want to create a design matrix using R. Can you explain the code which
creates the following please? I understand the first part.
b=g1(?) does what?
dd - data.frame(a = gl(3,4), b = gl(4,1,12)) # balanced 2-way
dd
a b
1 1 1
2 1 2
3 1 3
4 1 4
5 2 1
6 2 2
7 2 3
8 2 4
9 3 1
1. You need to learn to use R Help. It is there for a purpose.
help (help)
## or
?help
is where to start.
2. Before posting further, please read An Introduction to R. Ships with
every distro.
3. ?gl
4. This is not the best way to do this anyway.
dd - expand.grid(b=1:4,a=1:3) ## is preferable
On Jul 3, 2012, at 4:10 PM, mms...@comcast.net wrote:
Hello,
I want to create a design matrix using R. Can you explain the code
which creates the following please? I understand the first part.
What first part if the next part is ... this ?
b=g1(?) does what?
That's just generating a
On 03/07/2012 4:10 PM, mms...@comcast.net wrote:
Hello,
I want to create a design matrix using R. Can you explain the code which
creates the following please? I understand the first part.
b=g1(?) does what?
That is gl, not g1 (i.e. gee ell not gee one).See ?gl for a
description.
Hi,
Try this:
data1-data.frame(date,time,count)
dat1-data1[with(data1,rev(order(count))),]
data2-subset(dat1,rle(dat1$count)$lengths==1)
dat3-aggregate(data2$count,list(data2$date),max)
colnames(dat3)-c(date,count)
data4-merge(dat3,data2)
data4-data4[,c(1,3,2)]
data4
date
Marck,
A little late, but perhaps this will help someone in the future. I am
guessing that some of your integer fields contain scientific notation, and
for some reason read.table is not interpreting those as integers. Consider
changing the affected column classes from integer to numeric and I
If David's response is what you were seeking, I would then ask: homework??
In general, explicit model matrices are not needed for linear modeling in R.
Still not clear what you meant by reparametrization, but if David's
interpretation is correct, my impossible comment is clearly wrong.
-- Bert
try this:
myData = data.frame(Name = c('a', 'a', 'b', 'b'), length = c(1,2,3,4), type
+ = c('x','x','y','z'))
result - do.call(rbind, lapply(split(myData, myData$Name), function(.name){
+ data.frame(Name = .name$Name[1L]
+ , length = mean(.name$length)
+ , type = if (all(.name$type[1L] ==
On Tue, Jul 3, 2012 at 12:41 PM, jimmycloud jimmycl...@gmail.com wrote:
I have a general question about coefficients estimation of the mixed model.
I have 2 ideas for you.
1. Fit with lme4 package, using the lmer function. That's what it is for.
2. If you really want to write your own EM
Hi all,
Could you please help me?
I am looking for books/pointers/resources/tutorials on visualizing
complex/big data and on understanding multivariate relations in complicated
data.
More specifically, we have categorical variables and are interested in how
to visualize the categorical data and
On Tue, 3 Jul 2012, Michael wrote:
I am looking for books/pointers/resources/tutorials on visualizing
complex/big data and on understanding multivariate relations in
complicated data.
Michael,
You need to become familiar with the works of Edward Tufte, the dean of
complex data
I found this [1] book interesting. About big data It really depends from
a number of things... if can help, I know hdf5 work pretty Well with huge
dataset .
[1] http://www.ggobi.org/book/index.html
On Jul 3, 2012 7:14 PM, Michael comtech@gmail.com wrote:
Hi all,
Could you please help
Thank you all for providing various alternatives. They are all pretty fast.
Great help! Based on a test of a dataset with 800,000 rows, the time used
varies from 0.04 to 11.56 s. The champion is:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Regards,
Jin
Geoscience Australia Disclaimer: This e-mail (and
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty fast.
Great help! Based on a test of a dataset with 800,000 rows, the time used
varies from 0.04 to 11.56 s. The champion is:
a1$h2 - 0
a1$h2[a1$h1==H] - 1
Interesting. My
Le 04/07/2012 12:43, Peter Ehlers a écrit :
On 2012-07-03 17:23, jin...@ga.gov.au wrote:
Thank you all for providing various alternatives. They are all pretty
fast. Great help! Based on a test of a dataset with 800,000 rows, the
time used varies from 0.04 to 11.56 s. The champion is:
a1$h2 -
Hi
I have monthly data and the dates are in MM/YY Format
I need to convert them into DD/MM/YY format by pasting 01 in place of DD to
all the observations in my Year Column
ex:
YearStock Prices
01/2000 1
02/2000 2
03/2000 3
I need to convert them to
Thanks for your validation. Yes Peter's solution is the fastest, faster than
the previous one by saving 25% time. It was missed out in my previous testing.
Jin
-Original Message-
From: Pascal Oettli [mailto:kri...@ymail.com]
Sent: Wednesday, 4 July 2012 2:07 PM
To: Li Jin
Cc:
?paste
Please (re-) read the Introduction to R document supplied with the software
for faster answers.
Also, please read the Posting Guide mentioned at the bottom of every message on
this list. In particular, providing data in raw tabular form is often
ambiguous, and the use of the dput
Hi everyone I
have data on stock prices and market indices
and I need to run a seperate regression of every stock on market
so I want to write a for loop so that I wont have to write codes again
and again to run the regression...
my data is in the format given below
Date Stock1
-- Forwarded message --
From: Akhil dua akhil.dua...@gmail.com
Date: Wed, Jul 4, 2012 at 10:33 AM
Subject:
To: r-help@r-project.org
Hi everyone I
have data on stock prices and market indices
and I need to run a seperate regression of every stock on market
so I want to write a
Homework? (We don't do homework here).
-- Bert
On Tue, Jul 3, 2012 at 10:08 PM, Akhil dua akhil.dua...@gmail.com wrote:
-- Forwarded message --
From: Akhil dua akhil.dua...@gmail.com
Date: Wed, Jul 4, 2012 at 10:33 AM
Subject:
To: r-help@r-project.org
Hi everyone I
have
Dear R helpers,
I am using Beta distribution to generate the random no.s (recovery rates in my
example). However, each time I need to save these random no.s in a csv format.
To distinguish different csv files, one way I thought was use of Sys.time in
the file name. My code is as follows -
#
?lm
and note in particular the section beginning If response is a matrix...
-- Bert
On Tue, Jul 3, 2012 at 10:08 PM, Akhil dua akhil.dua...@gmail.com wrote:
-- Forwarded message --
From: Akhil dua akhil.dua...@gmail.com
Date: Wed, Jul 4, 2012 at 10:33 AM
Subject:
To:
On 3 July 2012 22:03, Akhil dua akhil.dua...@gmail.com wrote:
and I need to run a seperate regression of every stock on market
so I want to write a for loop so that I wont have to write codes again
and again to run the regression...
1. Do give a subject line -- a blank one is commonly used
Hello,
Try something like that:
lgd - format(Sys.time(), %Y_%m_%d_%H_%M_%S)
Regards,
Pascal
Le 04/07/2012 14:21, Vincy Pyne a écrit :
Dear R helpers,
I am using Beta distribution to generate the random no.s (recovery rates in my
example). However, each time I need to save these random
You forgot to follow the posting guide and tell us what operating system you
are using (sessionInfo), but I am going to guess that you are on Windows where
the colon (:) is an illegal symbol in filenames. Try formatting the time
explicitly in the conversion to character using the format string
Hi,
A few comments. First a for loop is probably not optimally efficient.
Consider instead (using a bulit in example dataset):
lm(cbind(mpg, hp) ~ cyl + vs, data = mtcars)
which gives:
Call:
lm(formula = cbind(mpg, hp) ~ cyl + vs, data = mtcars)
Coefficients:
mpg hp
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