Maybe efficiency is less of an issue than I thought, on 2 million
rows, it only took a bit over a minute, and my system only jumped up ~
600 MB of memory so it wasn't much of a strain there either.
> data.test <- matrix(
+ sample(seq(min(yourdata),max(yourdata)), size = 1000, replace = TRUE),
Hi,
Here is a little function that will do what you want and return a nice output:
#Function To calculate top two values and return
my.finder <- function(mydata) {
my.fun <- function(data) {
strongest <- which.max(data)
secondstrongest <- which.max(data[-strongest])
strongestantenna
try this:
> x <- read.table(textConnection(" value0 value60 value120 value180 value240
> value300
+ 1 -13007 -11707 -11072 -12471 -12838 -13357
+ 2 -12838 -13210 -11176 -11799 -13210 -13845
+ 3 -12880 -11778 -3 -12439 -13089 -13880
+ 4 -12805 -11653 -11071
try this:
> x <- read.table(textConnection(" GENEID col1 col2
> col3col4
+ G234064 1 0 0 0
+ G234064 1 0 0 0
+ G234064 1 0
Is there any function/way to merge/unite the following data
GENEID col1 col2 col3col4
G234064 1 0 0 0
G234064 1 0 0 0
G234064 1
I have a data frame with a couple million lines and want to retrieve the
largest and second largest values in each row, along with the label of the
column these values are in. For example
row 1
strongest=-11072
secondstrongest=-11707
strongestantenna=value120
secondstrongantenna=value60
Below
Dear community,
I'm currently attempting to predict the occurence of an event (factor)
having more than 2 levels with several continuous predictors. The model
being ordinal, I was waiting the glm function to return several intercepts,
which is not the case when looking to my results (I only have o
Hi, I'm looking for a way to use densityplot in the lattice package on Date
data or manipulate the labels for tick marks.
Here's an example:
#generate psuedo dates
dat.Date <- as.Date('2009-12-31')+as.integer(abs(rnorm(1))*100)
#convert to a julian date for densityplot
#because it doesn'
Hi everyone,
Is there any command for updating table withing a loop? For instance, at i, I
have a table as ZZ = table(data.raw[1:ind[i]]) where "ind" = c(10, 20, 30,
...). Then , ZZ will be as follow
"A" "B" "C"
3 10 2
At (i + 1), ZZ = table(data.raw[(ind[i]+1):ind[i+1]])
"A" "B" "D"
Try using odbcConnectExcel2007, instead of odbcConnectExcel.
That has worked for me.
Cheers, Mike.
> Message: 60
> Date: Thu, 22 Jul 2010 09:14:23 -0700 (PDT)
> From: "Jack T."
> To: r-help@r-project.org
> Subject: [R] 64 bit use of odbcConnectExcel
> Message-ID: <1279815263840-2298927.p...@
On 23/07/2010 7:39 AM, james.fo...@diamond.ac.uk wrote:
Dear RGL experts,
I haven't been able to add greek letters to my rgl plot3d.
I have tried "expression" with no success.
Here is the interested bit:
> library(rgl)
> cb <- cube3d()
> plot3d(cb,xlab=expression(alpha),ylab="",zlab="",box=FALS
Dear R list,
I use the constrOptim to maximize a function with four constriants but the
answer does not leave from the starting value and there is only one outer
iteration. The function is defined as follows:
tm<-function(p){
p1<-p[1]; p2<-p[2]; p3<-p[3];
p4<-1-p1-p2-p3;
p1*p2*p3*p4}
##the con
Seems to be a relevant and reasonable question to me.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Thanks in advance!
A=c(1, 2,3)
B=c (9, 10, 11, 12)
I want to get C=c(1*9, 1*10, 1*11, 1*12, ., 3*9, 3*10, 3*11, 3*12)?
C is still a vector with 12 elements
Is there a way to do that?
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-calculate-the-product-of-every-two-el
On Jul 23, 2010, at 10:29 PM, jim holtman wrote:
Well, I took you equation and put the following at the start:
Power <- function(x,y) x^y
EE <- function(x) x
alp <- 2
x <- 900*Power(-0.2030178326474623 + 0.23024073983368956*(1 -
alp) +
0.2807352820970084*(1 - alp)*(1 - alp*(1 + EE(1
Well, I took you equation and put the following at the start:
Power <- function(x,y) x^y
EE <- function(x) x
alp <- 2
x <- 900*Power(-0.2030178326474623 + 0.23024073983368956*(1 - alp) +
0.2807352820970084*(1 - alp)*(1 - alp*(1 + EE(1))) +
0.2145643524071315*(1 - alp)*
Power(1 - alp*(
As an alternative to Latent GOLD, I'm wondering if anyone knows of and R
package that can manage Latent Class Analysis with mixed variable types
(continuous, ordinal, and nominal/binary).
[[alternative HTML version deleted]]
__
R-help@r-project.
Dear UseRs!,
Everything about UseR! 2010 was terrific! I really mean "everything" - the
tutorials, invited talks, kaleidoscope sessions, focus sessions, breakfast,
snacks, lunch, conference dinner, shuttle services, and the participants. The
organization was fabulous. NIST were gracious hosts
I had addressed a problem similar to this only a few days ago. Please
see the following URL:
http://tolstoy.newcastle.edu.au/R/e11/help/10/07/1677.html
On Fri, 2010-07-23 at 08:45 -0400, nuncio m wrote:
> I have the following code to write the output from auto.arima function. The
> issue is not
Have you read AN INTRODUCTION TO R?
?%in%
Bert Gunter
Genentech Nonclinical Statistics
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of chipmaney
> Sent: Friday, July 23, 2010 3:39 PM
> To: r-help@r-project.org
> Subject: [R]
Try
Empirical.df[Empirical.df$ID %in% Samples.v,]
See ?"%in%" for more informartion.
HTH,
Jorge
On Fri, Jul 23, 2010 at 6:38 PM, chipmaney <> wrote:
>
> I have a dataframe:
>
> Empirical.df <- data.frame(ID=c("MCUP1-2","MCUP2-5", "MCUP3-3",
> "MCUP4-3","MCUP5-9"), Cover=c(60,40,45,68,72))
>
>
I have a dataframe:
Empirical.df <- data.frame(ID=c("MCUP1-2","MCUP2-5", "MCUP3-3",
"MCUP4-3","MCUP5-9"), Cover=c(60,40,45,68,72))
However, I only want to use a subset of Cover.df for my analysis. The
samples I want to use are:
Samples.v <- c("MCUP1-2", "MCUP4-3","MCUP5-9")
How do I use index
On Jul 23, 2010, at 6:18 PM, chipmaney wrote:
Here is an example dataset:
ZoneCover.df<- data.frame(Value=c(1,2))
row.names(ZoneCover.df) <- c("Floodplain1.Tree", "Floodplain1.Shrub")
I want to Export the Row.Names to a column in the dataframe:
ZoneCover.df$ID <- names(ZoneCover.df)
which
Here is an example dataset:
ZoneCover.df<- data.frame(Value=c(1,2))
row.names(ZoneCover.df) <- c("Floodplain1.Tree", "Floodplain1.Shrub")
I want to Export the Row.Names to a column in the dataframe:
ZoneCover.df$ID <- names(ZoneCover.df)
which yields this:
> ZoneCover.df
V
Here's one way - you can reorder the levels of a factor variable by specifying
the levels the way you want them.
> require("lattice")
Loading required package: lattice
> bwplot( conc ~ Type : Treatment, data = CO2 )
> str(CO2$Type)
Factor w/ 2 levels "Quebec","Mississippi": 1 1 1 1 1 1 1 1 1 1
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
anova(sr.reg.s4.nore)
Df Deviance Resid. Df-2*
The functions {summary,bandwidth}.rq() in the package quantreg looks
promising, but it is not really my area...
http://lmgtfy.com/?q=Hall-Sheater+sparsity+%22r-project%22
Hope this helps a little.
Allan
On 23/07/10 21:02, Nathalie Gimenes wrote:
Hi All,
I would like to estimate the "sparsit
Hannah: I am not sure if this is what you
need but you can use an array to do that.
Copy and paste the below code to your latex code.
\newpage
\begin{landscape}
\begin{figure}[h]
\begin{center}$
\begin{array}{cc}
\includegraphics[width=2in]{yourgraphicname} &
\includegraphics[width=2in]{yourgraphi
Hi All,
I would like to estimate the "sparsity function" using Hall Sheater
bandwidth.
Is there any command for this?
Thanks,
NGS
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https://stat.ethz.ch/mailman/listinfo
to make the desired within R's plotting device rather than in latex try:
par(mfrow = c(3, 2))
and you'll probably want to adjust other mar, parameters as well (e.g.
mar, cex.lab, etc).
Or, take advantage of the flexibility offered by the graphics package
by studying ?layout
hth,
Kingsford
On
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the following:
> anova(sr.reg.s4.nore)
Df Deviance Resid. Df-2*LL P(>|Chi|)
NULL NA NA45 33.89
I found the user guide available at Help > Main > User Guide > Html to
be quite useful for getting started. I don't think this was available
when first I looked at Tinn-R a couple years ago, at which point I too
felt the help files were very poor. But under 2.3.5.2 the file is
there, and Tinn-R i
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Hi,
I am trying to fit the following model:
sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Next time include a reproducible example. i.e. something we can run.
Now, Google "Hauck Donner Effect" to und
http://nixtricks.wordpress.com/2009/11/09/latex-multiple-figures-under-the-same-caption-using-subfigure/
"It will create two rows of subfigures with two subfigures on each row"
On Fri, Jul 23, 2010 at 6:43 AM, li li wrote:
> Hi all,
> I want to add 6 plots in the format of 2 columns and 3 rows
On Fri, Jul 23, 2010 at 2:07 PM, Dimitri Liakhovitski
wrote:
> David/Gabor,
>
> you helped me a lot.
> Gabor - I've run table(OrigData$Month) - and it looked weird.
> I went back to my file and changed the format of the date (Month) in
> Excel. I have no idea what it does - but after I saved it ag
David/Gabor,
you helped me a lot.
Gabor - I've run table(OrigData$Month) - and it looked weird.
I went back to my file and changed the format of the date (Month) in
Excel. I have no idea what it does - but after I saved it again, it
worked.
Thanks a lot!
I'll open another thread where I'll ask ho
On Jul 23, 2010, at 1:50 PM, Dimitri Liakhovitski wrote:
I am expecting to see the week names as row labels of z and the
corresponding values (like in the "monthly" example). I am pretty sure
- in order to get it one needs to install the latest version of zoo.
I've done it just a couple of days
On 2010-07-23 11:36, Deepayan Sarkar wrote:
On Fri, Jul 23, 2010 at 10:02 AM, Eck, Bradley J
wrote:
Dear list:
I'm using bwplot to compare concentrations by location and treatment as in:
# using built in data
bwplot( conc ~ Type : Treatment, data = CO2 )
I would like the order of the plots
Hi,
I am trying to fit the following model:
sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm),
data=bip.surv)
Where age_sym4 is the age that a subject develops clinical thought
problems; sym4 is whether they develop clinical thoughts problems (0 or
1); and lifedxm is mothe
On Fri, Jul 23, 2010 at 1:39 PM, Dimitri Liakhovitski
wrote:
> Very sorry - I mistunderstood and confused split with index.column -
> totally my fault.
> Ok, now I've run this line:
>
> z <- read.zoo(OrigData, index.column = 2, split = "Brand")
>
> And I am getting:
> Error in merge.zoo(` Plus` =
I am expecting to see the week names as row labels of z and the
corresponding values (like in the "monthly" example). I am pretty sure
- in order to get it one needs to install the latest version of zoo.
I've done it just a couple of days ago.
I am getting the error - and nothing is produced. Can i
On Jul 23, 2010, at 1:39 PM, Dimitri Liakhovitski wrote:
Very sorry - I mistunderstood and confused split with index.column -
totally my fault.
Ok, now I've run this line:
z <- read.zoo(OrigData, index.column = 2, split = "Brand")
And I am getting:
Error in merge.zoo(` Plus` = c(NA, 98L, 95L,
Very sorry - I mistunderstood and confused split with index.column -
totally my fault.
Ok, now I've run this line:
z <- read.zoo(OrigData, index.column = 2, split = "Brand")
And I am getting:
Error in merge.zoo(` Plus` = c(NA, 98L, 95L, 97L, NA, 98L, 97L, 98L, NA, :
series cannot be merged wit
On Fri, Jul 23, 2010 at 10:02 AM, Eck, Bradley J
wrote:
> Dear list:
>
> I'm using bwplot to compare concentrations by location and treatment as in:
>
> # using built in data
> bwplot( conc ~ Type : Treatment, data = CO2 )
>
> I would like the order of the plots to be: 3,4,1,2. I can't seem to f
Hello,
2010/7/23 jim holtman :
> It would be nice if you could post what the data looks like that you
> want to import. R can import any text file and then you have string
> manipulation that you can do to parse it. So the basic answer is
> probably yes, but we do need to understand the format o
But, but, but Did you read my message about the need to correctly
specify index columns?
The problem is that read.zoo is reading your first column as an index
and it's actually the second column that should be used for that
purpose.
--
David.
On Jul 23, 2010, at 1:01 PM, Dimitri Liak
Dear list:
I'm using bwplot to compare concentrations by location and treatment as in:
# using built in data
bwplot( conc ~ Type : Treatment, data = CO2 )
I would like the order of the plots to be: 3,4,1,2. I can't seem to figure
this out with index.cond or permc.cond.
Any help is appreciate
Strange, I did attach. Attaching again. Maybe the file just doesn't go through?
I have:
names(OrigData):
[1] "Brand" "Month" "Value"
I read ?read.zoo
According to that index should be the column number.
I thought it should be split = 1 in my case - because I am splitting by Brand.
But neither spl
Google on "tinn-r tutorial." However, I believe that you will find that the
answer is no, there isn't anything useful out there. Moreover, TINN-R's help
files are minimal and terrible, even though the package itself is quite
useful (I find). Please Do NOT post further TINN-R questions to this list
?read.zoo
You didn't specify the index column correctly.
On Jul 23, 2010, at 12:36 PM, Dimitri Liakhovitski wrote:
Hello!
I have a data set similar to the data set "monthly" in the example
below:
monthly<-
data
.frame
(month
=
c
(20090301,20090401,20090501,20100301,20100401,20090301,
On Fri, Jul 23, 2010 at 12:35 PM, wrote:
> David, Stephen,
> You're right - it's the time zone conventions that threw me as well. I tried
> those round() operations earlier, but inevitably ended up being either an
> hour behind. Even when I specified my time zone, it didn't make any
> differen
On 23/07/10 17:36, Duncan Murdoch wrote:
On 23/07/2010 12:10 PM, babyfoxlo...@sina.com wrote:
[...]
You probably won't get much faster than read.table with all of the
colClasses specified. It will be a lot slower if you leave that at
the default NA setting, because then R needs to figure ou
Hi,
does anyone know a tutorial for Tinn-R? During all my search I only found
R-Tutorials...
The problem now is: I would like to make Tinn-R an autosave. But since I had
several questions before concerning Tinn-R (for example, how to have Tinn-R
and R in one window or how do the new versions work
Hi,
Strange, there seems to be different behavior of "old style" classes and S4
classes.
This worked just like you expected (but it's not the same thing, no S4
classes)
f2=function(x,...) UseMethod("fxy")
fxy.default=function(x,...){
print("default")
}
fxy.X=function(x,...) {
read.table is not very inefficient IF you specify the colClasses=
parameter. scan (with the what= parameter) is probably a little more
efficient. In either case, save the data using save() once you have it
in the right structure and it will be much more efficient to read it
next time. (In fa
Hello!
I have a data set similar to the data set "monthly" in the example below:
monthly<-data.frame(month=c(20090301,20090401,20090501,20100301,20100401,20090301,20090401,20090501,20100301,20100401),monthly.value=c(100,200,300,101,201,10,20,30,11,21),market=c("Market
A","Market A", "Market A","M
On 23/07/2010 12:10 PM, babyfoxlo...@sina.com wrote:
Hi there,
Sorry to bother those who are not interested in this problem.
I'm dealing with a large data set, more than 6 GB file, and doing regression
test with those data. I was wondering are there any efficient ways to read
those data? Ins
David, Stephen,
You're right - it's the time zone conventions that threw me as well. I tried
those round() operations earlier, but inevitably ended up being either an hour
behind. Even when I specified my time zone, it didn't make any difference. So
there's something else that I'm missing. I'll
The arithmetic that David describes should work fine (POSIXct is
internally in UTC) unless you are in the Chatham Islands (which has a
UTC+12:45 time zone [1]) or Nepal (UTC+05:45 [2]) or some such place
with a funny idea about when the 1/2 hour mark is.
The formatting of the output may be tr
This truncates to quarter-hour rather than rounding to half-hour.
"stephen sefick" wrote:
>If you have a zoo series this should work. If it doesn't then please
>tell me because I think it works.
>
>snap2min <- function(zoo, min="00:15:00"){
>min15 <- times(min)
>a <- aggregate(zoo, trunc(time(z
By entering "trunc.POSIXt" at the R commandline, you can see the standard
truncate implementation. Riffing on this,
roundhalfhour <- function( x ) {
x <- as.POSIXlt( x + as.difftime( 15, units="mins" ) )
x$sec <- 0
x$min <- 30*(x$min %/% 30)
as.POSIXct(x)
}
The as.double approach ought to w
On Jul 23, 2010, at 12:04 PM, stephen sefick wrote:
If you have a zoo series this should work. If it doesn't then please
tell me because I think it works.
snap2min <- function(zoo, min="00:15:00"){
min15 <- times(min)
a <- aggregate(zoo, trunc(time(zoo), min15), function(x) mean(x,
na.rm=TR
Rkward is great.
If you are a programmer then have a look at the StatET plugin for eclipse
(Java). Not so simple to set up.
On Friday 23 July 2010 02:17:36 pm Alaios wrote:
> I would like to thank you for your immediate reply.
> Actually I do not like vim and emacs.. I am trying to find an ed
Hi there,
Sorry to bother those who are not interested in this problem.
I'm dealing with a large data set, more than 6 GB file, and doing regression
test with those data. I was wondering are there any efficient ways to read
those data? Instead of just using read.table()? BTW, I'm using a 64bit
Wow, that's even better!
Thanks to you both. I know both options will come in handy!
Jen
On Fri, Jul 23, 2010 at 11:58 AM, David Winsemius wrote:
>
> On Jul 23, 2010, at 11:32 AM, Jennifer Sabatier wrote:
>
> Hi R-List,
>>
>> I have a question regarding R-language formats, I think. I am pr
If you have a zoo series this should work. If it doesn't then please
tell me because I think it works.
snap2min <- function(zoo, min="00:15:00"){
min15 <- times(min)
a <- aggregate(zoo, trunc(time(zoo), min15), function(x) mean(x, na.rm=TRUE))
}
hth
Stephen Sefick
On Fri, Jul 23, 2010 at 11:00
Worked like a charm! Thanks a lot!
Jen
On Fri, Jul 23, 2010 at 11:48 AM, Alain Guillet
wrote:
> Hi,
>
> I am sure there is a better solution but a possibility is that you use
> sub("\\.",",",as.character(newdata)) instead of newdata in the text calls.
>
> HTH,
> Alain
>
>
>
>
> On 23-Jul-10
On Jul 23, 2010, at 11:35 AM, David Winsemius wrote:
On Jul 23, 2010, at 11:20 AM, > wrote:
Hi folks,
I've got a POSIXct datum as follows:
Sys.time()
[1] "2010-07-23 11:29:59 BST"
I want to convert this to the nearest half-hour, i.e., to
"2010-07-23 11:30:00 BST"
(If the time were
On Jul 23, 2010, at 11:32 AM, Jennifer Sabatier wrote:
Hi R-List,
I have a question regarding R-language formats, I think. I am
producing a
series of graphs (using plot, barplot, barchart, and bwplot, using
either
text or mtext to place values on the graphs) and tables for a
Francophone
On Jul 23, 2010, at 11:20 AM, > wrote:
Hi folks,
I've got a POSIXct datum as follows:
Sys.time()
[1] "2010-07-23 11:29:59 BST"
I want to convert this to the nearest half-hour, i.e., to
"2010-07-23 11:30:00 BST"
(If the time were "11:59:ss", I want to convert to "12:00:00").
How to a
Hi R-List,
I have a question regarding R-language formats, I think. I am producing a
series of graphs (using plot, barplot, barchart, and bwplot, using either
text or mtext to place values on the graphs) and tables for a Francophone
country. In fact, I have already done so. However, while they
In addition, there are 'theoretical' reasons for excluding intercept
from the model that must be considered. The reasons related to the
regressor(s) and depend on the phenomenon being modelled. For example,
whereas the intercept can be excluded in a bivariate model on the
expenditure of an individu
On Jul 23, 2010, at 11:16 AM, Brigid Mooney wrote:
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was hoping
On 23/07/2010 11:16 AM, Brigid Mooney wrote:
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was hoping someone
Hi folks,
I've got a POSIXct datum as follows:
> Sys.time()
[1] "2010-07-23 11:29:59 BST"
I want to convert this to the nearest half-hour, i.e., to "2010-07-23 11:30:00
BST"
(If the time were "11:59:ss", I want to convert to "12:00:00").
How to achieve this?
Thanks,
Murali
I'm working on a problem where I'm introducing random error and have
been using the built in function runif to provide that random error.
However, I realized that I seem to be getting some unexpected behavior
out of the function and was hoping someone could share some insight.
I don't know the ru
Hi Tom,
This seems to work like I'd expect:
HGlt10RawPerc2008 <- read.table(textConnection("
-5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
"))
recode(HGlt10RawPerc2008 ,"-100:0.0 = 10; 0:1.0 = 8; 1.001:3.0 = 6;
3.001:4.0 = 4; 4.001:5.0 = 2; else = 0")
I wonder if your data might n
Hi,
I am trying to recode the output from a matrix(here is a small snippet of it):
HGlt10RawPerc2008[1:20]
[1] -5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
H
Hi,
I am trying to recode the output from a matrix(here is a small snippet of it):
HGlt10RawPerc2008[1:20]
[1] -5
0
-1
-1
0
2
3
-5
-2
0
2
0
1
-2
3
0
4
1
4
2
Her
help("difftime") is probably what you want.
If not, please post a standalone example
Allan
On 23/07/10 15:18, Jim Hargreaves wrote:
Dear List
I have the POSIX timestamps of two events, A and B respectively. I
want an expression that tests if A occurs within 4 weeks of B.
Something like:
i
[cc'd to package maintainer]
This feels like a fairly straightforward question, so I'm surprised
not to have found an answer so far (RSiteSearch, package help,
vignette ...)
It seems fairly hard (without some fairly serious hacking) to
hard-code the axis limits in a hexbin plot from the hexb
Dear List
I have the POSIX timestamps of two events, A and B respectively. I want
an expression that tests if A occurs within 4 weeks of B.
Something like:
if (ts_A "is between" ts_B + 4(weeks) and ts_B - 4) ...
Can probably done by regex but the regex docu for R says nothing about
if state
Dear David and R users,
Thank you very much for the help. There is a function in the suggested
package called "midPoint" which does what I need.
df<- data.frame(lon1=c(-4.568,-4.3980), lat1=c(59.235,56.369),
lon2=c(-5.123,-4.698), lat2=c(60.258,59.197) )
library(geosphere)
p1 <- matrix(c(df$lo
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On 23/07/2010 15:56, Alaios wrote:
> I would like oto thank everyone for the replies.
> I instaled rkward.. It looks like an editor.. Is it possible to execute
Yes
> also your code from the text editor.
Look into the "Run" menu
> If there is a functi
I would like oto thank everyone for the replies.
I instaled rkward.. It looks like an editor.. Is it possible to execute also
your code from the text editor.
If there is a function call myfunc inside my file test.R
Is it possible to exectute the function from the rwkard?
Best Regards
Alex
___
On Jul 23, 2010, at 9:43 AM, li li wrote:
Hi all,
I want to add 6 plots in the format of 2 columns and 3 rows as one
figure in latex. The plots are in .eps file.
I know how to add 2 plots side by side, but could not figure out how
to do
multiple rows.
I know this may not be the right plac
I have been trying emacs with ess, Kate, plain gedit..nothing really
satisfied me..
but then I found a nice plugin for gedit: Rgedit !
it supports split screen with R terminal window (you can also personalize
colour if, like me, do like to
stick with old habits :) ), moreover gedit has a brackets
Hi all,
I want to add 6 plots in the format of 2 columns and 3 rows as one
figure in latex. The plots are in .eps file.
I know how to add 2 plots side by side, but could not figure out how to do
multiple rows.
I know this may not be the right place to ask such a question. But I do
not know who
Thank you all for your kind reply and help.
I changed the legend function as Mr. Peter Ehlers suggested and
the plots look good.
Thank you again.
Hannah
2010/7/22 Peter Ehlers
> On 2010-07-21 22:06, li li wrote:
>
>> Hi all,
>> I am have some difficulty with the legend
I like loops for this kind of thing so here is one:
df<- structure(list(start = c("15:00", "15:00", "15:00", "11:00",
"14:00", "14:00", "15:00", "12:00", "12:00", "12:00", "12:00",
"12:00", "12:00", "12:00", "12:00", "12:00", "12:00", "12:00",
"12:00", "12:00"), end = c("16:00", "16:00", "16:00",
On Jul 23, 2010, at 8:58 AM, Mafalda Viana wrote:
The arithmetic mean was my first approach and to nearby points it
doesn't make much difference. However, when the distance between the 2
points gets bigger this is no longer accurate enough. So yes, I was
thinking on spherical geometry, midpoint
The arithmetic mean was my first approach and to nearby points it
doesn't make much difference. However, when the distance between the 2
points gets bigger this is no longer accurate enough. So yes, I was
thinking on spherical geometry, midpoint considering the great circle
distance or similar.
Th
Hi,
I use heatmap.2 and heat.colors. Is it possible to specify the colors in
that way that
all values below, for instance, 1.5 should be coloured red, values between
1.5 and 1.7 green
and above 1.7 black?
Many thanks
--
View this message in context:
http://r.789695.n4.nabble.com/specify-heat-c
(1:length(a))[a != 0]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of vikrant
Sent: 23 July 2010 12:13
To: r-help@r-project.org
Subject: [R] How to get vector of poitions in a vector ?
I have a vector a = c(1,0,0,0,1,0,4,0,0,0)
Now
On 23/07/2010 7:14 AM, Duncan Murdoch wrote:
Nordlund, Dan (DSHS/RDA) wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Peter Dalgaard
>> Sent: Thursday, July 22, 2010 3:13 PM
>> To: Pat Schmitz
>> Cc: r-help@r-projec
The error message actually is
Error in inherits(object, "formula") : object 'frmlc' not found
and it is the omitted-by-you part after the colon that matters: the
body of model.tables does not have 'frmlc' in its scope.
Here is one approach to using dynamic formulae that works in a lot of
sim
I have the following code to write the output from auto.arima function. The
issue is not in finding the model but to divert its out put
fit to a file order_fit.txt. code runs but nothing is written to
order_fit.txt
where am I going wrong
library(forecast)
for (i in 1:2) {
filen = paste("file",i,"
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On 23/07/2010 14:17, Alaios wrote:
> I would like to thank you for your immediate reply.
> Actually I do not like vim and emacs.. I am trying to find an editor with a
> gui
> that will work fine in Linux.
Well - you are missing out with not using ema
It would be nice if you could post what the data looks like that you
want to import. R can import any text file and then you have string
manipulation that you can do to parse it. So the basic answer is
probably yes, but we do need to understand the format of the data to
give a more precise answer
try this:
> char2hr <- function(time){
+ mat <- do.call(rbind, strsplit(time, ":"))
+ mode(mat) <- 'numeric'
+ mat %*% c(1, 1/60) # convert to hours
+ }
> # convert to hours
> x.hr <- apply(x, 2, char2hr)
> # generate a set of sequences to set values
> x.seq <- apply(x.hr, 1, function
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