Maybe you can use ',=' as separators. ( I don't have R to check).
Otherwise, I would clean the file with an editor or tool like 'sed' to
replace the regular expression /key[0-9]=/ by nothing.
On Jan 18, 2013 8:05 AM, "Frank Singleton" wrote:
> Hi,
>
> Thanks for a great environmentfor statistical
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Mary
> Sent: Thursday, January 17, 2013 6:02 PM
> To: r-help@r-project.org
> Subject: Re: [R] Getting discrete colors on plot
>
> Hi,
>
> This is my first post; I'm new to
HI,
May be this helps:
mydata_long1<-within(mydata_long,{colorvar<-factor(colorvar,levels=1:3)})
require("ggplot2")
p <- ggplot(data=mydata_long1,
aes(x=variable, y=value,
group=id, colour = colorvar)) +
geom_line()
p
A.K.
- Original Message -
From: Mary
To: r-help@r-proje
Hi,
Thanks for a great environmentfor statistical computing :-)
I have some input data in a file ("input_kvpairs.csv") of the form
key1=23, key2=67, key3="hello there"
key1=7, key2=22, key3="how are you"
key1=2, key2=77, key3="nice day, thanks"
Now in my head I wish it was of the form ("input
On 01/18/2013 04:02 AM, Mary wrote:
Hi,
This is my first post; I'm new to R but am a senior statistical programmer. I
have done a lot of graphs using SAS Graph but now am trying to transition to
using graphs in R.
I'm trying to produce a graph where the colors have three categories- ideally
On Jan 17, 2013, at 7:00 PM, John Sorkin wrote:
Rolf
Perhaps the philosophy of the help system needs to change . . .
John
It's probably unwise to accept any one person's claim regarding the
"philosophy of the help system". If you have the time, energy and
skills to construct a help page
Rolf
Perhaps the philosophy of the help system needs to change . . .
John
Sent from my iPhone
On Jan 17, 2013, at 7:11 PM, "Rolf Turner "
wrote:
>
>
> The "help" facility is applicable to functions and data sets. It is not
> designed
> or intended to give "help" with respect to R syntax (w
The tables package may be of use to you for this.
On Fri, Jan 11, 2013 at 4:17 AM, Pancho Mulongeni <
p.mulong...@namibia.pharmaccess.org> wrote:
> Hi, I have a dataframe with n columns, but I am only looking at five of
> them. And lots of rows, over 700.
> So I would like to find frequencies fo
>But I don't want to plot random colors.
>...
> That's why I have this vector with length 24 - each one matches one line in
> the "npk" dataset.
... which is not what interaction.plot, or matplot, needs; it needs one per
line on the plot.
>How can I inform to the interaction.plot function the co
Sweave produces foo-concordance.tex from foo.Rnw, and writes
\input{foo-concordance.tex} in the LaTeX output. You can turn on the
concordance option in knitr as well. Since you do not use RStudio, you
have a couple of more steps to go:
1. borrow \Sconcordance from Sweave.sty;
2. manually \input{fo
Anybody know if it is possible to use texshop and knitr with the sync
working? I add a knitr engine but cannot sync.
PS, I am comfortable with texshop but not RStudio.
Huang
On Fri, Jan 18, 2013 at 7:33 AM, michele caseposta wrote:
> Hi,
> I just updated R and patchDVI (from CRAN).
> Now I can
Hello,
I would like to perform a Box-Cox (âbcPowerâ) transformation on my data.
For this, I am determining lambda using the âpowerTransformâ function.
However, with one of my variables I get the following Warning Message:
In estimateTransform(x, y, NULL, ...) :
Convergence failure: ret
The "help" facility is applicable to functions and data sets. It is not
designed
or intended to give "help" with respect to R syntax (with the exception of
the basic syntax of the operators --- unary and binary --- and the
associated
rules of precedence).
cheers,
Rolf Turner
On 01/18/2013 12:33 AM, Markku Karhunen wrote:
Hi again, R community.
I wonder how you do line breaks in \useage{} section in .Rd files. I
am sure there's some tutorial for this somewhere, but I just haven't
found it.
I have tried \\, \cr, \br and \newline, admittedly arbitrarily, but
all
Hi,
I just updated R and patchDVI (from CRAN).
Now I can reverse search from the pdf to the included.Rnw.
However, I cannot forward search from the included to the pdf. Is this how it
is expected to work?
Forward and inverse search work between main Rnw and pdf.
I am pasting below the code in the
Quoting Mat :
Hello togehter,
i have a data.frame like this one:
No. Date last change
1 1 2012-10-04 change settings
2 1 2012-10-20 bug fix
3 1 2012-11-05 final
4 2 2013-01-15new task
5 2 2013-01-16Bug fix
6 2 2013-01-17
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of mtb...@gmail.com
> Sent: Thursday, January 17, 2013 2:27 PM
> To: David Winsemius; r-help@r-project.org
> Subject: Re: [R] How to convert a string to the column it represents in
>
On 13-01-17 5:33 PM, C W wrote:
I was looking for the first answer.
In MCMC, at time t, when the candidate sample is rejected,
> candidate_sample[t] <- current_sample
say, at time t+1, the sample is rejected AGAIN, we have
> candidate_sample[t+1] <- current_sample
so, at time t, and t+1, w
Inline below.
-- Bert
On Thu, Jan 17, 2013 at 3:02 PM, David Winsemius wrote:
>
> On Jan 17, 2013, at 2:26 PM, mtb...@gmail.com wrote:
>
>> Hi David,
>>
>> I would like to have two objects, one containing the values in a column and
>> the other containing the column's name.
>
> You have not add
On Jan 17, 2013, at 2:26 PM, mtb...@gmail.com wrote:
> Hi David,
>
> I would like to have two objects, one containing the values in a column and
> the other containing the column's name.
You have not addressed the question ... why? Where are you going with this?
> Of course, that's easy to d
I was looking for the first answer.
In MCMC, at time t, when the candidate sample is rejected,
> candidate_sample[t] <- current_sample
say, at time t+1, the sample is rejected AGAIN, we have
> candidate_sample[t+1] <- current_sample
so, at time t, and t+1, we have the same value. When I calcu
Hi David,
I would like to have two objects, one containing the values in a column and
the other containing the column's name.
Of course, that's easy to do manually, but I don't want to have to type out
the name of the column more than once (thus, below, I have typed it once in
quotes, and I am tr
What answer is wanted for
c(1,1,1,2,3,1) ?
Note that Duncan's two suggestions below give different answers for this.
-- Bert
On Thu, Jan 17, 2013 at 1:59 PM, Duncan Murdoch
wrote:
> On 13-01-17 4:50 PM, C W wrote:
>>
>> Dear list,
>> How do you delete repeated samples? In MCMC, when your c
On Jan 17, 2013, at 1:50 PM, C W wrote:
> Dear list,
> How do you delete repeated samples? In MCMC, when your candidate value has
> been reject, so you remain on the same point, so you keep that value.
>
> Say I have this toy example,
>
>> c(1,6,6,6,3,5,4,4,2,3,5)
> c(1,6,6,6,3,5,4,4,2,3,5)[!
Exactly what I am looking for.
Thanks a lot!
Mike
On Thu, Jan 17, 2013 at 4:59 PM, Duncan Murdoch wrote:
> On 13-01-17 4:50 PM, C W wrote:
>
>> Dear list,
>> How do you delete repeated samples? In MCMC, when your candidate value
>> has
>> been reject, so you remain on the same point, so you ke
On Jan 17, 2013, at 1:36 PM, mtb...@gmail.com wrote:
> Hello R-helpers,
>
> I have run the following lines of code:
>
> x<-"cars$dist"
> y<-noquote(x)
>
>
> Now y is a string containing the characters "cars$dist"
>
> My questionis there an R function (or combination of functions) that I
On Thu, Jan 17, 2013 at 4:57 AM, vaseem shaikh wrote:
>
> But still, i am also trying to install the package by locally giving
> absolute path with repos= Null and type = "Source", will CRAN have any role
> to play here???
To compile a package on Windows, you need to install R tools (Rtools)
tha
On Jan 17, 2013, at 1:29 PM, mtb...@gmail.com wrote:
> Hi everyone, and thanks for your replies.
>
> Let me make this a little simpler. Please forget the plotting, that's not
> the issue.
>
> I have run the following line of code:
>
> x<-dat.col
>
> Now, is there a function (or combination of
Dear friends,
I have been trying out the C and the R codes in the fastICA package.
However, it turns out that these often give vastly different results,
especially when row.norm is set to T. This happens even though I have
initialized the input matrix to be exactly the same for both of them.
Here
On 13-01-17 4:50 PM, C W wrote:
Dear list,
How do you delete repeated samples? In MCMC, when your candidate value has
been reject, so you remain on the same point, so you keep that value.
Say I have this toy example,
c(1,6,6,6,3,5,4,4,2,3,5)
The 6 and 4 are repeated, I only want the index o
Dear list,
How do you delete repeated samples? In MCMC, when your candidate value has
been reject, so you remain on the same point, so you keep that value.
Say I have this toy example,
> c(1,6,6,6,3,5,4,4,2,3,5)
The 6 and 4 are repeated, I only want the index of the non-repeated values.
I thou
On Thu, Jan 17, 2013 at 1:29 PM, wrote:
> Hi everyone, and thanks for your replies.
>
> Let me make this a little simpler. Please forget the plotting, that's not
> the issue.
>
> I have run the following line of code:
>
> x<-dat.col
>
> Now, is there a function (or combination of functions) that
Hello R-helpers,
I have run the following lines of code:
x<-"cars$dist"
y<-noquote(x)
Now y is a string containing the characters "cars$dist"
My questionis there an R function (or combination of functions) that I
can apply to y that will cause y to contain the numbers in cars$dist? Even
be
Hello,
Try the following.
(I've named your data.frame 'dat')
do.call(rbind, lapply(split(dat, dat$`No.`), tail, 1))
Hope this helps,
Rui Barradas
Em 17-01-2013 10:50, Mat escreveu:
Hello togehter,
i have a data.frame like this one:
No. Date last change
1 1 2012-1
Hi everyone, and thanks for your replies.
Let me make this a little simpler. Please forget the plotting, that's not
the issue.
I have run the following line of code:
x<-dat.col
Now, is there a function (or combination of functions) that will let me
assign the character string "dat.col" to a new
?
But Pat...
The canonical way to do this is:
myPlotFin(Col2 ~ Col1, data = dat)
I have no idea what the OP wants, but my guess is that the right
answer is: Don't do that.
Cheers,
Bert
On Thu, Jan 17, 2013 at 12:25 PM, Patrick Burns
wrote:
> If you want the column names but not
> the data fra
Here's a link (on my local CRAN)...
http://cran.stat.auckland.ac.nz/doc/manuals/r-release/R-intro.html#The-three-dots-argument
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Friday, 18 January 2013 4:54a
To: iva
It's not really clear to me what you mean when you say that you want to
plot the hours, so it's hard to help. Regardless, take a look at looping
and plotting in any of the free documentation on CRAN.
http://cran.r-project.org/other-docs.html
I hope that this helps,
Andrew
On Fri, Jan 18, 2013
If you want the column names but not
the data frame name, then you could do:
with(dat, myPlotFun(Col1, Col2))
Pat
On 17/01/2013 20:07, Patrick Burns wrote:
You are thinking that 'names' does something different
than it does. What you seem to be after is the
deparse-substitute idiom:
dat <- d
Not sure if it would solve all your problems, but try to specify absolute
path first e.g. for "R", as *your* PATH is not available for the cronjob.
Best,
Gergely
On Thu, Jan 17, 2013 at 2:53 PM, Pinaki wrote:
> Could not figure out where I am going wrong. Following is my code in
> crontab -e:-
Hi Spyros,
I suggest that you borrow / buy the book that was written by the author of
that package, and study it. It's "Generalized Additive Models: An
Introduction with R". There's a lot of stuff going on in GAM fitting that
it would be worth paying close attention to.
I hope that this helps,
A
The ellipsis object is not listed in the base help pages!
help(`+`) # this works - help on arithmetic operators
help("+") # also works
help(`...`) # fails with Error: '...' used in an incorrect context
help("...") # fails also with No documentation for '...' in specified packages
and libraries: y
Hi,
This is my first post; I'm new to R but am a senior statistical programmer. I
have done a lot of graphs using SAS Graph but now am trying to transition to
using graphs in R.
I'm trying to produce a graph where the colors have three categories- ideally I
would like them to be Green for goo
thanks to your guys help I am closer to solving my problem but I have some
small problem. So let's say I start with
>data
number day hour
1 17 10
2 17 11
3 17 6
4 18 4
5 18 10
6 19 8
7 19 8
I want to
You are thinking that 'names' does something different
than it does. What you seem to be after is the
deparse-substitute idiom:
dat <- data.frame(Col1=1:10, Col2=rnorm(10))
myPlotFun <- function(x, y) {
plot(y ~ x, xlab=deparse(substitute(x)), ylab=deparse(substitute(y)))
}
myPlotFun(dat$Col1
hi guys
I need to interpolate values for the zero coupon yield curve. Following data
is given
Could not figure out where I am going wrong. Following is my code in
crontab -e:-
MAILTO: users
MAILTO= users
# m h dom mon dow command
3 19 * * * $HOME/users/REPORT/MAIL; time R --slave < report.R
[[alternative HTML version deleted]]
__
R
Hi,
I tried with kronecker()
do.call(rbind,lapply(1:4,function(i) t(kronecker(diag(4), x[i,]))[i,]))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] 1 5 9 13 17 0 0 0 0 0 0 0 0 0
#[2,] 0 0 0 0 0 2
Hi again, R community.
I wonder how you do line breaks in \useage{} section in .Rd files. I
am sure there's some tutorial for this somewhere, but I just haven't
found it.
I have tried \\, \cr, \br and \newline, admittedly arbitrarily, but
all of these produce warnings or errors.
br, M
Dear all,
I new to r and I would like your help.
I want to explore the patterns (unimodal, monotonically increased/decreased)
of species richness~altitude using GAM in R. Although I run the gam function
in mgcv package I do not know how to manually define knots and degrees of
freedom.
Any help wou
Hello Michael,
I have tried with different CRAN but still i am getting the same error.
But still, i am also trying to install the package by locally giving
absolute path with repos= Null and type = "Source", will CRAN have any role
to play here???
BR-
Vaseem Shaikh
On Thu, Jan 17, 2013 at 3:03
Hello togehter,
i have a data.frame like this one:
No. Date last change
1 1 2012-10-04 change settings
2 1 2012-10-20 bug fix
3 1 2012-11-05 final
4 2 2013-01-15new task
5 2 2013-01-16Bug fix
6 2 2013-01-17final
now i want
Hi Liu - I have been trying with assign() but it's not working. I don't
think that's what I'm looking forany other ideas? Many thanks, Mark
On Thu, Jan 17, 2013 at 1:11 PM, Wensui Liu wrote:
> are you looking for assign()?
> On Jan 17, 2013 1:56 PM, wrote:
>
>> Hello R-helpers,
>>
>> I
are you looking for assign()?
On Jan 17, 2013 1:56 PM, wrote:
> Hello R-helpers,
>
> I have run the following line of code:
>
> x<-dat$col
>
> and now I would like to assign names(x) to be "dat$col" (e.g., a character
> string equal to the column name that I assigned to x).
>
> What I am trying t
Hello R-helpers,
I have run the following line of code:
x<-dat$col
and now I would like to assign names(x) to be "dat$col" (e.g., a character
string equal to the column name that I assigned to x).
What I am trying to do is to assign columns in my dataframe to new objects
called x and y. Then I
I have 365 binary files:
https://echange-fichiers.inra.fr/get?k=oy3CN1yV1Um7ouRWm2U ,I want to
calculate the monthly average. So from the 365 files, I will get 12 files.I
would like also to tell R not to take into account the no-data value
(-32765).for example, for the first month, there are 31
Folks,
I run R on a early 2009 MacBook Pro running Mountain Lion.
I have a bunch of fonts in my user Library one of which is Garamond.
I have tried the ttf_import function to no avail. I played with this for a
couple of hours at least and I have gotten nowhere.
Here is a bit of one of my sess
> arun
> on Wed, 16 Jan 2013 19:20:46 -0800 writes:
> Hi,
> May be this helps:
> library(Matrix)
> res1<-lapply(split(x,1:nrow(x)),function(y)
sparseMatrix(i=rep(1:4,each=5),j=1:(4*5),x=y))
> do.call(rbind,lapply(seq_along(res1),function(i) res1[[i]][i,]))
>
it is easy to parse through yourself. if you don't care about the labels
and just want to import fixed-width file data, you can use the SAScii
package. if you do, run this code to get 'em :)
# load the stringr package to trim strings quickly
library(stringr)
# example proc format block--
# s
Iuri:
Your code as emailed reads:
##
data(npk, package="MASS")
fit <- by(npk, npk$block, function(bydata) fitted.values(lm(yield ~ N,
data=bydata)))
fit <- unlist(fit)
interaction.plot(npk$N, npk$block, fit, xlab="N", ylab="yield") # fake factor,
numeric fac <- c(rep(1,12),rep(2,12)) # plot
I've spent several days compiling the following code (I apologize in advance
- this code is very inelegant, and I'm sure could be written much more
efficiently, but I've stuck with whatever method I could get to work -
sometimes the more efficient code I just couldn't get to work without an
error,
Well..
On Thu, Jan 17, 2013 at 7:42 AM, Ivan Calandra
wrote:
> Ok, it is that simple... Actually I had tried it but messed up so that it
> didn't work.
> Do you know where I can find some documentation about it?
The "R language definition" manual would be the logical place to look,
no? And sure
On 17/01/2013 9:59 AM, Gabor Grothendieck wrote:
On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
wrote:
> On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
> wrote:
>> Hello:
>>
>> I'd like to know if R will run under Windows 8?
>>
>
> I am running R on Windows 8 with no apparent problems.
Ok, it is that simple... Actually I had tried it but messed up so that
it didn't work.
Do you know where I can find some documentation about it?
Regarding return(), I know that it's not necessary, but when the
function gets more complicated, I like to have it because it becomes
clearer to me.
Hello everybody,
I imported an SAS data-file into R. open.sas7bdat() did not work,
so I had to convert it to csv first. Now I would like to recode the
value values into factors. Unfortunately I only have a SAS
syntax file, having this form:
proc format;
value $resstatus
'B'= 'Jahresaufenthalt
William Dunlap wrote
>>> eval(parse(text=paste("dataset",IVcat[k],sep="$")))<-relevel(eval(parse(text=paste("dataset",IVcat[k],sep="$"))),ref="online")
>>This code returns the following error:
>>Error in eval(parse(text = paste("dataset", IVcat[k], sep = "$"))) <-
>>relevel(eval(parse(text = paste(
I also tried fitting a spline to the resulting survival curve and the result
was horrible.
maybe spline won't work or knots need special handling.
overall, I must have the final point of the smooth survival to be same as the
final point of the raw Cox survival and no flat days, the drops sho
On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
wrote:
> On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
> wrote:
>> Hello:
>>
>> I'd like to know if R will run under Windows 8?
>>
>
> I am running R on Windows 8 with no apparent problems.
Actually I found one tiny problem. If you ask whi
On 17-01-2013, at 15:36, Ivan Calandra wrote:
> Dear users,
>
> I'm trying to learn how to use the "...".
>
> I have written a function (simplified here) that uses doBy::summaryBy():
> # 'dat' is a data.frame from which the aggregation is computed
> # 'vec_cat' is a integer vector defining whi
On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
wrote:
> Dear users,
>
> I'm trying to learn how to use the "...".
>
> I have written a function (simplified here) that uses doBy::summaryBy():
> # 'dat' is a data.frame from which the aggregation is computed
> # 'vec_cat' is a integer vector defining
Hello users,
I would like to obtain a survival curve from a Cox model that is smooth and
does not have zero differences due to no events for those particular days.
I have:
> sum((diff(surv))==0)
[1] 18
So you can see 18 days where the survival curve did not drop due to no events.
Is there a wa
Dear users,
I'm trying to learn how to use the "...".
I have written a function (simplified here) that uses doBy::summaryBy():
# 'dat' is a data.frame from which the aggregation is computed
# 'vec_cat' is a integer vector defining which columns of the data.frame
should be use on the right side
On Thu, Jan 17, 2013 at 12:57 PM, vaseem shaikh wrote:
> Hello Michael,
>
> I have tried with different CRAN but still i am getting the same error.
>
Most likely you've got something blocking your network. Can you
download the source / binary from the CRAN pages (using your
web-browser) and then
> Stats is a base package, so that won't work. Base packages
> are built and installed differently from other packages.
Dang! Of course it is. Scratch previous response.
The irony is that my first thought was indeed "Read 'R Installation and
configuration' with special attention to 'Building f
On 13-01-17 8:31 AM, S Ellison wrote:
-Original Message-
I have changed some code in R file inside the stats package
(dendrogram.R).
That was brave. Others have already commented on its wisdom...
Now I wan to test and run the stats package
with the new updated code, what should I do
> -Original Message-
> I have changed some code in R file inside the stats package
> (dendrogram.R).
That was brave. Others have already commented on its wisdom...
> Now I wan to test and run the stats package
> with the new updated code, what should I do in detail?
1. Read and foll
Hi, Antti,
you should look at
?levels
(and particular its "Examples" section) to find out how to use
levels( X) <- c( "new1", ..., "newk")
to achieve what you want.
Regards -- Gerrit
On Thu, 17 Jan 2013, Antti Simola wrote:
Hi,
This is quite simple data manipulation task and I ne
Hi,
This is quite simple data manipulation task and I need help for it. I
want to make new factor variable that is an aggregation of an existing
factor.
This works as I intended:
X[Y == "original label"] <- " new label"
How to make following work then (to make coding more convenient):
orig
I was unable to run your code; 'fac' is missing and npk$fac in teh
interaction.plot returns NA.
> data(npk, package="MASS")
> fit <- by(npk, npk$block, function(bydata) fitted.values(lm(yield ~ N,
> data=bydata)))
> fit <- unlist(fit)
> interaction.plot(npk$N, npk$block, fit, xlab="N",
> ylab="y
Hi,
I am trying to plot an interaction.plot with different color for each
level of a factor. It has an erratic behavior.
For example, it works for the first interaction.plot below, with the
example from the ALDA book, but not with the other plots, from the NPK
dataset:
# from http://www.ats.ucla
Hello,
Em 17-01-2013 11:56, Jessica Streicher escreveu:
Rather unspecific.
Basically you'd need a loop to create the sets, and a way to write them into a
file.
You did not specify the format of your data. You might be able to use write,
write.table or write.csv and the like.
You could also ha
Rather unspecific.
Basically you'd need a loop to create the sets, and a way to write them into a
file.
You did not specify the format of your data. You might be able to use write,
write.table or write.csv and the like.
You could also have a look at ?save which allows you to save any R object.
Hello all,
Thanks a lot for you help!
Just in case someone else will have that same problem in the future:
Meanwhile I also found out that ggplot2 gives you the option to freely
swap x- and y-axes and apply a log-scale to whichever one you need:
> library(ggplot2)
> library(reshape2)
> molt
Hello
thank you for the fast and helpful answer! Now the following works fine
for me
x <- readLines(filename)
i <- grep("^year", x)
dlf <- read.table(textConnection(x[i:length(x)]),
header = T, stringsAsFactors=F,sep="\t")
Greetings
Christof
Am 16-01-2013 16:55, schrieb Rui Barradas:
Try a different CRAN mirror.
MW
On Jan 17, 2013, at 7:42 AM, vaseem shaikh wrote:
> Hi,
>
> Please find the snap shot attached for the error reported wile installing
> Hmisc Package.
>
> Is there any thing you can help me with.
>
> Currently using "R i386 2.15.2" Version of R on windows 7 p
you forgot to define the x and y variables that you provided as examples.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Hi,
Please find the snap shot attached for the error reported wile installing
Hmisc Package.
Is there any thing you can help me with.
Currently using "R i386 2.15.2" Version of R on windows 7 platform.
Ragards
Vaseem Shaikh
<>__
R-help@r-project.org m
Thank you for your reply!
When I copy and paste the code into the Console, I receive several errors!
1) Error in plot(x, y, type = "n") : object 'x' not found
> segments(x[-length(x)],y[-length(x)],x[-1],y[-length(x)])
Error in segments(x[-length(x)], y[-length(x)], x[-1], y[-length(x)]) :
HI,
May be this helps:
Example$Wi<-unlist(aggregate(Weight~ID,data=Example,function(x)
round(x/sum(x),2))[,2])
res<-do.call(rbind,lapply(split(Example,Example$Specie),function(x) with(x,
{aggregate(Wi,list(Food.item),function(y) sum(y)/length(unique(x[,1])))})))
names(res)<-names(Solution)[2:3]
Hi everyone, I have got an adjacency matrix here which gives my graph in R.
But is there any way to write Breadth First Search algorithm to obtain path
matrix? I'm really new to R, can anybody please help?
I know the idea but just don't know how to write it, basically I am trying
to look at only
On 01/17/2013 04:51 AM, David Arnold wrote:
Nice, worked very well. But because of the realignment, I now need to lower
by xlab a bit. Any suggestions?
Hi David,
This should give you an idea of how to do it:
par(mar=c(6,4,4,2))
plot(1:10,xlab="")
mtext("Index",side=1,line=4)
Jim
On 17/01/2013 07:59, e-letter wrote:
Readers,
Responding to an old post
(http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18850.html), and
using the example in the manual:
monthextract<-strptime("20/2/06 11:16:16.683", "%m")
monthextract
[1] NA
Why is the result 'NA' and not '2'?
Because you
Readers,
Responding to an old post
(http://tolstoy.newcastle.edu.au/R/e2/help/07/06/18850.html), and
using the example in the manual:
monthextract<-strptime("20/2/06 11:16:16.683", "%m")
monthextract
[1] NA
Why is the result 'NA' and not '2'?
--
r2151
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