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
__
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 have
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
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
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]
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,
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
attachment:
you forgot to define the x and y variables that you provided as examples.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Try a different CRAN mirror.
MW
On Jan 17, 2013, at 7:42 AM, vaseem shaikh vsma...@gmail.com 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
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:
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)
molten
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
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
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
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=yield) # fake
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):
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 need
-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 follow
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
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
On Thu, Jan 17, 2013 at 12:57 PM, vaseem shaikh vsma...@gmail.com 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
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
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
On Thu, Jan 17, 2013 at 2:36 PM, Ivan Calandra
ivan.calan...@u-bourgogne.fr 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
On 17-01-2013, at 15:36, Ivan Calandra ivan.calan...@u-bourgogne.fr 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
On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
claire.j.osw...@gmail.com 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
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
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(dataset, :
target of
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 $resstatus
'B'=
Ok, it is that simple... Actually I had tried it but messed up so that
it didn't work.
Do you know where I can find some documentation about it?
Regarding return(), I know that it's not necessary, but when the
function gets more complicated, I like to have it because it becomes
clearer to me.
On 17/01/2013 9:59 AM, Gabor Grothendieck wrote:
On Wed, Jan 16, 2013 at 12:39 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Jan 16, 2013 at 11:06 AM, Claire Oswald
claire.j.osw...@gmail.com wrote:
Hello:
I'd like to know if R will run under Windows 8?
I am running R on
Well..
On Thu, Jan 17, 2013 at 7:42 AM, Ivan Calandra
ivan.calan...@u-bourgogne.fr wrote:
Ok, it is that simple... Actually I had tried it but messed up so that it
didn't work.
Do you know where I can find some documentation about it?
The R language definition manual would be the logical
I've spent several days compiling the following code (I apologize in advance
- this code is very inelegant, and I'm sure could be written much more
efficiently, but I've stuck with whatever method I could get to work -
sometimes the more efficient code I just couldn't get to work without an
error,
Iuri:
Your code as emailed reads:
##
data(npk, package=MASS)
fit - by(npk, npk$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)) # plots
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--
#
arun smartpink...@yahoo.com
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)
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
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
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
are you looking for assign()?
On Jan 17, 2013 1:56 PM, mtb...@gmail.com 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
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 liuwen...@gmail.com wrote:
are you looking for assign()?
On Jan 17, 2013 1:56 PM, mtb...@gmail.com
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
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
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
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,
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
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 * * * $HOME/users/REPORT/MAIL; time R --slave report.R
[[alternative HTML version deleted]]
__
hi guys
I need to interpolate values for the zero coupon yield curve. Following data
is given
You are thinking that 'names' does something different
than it does. What you seem to be after is the
deparse-substitute idiom:
dat - data.frame(Col1=1:10, Col2=rnorm(10))
myPlotFun - function(x, y) {
plot(y ~ x, xlab=deparse(substitute(x)), ylab=deparse(substitute(y)))
}
myPlotFun(dat$Col1,
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
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
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: you
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,
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 pinakimah...@gmail.com wrote:
Could not figure out where I am going wrong. Following is my
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 -
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,
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:
?
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
pbu...@pburns.seanet.com wrote:
If you want the column
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
Hello,
Try the following.
(I've named your data.frame 'dat')
do.call(rbind, lapply(split(dat, dat$`No.`), tail, 1))
Hope this helps,
Rui Barradas
Em 17-01-2013 10:50, Mat escreveu:
Hello togehter,
i have a data.frame like this one:
No. Date last change
1 1
Hello R-helpers,
I have run the following lines of code:
x-cars$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
better,
On Thu, 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 functions)
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
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
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.
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
On Thu, Jan 17, 2013 at 4:57 AM, vaseem shaikh vsma...@gmail.com 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
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
can apply to y
Exactly what I am looking for.
Thanks a lot!
Mike
On Thu, Jan 17, 2013 at 4:59 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
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
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)
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
murdoch.dun...@gmail.com wrote:
On 13-01-17 4:50 PM, C W wrote:
Dear list,
How do you delete repeated samples? In
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
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
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 do
Inline below.
-- Bert
On Thu, Jan 17, 2013 at 3:02 PM, David Winsemius dwinsem...@comcast.net 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
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, we
-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
a
Quoting Mat matthias.we...@fnt.de:
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
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
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
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
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:
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 mic.c...@gmail.comwrote:
Hi,
I just updated R and patchDVI (from
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
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 color
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 for
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 rolf.tur...@xtra.co.nz
rolf.tur...@xtra.co.nz wrote:
The help facility is applicable to functions and data sets. It is not
designed
or intended to
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
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
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.csv)
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 mlhow...@avalon.net
To:
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 R but am
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 b17fly...@gmail.com wrote:
Hi,
Thanks for a great
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