Hi
see in line
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of David Carlson
Sent: Tuesday, October 23, 2012 6:47 PM
To: 'Macy Anonuevo'; r-help@r-project.org
Subject: Re: [R] plotting multiple variables in 1 bar graph
Hi
Hi
I want help for repeat because it takes a long time
repeat{
for (i in 1:n){
probb[i]=sum(Wc[z[,j]=yb[i]])
}
deltab[,b]=rbinom(n,1,probb)
if(length(which(is.na(deltab[,b])==T))==0){break}
}
This code works but takes much computation time.
What is this code supposed to do?
Dear all,
I am trying to create a list, where each list element is a vector of
different length arrays that contain 2by2 matrices. To be more specific
there are 11 treatments that are compared with placebo (we have 11
comparisons) and each comparison is studied by a different number of trials
and
On 2012-10-23 15:39, Rich Shepard wrote:
On Tue, 23 Oct 2012, Bert Gunter wrote:
I believe you are misunderstanding what a barchart is (or maybe I do,
since I never use 'em). I believe that there should be one quant value for
each tclass and stream, and you have several.
Bert,
My first
Hi
I'm not getting any results in R console when i run commands. I reinstalled
R but the results are same
-
Thanks in Advance
Arun
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HI,
May be this helps:
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm.D9 - lm(weight ~ group)
fun1-function(x){
Thanks, JeffThis is from a course, but the course is just needed to take some
slides for presentation. I just learn R by myself, and want this skill more
practical. So I try to use the innovative way to perform a more professional
presentation.hence, no worry about the homework support issue.
Hi,
Just a modification of David's method:
apply(dat1,2,function(x) names(which.max(x[!is.na(x)]))==Obama)
# AL AR CA NY
#FALSE FALSE TRUE TRUE
names(dat1)[apply(dat1,2,function(x) names(which.max(x[!is.na(x)]))==Obama)]
#[1] CA NY
A.K.
- Original Message -
From:
Hi,
I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it should be a simple thing, but I am unable to do it right
now !
The example re the structure of the spreadsheet is -
# Example:
Hi,
When I use untar command and I got an error.
untar(/public03/ara_gse_raw_data/GSE26679_RAW.tar,exdir=/public03/data/arabidopsis_data/GSE26679_RAW)
sh: /tmp/RtmpzQGJvh/file5f09b232: No such file or directory
I want to know how to solve this problem.
Thanks very much for any tips.
On Tue, Oct 23, 2012 at 11:52 PM, Luigi marongiu.lu...@gmail.com wrote:
Any tip on how to proceed?
You may want to do check rocplus package. Its vignette is pretty good.
http://cran.r-project.org/web/packages/rocplus/vignettes/rocplus.pdf
In your case, I think pairwise comparison would be one
On 2012-10-23 15:22, bwone wrote:
Hello all,
Is it possible to change the radii line type in radial plots? I wasn't able
to find anything online.
Do please be more specific: which package's radial plot function
are you using? Package plotrix has radial.plot() with an 'lty'
argument.
Peter
On Oct 23, 2012, at 11:36 PM, Loukia Spineli wrote:
Dear all,
I am trying to create a list, where each list element is a vector of
different length arrays that contain 2by2 matrices. To be more specific
there are 11 treatments that are compared with placebo (we have 11
comparisons) and
Hello,
And what do you expect from us, without example?
Regards,
Pascal
Le 12/10/24 15:44, arunkumar a écrit :
Hi
I'm not getting any results in R console when i run commands. I reinstalled
R but the results are same
-
Thanks in Advance
Arun
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On Oct 23, 2012, at 7:43 PM, zhaohu44 wrote:
Hi,
When I use untar command and I got an error.
untar(/public03/ara_gse_raw_data/GSE26679_RAW.tar,exdir=/public03/data/arabidopsis_data/GSE26679_RAW)
sh: /tmp/RtmpzQGJvh/file5f09b232: No such file or directory
I want to know how to solve
wow. That sounds like a problem.
For a more constructive response, read the posting guide and post specific
examples and explain how the output you get is not what you expect. Also, the
output of sessionInfo() is likely to be helpful, and if you use a Mac, or
Debian/other Linux, one of the
You are absolutely right. I read the guidelines! Mistake to attach 'word'
file! The comfort is that at least I got an A :).
I revise my question approapriately!
On Wed, Oct 24, 2012 at 10:36 AM, David Winsemius dwinsem...@comcast.netwrote:
On Oct 23, 2012, at 11:36 PM, Loukia Spineli wrote:
Dear all,
I am trying to create a list, where each list element is a vector of
different length arrays that contain 2by2 matrices. To be more specific
there are 11 treatments that are compared with placebo (we have 11
comparisons) and each comparison is studied by a different number of trials
and
Le mardi 23 octobre 2012 à 20:36 -0700, Shane2012 a écrit :
Thanks, Jeff
This is from a course, but the course is just needed to take some
slides for presentation. I just learn R by myself, and want this skill
more practical. So I try to use the innovative way to perform a more
professional
Hello,
testclass - setRefClass(
testclass,
fields = list(testfield = logical),
methods = list(validate=function(){testfield-TRUE}))
test - testclass$new()
test$testfield
logical(0)
test$validate()
test$testfield
[1] TRUE
Works just fine for me.
I would love to be able to do something
On 10/23/2012 07:22 PM, Macy Anonuevo wrote:
I'd greatly appreciate your help in making a bar graph with multiple
variables plotted on it. All the help sites I've seen so far only plot 1
variable on the y-axis
Data set:
I have 6 sites, each measured 5 times over the past year. During each
Dear Maya,
You and I corresponded about this problem yesterday, so I'll respond on the
r-help list only briefly:
On Tue, 23 Oct 2012 23:12:23 +0300
Maya Abou Zeid ma...@aub.edu.lb wrote:
Hello,
I am using the SEM package in R to fit a multigroup latent variable model and
ran into some
Hello,
Try the following.
apply(MyFile, 1, scale)
Hope this helps,
Rui Barradas
Em 24-10-2012 07:17, Vedant Sharma escreveu:
Hi,
I need to find the z-score of the data present in a speardsheet. The values
needs to be calculated for each gene across the samples (refer the
example). And, it
On 10/24/2012 03:21 AM, Thiho Jules wrote:
Hi,
I want to make a barplot with the following datasets:
I have a file as following:
name chr position A1 A2 pop1 pop1 pop2 pop2
I have calculated a measure using all the values in columns pops, the
values are saved in a vector.
Now I want to
Hi,
Try this:
res-do.call(rbind,lapply(lapply(apply(MyFile,1,function(x)
x[!is.na(x)]),function(x) (x-mean(x))/sd(x)),function(x)
x[c(Sample_1,Sample_2,Sample_3)]))
res
# Sample_1 Sample_2 Sample_3
#Gene_1 0.4931970 -1.1507929 0.6575959
#Gene_2 1.1421818 -0.7179429 -0.4242390
Hi,
In cases, with more sample columns, you could also use this:
res2-t(sapply(lapply(apply(MyFile,1,function(x) x[!is.na(x)]),function(x)
(x-mean(x))/sd(x)),function(x) x[colnames(MyFile)] ))
res2
# Sample_1 Sample_2 Sample_3
#Gene_1 0.4931970 -1.1507929 0.6575959
#Gene_2
Hello,
Sorry but, what doesn't work? What is the error message or result?
As for stringsAsFactors not keeping character strings as character
strings maybe this is from calling data.frame from within the
environment used by with(), but you can use other ways of converting to
character strings,
Hello,
I was studying several packages related to time series analysis (urca,
vars, tseries). I understand that we can estimate a VECM and also test
restrictions on alphas and betas. However, I couldn't find a function that
allows me to specify the five cases of VECM (restricted constant,
Dear all,
I have been trying the following without avail and would be very grateful for
any help. From a dendrogram (recursive list of lists with some structure), I
would like to obtain some information of the component lists and of the
enclosing list at the same time. In dendrogram-speech I
Ai...I see, i didn't know I can sort rowname.
more study to do :)
cheers,
martin
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On 24/10/12 19:44, arunkumar wrote:
Hi
I'm not getting any results in R console when i run commands. I reinstalled
R but the results are same.
I *suspect* that you are displaying the results inside some sort of
function
(e.g. a for loop, or source() ) whence the results are invisible
Hi,
I'm Pina and I'm a student in geology. I'm working with spectral profile of
sand and I have to find the similarity between one spectral profile selected
by hyperspectral image anche one that I created to mix different percentage
of 4 mineral component. I have to find the best mix of
Hi Petr
Thanks for reply.
On 24 October 2012 07:26, PIKAL Petr [via R]
ml-node+s789695n464726...@n4.nabble.com wrote:
Hi
It is a piece of my code. At this, i create a matrix(deltab) with each
column is a binomial, but in there are several NAs, and i must restart
the cycle the number of
Hi all,
I am attempting to create a new variable based on values of other variables.
The variable is called pharm. It basically takes the numeric code of 1 as yes
and 0 to be No from the variable B20_C1 (a question on a survey). However, I
would also like to have a level for non-respondents and
Hi,
still doesn't work.
Hello,
I've just seen the error, you are _not_ searching for colnames in
mod5.sig$snps. Corrected:
Selec = todos[ , colnames(todos) %in% mod5.sig$snps]
Hope this helps,
Rui Barradas
Em 23-10-2012 21:17, Silvano Cesar da Costa escreveu:
Hi Rui,
it doesn't
Sorry, I have been away. I'll have a look at this and get back to you
tomorrow.
Tim
On Thu, Oct 18, 2012 at 10:35 PM, ahs [via R]
ml-node+s789695n4646600...@n4.nabble.com wrote:
Hello,
I am trying to use this fix for the convergence problem in polr, but I
don't seem to get the change of
On 24-10-2012, at 10:04, pina pina@hotmail.it wrote:
Hi,
I'm Pina and I'm a student in geology. I'm working with spectral profile of
sand and I have to find the similarity between one spectral profile selected
by hyperspectral image anche one that I created to mix different percentage
Hannes,
A bit inelegant, but it works. Try this:
mtext(Overall Title Row 2, outer=TRUE, line=-17)
mtext(Overall Title Row 3, outer=TRUE, line=-34)
Jean
capy_bara hettl...@few.vu.nl wrote on 10/23/2012 09:08:39 AM:
Dear all,
I have a 3x2 plot and in addition to the title of the
On Tue, 23 Oct 2012, Peter Ehlers wrote:
As Bert correctly points out, you have more than one quant value per
tclass/stream combination. You need to summarize your 1987-case data.frame
to a 56-case data.frame (8 levels of tclass and 7 levels of stream). The
easy way to do that is:
benthos2 -
Hi,
May be this helps:
set.seed(1)
dat1-data.frame(B20_C1=c(NA,sample(0:1,4,replace=TRUE),NA),B20_C2=c(NA,sample(0:1,3,replace=TRUE),sample(0:1,2,replace=TRUE)),nrB20C=c(1,NA,NA,NA,NA,NA))
dat1
# B20_C1 B20_C2 nrB20C
#1 NA NA 1
#2 0 0 NA
#3 0 1 NA
#4
I **think** he is saying that he is not getting anything at all back
on the console. If so, he needs to *follow the posting guide* and
tell us what OS and setup he is using and **exactly** what is meant by
the R console (GUI, command line, ...?) . Your last suggestion to
post elsewhere may also be
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Wednesday, October 24, 2012 3:31 PM
To: Pancho Mulongeni
Cc: R help
Subject: Re: [R] Recode function car package erases previous values
Hi,
May be this helps:
set.seed(1)
I'm used to run that on Matlab
for indice=1:23,
V=vector(1:indice)
end
that give me 23 vectors as output.
Why the same command on R for ( i in 23:0 ) { V - vector [ 1 : i ] }
returns one only vector?
How can I obtain the 23 vectors that I need ?
Thank you.
--
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Great!
You can skip my question about s0 though, I found where it is being used,
but I still struggle with the code and convergence problem.
I also found something here
http://biostat.mc.vanderbilt.edu/wiki/pub/Main/CharlesDupontStuff/newPolr.R
that seems like someone tried to fix it, but with
Hello Rmillan,
I was looking for the same issue and I found that the variables ECT1,
ECT2... are the loading matrix, as you can check in the example of the paper
found at http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf
http://cran.r-project.org/web/packages/vars/vignettes/vars.pdf
A perhaps more elegant way to do it would be to use the ?layout
function rather than the ancient mfrow() to create space for both
titles and graphs. You could then e.g. draw your titles in the space
you created for titles.
-- Bert
On Wed, Oct 24, 2012 at 5:33 AM, Jean V Adams jvad...@usgs.gov
Quoting Rolf Turner:
(1) Learn something about R; don't just hammer and hope. Read the
introductory manuals and scan the FAQ.
-- Bert
On Wed, Oct 24, 2012 at 5:25 AM, colaiutachambers
gabriele.carrar...@gmail.com wrote:
I'm used to run that on Matlab
for indice=1:23,
V=vector(1:indice)
end
I am looking at your pdf file.
it doesn't match your description.
There are no missing values in the vectors
x, y, t.
Each vector has length 55 which is not a multiple of 4,
so we don't know where the 2x2 matrices come from.
The code doesn't run. There are are too many }.
PDF files are
Hi Pancho,
I tried ur method:
pharm-as.factor(recode(dat1$B20_C1,1='Yes';0='No'))
pharm
#[1] NA No No Yes Yes NA
#Levels: No Yes
pharm[dat1$nrB20C==1]-'no resp'
#Warning message:
#In `[-.factor`(`*tmp*`, dat1$nr.B20C == 1, value = no resp) :
# invalid factor level, NAs generated
pharm
If you provide some data we can work with (a reproducible example), then we
can help you.
Now we can only guess?
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Hi Arun,
I also used to get that error, but what class is your nr.B20C, mine is not a
factor, it is numeric and perhaps that's why it works
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Wednesday, October 24, 2012 4:35 PM
To: Pancho Mulongeni
Cc: R help
Subject:
On Oct 24, 2012, at 2:14 AM, Johannes Graumann wrote:
Hello,
testclass - setRefClass(
testclass,
fields = list(testfield = logical),
methods = list(validate=function(){testfield-TRUE}))
test - testclass$new()
test$testfield
logical(0)
test$validate()
test$testfield
[1] TRUE
Works
On Oct 24, 2012, at 2:27 AM, martiny wrote:
Ai...I see, i didn't know I can sort rowname.
more study to do :)
You deleted the context of this thread, but while you are looking at
the rownames documentation you might also keep your eyes open for the
zoo and xts packages which could
I'd greatly appreciate your help in making a bar graph with multiple
variables plotted on it. All the help sites I've seen so far only plot 1
variable on the y-axis
...
I've spent several hours looking for code to do this but didn't find
anything. I'd use the Excel graph except that it
Hello all.
Steven's approach didn't work for me. I'm running 64 bit R
R.Version()
$platform
[1] x86_64-pc-mingw32
$arch
[1] x86_64
$os
[1] mingw32
$system
[1] x86_64, mingw32
and downloaded and installed the 64 bit version of Java directly from
Oracle. I then tried to load the rJava
The examples I gave--Null, Empty string, white space, etc where just examples
based on SPSS Modeler's Data Audit node.
I just want something that both identifies the columns having missing values--
regardless of what they technically are stored as(NA or a field with space bar
hit a couple of
Dear Pancho,
I'm not going to respond to the subsequent messages in this thread, since you
appear to have solved your problem, just explain that what you did originally
was to recode the variable B20_C1, creating the new variable pharm. Then you
recoded another variable, nr.B20C, replacing the
HI Stuart,
Just a doubt:
When I run Rui's function:
I got results as:
g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]
# ID DATE DG
#22 841 20050421 1
#23 841 20050421 2
#38 1019 19870508 2
#39 1019 19870508 1
But, when I look at the id.d
16 814 20020814 2
17 814
tmuman mumantariq at gmail.com writes:
Hi, am very new to R and I've written an optim function, but can't
get it to work
least.squares.fitter-function(start.params,gr,
low.constraints,high.constraints,model.one.stepper,data,scale,ploton=F)
{
result-optim(par=start.params,
Hello,
Try the following.
lapply(1:23, function(i) 1:i)
As for why yuor code doesn't work, you're assigning 24 times a value to
V, on loop exit only the last one is there. A correct way using loops
would be
V - list()
for(i in 1:23)
V[[i]] - 1:i
Beware that the last vector goes from
Hi,
With the dataset that you provided, and considering that my solution works (ID
323 included), the below code outputs a variable INCLUDE.
res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x)
That's simple and appears to work. Thanks for the prompt response.
Ben
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__
Hi,
Try this:
test1-read.table(text=
id x1 x2 x3 x4 x5 x6 x7
1 1 36 26 21 32 31 27 31
2 2 45 21 46 50 22 36 29
3 3 49 47 35 44 33 31 46
4 4 42 32 38 28 39 45 32
5 5 29 42 39 48 25 35 34
6 6 39 31 30 37 46 43 44
7 7 41 40 25 23 42 40 24
8 8 27 29 47 34 26 38 28
9 9
Hello,
I have a time series of 500 returns. I am fitting a garch(1,1) model to this
series using
fit - garchFit(~garch(1,2), data=y)
Then I run: predict(fit,n.ahead=100,plot=TRUE)
To get the plot.
My question, how can I access the data that is being plotted with the black
line. Basically I
Hello,
Using one of Arun's ideas, some post ago, this new function returns a
logical index into id.d of the rows that should be _removed_, hence rm1
and rm2. I think
getRepLogical - function(x, first = TRUE){
fun - if(first) head else tail
dte - tapply(x[,2], x[,1], FUN =
Do you care about local topography/terrain? I think most of the
calculators/tables that are commonly used assume that you are at a
fairly flat place on the earth's surface, but that is not always true.
The area where my wife grew up had its longest day closer to the
equinox than the summer
Hi Rui,
I think now our results are matching except in the INCLUDE column
id.d[c(11:13,22:24,38:40),]
# ID DATE DG INCLUDE
#11 323 20080407 1 TRUE
#12 323 20080521 2 FALSE
#13 323 20080521 3 TRUE
#22 841 20050421 1 TRUE
#23 841 20050421 2 FALSE
#24 841 20060428
On Oct 24, 2012, at 8:32 AM, Lopez, Dan wrote:
The examples I gave--Null, Empty string, white space, etc where just examples
based on SPSS Modeler's Data Audit node.
I just want something that both identifies the columns having missing
values-- regardless of what they technically are
Hello,
I just realized that function getRepLogical marks the second, not the
first (eventually from last) to be removed. The first tapply should be
dte - tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2),
fromLast = TRUE))
in order to remove the first (or last).
Rui Barradas
Hello,
Inline.
Em 24-10-2012 19:05, arun escreveu:
Hi Rui,
I think now our results are matching except in the INCLUDE column
id.d[c(11:13,22:24,38:40),]
# ID DATE DG INCLUDE
#11 323 20080407 1TRUE
#12 323 20080521 2 FALSE
#13 323 20080521 3TRUE
#22 841 20050421 1
I am using the package plotrix radial.plot(). Yes, radial.plot() has a line
type argument, lty, but that is for the polygons or the radial lines, not
the radii or axes of the radial plot.unless I am doing something wrong.
Thanks!
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HI Stuart,
Just a small comment:
id.d1$id.d_INCLUDE-TRUE #is not needed.
res1- data.frame(flag=tapply(id.d[,2],id.d[,1],FUN=function(x)
head(duplicated(x)|duplicated(x,fromLast=TRUE),1)|tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
Hello,
I am trying to create a function that will move through each record of a data
frame, find the value in the HUC column, then
randomly select another observation from the dataframe with the same value in
HUC column, as well as the same value in Yr column as the first
observation. I want
Hi,
According to the OP So the function should only exclude an ID, having
identified a first (or last) DATE duplicate, the DGs for these two dates are
different.
Rui:
By running your modified function (using dte - tapply(x[,2], x[,1], FUN =
function(x) duplicated(fun(x, 2),fromLast = TRUE))),
Hi,
M1 and M2 are extreme in that all or none of the variables have
parallel lines on the logit scale. One can try fitting a partial
POM, which remains fraught (but not as much as M2) because if
the lines intersect for a particular variable where the data lie
then there will be numerical
Hi,
I think the as.factor in ur code created trouble.
library(car)
pharm-as.factor(recode(B20_C1,1='Yes';0='No'))
pharm-recode(dat1$B20_C1,1='Yes';0='No')
pharm[dat1$nrB20C==1]-no resp
pharm
#[1] no resp No No Yes Yes NA
table(pharm)
#pharm
#
Thank you very much Michael and Berend for your ideas and feddback.
I apologize for my mistakes. It's true that I still have much to learn (and
I sometimes forget what I read). I'll surely use the dput() command next
time I post something and I'll try to be more clear with my questions too.
This is more of a general question without data. After doing 'survdiff',
from the 'survival' package, on strata including four groups (so 4 curves
on a Kaplan Meier curve) you get a chi squared p-value whether to reject
the null hypothesis or not. Is there a method to followup with pairwise
I mis-typed, missing an if. I think you've got it, but let me try again:
The function should:
- put FALSE in a column for every instance of an ID
IF ( that ID has a first (or last) DATE duplicated )
AND
IF (the DGs for the duplicated dates are different).
So for the earliest/first date
Hello,
Inline.
Em 24-10-2012 22:40, Stuart Leask escreveu:
I mis-typed, missing an if. I think you've got it, but let me try again:
The function should:
- put FALSE in a column for every instance of an ID
IF ( that ID has a first (or last) DATE duplicated )
AND
IF (the DGs for the duplicated
On Oct 24, 2012, at 4:33 PM, Charles Determan Jr deter...@umn.edu wrote:
This is more of a general question without data. After doing 'survdiff',
from the 'survival' package, on strata including four groups (so 4 curves
on a Kaplan Meier curve) you get a chi squared p-value whether to reject
Hello R user,
Data below represent year in decimal. I would like to catagorize it
in such a way that any valye [0,1] goes to catagory 1 , (1,2] goes to
catagory 2 and so on..
Any suggestion how it can be done with if else statement or any other way?
2.880556
0.616667
5.08
0.858333
0.47
Hi,
May be this:
dat1-read.table(text=
2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
4.116667
1.63
2.147222
,sep=,header=FALSE)
dat1$category-ifelse(dat1$V1=1 dat1$V10,1,ifelse(dat1$V11
See ?cut for a simpler way of doing this.
HTH,
Jorge.-
On Thu, Oct 25, 2012 at 10:02 AM, arun wrote:
Hi,
May be this:
dat1-read.table(text=
2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
I run system of equations. The system package gives variance
covariance matrix for each equation. I want to extract the variance covariance
matrix
(of the coefficients) for each equation. How do we extract it?.
Y1~X1+X2
Y2~X1+X2
fit.ols-system (system, ols,
data)â¦â¦. To extract it I wrote as
Hi Ved,
Sorry, I didn't test it well enough at that time.
In your example file,
#there were NAs
MyFile1 - read.csv( text=
Names,'Sample_1','Sample_2','Sample_3'
Gene_1,87,77,88
Gene_2,98,22,34
Gene_3,33,43,33
Gene_4,78,,81
, header=TRUE, row.names=1, as.is=TRUE, quote=', na.strings= )
Hi,
(Jorge: Thanks for the suggestion.)
cut? will be much easier.
dat1-read.table(text=
2.880556
0.616667
5.08
0.858333
0.47
2.936111
4.258333
0.258333
2.03
2.58
1.09
0.447222
1.87
0.080556
4.03
4.116667
1.63
2.147222
,sep=,header=FALSE)
On 2012-10-24 11:06, bwone wrote:
I am using the package plotrix radial.plot(). Yes, radial.plot() has a line
type argument, lty, but that is for the polygons or the radial lines, not
the radii or axes of the radial plot.unless I am doing something wrong.
Thanks!
First, even if you must
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