plot(1:5, main = paste0(Figure 1: x=, x))
Michael
On Jul 12, 2012, at 3:30 PM, JeffND zuofeng.shan...@nd.edu wrote:
Hi folks,
To simplify my question, consider an example:
x=runif(1,0,1)
plot(1:5,1:5)
Now I want to add a title to the above plot showing the vaue of x, so if the
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal distribution with mean 60 and standard deviation 3. The second is
drawn from a normal distribution with mean 65 and standard deviation 3. The
Hi, Arnold,
looking at the example section of
?stripchart
may help you.
Hth -- Gerrit
On Thu, 12 Jul 2012, darnold wrote:
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal
Dear R users,
I am struggling with the colorkey on a levelplot lattice graphic.
I want that no ticks are printed on the colorkey. That is, I want their size
tck=0.
Merely setting tck=0 t in the colorkey parameter does not work. Setting it in
the lattice.par.set()
removes the ticks from the
On Jul 13, 2012, at 08:18 , darnold wrote:
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal distribution with mean 60 and standard deviation 3. The second is
drawn from a normal
On Thu, Jul 12, 2012 at 9:39 PM, Martin Ivanov tra...@abv.bg wrote:
Dear R users,
I have a lot of experience with traditional R graphics, but I decided to turn
to trellis as
it was recommended for spatial graphs by the sp package. In traditional R
graphics
I always first set the size of
Hello Veronica,
what makes you think that this is an error? It is a warning that your specified
SVAR-model is **just** identified and hence an over-identification test cannot
be conducted. You can suppress this warning by not asking for an
over-identification in the first place, by setting
Hello,
Works unchanged with me.
Yesterday it could have worked for some other reason, like having other
variables in my environment, which I had, but this time I have started
anew. Try including
tUnitsmall - tUnitsort[, cols]
and then use this data.frame to see what happens.
Rui Barradas
Hi Peter,
I had never heard of this 'frame' argument and it's a breakthrough for
me to be finally able to get rid of this frame!
But where is this argument explained? I couldn't find it in plot(),
boxplot(), bxp() or par().
Thank you for your answer :)
Ivan
--
Ivan CALANDRA
Université de
Dear R users,
I need to add minor axis ticks to my graph. In traditional R this is easily
achievable by simply
adding a second axis with the minor ticks. But how to do that in trellis? I am
already out of ideas.
Any suggestions will be appreciated.
Best regards,
Martin
Hello, Ivan,
you'll find argument frame (actually frame.plot) explained in
?plot.default
Regards -- Gerrit
Hi Peter,
I had never heard of this 'frame' argument and it's a breakthrough for me to
be finally able to get rid of this frame!
But where is this argument explained? I couldn't
To use variables in mathematical expressions, bquote can be used:
#Term in in .( ) is evaluated
plot(1:10,main=bquote(frac(alpha,.(2+2
--
GnuPG Key: 0x7340821E
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
two things:
- R always counts from 1, not from 0
- listmembers are accessed by using [[ ]] , not [ ]
try
t1[t==ll[[1]], v] - 99
greetings Jessi
On 11.07.2012, at 15:47, Charles Stangor wrote:
I can't seem to determine how to get the name of a list member to
substitute:
ll - list(a1 =
On Thu, Jul 12, 2012 at 03:51:54PM -0500, Vineet Shukla wrote:
I have independent event sequences for example as follows :
Independent event sequence 1 : A , B , C , D
Independent event sequence 2 : A, C , B
Independent event sequence 3 :D, A, B, X,Y, Z
Independent event sequence 4
You could probably make them numeric, like
v-c(a,a,b,c)
f-factor(v)
as.numeric(f)
[1] 1 1 2 3
to get a numeric rock_id, but i wouldn't per se recommend it.
You should ask someone who knows more about the scientific side of this method
to tell you how factorial data is properly treated.
On Fri, Jul 13, 2012 at 12:21 PM, Martin Ivanov tra...@abv.bg wrote:
Dear R users,
I am struggling with the colorkey on a levelplot lattice graphic.
I want that no ticks are printed on the colorkey. That is, I want their size
tck=0.
Merely setting tck=0 t in the colorkey parameter does not
Thank you for this useful code!
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Hello,
This information is new. Are you able to import data from access?
A.K.
- Original Message -
From: imnew jubil...@live.com.sg
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 9:52 PM
Subject: Re: [R] plot graph by first letter
Hi, for my dataset is actually
Thanks to both of you, you are probably right that memory is the limiting
factor. I have no knowledge about the available memory on the lab machines,
but I will find out and make sure that this is the explanation.
Best,
Ulrike
--
View this message in context:
Hi, I'm currently working on the below codes however whenever I run it in
openbugs it gives an error message saying: unknown type of logical function
error pos 76. Any help would be appreciated.
## bugs code
library(R2OpenBUGS)
sink(C:/Users/CCF/Documents/Suzie Work/PTY Project/Waterhole
Hi,
here i have a Max and Min values
Min -3
Max -6
and also a matrix like this,
ABCXYZ PQR
-- ------
2 43
5 48
7 13
In this i need to check each particular
I have the following dataframe with the first column being of type datetime:
dateTime - c(10/01/2005 0:00,
10/01/2005 0:20,
10/01/2005 0:40,
10/01/2005 1:00,
10/01/2005 1:20)
var1 - c(1,2,3,4,5)
var2 - c(10,20,30,40,50)
df -
Dear all, I am analyzing data from a survey question where I have many Don'k
know responses. Respondents could either say Don't know or give a rating
from 0 to 10. I could analyse the Don't know responses separately using a
logistic regression to see who is the typical person who doesnt know
Hello,
Within my department I would like to share the latest version(s) of my R
scripts to my colleagues using subversion. This repository is password
protected (WebDAV) as it should not be accessed by external persons.
In the majority of my scripts I load scripts using the source() function,
thanks!
On Fri, Jul 13, 2012 at 4:35 AM, Jessica Streicher j.streic...@micromata.de
wrote:
two things:
- R always counts from 1, not from 0
- listmembers are accessed by using [[ ]] , not [ ]
try
t1[t==ll[[1]], v] - 99
greetings Jessi
On 11.07.2012, at 15:47, Charles Stangor wrote:
Hello,
Check the structure of what you have, df and newdf. You will see that in
df dateTime is of class POSIXlt and in newDf newDateTime is of class
POSIXct.
Solution:
[...]
df$dateTime - strptime(df$dateTime,%m/%d/%Y %H:%M)
df$dateTime - as.POSIXct(df$dateTime)
[...]
Hope this helps,
Rui
On Thu, Jul 12, 2012 at 8:17 AM, Bharat Warule bwar...@gmail.com wrote:
Hello,
I am using read.csv.sql first time for reading the large data file.If I am
ran this code that showns warning “closing unused connection”.
Is it I am missing any argument from my command or how to comeout from this
On Jul 13, 2012, at 12:35 AM, sanchez ana wrote:
Dear All,
I am using the function vuong from pscl package to compare 2 non nested
models NB1
(negative binomial I ) and Zero-inflated model.
NB1 - glm(, , family = quasipoisson), it is an
object of class: glm lm
zinb -
zeroinfl(
Thank you for the explanation. Good to know about the issue how double values
are constructed by a bit system. This makes me handling double values with
care in using it in R or aother languages control structures etc.
Thank you also for the hint concerning the Null vs. Zero vs.. issue. Yes,
the
Someone pointed me to this paper:
http://www.validlab.com/goldberg/paper.pdf
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__
Hello,
hiere is a small reproducible example.
All z.i which are NA should be transparent at the plot, but they are white
colored.
### Example image.plot regular x,y grid ###
x - seq(2,2.9,0.1)
y - seq(42,42.9,0.1)
z - matrix(seq(-5,4.9,0.1),nrow=10)
image.plot(x,y,z)
### overplotting by
Hi,
I need to generate HTML index file from Rd file.
example
Module Class Function
Description
Csvfunction csvoperations Module Description
http://test Read CSV Function
Jessica:
On Fri, Jul 13, 2012 at 1:35 AM, Jessica Streicher j.streic...@micromata.de
wrote:
two things:
- R always counts from 1, not from 0
- listmembers are accessed by using [[ ]] , not [ ]
FALSE! -- or at least not clearly stated:
x - list(a=letters[1:3],b=1:4)
x[[2]]
[1] 1 2 3 4
On Jul 13, 2012, at 04:27 , arun wrote:
Hello,
I saw your reply in nabble. Sorry about that. I thought the dataset had
only few columns.
#You can read first line of a file using:
readLines(foo.txt,n=1)[1]
#The more generic colname substitution
dat1-read.table(text=
2.5 3.6
On 13/07/2012 9:50 AM, Bert Gunter wrote:
Jessica:
On Fri, Jul 13, 2012 at 1:35 AM, Jessica Streicher j.streic...@micromata.de
wrote:
two things:
- R always counts from 1, not from 0
- listmembers are accessed by using [[ ]] , not [ ]
FALSE! -- or at least not clearly stated:
x -
Hello All,
Does anyone know where I can find information about how to do a power analysis
for Cox regression with a time-varying covariate using R or some other readily
available software? I've done some searching online but haven't found anything.
Thanks,
Paul
Try this:
x - read.table(text = 2.5 3.6 7.1 7.9
+ 100 3 4 2 3
+ 200 3.1 4 3 3
+ 300 2.2 3.3 2 4, header = TRUE, check.names = FALSE)
x
2.5 3.6 7.1 7.9
100 3.0 4.0 2 3
200 3.1 4.0 3 3
300 2.2 3.3 2 4
names(x)
[1] 2.5 3.6 7.1
BTW, is there an R command to read just the first line of the file?
scan() or readLines() will read as many lines of the file as you want.
Use the file() function to open a file connection so a subsequent
read.table() will start where scan() or readLines() finished. E.g.,
tfile - tempfile()
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I wanted. I appreciate the
clarification.
I think that multiple imputation is more flexible in some ways because
you can easy create different models for every variable. At the
Why does the subset not work in the which() version below?
Thank you
v1 - subset(t1,
version_1==as.character(100-1)
| version_1==as.character(100-2))
a-c(100-1, 100-2)
v1 - subset(t1, which(a==as.character(version_1)) != 0)
[[alternative HTML version deleted]]
Dear community,
I'm using rpart and would like to extract 95% prediction interval at each
terminal node from a regression tree (built with rpart, method= anova) Is
that possible?
Thanks in advance, u...@host.com as crossp...@host.com
--
View this message in context:
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g - length(table(G))
TP
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Dear R users,
I have question concerning box plot and it's whiskers. As I understood from
the description of the boxplot() function, if the range value is positive
the plot whiskers extend out from the box to the most extreme data points
defined by the values of the IQR times range (default
Thank you a ton! This is perfect!
From: Jean V Adams [via R] [mailto:ml-node+s789695n4636370...@n4.nabble.com]
Sent: Thursday, July 12, 2012 4:59 PM
To: Lauren Vogric
Subject: Re: Grabbing Indexes of a certain standard deviation
I wrote a little function called first() to help with situations
Thanks, I see that it is working in the sample data. My data, however, gives
me an error message:
data - strapplyc(text, batch[[l]])
Error in structure(.External(dotTcl, ..., PACKAGE = tcltk), class =
tclObj) :
[tcl] couldn't compile regular expression pattern: parentheses () not
balanced.
Thanks all you guys' help!
--
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__
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Hi,
I'm trying to run an R script from the command line and I do it in the
following way:
R CMD BATCH -q something.R output file*
*I write the R script from inside a Perl program (on UNIX of course!)
and execute the shell command using the function system.**What
intrigues me is the strange
They look fine to me.
luke
On Fri, 13 Jul 2012, Joshua Wiley wrote:
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I wanted. I appreciate the
clarification.
I think that multiple imputation is more flexible in some ways
Apologies -- replied to the wrong message.
luke
On Fri, 13 Jul 2012, luke-tier...@uiowa.edu wrote:
They look fine to me.
luke
On Fri, 13 Jul 2012, Joshua Wiley wrote:
Dear John,
Thanks very much for the reply. Looking at the optimizers, I had
thought that the objectiveML did what I
The whiskers will not be symmetrical if 1.5*IQR extends beyond the maximum or
minimum values. In that case, the whisker stops at the maximum or minimum.
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
On Fri, Jul 13, 2012 at 9:40 AM, mdvaan mathijsdev...@gmail.com wrote:
Thanks, I see that it is working in the sample data. My data, however, gives
me an error message:
data - strapplyc(text, batch[[l]])
Error in structure(.External(dotTcl, ..., PACKAGE = tcltk), class =
tclObj) :
[tcl]
Hi,
I have two non-normal distributions and use interquartile ranges as a
dispersion measure.
Now I am looking for a test, which tests whether the interquartile ranges from
the two distributions are significantly different.
Any idea?
Thanks,
joerg
[[alternative HTML version deleted]]
Hi,
I have been using lm in R to do a linear regression and find the slope
coefficients and value for R-squared. The R-squared value reported by R
(R^2 = 0.9558) is very different than the R-squared value when I use the
same equation in Exce (R^2 = 0.328). I manually computed R-squared and the
Hello,
To know why, just evaluate the condition, with 't1$' before 'version_1':
which(as.character(t1$version_1) %in% a) != 0
[1] TRUE TRUE
It allways evaluates to TRUE, therefore, subset() returns all rows.
See if this isn't simpler than both of your forms.
v2 - subset(t1, version_1 %in%
?plot.default
## documents the frame.plot arg and
?boxplot
## documents miscellaneous graphical parameters that go into ...
I would agree that the documentation is relatively poor, a legacy of the
original graphics system, which splits up parameter documentation among
par(), plot(), and specific
-Original Message-
I have question concerning box plot and it's whiskers. As I
understood from the description of the boxplot() function, if
the range value is positive the plot whiskers extend out from
the box to the most extreme data points defined by the values
of the IQR
Thanks Dennis and Mike... I'm getting it!!!
Sent from my Android
Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
To know why, just evaluate the condition, with 't1$' before 'version_1':
which(as.character(t1$version_1) %in% a) != 0
[1] TRUE TRUE
It allways evaluates to TRUE, therefore,
On 13.07.2012 12:51, Tom Porteus wrote:
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g - length(table(G))
Additional hint:
For debugging purposes, it is advisable to open
On 13.07.2012 12:51, Tom Porteus wrote:
It seems to not be recognising a function you are calling within your model -
taking a quick look you might want to check that table() is a function in
OpenBUGS.
g - length(table(G))
[This time also CCing the OP]
Additional hint:
For debugging
On 13.07.2012 03:18, elong zhang wrote:
Dear All:
Could anybody help me figure out why I get the Error message below while I
running the example code of bugs() function in R2OpenBUGS packages? I have
tried the code both in Win 7 and Ubuntu 12.04, but they show the same
message. My R version
First of all, although the question uses the term list, the object named
'a' is clearly not a list as R defines the term. It is a vector.
Thus a better example answer is
a - c('abc', 'def')
paste(a , collapse=' ')
and this works for however many elements there are in the vector a.
The
OK, got this far:
x1 - round(rnorm(10,60,3))
x2 - round(rnorm(10,65,3))
x3 - round(rnorm(10,70,3))
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method=stack,
pch=4,
offset=1/2,
col=blue,
lwd=2,
las=1)
Any ideas on how to get
On 13/07/2012 9:46 AM, Abhi Raghavan wrote:
Hi,
I'm trying to run an R script from the command line and I do it in the
following way:
R CMD BATCH -q something.R output file*
*I write the R script from inside a Perl program (on UNIX of course!)
and execute the shell command using the function
What does Excel give for the following data, where the by-hand formula
you gave is obviously wrong?
x - c(1, 2, 3)
y - c(13.1, 11.9, 11.0)
M1 - lm(y~x+0)
sqerr - (y- predict(M1)) ^ 2
sqtot - (y - mean(y)) ^ 2
1 - sum(sqerr)/sum(sqtot)
[1] -37.38707
Bill Dunlap
Spotfire,
You might want to look at
http://support.microsoft.com/kb/214230
entitled
Incorrect output is returned when you use the Linear Regression (LINEST)
function in Excel
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
Hello,
I am using the locfit to fit a non parametric glm model to data with a gamma
distributed response variable. In the parametric glm regression the diagnostics
were based on the study of the standardized deviance or pearson residuals. How
can I estimate the the standardized Pearson
Pamela
R squared with a non-zero, and with a zero intercept can be very different as
the regression line that you get with and without a zero intercept can be very
different. Have you plotted your data plot(k[,2],k[,1]) to see if a zero
intercept is reasonable for your data? Have you drawn the
While excluding the intercept may make sense, your formula for r^2 assumes
that there was an intercept (that is why mean(y) is in your expression for
sqtot).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: Pamela Krone-Davis [mailto:pkrone-da...@csumb.edu]
Sent: Friday, July 13,
On Jul 13, 2012, at 17:59 , arun wrote:
Hi Peter,
I copied the data from your email and run it again.
dat1-read.table(text=
2.5 3.6 7.1 7.9
100 3 4 23
200 3.1 4 3 3
300 2.2 3.3 24
,sep=,header=TRUE)
dat1
X2.5 X3.6 X7.1 X7.9
100
try something like
abline(h=1.9)
John Kane
Kingston ON Canada
-Original Message-
From: dwarnol...@suddenlink.net
Sent: Fri, 13 Jul 2012 09:54:35 -0700 (PDT)
To: r-help@r-project.org
Subject: Re: [R] Side by side strip charts
OK, got this far:
x1 - round(rnorm(10,60,3))
x2 -
On Jul 13, 2012, at 18:54 , darnold wrote:
OK, got this far:
x1 - round(rnorm(10,60,3))
x2 - round(rnorm(10,65,3))
x3 - round(rnorm(10,70,3))
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method=stack,
pch=4,
offset=1/2,
col=blue,
What I'm trying to do is create best fit line in R for a set of data points
and then remove all the outliers to re-create a best fit. I can't use IQR
because the outliers I have in mind are easily within the range, but way out of
line for the best fit, which is ruining the fit. I'd rather
Hello,
Or maybe the argument 'pos' of axis().
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method=stack,
pch=4,
offset=1/2,
col=blue,
lwd=2,
las=1,
xlim=c(53, 77),
xaxt=n)
axis(1, at = seq(55, 75,
I didn't actually see any question in this posting, but instead of removing the
outliers consider using a robust linear model.
library(MASS)
?rlm
The TeachingDemos package has a data set called outliers to show what can
happen when you iteratively remove outliers in the way you suggest.
Do you have a good reason to throw these points out?
On Fri, Jul 13, 2012 at 2:17 PM, David L Carlson dcarl...@tamu.edu wrote:
I didn't actually see any question in this posting, but instead of removing
the outliers consider using a robust linear model.
library(MASS)
?rlm
The
Hi Petr,
Yes, that's really very helpful.
Petr : Using this interpretation, AB occurs at lines 1,3,4 and not 1,3,5.
Is this correct?
Vineet : Yes , thats right sorry for the typo.
Petr: If some sequence contains several ocurrences of a pattern, for
example,
the sequence
A, B, A, B
How do I read/write liblinear models to files?
E.g., if I train a model using the command line interface, I might want
to load it into R to look the histogram of the weights.
Or I might want to train a model in R and then apply it using a command
line interface.
--
Sam Steingold
S+ and, I assume, R compute r^2 when there is no intercept as
sum(fitted(M1)^2) / sum(y^2)
where M1 is the fitted model and y the response.
See, for example,
http://web.ist.utl.pt/~ist11038/compute/errtheory/,regression/regrthroughorigin.pdf
for the derivation of this formula.
Bill Dunlap
They are due to measurement error, sample of a different population, or
... ? What is the unusual event? Does it explain something important
about the system that you are working on? I am not telling you not to
do what you are doing, but just writing things that I consider when I am
doing
For something like this the best (and possibly only reasonable) option
is to use simulation. I have posted on the general steps for using
simulation for power studies in this list and elsewhere before, but
probably never with coxph.
The general steps still hold, but the complicated part here will
Hi Joerg,
Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
which tests differences anywhere in two distributions, e.g. tails,
interquartiles and center.
Weidong Gu
On Fri, Jul 13, 2012 at 7:32 AM, Schaber, Jörg
joerg.scha...@med.ovgu.de wrote:
Hi,
I have two non-normal
A permutation test may be appropriate:
1. compute the ratio of the 2 IQR values (or other comparison of interest)
2. combine the data from the 2 samples into 1 pool, then randomly
split into 2 groups (matching sample sizes of original) and compute
the ratio of the IQR values for the 2 new
If not linear, then perhaps nlrob() in package robustbase.
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
- Original Message -
From: Stephen Sefick ssef...@gmail.com
To: Lauren Vogric
On 2012-07-13 13:33, Weidong Gu wrote:
Hi Joerg,
Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
which tests differences anywhere in two distributions, e.g. tails,
interquartiles and center.
The ks.test() function refers to the Kolmogorov-Smirnov test,
not the Wilcoxon test.
I just want to integrate a 3D data set along one dimension to obtain a
2D data set. Something like:
(given array d with dim nx,ny,nz ...)
data_int-array(dim=c(nx,ny))
for (n in 1:ny) {
for (m in 1:nx) {
data_int[m,n]-sum(d[m,n,])
}
}
The thing is, given R's facility with
On 2012-07-13 06:07, Chris82 wrote:
Hello,
hiere is a small reproducible example.
All z.i which are NA should be transparent at the plot, but they are white
colored.
### Example image.plot regular x,y grid ###
x - seq(2,2.9,0.1)
y - seq(42,42.9,0.1)
z - matrix(seq(-5,4.9,0.1),nrow=10)
Sorry, I meant Kolmogorov-Smirnov test.
Thanks Peter for correction.
Weidong
On Fri, Jul 13, 2012 at 4:56 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2012-07-13 13:33, Weidong Gu wrote:
Hi Joerg,
Seems Mann-Whitney-Wilcoxon test (ks.test in R) would do the work
which tests differences
I have a function that requires a distance matrix of class dist with
species names as row names. For the life of me, I cannot figure out how to
get dist() to include species names.
I am sure this must be easy, because a lot of packages and functions out
there require dist objects to have row
On 2012-07-13 15:55, Katharine Miller wrote:
I have a function that requires a distance matrix of class dist with
species names as row names. For the life of me, I cannot figure out how to
get dist() to include species names.
I am sure this must be easy, because a lot of packages and functions
On 2012-07-13 11:37, Rui Barradas wrote:
Hello,
Or maybe the argument 'pos' of axis().
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method=stack,
pch=4,
offset=1/2,
col=blue,
lwd=2,
las=1,
set.seed(42)
d - array(as.integer(round(runif(125)*10, 0)), dim=c(5, 5, 5))
data_int - apply(d, c(1,2), sum)
-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77840-4352
- Original Message -
From: Sam B
Very nice suggestion. I am getting some very kind help here.
x1 - round(rnorm(10,60,3))
x2 - round(rnorm(10,65,3))
x3 - round(rnorm(10,70,3))
stripchart(list(sample1=x1,sample2=x2,sample3=x3),
method=stack,
pch=4,
offset=1/2,
col=blue,
lwd=2,
Hi,
I hope that folks can give me some simple approaches to taking the data set
below, which is accumulated in two columns called long and group, then
arrange the data is the long column into a data frame containing five
variables: Group 1, Group 2, Group 3, Group 4, and Group 5. I am
hoping for
On 7/13/2012 8:37 PM, darnold wrote:
Hi,
I hope that folks can give me some simple approaches to taking the data set
below, which is accumulated in two columns called long and group, then
arrange the data is the long column into a data frame containing five
variables: Group 1, Group 2, Group 3,
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