Dear Any
Thanks for your response. Maybe I did not explain the behavior well. I am
aware that the Not Responding is a windows default. What I was trying to
explain is that once the process that generated the Not Responding is
finished and I can use R for othe computations the Not
Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro).
No idea why...
Andy
From: Francisco J. Zagmutt
Dear Any
Thanks for your response. Maybe I did not explain the
behavior well. I am
aware that the Not Responding is a windows default. What I
was trying to
How do I convert the output of cor(x) to a columnar format?
Ex. from format below
XYZ
X 1.0 0.9 0.5
Y 0.9 1.0 0.1
Z 0.5 0.1 1.0
to format below
X X 1.0
X Y 0.9
X Z 0.5
Y X 0.9
Y Y 1.0
Y Z 0.1
Z X 0.5
Z Y 0.1
Z Z 1.0
Thanks!
Omer
Liaw, Andy wrote:
Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro).
No idea why...
I guess this is a Windows bug, because I have seen it in other
applications as well. Hence I don't think we should waste our time here ...
Uwe Ligges
Andy
From: Francisco J.
Omer Bakkalbasi wrote:
How do I convert the output of cor(x) to a columnar format?
Ex. from format below
XYZ
X 1.0 0.9 0.5
Y 0.9 1.0 0.1
Z 0.5 0.1 1.0
to format below
X X 1.0
X Y 0.9
X Z 0.5
Y X 0.9
Y Y 1.0
Y Z 0.1
Z X 0.5
Z Y 0.1
Z Z 1.0
See, e.g.,
Dear Uwe
I have not seen this behavior in other windows applications but I
definitivelly agree with you that it is probably not worth spending time on
this trivial issue.
Thanks
Francisco
From: Uwe Ligges [EMAIL PROTECTED]
To: Liaw, Andy [EMAIL PROTECTED]
CC: 'Francisco J. Zagmutt'
[EMAIL
Yes, it is a Windows bug: the frame is controlled by Windows and not by R.
On Fri, 17 Jun 2005, Uwe Ligges wrote:
Liaw, Andy wrote:
Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro).
No idea why...
I guess this is a Windows bug, because I have seen it in other
On Thu, 16 Jun 2005 [EMAIL PROTECTED] wrote:
This question is partly about R and partly out of my ignorance about
time series.
I want to regress one time series on another, taking into account the
autocorrelation (in an AR1 model) within each series. I am interested in how
the standard
What version of R is this (please do see the posting guide)?
In both 2.1.0 and 2.1.1 beta I get
all
Promoter ip.x ip.y ip
130 40 40
240 40 40
3a 10 NA NA
4c 20 20 20
5b NA 15 15
6d NA 30 30
so cannot reproduce your
On Thu, 16 Jun 2005, Piotr Majdak wrote:
I'm looking for a solution to analyse data, which consists of
dichotomous responses (yes/no) for 2 multinomial ordinal variables.
Please explain how you get a binary response for a `multinomial ordinal
variables'? If you intend these variables to be
apply() is just a for() loop internally so why do you expect it to be
faster?
Some comments:
1) Here predict() is just extracting the fitted values.
2) Using lm.fit will be faster if fitted values is all you want.
3) You are actually regressing each column on all other columns plus an
On Thu, 16 Jun 2005, Joshua Gilbert wrote:
I need to compute generalized eigenvalues. The eigen function in base
doesn't do it and I can't find a package that does.
They are very rarely used in statistics, so this is not surprising.
I presume you mean solving Ax = lambda B x: if B is
Paul,
thanks! An elegant solution.
Andrew
--
Andrew Robinson Ph: 208 885 7115
Department of Forest Resources Fa: 208 885 6226
University of Idaho E : [EMAIL PROTECTED]
PO Box 441133W : http://www.uidaho.edu/~andrewr
Moscow ID
Dear R-users,
Earlier this year I posted a message to this list regarding
negative binomial mixed models in R. It was suggested that
the program I had written should be turned into an R-package.
This has now been done, in collaboration with David Fournier
and Anders Nielsen.
The R-package
Hello
I would like to generate covariance matrix with autoregressive
structure. I saw some functions in nlme such as corAR1 for example but I
don't know how to use it for my goal. Could someone help me or advise me
another function?
Thank you in advance
Caroline
Hi,
I have a plot and a custom axis labeling, e.g.
x-c(...)
plot(x,axes=FALSE)
axis(2)
axis(1,1:50,c(label1,...,label50))
now since the labels are quite long, only a few fit on the page.
Can I rotate each label by 90 degree counterclockwise
(so that they are vertical)
ARMAacf() will give you the acf for an autoregression, and toeplitz()
wil turn this into a correlation matrix. Then just multiply by the
desired variance.
I am not sure what this has to do with your subject line, and ?arima.sim
might be helpful for that.
On Fri, 17 Jun 2005, Caroline TRUNTZER
Prof Brian Ripley wrote:
On Thu, 16 Jun 2005, Piotr Majdak wrote:
I'm looking for a solution to analyse data, which consists of
dichotomous responses (yes/no) for 2 multinomial ordinal variables.
Please explain how you get a binary response for a `multinomial ordinal
variables'? If
Prof Brian Ripley [EMAIL PROTECTED] writes:
On Thu, 16 Jun 2005, Joshua Gilbert wrote:
I need to compute generalized eigenvalues. The eigen function in base
doesn't do it and I can't find a package that does.
They are very rarely used in statistics, so this is not surprising.
An aside,
As I suggested before, a binomial logistic model is appropriate here, not
a Poisson log-linear one. (They are equivalent, but the binomial version
is easier to interpret and less wasteful to fit.)
You have still not defined v' and w', nor the scores (are they estimated
or not). But the model
Hi,
I am new to this list as a poster, but a reader for some time.
I've using R for several weeks now, and I have a lot of questions about
certain procedures. Here I go:
I want to test if there are differences in the time spent by pollinators
visiting flowers of a given plant species,
Hi,
I am new to this list as a poster, but a reader for some time.
I've using R for several weeks now, and I have a lot of questions about
certain procedures. Here I go:
I want to test if there are differences in the time spent by pollinators
visiting flowers of a given plant species,
BoM DS wrote:
Hi,
I have a plot and a custom axis labeling, e.g.
x-c(...)
plot(x,axes=FALSE)
axis(2)
axis(1,1:50,c(label1,...,label50))
now since the labels are quite long, only a few fit on the page.
Can I rotate each label by 90 degree counterclockwise
(so that they
Maybe like:
dat
X Y Z
X 1.0 0.9 0.5
Y 0.9 1.0 0.1
Z 0.5 0.1 1.0
datrow - stack(as.data.frame(dat))
datrow$X=rownames(dat)
datrow
values ind X
11.0 X X
20.9 X Y
30.5 X Z
40.9 Y X
51.0 Y Y
60.1 Y Z
70.5 Z X
80.1 Z Y
91.0 Z Z
BoM DS wrote:
Hi,
I have a plot and a custom axis labeling, e.g.
x-c(...)
plot(x,axes=FALSE)
axis(2)
axis(1,1:50,c(label1,...,label50))
now since the labels are quite long, only a few fit on the page.
Can I rotate each label by 90 degree counterclockwise
(so that they are
Prof Brian Ripley wrote:
You have still not defined v' and w', nor the scores (are they estimated
or not). But the model I suggested is such a model with scores 1,2,...
Sorry for that, here it is:
scores v and w: integer scores, reflecting the ordering of columns/rows.
Agresti suggests to
2005/6/17, Sundar Dorai-Raj [EMAIL PROTECTED]:
You should (re-)read ?axis, which points you to the las parameter:
x - 1:50
plot(x, axes = FALSE)
axis(1, x, paste(label, x), las = 2, cex.axis = 0.5)
axis(2)
box()
thank you very much. This answer helps me in several ways:
1. it solves my
Hi,
I got the fowllowing error message when I run pamr. Could you please advise me
what does this error mean?
Many thanks
--
mydata - pamr.from.excel(datgrp4, 352, sample.labels=TRUE)
Read 812768 items
Read in 2307 genes
Read in 350 samples
Read in 350 sample labels
Make
HI!
I have 2D feature vectors and 2 classes. I want to solve this
classification problem with SVM. I create an object XOR of the mlbench
library.
Then I replace the values of the XOR object with my values. Then I do a
plot and I have my data plotted and coloured with the labels from the
XOR
Dear list,
I am trying to plot the local variance in a moving window on a
dataset. The function that I am trying to use for this is wapply from
gtools.
However, from the lattice panel function code:
snip
cat(x)
cat(y)
wapply(x,y,method=range,width=1/10,fun=sd,na.rm=TRUE)
I just realised that I had forgotten to copy this to the list.
-- Forwarded message --
From: Deepayan Sarkar [EMAIL PROTECTED]
Date: Jun 15, 2005 12:13 PM
Subject: Re: [R] lattice, panel.grid, and scales=list(tick.number=XXX)
To: Berton Gunter [EMAIL PROTECTED]
On 6/14/05,
Thorstensen Nicolas wrote:
HI!
I have 2D feature vectors and 2 classes. I want to solve this
classification problem with SVM. I create an object XOR of the mlbench
library.
Then I replace the values of the XOR object with my values. Then I do a
plot and I have my data plotted and
luk wrote:
Hi,
I got the fowllowing error message when I run pamr. Could you please advise
me what does this error mean?
Many thanks
--
mydata - pamr.from.excel(datgrp4, 352, sample.labels=TRUE)
Read 812768 items
Read in 2307 genes
Read in 350 samples
On Fri, 17 Jun 2005 13:28:21 +0200,
Thorstensen Nicolas (TN) wrote:
HI!
I have 2D feature vectors and 2 classes. I want to solve this
classification problem with SVM. I create an object XOR of the mlbench
library.
Then I replace the values of the XOR object with my values. Then
This is a rather obscure question, I realize. I have written to the
package author but have not heard back as of yet. I have read the
README in the package, as well, but it didn't give me enough detail to
diagnose the problems that I am having. (I did edit out the LAM-MPI
check in
Something like:
dat - data.frame(x=runif(10), y=runif(10), z=runif(10))
m - cor(dat)
m
x y z
x 1.000 0.1183305 0.1096394
y 0.1183305 1.000 -0.2819285
z 0.1096394 -0.2819285 1.000
mat2col - function(m) {
+ m2 - matrix(m, ncol=1)
+
Excellent! This is the most flexible and intuitive option. Thanks!
Omer
Cell: (914) 671-7447
-Original Message-
From: Liaw, Andy [mailto:[EMAIL PROTECTED]
Sent: Friday, June 17, 2005 8:40 AM
To: '[EMAIL PROTECTED]'; r-help@stat.math.ethz.ch
Subject: RE: [R] CORRELATION MATRIX
Thanks for your quick responses, Gabor and Brian.
I'm currently running R version 1.9.1 on Linux. Actually, I have just
tested this on R v.2.1.0 running under Windows XP, and indeed, as you both
indicate, the problem does not exist on that version for that OS. So, at an
appropriate time I'll
I am trying without success to build the R-2.1.0 sources on Windows
2003 server.
I have MinGW/bin in front of cygwin/bin in the Windows path.
However, I try to build, I get failures trying to include headers
which are not part of MinGW. I am definitely using the MinGW
compilers, so why
Hi!
I have had this problem for a long time. I have tried to study the manuals and
search the mailing lists, but I can not solve this. I think there has to be one
simple solution to this, but I just can not find it.
I have saved the data in excel (csv-format). Then I read the data in R e.g.
We're happy to announce the CRAN release of sp, an R package which
has new-style classes and methods for spatial data, version 0.7-9.
Spatial data types that sp implements are: points, grids, lines, and
polygons (i.e., rings) optionally with holes. Methods include
+ the usual print, summary,
i'm trying to build a little summary table for the contents of a data frame.
t-sapply(macro, data.class)
c-sapply(macro, length)
m-sapply(macro, mean, na.rm=T, digits=2)
cbind(type=t, n=c , mean=m)
I want to drop the variables that are factors so I can include -max- and
-min- in my table.
On Fri, 17 Jun 2005, Johanna Sundvik wrote:
However, this Mean1 is categorical when it should be real numbers.
Mean1
[1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364
Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693
Why R does not understand
Hi R users,
I am trying to calculate MonteCarlo from the multivariate normal
distribution. I am utilizing the parameters vector (how mean) and
covariance matrix (or 1/hessian) how input.
Can anyone provide guidance on how I could do this?
Thank you.
Hi,
To obtain estimates for some missing values in my data I fitted a
linear regression and then used the command fitted(model) to get the
fitted values from the model, but R doesn't return any values for the
NA's. I can calculate the fitted values from the estimates obtained
from the summary
Hi Brian (and the list of course!),
I still have problems analysing data in R, because I don't know how to
tell glm() use row-effect model, please. The models are well defined
by Agresti, but can't get the link from the theory to the
implementations in R. Different names, definitions and no
Thomas Lumley wrote:
On Fri, 17 Jun 2005, Johanna Sundvik wrote:
However, this Mean1 is categorical when it should be real numbers.
Mean1
[1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364
Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693
In my experience, this has always been due to the presence of
non-numeric values in the input.
In the example you show, it is not obvious that there is any. I would
start by first inspecting the input file very carefully, using a text
editor outside of R. Since your example appears to have
On 6/17/2005 11:03 AM, Juan Carlos Quiroz Espinosa wrote:
Hi R users,
I am trying to calculate MonteCarlo from the multivariate normal
distribution. I am utilizing the parameters vector (how mean) and
covariance matrix (or 1/hessian) how input.
Can anyone provide guidance on how I could
try this
# to get only the factors
macro.f - macro[sapply(macro, is.factor)]
to drop the factors
macro.nf - macro[sapply(macro, !is.factor)]
# to get only numerics
macro.n - macro[sapply(macro, is.numeric)]
and so on.
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
RSiteSearch(random multivariate normal) produced 82 hits, the
fourth of which was for multivariate normal distribution, function
pmvnorm in library mvtnorm, which also includes a function rmvorm, that
should do what you want.
Buena suerte. spencer graves
Juan Carlos
a - data.frame(a = 1:2, b = c(a, b), c = I(c(a, b)))
a - a[ , !sapply(a, class) %in% factor]
-Original Message-
From: E. Michael Foster [mailto:[EMAIL PROTECTED]
Sent: Friday, June 17, 2005 10:58 AM
To: r-help@stat.math.ethz.ch
Subject: [R] drop elements of vector by class
i'm trying
Have you also tried lmer in library(lme4)? This is newer and
better in many ways. Unfortunately, I see only one example in the Help
file, but you might be able to figure out how to use lmer from the help
file and from the 125 hits on RSiteSearch(lmer).
The definitive work
Have you considered:
set.seed(1)
tstDF - data.frame(x=1:4, y=c(NA, 2:4+rnorm(3)))
fit - lm(y~x, tstDF)
predict(fit)
234
1.678441 2.573854 3.469266
predict(fit, tstDF)
1 2 3 4
0.7830284 1.6784410 2.5738536 3.4692662
I've struggled with this myself in the past. I've recently started
using the following:
File - pair.txt
# File name with path if different from getwd()
readLines(File, n=9)
The function readLines reads the first n lines as n individual
character strings. From this, you
Question : Is it possible to create a function, using a for ifelse
function, inside sapply, to compare the values in one data frame to a set
of upper and lower limits in another data frame, same number of columns.,
Take the values which meet the requirements TRUE and create a new data
frame
It depends on the na.action used in lm(), which defaults to na.omit. Here's
an example:
x - c(1:5, NA, 7:10)
y - rnorm(x)
fitted(lm(y ~ x))
1 2 3 4 5 7
8
0.68680104 0.47913875 0.27147646 0.06381417 -0.14384812 -0.55917271
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for each
education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
I posted a bad example. The question would only make sense if NAs are all
in the response. Here's try #2:
I believe you can not get fitted values for cases where y is missing
directly from the model object. You can, however, use predict(lm.object,
newdata=mydata[!complete.cases(mydata)]) to
On Fri, 17 Jun 2005 07:39:30 +1000
Stephen Choularton [EMAIL PROTECTED] wrote:
Hi
I can get all my features by doing this:
logistic.model = glm(similarity ~ ., family=binomial, data =
cData[3001:3800,])
I can get the product of all my features by this:
logistic.model =
How to trim the leading and trailing white space off of a string?
If the variable is E I need to convert it to E.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
and the newest R-new in the www.r-project.org has an article about how to use
the lmer function.
On Fri, 17 Jun 2005 08:43:41 -0700
Spencer Graves [EMAIL PROTECTED] wrote:
Have you also tried lmer in library(lme4)? This is newer and
better in many ways. Unfortunately, I see only
Hi
I want to get the values of a vector which I have its indices. How it is
possible?
For example after clustering , I can access to the indices of the first cluster
using:
first- which(clusters$clustering==1)
first give me the indices, but how can I access to the values?
Thanks a lot and
Here I go again with ave():
mydata$md - ave(mydata$income, mydata$education, FUN=median, na.rm=TRUE)
IMHO it's one of the most under-rated helper functions in R.
Andy
From: Dimitri Joe
Hi there,
I have a data frame (mydata) with 1 numeric variable (income)
and 1 factor (education). I
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/14493.html
Andy
From: Omar Lakkis
How to trim the leading and trailing white space off of a string?
If the variable is E I need to convert it to E.
__
R-help@stat.math.ethz.ch
You can use tapply() to compute the medians, as in
meds - tapply(mydata$inc,INDEX=mydata$ed,FUN=median)
then create a new column with the medians as
medianEd - meds[mydata$ed]
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitri
These two lines worked for me:
rst - tapply(mydata$income, mydata$education, median)
mydata$md - rst[mydata$education]
Here's my cheesy example:
mydata - data.frame(income= round(rnorm(3, 55000, 1)),
+ education = letters[rbinom(3, 4, 1/2)+1])
rst -
try this:
x.1 - data.frame(income=runif(100)*1,
educ=sample(c('hs','col','none'),100,T))
x.1
income educ
1 5930.30882 col
2 5528.83222 hs
3 5967.04041 hs
4 3926.30682 hs
5 2603.75924 none
...
x.2 - tapply(x.1$income, x.1$educ, mean)
x.2
col
RSiteSearch(trim) will give you a lot of answers. You cal also use the
higher level function trim{R.oo} i.e.:
library(R.oo)
x=e
trim(x)
[1] e
From: Omar Lakkis [EMAIL PROTECTED]
Reply-To: Omar Lakkis [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] trim a string
Date: Fri, 17
Better code for this purpose is in example(grep).
`white space' and `a blank' are not necessarily the same thing.
On Fri, 17 Jun 2005, Liaw, Andy wrote:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/14493.html
Andy
From: Omar Lakkis
How to trim the leading and trailing white space off
Hi,
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitri Joe
Sent: Friday, June 17, 2005 7:01 PM
To: R-Help
Subject: [R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income)
and 1 factor (education). I
How to trim the leading and trailing white space off of a string?
If the variable is E I need to convert it to E.
gsub('^[[:space:]]+', '',E )
gsub('[[:space:]]+$', '',E )
as in R-2.1.0/library/base/html/grep.html
On 6/17/2005 12:33 PM, [EMAIL PROTECTED] wrote:
Question : Is it possible to create a function, using a for ifelse
function, inside sapply, to compare the values in one data frame to a set
of upper and lower limits in another data frame, same number of columns.,
Take the values which
I commend you to (a) the recent article by Doug Bates on Fitting
nonlinear mixed models in R pp. 27-30 in the latest issue of R News
available from www.r-project.org - Newsletter and (b) Doug's book
with Pinheiro (2000) Mixed-Effects Models in S and S-PLUS (Springer). I
suggest you
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained) }
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
Yes, %in% or is.element().
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mike R
Sent: Friday, June 17, 2005 4:32 PM
To: r-help@stat.math.ethz.ch
Subject: [R] an operator for contains
k = c(1:9)
if( length( which(k==3) ) ){
Does '?%in%' or '?match' meet your needs?
spencer graves
Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along
See match(). Also intersect().
The documentation for operators is, I think mostly at the top of
the index page for the base package, the one you searched. The
relevant one is %in%. I guess Value matching didn't ring the
right bell.
On 06/17/05 13:31, Mike R wrote:
k = c(1:9)
if( length(
%in%
R k - 1:9
R 3 %in% k
[1] TRUE
R 33 %in% k
[1] FALSE
Arne
On Friday 17 June 2005 22:31, Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not
contained) }
is therre a simple way to test if a vector/list contains a particular
value?
for example an
On 6/17/05, Mike R [EMAIL PROTECTED] wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
Yes, several. Here's one:
k - 1:9
if(any(k == 3)) {
cat(is an element\n) } else {
On Fri, 17 Jun 2005, Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
value %in% vector
more generally, is the a documentaion page that lists/describes all
I thought the point of adjusting the R^2 for degrees of
freedom is to allow comparisons about goodness of fit between
similar models with different numbers of data points. Someone
has suggested to me off-list that this might not be the case.
Is an ADJUSTED R^2 for a four-parameter, five-point
Hello:
I would like to be able to do a 3D scatter plot from 3 variables, 2
independent and 1 dependent. The closest R function I could find for
this is cloud. However cloud uses, as input, a matrix where the value
of each matrix element is the dependent variable value at that matrix
wow. thanks everyone for the multitude of suggestions !
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
James Salsman [EMAIL PROTECTED] writes:
I thought the point of adjusting the R^2 for degrees of
freedom is to allow comparisons about goodness of fit between
similar models with different numbers of data points. Someone
has suggested to me off-list that this might not be the case.
Is an
2. I wasn't aware that there is an online-help. (was using the
tutorial on the web, which is nat at all detailed.
help(), help.search(), and apropos() may all be useful.(thanks Sarah !!)
and RSiteSearch()
You might find this handy too
example(axis)(thanks Brian !!)
2. I wasn't aware that there is an online-help. (was using the
tutorial on the web, which is nat at all detailed.
help(), help.search(), and apropos() may all be useful.(thanks Sarah !!)
and RSiteSearch()
You might find this handy too
example(axis)(thanks Brian !!)
I am having trouble figuring out this one.
I want to do a one way anvoa with 13 levels. mydata is in a vector with
length 65. each level has 5 repeats. but it contains NA.
I made mygroup-gl(13, 5, 65, labels=(...))
anova(lm(mydata ~ mygroup))
it gives following error:
Error in
Dear all:
Here is my problem:
Example data:
dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17))
If I wanted to aggregate each level of column dat$x I
could use:
aggregate(dat$y,list(x=dat$x),sum)
But I just want to aggregate two levels (?c? and ?d?)
to obtain a new level ?e?
I am expecting
# CASE 1
# The following plots a single cell or block for all three location 0,1,2.
y - rep(2,8)
hist(y)# why is this a single block?
hist(y,xlim=c(0,2))# same thing
hist(y,breaks=2) # same
# CASE 2
# adding a different value, plots as expected
y - append(y,0)
hist(y)
Test that:
mydata - as.numeric(mydata)
anova(lm(mydata ~ mygroup))
2005/6/18, Lei Jiang [EMAIL PROTECTED]:
I am having trouble figuring out this one.
I want to do a one way anvoa with 13 levels. mydata is in a vector with
length 65. each level has 5 repeats. but it contains NA.
I made
On 6/17/05, Lauren, Peter [EMAIL PROTECTED] wrote:
Hello:
I would like to be able to do a 3D scatter plot from 3 variables, 2
independent and 1 dependent. The closest R function I could find for
this is cloud. However cloud uses, as input, a matrix where the value
of each matrix
In case #1, the argument 'breaks' can break the histogram cells:
hist(y, breaks=c(0, 0.5, 1.0, 1.5, 2.0)) # not the same ^_^
2005/6/18, ap [EMAIL PROTECTED]:
# CASE 1
# The following plots a single cell or block for all three location 0,1,2.
y - rep(2,8)
hist(y) # why is this a single
On 6/17/05, alex diaz [EMAIL PROTECTED] wrote:
Dear all:
Here is my problem:
Example data:
dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17))
If I wanted to aggregate each level of column dat$x I
could use:
aggregate(dat$y,list(x=dat$x),sum)
But I just want to aggregate two levels (c
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