On 24/04/2014 22:33, Greg Snow wrote:
Many, probably even most (but I have not checked) of the datasets
available in R packages have help files with a references section.
That section should lead you to an original source that may have the
copyright information and is what should be referenced.
On 24 Apr 2014, at 23:56, Greg Snow 538...@gmail.com wrote:
library(Matrix)
adjM - Matrix(0,nrow=10,ncol=10)
locs - cbind( sample(1:10), sample(1:10) )
vals - rnorm(10)
adjM[ locs ] - vals
... and once you've got your data in this format, why not construct the sparse
matrix
Stefan Evert stefa...@collocations.de
on Fri, 25 Apr 2014 09:09:31 +0200 writes:
On 24 Apr 2014, at 23:56, Greg Snow 538...@gmail.com wrote:
library(Matrix)
adjM - Matrix(0,nrow=10,ncol=10)
locs - cbind( sample(1:10), sample(1:10) )
vals - rnorm(10)
Hello list,
I would be very grateful if somebody could suggest me which kind of call
may be used to perform a manova with unbalanced data avoiding NA values.
Thanks a lot
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__
R-help@r-project.org mailing list
Hi,
Im trying to plot a linear line on the scatter plot using the pairs()
function. At the moment the line is non linear. However, I want a linear
line and the associated R value.
Here is my current code:
panel.cor.scale - function(x, y, digits=2, prefix=, cex.cor)
{
usr - par(usr);
Hi
Have a look on how panel.smooth is defined:
panel.smooth
function (x, y, col = par(col), bg = NA, pch = par(pch),
cex = 1, col.smooth = red, span = 2/3, iter = 3, ...)
{
points(x, y, pch = pch, col = col, bg = bg, cex = cex)
ok - is.finite(x) is.finite(y)
if (any(ok))
Hi, Can someone explain why it says the cases are complete. It seems they
are not.
head(x)
rnumExam.Title Patient.ID Age Sex Impression2
11 Temporal bone3529243 009Y M
22 Temporal bone3529243 010Y M
33 Head3508570 002M M
44Sacrum
Great, thanks Frede,
This works perfectly. Ive tested these correlations with ones in sigma plot
and excel and for some reason the r squared value is different. Any idea
why this is? Thanks again for your help.
Cheers
On Fri, Apr 25, 2014 at 11:57 AM, Frede Aakmann Tøgersen
At 13:15 24/04/2014, Verena Weinbir wrote:
Hello!
I am using the metafor package for my master's thesis as an R-newbie. While
calculating effectsizes from my dataset (mean values and
standarddeviations) using escalc shouldn't be a problem (I hope ;-)), I
wonder how I could at this point
No it is difficult to say when I donât know what you did in Excel and Sigma
Plot. The latter I donât know and do not use it my self.
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730
You have no missing data. Note that the string is not missing! You need to
code missings as NA. Have look at ?is.na
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Hi R users,
is there any possibility that the file created by:
dump.frames(dumpto = file.name, to.file = TRUE)
has any upper size limitations? I use the command above to debug R Code
run on a cluster environment. Looking at the logs, I get an indication
that an error occoured but the
Dear R users;
I have a question about Cramer Rao upper/lower bounds
Is it possible to compute Crammer Rao upper/lower bounds from residuals
and corresponding covariance matrices ?
Any suggestions will be appreciated, thanks in advance.
M.O
__
I just plotted one variable against the other, added a trend line and got
my R squared value. I presumed the absolute correlation from pairs() would
be the same?
On Fri, Apr 25, 2014 at 12:24 PM, Frede Aakmann Tøgersen
fr...@vestas.comwrote:
No it is difficult to say when I donât know what
Hey Frede,
I found the answer,
thanks for your help.
Shane
On Fri, Apr 25, 2014 at 12:42 PM, Shane Carey careys...@gmail.com wrote:
I just plotted one variable against the other, added a trend line and got
my R squared value. I presumed the absolute correlation from pairs() would
be the
Do you know how I would put in the R squared value rather than the
correlation by any chance?
Cheers
On Fri, Apr 25, 2014 at 12:56 PM, Shane Carey careys...@gmail.com wrote:
Hey Frede,
I found the answer,
thanks for your help.
Shane
On Fri, Apr 25, 2014 at 12:42 PM, Shane Carey
Dear Sergio,
The Anova() function in the car package can perform MANOVA with a multivariate
linear model fit to unbalanced data by lm() -- see the examples in ?Anova. I'm
not sure what you mean by avoiding NA values, however. With the default
na.action, which is na.omit, lm() will perform a
Well, you could upgrade to the latest version of R, as the R-Forge
page states, R-Forge provides these binaries only for the most recent
version of R, but not for older versions.
If there's no Windows binary, it' s possible that some necessary
requirement isn't available for Windows. You'll need
Hello R-Community,
I have quarterly panel-data and not surprisingly missing values, in a few
variables, which are differently distributed around the panel.
Now I want to run different unit-root tests.
For ADF on the pooled data set, I chose CADF-package, which can determine the
number
At 12:33 25/04/2014, you wrote:
Thank you very much for your reply and the book recommendation, Michael.
Yes, I mean Cohen's d - sorry for the typo :-)
Just to make this sure for me: There is no
possibility to integrate stated Cohens' ds in an
R-Metaanalysis (or a MA at all), if there is no
Because, my inbox shows a cutted-email-version, once again:
Hello R-Community,
I have quarterly panel-data and not surprisingly missing values, in a few
variables, which are differently distributed around the panel.
Now I want to run different unit-root tests.
For ADF on the pooled data set, I
Hi,
May be this helps:
library(plm)
data(Grunfeld)
Grunfeld$inv[c(2,8)] - NA
GrunfeldNew - subset(Grunfeld, !year %in% year[is.na(inv)])
x - data.frame(split(GrunfeldNew$inv, GrunfeldNew$firm))
purtest(x, pmax = 4, exo = none, test = levinlin,lags=SIC)
# Levin-Lin-Chu Unit-Root Test (ex.
Dear Rs,
I am re-executing some older code. It does work in the ancient R 2.12.0 which I
still have on my PC but with the new version R 3.1.0 it does not work any more
(but some other new stuff, which won't work with 2.12).
The problem arises in context with the systemfit package using the
On Apr 24, 2014, at 6:40 PM, christian millan wrote:
Hi,I have been trying to make my axis fonts and axis labels fonts in bold
even when the I write the right command. I writing font.lab=2, font.axis=2
but the bold fonts don't show up. Any help?
There are three different plotting systems
Hi guys,
I will be very grateful if you guys can do me a little favor on R. I am
calculating the sensitivity and specificity for a 2*2 matrix, such as
t
01
0 1427 110
1 271 166
My codes are: sens - function(ct) { ct[2,2] / sum(ct[,2]) }
spec -
On Apr 25, 2014, at 10:58 AM, Si Qi L. wrote:
Hi guys,
I will be very grateful if you guys can do me a little favor on R. I am
calculating the sensitivity and specificity for a 2*2 matrix, such as
t
01
0 1427 110
1 271 166
My codes are: sens - function(ct) {
On Apr 25, 2014, at 9:17 AM, Werner W. wrote:
Dear Rs,
I am re-executing some older code. It does work in the ancient R 2.12.0 which
I still have on my PC but with the new version R 3.1.0 it does not work any
more (but some other new stuff, which won't work with 2.12).
The problem
Try this:
sens - function(ct) { ct[2,2] / sum(ct[,2]) }
spec - function(ct) { ct[1,1] / sum(ct[,1]) }
myt - matrix( c(1427,271,110,166), ncol=2)
sens(myt)
[1] 0.6014493
spec(myt)
[1] 0.8404005
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
If you know the d-value and the corresponding group sizes for a study, then
it's possible to add that study to the rest of the dataset. Also, if you only
know the test statistic from an independent samples t-test (or only the p-value
corresponding to that test), it's possible to back-compute
Hello all,
I looking for confirmation of what I'm seeing. The qq plot in gam.check
and qq.gam is not standardizing the residuals.
The current help doesn't suggest they're standardized.
Somehow I found, online, a help for gam.check from version [Package
mgcv version 1.3-23 Index]:
If the fit
Hi Michael,
You seem to have a quite recent mgcv if you are using qq.gam, so a help
file for version 1.3 is probably not going to be much help (gam.fit2 no
longer exists, for example).
By default qq.gam plots deviance residuals (see ?qq.gam). So the default
is standardization. When possible
Hi all,
I recently upgraded to R-3.1.0 from R-3.0.2. Things seem to be fine, but
when trying to plot a ggplot image with transparency, I get the following
issue:
Warning message:
In grid.Call.graphics(L_polygon, x$x, x$y, index) :
semi-transparency is not supported on this device: reported
On 04/25/2014 07:26 PM, Fong Chun Chan wrote:
Hi all,
I recently upgraded to R-3.1.0 from R-3.0.2. Things seem to be fine, but
when trying to plot a ggplot image with transparency, I get the following
issue:
Warning message:
In grid.Call.graphics(L_polygon, x$x, x$y, index) :
So, I know that's a confusing Subject header.
Here's similar data:
tmp - data.frame(matrix(
c(rbinom(1000, 1, .03),
array(1:127, c(1000,1)),
array(format(seq(ISOdate(1990,1,1), by='month',
length=56),
Dear all,
When I try to run IDEAL it gives me an error saying: Error in
ideal(mydatarc.working, codes = mydatarc.working$codes, dropList = list(lop
= 0), : object 'normalize' not found.
Below is what I did
require(pscl)
mydata-read.csv(file.choose())
summary(mydata)
legNames-mydata[,1]
Hi Arun,
Thanks a lot for your script. Ill work on it tomorrow.
Cheers
On 24/04/2014, at 11:54 AM, arun kirshna [via R]
ml-node+s789695n4689356...@n4.nabble.com wrote:
HI,
I guess you got an output like this using my script:
##Please use ?dput() to show the example data.
FA -
Dear R users,
is there any way to access/print/save the content of an environment in
which an error occoured? Image the following testcase:
test = function() {
b = 3
plot(notavailable)
}
dump.frames.mod = function() {
save(list=ls(parent.frame(1)), file='dummy.RData')
}
Hi,
I am using hclust and cutree to cluster a data frame y and cut it into few
clusters as follows
y
V1 V2 V3 V4
A 1 2 3 4
B 5 6 7 8
C 9 10 11 12
D 13 14 15 16
E 17 18 19 20
clu-hclust(dist(y),method=complete)
clu-hclust(dist(y),method=complete)
clu
Call:
hclust(d = dist(y),
Hello
I am running some k-medoids analysis to see if that algorithm will work better
on my dataset compared to the k-means I have been using so far.
When running k-means I used to specify the nstart option, equal to a certain
number, depending on how many random sets of clusters I wanted.
Could
Hello,
When using hist() with a POSIXt date that has 00 seconds an error is
produced. For example
hist(as.POSIXlt(2010-07-01 00:00:00), breaks=mins)
Error in seq_len(1L + max(which(breaks maxx))) :
argument must be coercible to non-negative integer
In addition: Warning message:
In
Greetings --
For a host of reasons, I need to use/embed Garamond font with various
graphics for a particular publication. I've figured out how to more or
less get there from here, using the following sequence:
library(extrafont)
Then, I need to import the system fonts (Windoze box).
So, I
On 04/26/2014 12:42 PM, Jennifer Sabatier wrote:
So, I know that's a confusing Subject header.
Here's similar data:
tmp- data.frame(matrix(
c(rbinom(1000, 1, .03),
array(1:127, c(1000,1)),
On Fri, Apr 25, 2014 at 6:35 PM, chris Jhon cjhon...@gmail.com wrote:
Hi,
I am using hclust and cutree to cluster a data frame y and cut it into few
clusters as follows
y
V1 V2 V3 V4
A 1 2 3 4
B 5 6 7 8
C 9 10 11 12
D 13 14 15 16
E 17 18 19 20
On 26 Apr 2014, at 02:35, chris Jhon cjhon...@gmail.com wrote:
The question is how to plot the cluster number on the dendrogram plot?
Besides Peter Langfelder’s suggestion, you may also want to take a look at CRAN
packages dendroextras or dendextend
For example using the dendroextras
#
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