Hi,
I cannot get rid of this error message:
Warning messages:
1: In if (data$SP == 1) { :
the condition has length 1 and only the first element will be used
2: In if (data$SP == 1) { :
the condition has length 1 and only the first element will be used
3: In if (data$SP == 1) { :
the
Hi Tiff
data1 - data.frame(SP = c(2, 2, 1), A1 = 1:3)
data1
SP A1
1 2 1
2 2 2
3 1 3
i - 1
data1$SP
[1] 2 2 1
data1$SP[i]
[1] 2
# a warning is generated when
# the length of the argument is
# greater than 1
if(c(TRUE, TRUE, TRUE)) print(TRUE)
[1] TRUE
Warning message:
In if
Hi
you are coming from different language paradigm?
Although you did not provide your data I presume you have data frame
called data with columns SP, A1-A5
Your construction
data[1:n, A1]- data$A1 -1
seems to me rather strange and basically your cycle shall do
Dear All,
It may be a trivial question but how to determine the number of days between
two dates? What I want to do is to subtract two dates by a function which
returns the number of days between these two dates.
11.11.2008-11.11.2006 ~= 730 days
Look forward to your reply,
Carol
On 09/03/12 10:04, carol white wrote:
Dear All,
It may be a trivial question but how to determine the number of days between
two dates? What I want to do is to subtract two dates by a function which
returns the number of days between these two dates.
11.11.2008-11.11.2006 ~= 730 days
On 09-03-2012, at 10:04, carol white wrote:
Dear All,
It may be a trivial question but how to determine the number of days between
two dates? What I want to do is to subtract two dates by a function which
returns the number of days between these two dates.
11.11.2008-11.11.2006 ~= 730
Thanks, Peter.
I did the following: restart R and run the same code with and without a minimal
system pause (Sys.sleep) after the line that adds the device to the GTK window.
Adding a pause will make it work, though I do not understand what is happening
here. Note the different settings of
I need confint() for glm() to supress the messages
Waiting for profiling to be done...
because they mess up the caching mechanism of pgfSweave (see
https://github.com/cameronbracken/pgfSweave/issues/40).
I have read the help page of confint(), but I do not know how to get
the help page for the
Hi,
according to the help file rtags does not support VI(M) yet. Is there any known
hack to ctags to get tags for R in VI(M)?
BW
F
--
Federico C. F. Calboli
Neuroepidemiology and Ageing Research
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax
Yes, please upload some code / data (minimal working example). The
easiest way to upload data is to use dput() for a plain text
representation. Also remember that the majority of R-Helpers don't use
Nabble (rather they use an email client directly) so it's easier for
us if you put things in the
library(stringr)
x$Name - str_trim(gsub([ABC] Branch, ,x$Name))
Michael
On Fri, Mar 9, 2012 at 12:32 AM, Sichong Chen csc...@gmail.com wrote:
Dear Community
I have a large dataframe x as follows with common ids but different names:
x - data.frame(ID = c(1,1,2,2,2,3,3),
+ Name = c(B Branch
Hi Everyone,
I ran the same code in R and in R-studio, but got two different results.
Does anybody know why this is occurring, and if there is a fix for this?
and which is the correct program to use ?
Some information about the code I am running: I am running the fisher test
and it seems that
Reproducible code please. (I'm quite surprised this would happen --
are you sure there's no stochastic element to your calculation that
explains the differences?)
But the canonical answer is the CLI R and the CRAN binaries.
Michael
On Fri, Mar 9, 2012 at 7:11 AM, Aayush Raman
Hi
Hi,
Ok I think I have to be more precise. Here is some example:
The data includes tree samples (one of each species: 1, 2, 3 ) with
different properties that are on rows.
It is difficult to reproduce your code as the data can not be easily
transferred to R. Better to use
Dear Duncan,
thanks for the quick answer. However, I already sorted one error out - turned
out that I was just stupid: The security token worked fine once I filled all
the additional required information on the ChemSpider profile (which means
the token showing up on your profile is
On Mar 9, 2012, at 13:15 , R. Michael Weylandt wrote:
Reproducible code please. (I'm quite surprised this would happen --
are you sure there's no stochastic element to your calculation that
explains the differences?)
Notice that the code is running on two separate platforms.
It's not
Missed that...thank you.
Michael
On Fri, Mar 9, 2012 at 8:21 AM, peter dalgaard pda...@gmail.com wrote:
On Mar 9, 2012, at 13:15 , R. Michael Weylandt wrote:
Reproducible code please. (I'm quite surprised this would happen --
are you sure there's no stochastic element to your calculation
... Is there a way to extend all the lines to make them end at a
certain time point? (i.e outcome.[,1] is a time to event variable and
I would like thesurvival lines on the plot to extend out to say
5(years) )
No, there is no option in the plot.survival function to do this. No one
has
Hello,
the idea is to copy the d for df, with new results.
x-data.frame(name=x1,pos=4,age=20)
x-rbind(x,data.frame(name=x2,pos=5,age=20))
x-rbind(x,data.frame(name=x3,pos=6,age=21))
x-rbind(x,data.frame(name=x4,pos=7,age=24))
x-rbind(x,data.frame(name=x5,pos=8,age=27))
May i please update my question, and I understand now something that i did
not yesterday.
In this example:
table - structure(c(4, 7, 0.2, 3, .1, 7, 222, 3, 10, 5, 11,
8, 8, 10, 7), .Dim = c(5L, 3L), .Dimnames = list(c(gene1,
gene2, gene3, gene4, gene5), c(codon1, codon2,
codon3)))
Dear All,
I have a data set with variables x1, x2, x3, ..., x20 and I want to
create z1, z2, z3, ..., z20 with the following formula:
z1 = 200 - x1
z2 = 200 - x2
z3 = 200 - x3
.
.
.
z20 = 200 - x20.
I tried using a for loop and its index as:
for (i in 1:20) {
z(i) = 200 - x(i)
}
But R gives
May I ask, is it possible using plotrix to shade a group of variables
differentially from the rest of a graph, eg so the output looks similar to
this, where the nodes of open circles are my nodes of interest:
http://r.789695.n4.nabble.com/file/n4459137/Screen_shot_2012-03-08_at_12.18.34.png
Hi Michael,
Thank you for your reply. I have uploaded the minimum, I have left out the
formulas for calculating the amounts as they are not important to the loop.
Basically I have a while loop running that adds to the list of values and
then outside this loop I have a list called sis, this is the
Thanks, rle is what I was looking for.
Jorge
From: R. Michael Weylandt [michael.weyla...@gmail.com]
Sent: 08 March 2012 16:29
To: Jorge Molinos
Cc: r-help@R-project.org
Subject: Re: [R] Calculating length of consecutive sequences within a vector
rle
In stats:::confint.glm (actually it is MASS:::confint.glm) you see that the
message() function is used to produce waiting for To avoid this
message you can just use
suppressMessages(confint(my.fit))
At least in R it works, hope it works with Sweave as well.
Regards,
Sina
--
View this
Here is one way of doing it; it reads the file and create a 'long' version.
##
input - file(/temp/ClinicalReports.txt, 'r')
outFile - '/temp/output.txt' # tempfile()
output - file(outFile, 'w')
writeLines(ID, Date, variable, value, output)
ID - NULL
dataSw - NULL
repeat{
line -
Hi Mark,
This comes down to the way that GTK+ allocates size to its widgets. The
allocation of a widget is initialized to have a width and height of 1. When
a child is added to a visible parent, the parent will execute its lay out
algorithm and allocate a certain amount of space to the widget.
Hello,
How is it possible to sort dates in R?
Cheers,
Carol
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal,
Hi
Hello,
the idea is to copy the d for df, with new results.
x-data.frame(name=x1,pos=4,age=20)
x-rbind(x,data.frame(name=x2,pos=5,age=20))
x-rbind(x,data.frame(name=x3,pos=6,age=21))
x-rbind(x,data.frame(name=x4,pos=7,age=24))
x-rbind(x,data.frame(name=x5,pos=8,age=27))
Hi
Dear All,
I have a data set with variables x1, x2, x3, ..., x20 and I want to
create z1, z2, z3, ..., z20 with the following formula:
z1 = 200 - x1
z2 = 200 - x2
z3 = 200 - x3
.
.
.
z20 = 200 - x20.
I tried using a for loop and its index as:
for (i in 1:20) {
z(i) =
? sort
x - c(Sys.Date(), Sys.Date() + 1, Sys.Date() - 1)
print(x)
print(sort(x))
Michael
On Fri, Mar 9, 2012 at 8:35 AM, carol white wht_...@yahoo.com wrote:
Hello,
How is it possible to sort dates in R?
Cheers,
Carol
__
Hi
Hello,
How is it possible to sort dates in R?
You mean sort? Or order? Something like:
dd -sample(Sys.Date()-1:10)
dd
[1] 2012-03-04 2012-03-05 2012-02-29 2012-03-01 2012-03-02
[6] 2012-03-08 2012-03-03 2012-02-28 2012-03-06 2012-03-07
sort(dd)
[1] 2012-02-28 2012-02-29 2012-03-01
How is it possible to sort dates in R?
Try this:
a = sample(as.Date(1:100, origin = '2012-01-01'),15)
a
[1] 2012-01-31 2012-01-22 2012-03-18 2012-03-05 2012-03-17
[6] 2012-03-08 2012-01-08 2012-01-20 2012-03-01 2012-03-21
[11] 2012-02-17 2012-01-17 2012-02-12 2012-02-28 2012-04-01
sort(a)
Le vendredi 09 mars 2012 à 13:24 +0330, Hassan Eini Zinab a écrit :
Dear All,
I have a data set with variables x1, x2, x3, ..., x20 and I want to
create z1, z2, z3, ..., z20 with the following formula:
z1 = 200 - x1
z2 = 200 - x2
z3 = 200 - x3
.
.
.
z20 = 200 - x20.
I tried
On Fri, Mar 9, 2012 at 8:35 AM, carol white wht_...@yahoo.com wrote:
Hello,
How is it possible to sort dates in R?
Your question has already been answered but note that if your data is
a time series and you represent it using zoo it will automatically be
sorted. Here dates is in reverse
Perhaps ?polygon
Michael
On Fri, Mar 9, 2012 at 6:10 AM, aoife aoife.m.dohe...@gmail.com wrote:
May I ask, is it possible using plotrix to shade a group of variables
differentially from the rest of a graph, eg so the output looks similar to
this, where the nodes of open circles are my nodes
First, be sure your R and the R associated with Rstudio are the same R
versions. In Rstudio, check Tools - Options - R version. It looks as
if your R-studio is running the 32-bit version of R.
Aayush Raman ayushra...@gmail.com
Sent by: r-help-boun...@r-project.org
03/09/2012 07:11 AM
To
On 3/8/2012 1:08 PM, Gabriel Yospin wrote:
I would like to make a legible boxplot of tree growth rates for each of
seven tree species at each of seven different sites. It's a lot of data to
put on one figure, I know. I made a beautiful, interpretable figure using
color, but my target journal
On Mar 8, 2012, at 7:37 PM, Jeff Garcia wrote:
I have a simulated matrix of dates that I generated from a
probability function. Each column represents a single iteration.
I would like to bin each run _separately_ by decades and dump them
into a new matrix where each column is the length
On Fri, 2012-03-09 at 12:00 +0100, r-help-requ...@r-project.org wrote:
A note on standard errors: ?S(t) +- std is a terrible confidence
interval. ?You will be much more accurate if you use log
scale. ?(Some
argue for logit or log-log, in truth they all work well.) ? If n is
large
Hello
Thanks for the reply. As yet I have not much experience in r, although I
make some mistakes.
Any tips to solve the problem of a more effective way.
Regards
--
View this message in context:
http://r.789695.n4.nabble.com/Copy-dataframe-for-another-tp4456893p4459526.html
Sent
You should send this to r-h...@stat.math.ethz.ch.
On 03/09/2012 09:21 AM, Andrea Sica wrote:
Hello everybody, I'm looking for someone who is able with MCA and
would like to gives some help.
If what I'm doing is not wrong, according to the purpose I have, I
need to understand how to create
You could also have a look at the LaF package which is written to
handle large text files:
http://cran.r-project.org/web/packages/LaF/index.html
Under the vignettes you'll find a manual.
Note: LaF does not help you to fit 9GB of data in 4GB of memory, but
it could help you reading your
On Mar 9, 2012, at 6:14 AM, Hans Ekbrand wrote:
I need confint() for glm() to supress the messages
I'm wondering if suppressMessages would be helpful? Which in turn
suggests that you do not know how to use ??, so firt you should get
in the habit of doing a helpSearch before posting.
Hi Robert,
If you type
?loess
It pulls up the documentation. What about that function do you not like? As
you said, it needs two variables, but typically the second is just your
time index. Try this:
n - 50
x - rep(0,n)
for(i in 2:n){
x[i] - rnorm(1,x[i-1])
}
loess(x ~ seq(1,n))
Hello everybody, I'm looking for someone who is able with MCA and would
like to gives some help.
If what I'm doing is not wrong, according to the purpose I have, I need to
understand how to create a dependence matrix, where I can analyze the
dependence between all my variables.
Till now this is
On Fri, Mar 09, 2012 at 01:24:00PM +0330, Hassan Eini Zinab wrote:
Dear All,
I have a data set with variables x1, x2, x3, ..., x20 and I want to
create z1, z2, z3, ..., z20 with the following formula:
z1 = 200 - x1
z2 = 200 - x2
z3 = 200 - x3
.
.
.
z20 = 200 - x20.
I tried
On 2012-03-09 15:30, David Winsemius wrote:
On Mar 9, 2012, at 6:14 AM, Hans Ekbrand wrote:
I need confint() for glm() to supress the messages
I'm wondering if suppressMessages would be helpful? Which in turn
suggests that you do not know how to use ??, so firt you should get
in the habit
Hello.
I'm looking for a method to simulate n different 2x2 tables having all the
same odds ratio.
For example.
I need
100 tables with odds ratio 1
100 tables with odds ratio 2
100 tables with odds ratio 3
and so on.
All tables should have the same marginal frequencies.
Thank you
--
View
Thanks Pascal,
The good new is the code works, but it doesn't produce the result that I
expected, or at least it doesn't match what matlab does. I did a bit of search
myself and came across the following post on stackoverflow
http://stackoverflow.com/questions/7746529/smoothing-with-lowess
On Fri, Mar 09, 2012 at 05:37:42AM -0800, hubinho wrote:
Hello.
I'm looking for a method to simulate n different 2x2 tables having all the
same odds ratio.
For example.
I need
100 tables with odds ratio 1
100 tables with odds ratio 2
100 tables with odds ratio 3
and so on.
All
Hi,
what is the proper of of passing a missing value so I can extract
the entire i-th row of a matrix (in a list of lists) without
pre-computing the number of cols?
For example, if I know that the matrices have 2 columns, I can do the following:
set.seed(1)
x0 - lapply(1:10, function(i)
Folks,
This is more of a stat question than an R question.
Apologies in advance!
Suppose I fit an AR(1) to a time-series and also fit an AR(1) to the logs of
the same time-series and then simulate future paths.
In my case I see a big difference in the resulting paths. If I simulate
Dear Federico,
I am using the r-plugin provided for VIM on unix. This has an option of
building tag files.
http://www.vim.org/scripts/script.php?script_id=2628
I 've just tried the command, and albeit some error messages showed up,
a tag file for the current dir was build. Not sure how the
Dear R-users,
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
On Mar 9, 2012, at 5:31 AM, RMSOPS wrote:
Hello,
the idea is to copy the d for df, with new results.
x-data.frame(name=x1,pos=4,age=20)
x-rbind(x,data.frame(name=x2,pos=5,age=20))
x-rbind(x,data.frame(name=x3,pos=6,age=21))
x-rbind(x,data.frame(name=x4,pos=7,age=24))
Hello.
I have 2 plots.
plot1 -plot(table1)
plot2 -plot(table2)
How may i plot these both on the same graph, i.e. layer one graph on top of
the other one.
The result should look similar to this the image below, where the black
lines indicate one plot, and the red dots indicate the second plot.
I would like to read hdf5 or h5 files in R .
I found two packages hdf5 and h5r , but in both packages they have
information how to get attribute from data.
In my case I would like see what kind of parameters are saved in every file.
I have tried
h5 -
Dear all,
I'm trying to create a list of point patterns ppp.object {spatstat} in a loop.
My dataset looks like this:
names(OT1);head(OT1);dim(OT1)
[1] EID latitude longitude month year CPUE
TSUM
[8] fTSUM
EID latitude longitude
On Mar 9, 2012, at 10:41 AM, David Winsemius wrote:
On Mar 9, 2012, at 5:31 AM, RMSOPS wrote:
Hello,
the idea is to copy the d for df, with new results.
x-data.frame(name=x1,pos=4,age=20)
x-rbind(x,data.frame(name=x2,pos=5,age=20))
x-rbind(x,data.frame(name=x3,pos=6,age=21))
No idea what table1, table2 are
plot(1:5, type = l)
points(5:1, col = 2)
should get you started.
Michael
On Fri, Mar 9, 2012 at 10:17 AM, aaral singh aaral.si...@gmail.com wrote:
Hello.
I have 2 plots.
plot1 -plot(table1)
plot2 -plot(table2)
How may i plot these both on the same
Hi Aurelie,
Please give this a look:
http://www.nealgroothuis.name/introduction-to-data-types-and-objects-in-r/
And see if this resolves most, or all, of your questions...
Contact
Details:---
Contact me: tal.gal...@gmail.com |
You are overriding b at each loop iteration and consequently only
keeping the last one.
Perhaps
b - list()
for(i in sort(unique(OT1$month))){
a-OT1[OT1$month==i,]
b[[i]]-ppp(a$longitude,a$latitude,marks=a$fTSUM,window=newW)
plot(b[[i]],main=i)
}
Generally it's bad practice
Many thanks for reply.
I have trouble understanding how to use response, i am sorry.
My question is i have two matrices. I then plot two matrices. Then I have 2
seperate plots. I can color the nodes in the plots in two different colors.
Then, how do i merge the two plots to view one overlapping
Thank you Tal, useful link.
Best,
Aurelie
On 2012-03-09, at 11:53 AM, Tal Galili wrote:
Hi Aurelie,
Please give this a look:
http://www.nealgroothuis.name/introduction-to-data-types-and-objects-in-r/
And see if this resolves most, or all, of your questions...
Contact
Thank you very much Michael!
Best,
Aurelie
On 2012-03-09, at 11:56 AM, R. Michael Weylandt wrote:
You are overriding b at each loop iteration and consequently only
keeping the last one.
Perhaps
b - list()
for(i in sort(unique(OT1$month))){
a-OT1[OT1$month==i,]
The chi-squared test is one option (and seems reasonable to me if it
the the proportions/patterns that you want to test). One way to do
the test is to combine your 2 matrices into a 3 dimensional array (the
abind package may help here) and test using the loglin function.
On Thu, Mar 8, 2012 at
Why do you want to do this? Lattice was not really designed to put
just part of the graph up, but rather to create the entire graph using
one command.
If you want to show a process, putting up part of a graph at a time,
it may be better to create the whole graph as a vector graphics file
(pdf,
Thank you. Can the chi-squared test compare two matrices that are not the
same size, eg if matrix 1 is a 2 X 4 table, and matrix 2 is a 3 X 5 matrix?
On Fri, Mar 9, 2012 at 4:37 PM, Greg Snow 538...@gmail.com wrote:
The chi-squared test is one option (and seems reasonable to me if it
the the
R tends to see the ordering of factor levels as a property of the data
rather than a property of the table/graph. So it is generally best to
modify the data object (factor) to represent what you want rather than
look for an option in the table/plot function (this will also be more
efficient in
On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
Dear list members,
Within a loop, I need to create an xyplot with only a legend, not even
with the default external box drawn by lattice.
I already managed to remove the axis labels and tick marks, but I
couldn't find in the
On Mar 9, 2012, at 12:10 PM, David Winsemius wrote:
On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
Dear list members,
Within a loop, I need to create an xyplot with only a legend, not
even
with the default external box drawn by lattice.
I already managed to remove the
You should probably read up on what the chi-squared test actually
tests: in one form, it asks whether some set of observations could
have come from a given multinomial distribution. Concretely, it asks
whether it is reasonable to get 3 blues, 4 reds, and 2 whites from a
uniform distribution over
Do your matrices match up with each other in any meaningful way or
do you just want two independent plots on a single page?
You should probably provide the dput() output of each table object so
we can see what you've got.
Michael
On Fri, Mar 9, 2012 at 11:07 AM, aoife doherty
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of aoife doherty
Thank you. Can the chi-squared test compare two matrices that
are not the same size, eg if matrix 1 is a 2 X 4 table, and
matrix 2 is a 3 X 5 matrix?
No.
Hi all,
In using rlm I've got a bunch of warnings... failed to converge in 20
steps, etc.
My question is:
what are the results then after the failure?
Will rlm automatically downgrade back to lm upon failure?
Thanks a lot!
[[alternative HTML version deleted]]
Hi All,
s = Surv(outcome.[,1], outcome.[,2])
survplot= (survfit(s ~ person.list[,1]))
summary(survplot)
This prints a summary of all the curves at specified time intervals of
events. Is there a way to suppress this summary to only display a
The response much appreciated. They do match up, one is a small subset of
the other.
I have this:
dput(table1)
structure(list(var1 = c(2L, 4L, 4L, 1L, 423L), var2 = c(3L, 5L,
6L, 342L, 3L)), .Names = c(var1, var2), class = data.frame, row.names
= c(node1,
node2, node3, node4, node5))
Hassan,
Others have provided you with better solutions, but I hope this allows you
to see why yours didn't work.
# first (going with your code) you needed a data.frame called x
# here is an example:
x - structure(list(x1 = c(0.0986048226696643, -0.445652024980979,
0.0893989676314604,
I'm accessing R via a socket connection. I set up a connection using
socketConnection and then use readLines inside of a while(TRUE) loop to
listen for activity. Is that the best way of doing this sort of activity?
It works, that's not the issue, I am just wondering if there's a better
way.
Benilton Carvalho beniltoncarva...@gmail.com writes:
Hi,
what is the proper of of passing a missing value so I can extract
the entire i-th row of a matrix (in a list of lists) without
pre-computing the number of cols?
For example, if I know that the matrices have 2 columns, I can do the
On 09-03-2012, at 20:00, Michael wrote:
Hi all,
In using rlm I've got a bunch of warnings... failed to converge in 20
steps, etc.
My question is:
what are the results then after the failure?
They haven't converged. So inaccurate. Maybe your model is badly formulated or
ill
For a paper dealing with generalized ellipsoids, I want to illustrate in
3D an ellipsoid that is unbounded
in one dimension, having the shape of an infinite cylinder along, say,
z, but whose cross-section in (x,y)
is an ellipse, say, given by the 2x2 matrix cov(x,y).
I've looked at
Thanks, we will try to get it updated soon.
Uwe
On 08.03.2012 09:40, Caitlin wrote:
Hi all.
I just noticed that the release of version 2.14.2 was not announced on the
R home page.
Thanks,
~Caitlin
[[alternative HTML version deleted]]
__
I think the main issue of the OP is that he geneartes a 55000x55000
distance matrix and has to calculate on it. Beside immense main memory
consumption this may take ages to complete with hierarchical clustering.
Uwe Ligges
On 08.03.2012 15:02, Sarah Goslee wrote:
See inline:
On Thu, Mar 8,
Seems to be fine for me.
Uwe Ligges
On 07.03.2012 03:24, Robert King wrote:
Where should I report mirror problems? There doesn't seem to be anywhere on
http://cran.r-project.org/mirrors.html listing contact emails for mirror
admins.
There is some problem with the debian binaries on
HI All:
Does anyone know the code behind the qbeta function in R?
I am using it to calculate exact confidence intervals and I am getting
'NaN' at places I shouldnt be. Heres the simple code I am using:
k-3
x-NULL
p-rbeta(k,3,3)# so that the mean nausea rate is alpha/(alpha+beta)
min-10
On Mar 9, 2012, at 2:48 PM, Anamika Chaudhuri wrote:
HI All:
Does anyone know the code behind the qbeta function in R?
Well, yes, but don't you think it would be wise to question whether
your code might be the problem rather than the R code?
I am using it to calculate exact confidence
Take a look at n-x+1, the second parameter to the beta distribution:
n - c(10, 45, 38)
x - rbind(c( 7, 45, 31),
+c(10, 40, 35),
+c( 9, 44, 33),
+c( 8, 44, 31),
+c( 8, 45, 36))
n - x + 1
[,1] [,2] [,3]
[1,]4 -6 15
[2,] 36 -294
Here is my latest. I kind of changed the problem (for speed). In real life
I have over 300 uadata type matrices, each having over 20 rows and over
11,000 columns. However the rddata file is valid for all of the uadata
matrices that I have (300). What I am doing now: I'm creating a matrix of
row
2012/3/9 Uwe Ligges lig...@statistik.tu-dortmund.de:
I think the main issue of the OP is that he geneartes a 55000x55000 distance
matrix and has to calculate on it. Beside immense main memory consumption
this may take ages to complete with hierarchical clustering.
Indeed. I missed that in the
On Thu, Mar 8, 2012 at 4:41 AM, Massimo Di Stefano
massimodisa...@gmail.com wrote:
Hello All,
i've a set of observations that is in the form :
a, b, c, d, e, f
67.12, 4.28, 1.7825, 30, 3, 16001
67.12, 4.28, 1.7825, 30, 3, 16001
66.57, 4.28,
It's hard to help if you keep changing the framework of your problem,
first two matrices - now it's a data.frame and a list of subset row
names in a plotting method from whatever package suprow comes from.
Regardless, Michael's original answer already gave you a solution:
#The code of rank 1 in the previous post should have read
#rank1-apply(iterator1,1,function(x) x+base1)
#corrected code below
siegel.tukey=function(x,y,id.col=TRUE,adjust.median=F,rnd=-1,alternative=two.sided,mu=0,paired=FALSE,exact=FALSE,correct=TRUE,conf.int=FALSE,conf.level=0.95){
Hi Chuck, thank you *very* much! That really helped! b
On 9 March 2012 17:15, cbe...@tajo.ucsd.edu wrote:
Benilton Carvalho beniltoncarva...@gmail.com writes:
Hi,
what is the proper of of passing a missing value so I can extract
the entire i-th row of a matrix (in a list of lists) without
Dear all,
I have been running some tests of my package RSiena on different
platforms and trying to reconcile the results.
With Mac, the commands
options(digits=4)
round(1.81652, digits=4)
print 1.817
With Windows, the same commands print 1.816
I am not bothered which answer I get, but it
Can anyone recommend a good nonparametric density approach for data bounded
(say between 0 and 1)?
For example, using the basic Gaussian density approach doesn't generate a
very realistic shape (nor should it):
set.seed(1)
dat - rbeta(100, 1, 2)
plot(density(dat))
(note the area outside of
Peter,
really thanks for your answer.
install.packages(flashClust)
library(flashClust)
data - read.csv('/Users/epifanio/Desktop/cluster/x.txt')
data - na.omit(data)
data - scale(data)
mydata
a bc d e
1 -0.207709346 -6.618558e-01
With coordination with the code's author (Daniel),
The updated code has been uploaded to github here:
https://github.com/talgalili/R-code-snippets/blob/master/siegel.tukey.r
And also the following post was updated with the code:
Is there any way to issue operating system commands and geting back the
results,
in R?
I mean, for instance, in Linux, to execute from R the 'ls' command and getting
back a list of files in the current directory, or, equivalently, in
Windows/DOS,
the 'dir' command?
I'm not interested in the
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