Hi Dan,
I am sorry about that it was typo mistake but the both values are -999.99 .
your solution works.
Thanks
Uday
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On 22.02.2012 03:33, zheng wei wrote:
Thanks.
I was just reminded by the tech support in my university that cplex is an independent
software owned by ILOG, which in turn is now owned by IBM. I suceeded in installing the
software cplex under the directory of C:/Program
On 22/02/2012 07:53, Martin Maechler wrote:
LS == Libo Sunlibo...@rams.colostate.edu
on Tue, 21 Feb 2012 21:09:54 -0700 writes:
Thanks. Shall I sum the user time and system time, which
roughly equals to the elapsed time?
No. Rather just use the user time,
i.e.
Shi, Tao shidaxia at yahoo.com writes:
Hi list,
I want to draw a bar plot with color indicating one grouping and different
shading on top of the color
indicating another grouping. How should I proceed?
Thanks!
...Tao
Does this help?
Dear R-Core team,
Dear Rcpp team and other package teams,
Dear R users,
The new package 'bit64' is available on CRAN for beta-testing and
code-reviewing.
Package 'bit64' provides fast serializable S3 atomic 64bit (signed)
integers that can be used in vectors, matrices, arrays and
Dear all,
I am using the package igraph and try to use the neighbors() function. I am
a bit confused, see the following :
Given a graph g with 18 vertices. Names of the vertices are indicated by :
V(irt.graph)$name
[1] AT2G41240 AT3G13610 AT3G02470 AT3G30775 AT3G18290
AT4G13770 AT1G01580
Hi,
I have created two separate overlapping density plots- see example code
below.
What I wish now to do is combine them into one figure where they sit side
by side.
Any help would be great!
many thanks in advance, josh.
#
thedataA -
Dear all,
I have a (probably very basic) question. I am imputing data with the mice
package, using 10 chains. I can then write out the 10 final values of the
chains simply by
name1 - complete(imp, 1)
:
:
name10 - complete(imp,10)
Not a big deal, I
i want hash like
A : 1.2, 3.4, 4.5
B : 9.7, 5.6, 4.8
C : 3.4 ,5.7, 4.6
where A key contain all three value at right sight
how to append single keys values if i have predefined hash with known
length
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Hi,
I'm a bit puzzled by the gamlss fitting of exponential and lognormal data.
Gamlss seems to think that exponentially distributed data fits better with a
lognormal distribution, and vice versa.
For example,
X - rexp(1000)
Gexp - gamlss(X~1,family=EXP) # X~1 is X tilde 1
GAMLSS-RS
This finally worked- grid.draw(
my.sgpd[[Goodness_of_Fit]][[READ.2006]][[i]] )
dev.off()
On 2/21/2012 2:04 PM, David Winsemius [via R] wrote:
On Feb 21, 2012, at 10:39 AM, kmittapalli wrote:
should i keep the first line of the code like-
pdf(file=Biology_2012_GOF.
pdf, width=8.5,
Hi,
I have a dataset
X Y
A= 10 15
20 30
40 50
B = X Z
10 30
20 50
I have a table containing X Y Z columns
How to insert a data frame into the table without using for loop ?
now currently i'm using the for loop
Hey guys,I'm working with R version 2.14.1 (2011-12-22) on a unix machine and
I'm having a similar problem. I have a matrix and i want to do a simple
correspondance analysis. I tried to install the ca package i get this error:
library(ca)
Loading required package: rgl
Warning messages:
1: In
Dear,
I'm a master student mathematics at university Gent, who's writing a thesis
about vines and copula's.
I'm in trouble with the package 'fCalendar' which I need for running 'QRMlib'.
The problem is that 'fCalendar' doesn't have a namespace. I need to use
R.2.14.1 because
I also need the
Hi,
I have the following equation:
x1 + x2 + x3 - 2(x4 + x5 + x6) + 3(x7) = N
each x_i can take any value: 1, 2, 3, 5, 6, 10, 15 or 30 and
each one is different from each other.
Which combination of values ??in the formula which leads to
the smallest value of N?
How can I program this
--
Helios de Rosario Martínez
Researcher
El día 22/02/2012 a las 11:32, Florian Weiler fweile...@jhubc.it
escribió:
Dear all,
I have a (probably very basic) question. I am imputing data with the
mice
package, using 10 chains. I can then write out the 10 final values of
the
chains
On 22-02-2012, at 13:49, Silvano wrote:
Hi,
I have the following equation:
x1 + x2 + x3 - 2(x4 + x5 + x6) + 3(x7) = N
each x_i can take any value: 1, 2, 3, 5, 6, 10, 15 or 30 and
each one is different from each other.
Which combination of values ??in the formula which leads to the
Because for example I've done this:
I made a simple table:
A B C
G1 1 34 231
G2 231 1 0.1
G3 12 0.0223
and ran a correspondance analysis on it using the ca package:
library(ca)
Loading required package: rgl
Warning messages:
1:
Hi Alok
Yes, last_insert_rowid() will always look for the last row inserted in this
connection, so you will always get 0 before inserting a new row in a new
connection. Maintaining the connection may cause other problems as the last
inserted rowid might not be relevant to your query.
Getting
Hi
My data looks like this
startDate=2008-06-01
dateRange =c( 2008-10-01,2008-12-01)
Is there any method to find the week number from the startDate range
-
Thanks in Advance
Arun
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On Feb 22, 2012, at 12:03 AM, Subha P. T. wrote:
Hi David
My data set has about 20 significant variables and step function
with logistic regression is working fine( in R-commander). I tried
to get conditional logistic by introducing the stratum variable and
clogit. The clogit is not
Stepwise variable selection is an invalid statistical method. Who or which
book recommended it?
Frank
Subha P. T. wrote
Hi,
Is there any function available to do stepwise selection of variables in
Conditional(matched) logistic regression( clogit)? step, stepwise etc are
failing in case
Hi,
I have created two separate overlapping density plots- see example code
below.
What I wish now to do is combine them into one figure where they sit side
by side.
Any help would be great!
many thanks in advance, josh.
#
thedataA -
?strptime is a good place to start
hth, Ingmar
On Wed, Feb 22, 2012 at 2:09 PM, arunkumar akpbond...@gmail.com wrote:
Hi
My data looks like this
startDate=2008-06-01
dateRange =c( 2008-10-01,2008-12-01)
Is there any method to find the week number from the startDate range
-
Thanks for the answer, and sorry if I was not clear.
So I run the data imputation using mice with 10 chains and then I get a
mids-object. From that object I can then extract 10 data sets using the
complete(imp, n) command (with n=c(1:10)).
Now I can type this out 10 times:
set1 - complete(imp,
On 22-02-2012, at 14:40, Florian Weiler wrote:
Thanks for the answer, and sorry if I was not clear.
So I run the data imputation using mice with 10 chains and then I get a
mids-object. From that object I can then extract 10 data sets using the
complete(imp, n) command (with n=c(1:10)).
Le mercredi 22 février 2012 à 04:51 -0800, aoife a écrit :
Because for example I've done this:
I made a simple table:
A B C
G11 34 231
G2231 1 0.1
G312 0.0223
and ran a correspondance analysis on it using the ca package:
Le mercredi 22 février 2012 à 05:40 -0800, Florian Weiler a écrit :
Thanks for the answer, and sorry if I was not clear.
So I run the data imputation using mice with 10 chains and then I get a
mids-object. From that object I can then extract 10 data sets using the
complete(imp, n) command
On Feb 22, 2012, at 5:28 AM, josh rosen wrote:
Hi,
I have created two separate overlapping density plots- see example
code
below.
What I wish now to do is combine them into one figure where they sit
side
by side.
Any help would be great!
many thanks in advance, josh.
To give a little more detail, you can convert your character strings
into POSIX objects, then extract from it virtually anything you would
want using strftime. In particular, %W is how you get the week number:
dateRange - c(2008-10-01,2008-12-01)
x - as.POSIXlt(dateRange)
Hi Thomas,
I've been out of office for a time, but maybe you are still waiting ...
As far as I see your model is a correctly implemented ODE (!!!) system
and I don't understand what you mean with unexpected results.
Would it be possible that your original intention was not a *continuous*
Hi,
On Wed, Feb 22, 2012 at 4:47 AM, arunkumar akpbond...@gmail.com wrote:
Hi,
I have a dataset
X Y
A= 10 15
20 30
40 50
B = X Z
10 30
20 50
I have a table containing X Y Z columns
A table, or is this
--- begin included message ---
I have a left truncated, right censored cox model:
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to know how much of the observed variance (as a number
between 0 and 1) is explained by each variable. How could I do that?
Adding terms sequentially
On 22 Feb 2012, at 14:01, Terry Therneau wrote:
--- begin included message ---
I have a left truncated, right censored cox model:
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to know how much of the observed variance (as a number
between 0 and 1) is explained by each
I believe fCalendar was replaced by timeDate which does have a
namespace and can be acquired from CRAN.
Michael
On Wed, Feb 22, 2012 at 5:41 AM, Britt Grt britt...@hotmail.com wrote:
Dear,
I'm a master student mathematics at university Gent, who's writing a thesis
about vines and copula's.
If a variable y has (approximately) constant CV, then log(y) has
(approximately) constant variance. So, use the standard deviation of
the data.
begin included message ---
Hi, I have a microarray dataset from Agilent chips. The data were really
log ratio between test samples and a universal
Thanks a lot all of you!
@ Berend, your code works fine, thanks.
@ Milan, you have a point there, makes sense to create 1 instead of 10
objects!
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thank you very much David! Can I follow up with a slight complication of
this?
A 2 by 2 case, where the abline is different in each plot.
A relevant input code would be
thedataA1 - data.frame(x1=rnorm(100,1,1),x2=rnorm(100,3,1)) #create data
thedataB1 -
The easiest way to get a hash structure is to use environments.
hash1 - new.env()
hash1$A - c(1.2, 3.4, 4.5) # etc.
If I remember right there is a package on CRAN that abstracts some of
this interface away (it's not the most intuitive) but this would
likely be the core of it.
Michael
On Wed,
Dear R users,
I have difficulty to create a list within a list.
Example: with
A - vector(mode=list, 4)
I create a list of 4 elements:
A
[[1]]
NULL
[[2]]
NULL
[[3]]
NULL
[[4]]
NULL
In each element of this list I can store, for example, a matrix:
A[[1]] - matrix ...
I need each element of A
Le mercredi 22 février 2012 à 16:17 +0100, Stefano Sofia a écrit :
Dear R users,
I have difficulty to create a list within a list.
Example: with
A - vector(mode=list, 4)
I create a list of 4 elements:
A
[[1]]
NULL
[[2]]
NULL
[[3]]
NULL
[[4]]
NULL
In each element of
It's certainly possible and I don't think you get any grueling
inefficiencies. along the way.
Another way that might or might not make sense for you is to use a
nifty trick I saw Gabor and Duncan M use a few weeks ago: you can give
dimensionality to a list and then subset it in the normal ways:
On Feb 22, 2012, at 9:13 AM, josh rosen wrote:
thank you very much David! Can I follow up with a slight
complication of this?
A 2 by 2 case, where the abline is different in each plot.
A relevant input code would be
thedataA1 - data.frame(x1=rnorm(100,1,1),x2=rnorm(100,3,1)) #create
data
The suggestion below gives you week numbers with week 1 being the week
containing the first monday of the year and weeks going from monday to
sunday. There are other conventions. The ISO convention is that week 1
is the first week containing at least 4 days in the new year (week 1
of
On Wed, Feb 22, 2012 at 8:09 AM, arunkumar akpbond...@gmail.com wrote:
Hi
My data looks like this
startDate=2008-06-01
dateRange =c( 2008-10-01,2008-12-01)
Is there any method to find the week number from the startDate range
Is the question how many weeks are from the startDate to
Hi Florian,
'yet that doesn't work' is an improper question on this list, see the
posting guide.
Besides that, something like
set-vector(mode = list, length = 10)
for (i in 1:10){
set[[i]] - complete(imp,i)}
or saving some typing
set-lapply(1:10,function(i)complete(imp,i))
should work.
Thanks so much for the good advice. I was able to get the function to work.
Cloe
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i did it, but using hash package ,i got error like
hash1 - new.env()
hash1$A - c(1.2, 3.4, 4.5)
hash1$A
[1] 1.2 3.4 4.5
h-hash(keys=c(A,B),values=c(hash1$A,hash1$A) )
Error in .set(h, ...) : Keys of length 2 do not match values of length 6
how should i proceed
--
View this message in
Thank you for the reply, my problem is that i don't understand the error that
plot(ca) is giving me:
plot(ca)
Error in matrix(as.matrix(obj), nrow = I, ncol = J) :
non-numeric matrix extent
so just to put this in context: my workscreen looks like this:
table
A B C
G1 1 34.00
sagarnikam123 wrote
i did it, but using hash package ,i got error like
hash1 - new.env()
hash1$A - c(1.2, 3.4, 4.5)
hash1$A
[1] 1.2 3.4 4.5
h-hash(keys=c(A,B),values=c(hash1$A,hash1$A) )
Error in .set(h, ...) : Keys of length 2 do not match values of length 6
how should i
Thank you both, Michael and Jim, for the answers!
- Original Message -
From: Michael Bibo michael_b...@health.qld.gov.au
To: r-h...@stat.math.ethz.ch
Cc:
Sent: Wednesday, February 22, 2012 12:46 AM
Subject: Re: [R] barplot with both color and shading
Shi, Tao shidaxia at
Hello,
Just look at your second 'apply': the 'function(x)' is not using the 'x' (!)
Solution:
function(x) hist(x, breaks=0:nrow(test), plot=FALSE)$counts)
Note that instead of 'nrow' you could also use 'length(x)'.
Hope this helps,
Rui Barradas
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View this message in context:
Hello,
I'm trying to maximize a likelihood function for a HMM with the optim()
function using the Nelder-Mead Method. The LLF has 20 Parameters which
are to be estimated. We found out that R changes some variables but not
all of them, especially the last 2 parameters aren't changed. I already
Thanks.
I also did some parallel computing. Here are some time I spent by using
'snowfall' package on intel centrino 2 laptop (2 cores) on windows 7.
user system elapsed
0.240.05 2442.77
Libo
On Wed, Feb 22, 2012 at 12:53 AM, Martin Maechler
maech...@stat.math.ethz.ch wrote:
LS
On Wed, Feb 22, 2012 at 8:49 AM, David Winsemius dwinsem...@comcast.net wrote:
After going back and constructing a proper dataset, you should be passing
'groups' into the panel function and picking it up inside panel.abline.
Close, but unfortunately things get more complicated when using
Dear R developers,
The following command produces an interaction plot with lwd=2.
interaction.plot(c(1, 2, 1, 2), c(1, 1, 2, 2), 1:4, lwd=2)
In the legend, however, lwd seems to be 1, which does not seem
to be intended behavior. Probably the lwd is not correctly forwarded
to legend:
from the
I haven't been following this thread so I may be off base, but are you
sure you don't mean plot(ca(table))?
ca is a function from the ca package -- you want to plot the output of
the function, not the function itself.
Sorry if this is unhelpful,
Michael
On Wed, Feb 22, 2012 at 10:02 AM, aoife
On Wed, Feb 22, 2012 at 3:23 AM, aoife aoife.m.dohe...@gmail.com wrote:
Hey guys,I'm working with R version 2.14.1 (2011-12-22) on a unix machine
You may be missing some openGL libraries (mesa ? )
On Ubuntu, this solved my problem of installing rgl:
sudo apt-get build-dep r-cran-rgl
Can't vouch
Dear Max and Greg
Thank you for your help. Unfortunately I was not able in getting what I need
using the functions you suggested. I believe it can be a result of my
inexperience with the packages caret and rms. Therefore, I provide more
information about my problem and wish you can again provide
Hi Petr,
Thanks for reply
sorry for late message there was typo error the both values are -999.99
a[rowSums(a == -999.99) == 0, ], this solution works only if we have to
remove certain value from matrix.
but if
a-matrix(c(1,2,3,5,-999.99,5,-999.99,6,1,5,9,1),nrow=4)
a
[,1][,2]
My data set consist of number of calls (lcin) across Day. I am looking for
activity differences between three features (4 sites per feature). I am also
looking for peaks of activity across time (Day). I am using a gamm since I
believe these are nonlinear trends with nested data.
On Wed, Feb 22, 2012 at 11:37 AM, Matthias Gondan
matthias-gon...@gmx.de wrote:
Dear R developers,
The following command produces an interaction plot with lwd=2.
interaction.plot(c(1, 2, 1, 2), c(1, 1, 2, 2), 1:4, lwd=2)
In the legend, however, lwd seems to be 1, which does not seem
to be
all you need is rowSums(a == -999.99) -- this will check for -999.99
in *any* spot. If you do only want to check a certain column/row, add
drop=FALSE to your subscripting.
Michael
On Wed, Feb 22, 2012 at 1:50 PM, uday uday_143...@hotmail.com wrote:
Hi Petr,
Thanks for reply
sorry for late
Hi Folks,
I just discovered that my dataset (coming from QuestionPro platform) has got
multiple lines for each respondent id, but what I would really need is a
regular data matrix where each respondent's data is shown on a single line.
Does anyone has already develop a procedure that
If you provide a small reproducible example of your data format and
expected output, I'm sure someone here can offer a useful solution.
Without knowing what your data look like, not so easy.
Sarah
On Wed, Feb 22, 2012 at 2:22 PM, Luca Meyer lucam1...@gmail.com wrote:
Hi Folks,
I just
Is this not what you want:
a[a[,2] != -999.99,]
I didn't see the earlier message so I'm not sure how rowSums
was involved.
Sarah
On Wed, Feb 22, 2012 at 1:50 PM, uday uday_143...@hotmail.com wrote:
Hi Petr,
Thanks for reply
sorry for late message there was typo error the both values are
Le mercredi 22 février 2012 à 14:00 -0500, R. Michael Weylandt a écrit :
I haven't been following this thread so I may be off base, but are you
sure you don't mean plot(ca(table))?
ca is a function from the ca package -- you want to plot the output of
the function, not the function itself.
Hi all,
I am looking for a good and modern Kernel Regression package in R, which
has the following features:
1) It has cross-validation
2) It can automatically choose the optimal bandwidth
3) It doesn't have random effect - i.e. if I run the function at different
times on the same data-set, the
Sure, I am sorry I have not done that in the first place.
The datasets I have looks like:
id - c(1,1,2,2,2,3)
v1 - c(NA,1,NA,1,NA,1)
v2 - as.character(c(yes,,no,,,yes))
v3 - as.factor(c(NA,1,NA,NA,3,2))
d0 - data.frame(id,v1,v2,v3)
d0
What I would need is to derive a dataset that looks like:
Hi Luca,
Thank you for the example. Here is one way of doing what you want (of
course there are many of them!):
# data
d0 - structure(list(id = c(1, 1, 2, 2, 2, 3), v1 = c(NA, 1, NA, 1,
NA, 1), v2 = structure(c(3L, 1L, 2L, 1L, 1L, 3L), .Label = c(,
no, yes), class = factor), v3 =
savehistory() gives me the option of saving the executable lines only. I'd like
to save everything.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Based on my understanding of the manual, I moved upziped the file and put the
folder of Rcplex under the directory of c:/temp
Then I use cmd under windows to go to the director of C:\Program
Files\R\R-2.13.0\bin, where my R is installed
and typed R CMD INSTALL
c:/temp/Rcplex
I got the error of
Hello,
What is the best way to get ranks for a vector of values, limit the range
of rank values and create equal count in each group? I call this uniform
ranking...uniform count/number in each group.
Here is an example using three groups:
Say I have values:
x = c(3, 2, -3, 1, 0, 5, 10, 30, -1,
Hi,
On Wed, Feb 22, 2012 at 4:01 PM, Ben quant ccqu...@gmail.com wrote:
Hello,
What is the best way to get ranks for a vector of values, limit the range
of rank values and create equal count in each group? I call this uniform
ranking...uniform count/number in each group.
Here is an example
On Feb 22, 2012, at 4:01 PM, Ben quant wrote:
Hello,
What is the best way to get ranks for a vector of values, limit the
range
of rank values and create equal count in each group? I call this
uniform
ranking...uniform count/number in each group.
Here is an example using three groups:
Thank you everyone! We already use the Hmisc package so I'll likely use
cut2.
Ben
On Wed, Feb 22, 2012 at 2:22 PM, David Winsemius dwinsem...@comcast.netwrote:
On Feb 22, 2012, at 4:01 PM, Ben quant wrote:
Hello,
What is the best way to get ranks for a vector of values, limit the range
Hi all,
I'm trying to generate a Weibull distribution including four covariates in
the model. Here is the code I used:
# Generate survival time
T = rweibull(200, shape=1.3,
scale=0.004*exp(-(-2.5*b1+2.5*b2+0.9*x1-1.3*x2)/1.3))
C = rweibull(n, shape=1.5, scale=0.008) #censoring time
time =
Hi Jorge,
The method you suggest is indeed working fine on the small sample data set.
When I apply to a larger dataset (714 rows by 160 columns) it transforms some
variables from factor to list, how can I change it back to their original
class in an automatic way?
Thanks,
Luca
Il giorno
Hi, I need to export to LaTex the summary of a PCA. So:
myPCA - prcomp(myDF)
mySummary - summary(myPCA)
#
print(xtable(mySummary))
How can I export to LaTeX not all the summary but only the first nPCs??
Best
Riccardo
__
R-help@r-project.org mailing
Hi all,
Is there a problem when accessing/writing to global variable in using
doSNOW package on multiple cores?
In the below program, each of the MyCalculations(ii) writes to the ii-th
column of the matrix globalVariable...
Do you think the result will be correct? Will there be hidden catches?
On Tue, Feb 21, 2012 at 4:04 PM, Matthew Keller mckellerc...@gmail.com wrote:
X - read.big.matrix(file.loc.X,sep= ,type=double)
hap.indices - bigsplit(X,1:2) #this runs for too long to be useful on
these matrices
#I was then going to use foreach loop to sum across the splits
identified by
Would you like it to do your your taxes for you too? :-)
Bert
Sent from my iPhone -- please excuse typos.
On Feb 22, 2012, at 11:46 AM, Michael comtech@gmail.com wrote:
Hi all,
I am looking for a good and modern Kernel Regression package in R, which
has the following features:
1)
Dear Max
Thank you for your attention. The train function in the caret package realy
does what I need.
Best regards,
-
Bc.Sc.Agri. Alessandro Samuel-Rosa
Postgraduate Program in Soil Science
Federal University of Santa Maria
Av. Roraima, nº 1000, Bairro Camobi, CEP 97105-970
Santa Maria,
I have a problem with the textmatrix() function of the LSA package whenever I
specify 'removeNumbers=TRUE'. The data for the function are stored in a
directory LSAwork which consists of a series of files that houses the text in
column form. As long as removeNumbers = FALSE or it is not
Hello,
I am try to make a density plot where plots are stacked like the one
found here:
http://dsarkar.fhcrc.org/lattice/book/images/Figure_14_03_stdBW.png
I am facing problems, however. Using the code example below, I'd like
to generate a separate panel for each val of id2. Within each panel,
As far as I know, you must convert your main program into a subroutine.
In my experience, this consist mostly if not entirely of converting user
input from prompts to subroutine arguments.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
Are you absolutely certain that the data must be stored in Excel?
In the long run I believe you will find it easier if the data is stored in
an external database, or some other data repository that does not require
you to read so many separate files.
Probably the best you can hope for as it is
Hello,
I have current data from a nortek ADP, which is basically current speed and
direction data in a 3 dimensional X Y Z format
http://www.nortekusa.com/usa/products/current-profilers/aquadopp-profiler-1
The instrument logs data in a complex way and I was wondering if anyone has
had any
Hi,
I am new to R and am a very basic user. I'm importing SHP files, adding
plots of random locations within my polygon (these files have GPS data), and
then want to save these plots (intact with added points) as KML files to
look at in GoogleEarth (or possibly as SHP files which I can then
hi
I have a problem with Median in Survival.
when I use
S1=survfit(Surv(Time,Status))
the result shows the median but I cannot use it as numeric!
S1$median in Null
Could u pleas help me.
Many Many Tank You.
Niloofar.
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Hi,
why my script iss always generating the same graph?when I change the parameters
and the name of text file?
library(MASS)
dados-read.table(inverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)
png(filename=invernoRG.png,width=800,height=600)
hist(dados[[1]], seq(0, 18, 0.5),
Good morning
In case control study
I do have 35 covariates (personal and environmental details about the
patients)
and genotype data for 50 SNPs
I need to apply all the covariates and one SNP at a time in the logistic
regression model
How can i do this in R ?
I have the Covariate file and
Hi team,
My RODBC connection times out after certain period (while creating a
new channel). Is there a way to set a different value for timeout
property? I looked into RODBC document and I didn't find any such
property. I am trying to connect to a Sybase database.
Maybe I can use something
How to make the gwaa.data file
I have the genotype data in plink format
As binary and as text file
Can anyone assist me ?
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On Feb 22, 2012, at 6:59 PM, niloo javan wrote:
hi
I have a problem with Median in Survival.
when I use
S1=survfit(Surv(Time,Status))
the result shows the median but I cannot use it as numeric!
S1$median in Null
That does not look like a formula.
Furthermore I do not think there is an
On Feb 22, 2012, at 7:08 PM, Vanúcia Schumacher wrote:
Hi,
why my script iss always generating the same graph?when I change the
parameters and the name of text file?
library(MASS)
dados-read.table(inverno.txt,header=FALSE)
vento50-fitdistr(dados[[1]],densfun=weibull)
Thanks for the information. In my case I know that the rows will never
be deleted so logically I could use max(rowid), but I am going to stick
to my original solution of using a table to get the max id. But there is
no in built table to maintain this, I will have to create a new table
and maintain
Bert's question aside (I was going to ask about laundry, but that's much harder
than taxes...), my understanding of the situation is that optimal is in the
eye of the beholder. There were at least two schools of thought on which is
the better way of automatically selecting bandwidth, using
Gabor,
Thanks very much.
I have all the zoo functions to get 1 minute aggregation and 15 min. means
working and now able to write out to a file/etc.
One question on the 15 min. mean results.
m1 - times(00:01:00)
g - seq(trunc(start(z),m1),end(z),by = m1)
z1-na.approx(z,xout = g)
m15 -
Hello,
I'm very new to R so my apologies if I'm making an obvious mistake.
I have a data frame with ~170k rows and 14 numeric variables. The first 2
of those variables (let's call them group1 and group2) are used to define
groups: each unique pair of (group1,group2) is a group. There are roughly
1 - 100 of 111 matches
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