Hi
x$dat - as.Date(paste(x$Period,0,1, sep=), format=%Y%m%d)
shall do it.
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of eric
Sent: Friday, November 16, 2012 12:46 AM
To: r-help@r-project.org
Subject: [R] Creating
On Thu, 15-Nov-2012 at 07:29PM -0700, ilai wrote:
| dotplot(variety ~ yield | year+ site, barley,
| strip = function(...,which.given,factor.levels) {
| if(which.given==2){
| strip.default(which.given,factor.levels=substr(levels(barley$site), 1,
| 1),style=4,...)
| }
| else{
|
Hi Nick,
Have you tried this:
http://www.bioconductor.org/packages/release/bioc/html/RBGL.html
There is a function there called 'lambdaSets'
Best,
Mehmet
On Thu, Nov 15, 2012 at 10:45 AM, Nick Duncan nickd...@gmail.com wrote:
Dear All,
I would like to extract Lambda Sets from a binary
Hi
Please include context.
Your numbers are not numbers. They are strings in csv file e.g. 1,200,300
and are converted to factors during reading.
First do not convert them to factors by stringsAsfactors=FALSE option in
read.table.
If you are sure that all commas are thousands separators (in
On 16/11/2012, Rolf Turner rolf.tur...@xtra.co.nz wrote:
Your question makes little sense. Functions have derivatives --- at
least some of them do. Data sets do not have derivatives. The
functions D(), deriv() etc. work on specified analytic expressions
for functions --- data sets do not
Given my reproducible example:
new.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.007,
0.21, 0.077, 0.01, 0.033, 0.05, 0.014), LR = c(FALSE, FALSE,
TRUE, FALSE, TRUE, FALSE, FALSE, FALSE, TRUE, FALSE, TRUE, TRUE
), group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L,
Hello Eric,
strptime(201206,format=%Y%m) gives NA (I don't know why but I'm a
newbie).
The only solution I found is to add 01 as a day to have a complete date
(Year Month Day):
x$Period-as.Date(paste(x$Period,01,sep=),format=%Y%m%d)
Then you get a Date format you can use to plot graph.
I think
Marvelous! That is exactly what I have been looking for.
Many thanks,
Nick
On 16 November 2012 08:33, Suzen, Mehmet msu...@gmail.com wrote:
Hi Nick,
Have you tried this:
http://www.bioconductor.org/packages/release/bioc/html/RBGL.html
There is a function there called 'lambdaSets'
Best,
On 16-11-2012, at 09:43, e-letter wrote:
On 16/11/2012, Rolf Turner rolf.tur...@xtra.co.nz wrote:
Your question makes little sense. Functions have derivatives --- at
least some of them do. Data sets do not have derivatives. The
functions D(), deriv() etc. work on specified analytic
This question has been solved off-list. Please find the solution below:
a - list(var1=c(100:1), var2=c(1:100), var3=rnorm(100))
foo - function(x) {
print(names(x))
print(==)
x - x[[1]]
print(x)
}
for(i in 1:length(a)) {
foo(a[i])
}
The main trick here was to send a[i] into the
Hi @ all,
thanks for the solutions! Now I can go on.
Greetz
GeoPhagUS
--
View this message in context:
http://r.789695.n4.nabble.com/merge-dataframes-with-condition-tp4649605p4649712.html
Sent from the R help mailing list archive at Nabble.com.
Refering the discussed issue: As a result I have dataframe like the
following.
animal-c(bear,lion,monkey,fish,zebra)
S-c(10,20,40,5,12)
N-c(5,8,15,26,1)
Z-c(24,12,8,7,2)
R-c(21,14,2,5,3)
Q-c(13,9,9,16,1)
df1-data.frame(animal,S,N,Z,R,Q)
Now, I want to plot it as a stacked barchart so that it
I would like to create a function to convert data based on a class using
as.numeric
Similarly, using a plot, I do:
plot.essai - function(x, ...) {return(x*2)}
d - 10
class(d) - essai
plot(d)
It works:
[1] 20
attr(,class)
[1] essai
Now same with as.numeric:
as.numeric.essai - function(x,
I just discover the deriv function but I have a minor problem at the end
when using its result:
For example:
dx2x - deriv(~ A*x^2, x) ; dx2x
# it works fine:
# expression({
# .value - A * x^2
# .grad - array(0, c(length(.value), 1L), list(NULL, c(x)))
# .grad[, x] - A * (2 * x)
# attr(.value,
Hello,
Actually, it's working as expected. 'numeric' is a class and the
function as.numeric() is doing what it should:
as.numeric.essai - function(x, ...) {return(x*2)}
d2 - as.numeric(d)
class(d2)
[1] numeric
Maybe you want an as.essai function.
as.essai - function(x, ...) {x -
On 16/11/2012 10:01, Marc Girondot wrote:
I would like to create a function to convert data based on a class using
as.numeric
Similarly, using a plot, I do:
plot.essai - function(x, ...) {return(x*2)}
d - 10
class(d) - essai
plot(d)
It works:
[1] 20
attr(,class)
[1] essai
Now same with
I have a data set (data.txt) containing information on affection status
(1=affected; 0=not affected) for some subjects identified by an ID. Then I
have a sort of correlation matrix of the same subjects (matrix.txt). For
each affected subject in the data set I have to retrieve the IDs of
unaffected
Hi all,
I am running 4 series of quantile regressions with tau=10:90/100, each
series corresponding to a different year.
I would like to restrict ylim for each coefficient to be the same across
years in order to help comparing coeff across years. Therefore, I need to
specify ylim for each coef.
Dear Katherine,
function flexmixedruns in package fpc may do what you want; it fits mixtures
with continuous and categorical variables, can use the BIC for giving you the
number of mixture components and also gives you posterior probabilities for
cases to belong to components.
Note that
thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/no-drawn-y-axis-but-values-tp4649298p4649718.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hi,
I'm trying to build a modified version of the package RSvgDevice,
without warnings.
I'm having troubles with the description file. Here's the logs :
* checking Rd files ... WARNING
prepare_Rd: RSvgDevice.Rd:1: All text must be in a section
But the RSvgDevice.Rd file seems OKto me, what
Hello All,
i would fit my data with a function like this :
y = a0 + a1 * exp(-x/a2) + a3 * exp(-x/a4) + a5 * exp(-x/a6) + a7 *
exp(-x/a8) + a9 * exp(-x/a10)
plus i have to impose that
a1 + a3 + a5 + a7 + a9 = 1
a1 , a3 , a5 , a7 , a9 = 0
i try with nls, but i do not know how to impose the
Hi Ramoss,
There are a few solutions to this - probably the best solution involves proper
handling of time series/date objects.
But, for an answer try:
# Build a data frame ...
set.seed(123)
someTimes - sapply(1:5, function(i) sort(round(runif(6, 0, 24
myDf - data.frame(Day = rep(1:5, each
On 12-11-16 5:14 AM, Matthieu Decorde wrote:
Hi,
I'm trying to build a modified version of the package RSvgDevice,
without warnings.
I'm having troubles with the description file. Here's the logs :
* checking Rd files ... WARNING
prepare_Rd: RSvgDevice.Rd:1: All text must be in a section
But
On 15/11/12 21:22, David Winsemius wrote:
On Nov 15, 2012, at 5:38 AM, Matthias Ziehm wrote:
Hi all,
Sorry if this has been answered already, but I couldn't find it in the archives
or general internet.
A Markmail/Rhelp search on: gompertz survreg ...brings this link to a reply
by Terry
HI,
Sys.setenv(TZ=GMT)
c(d)
#[1] 2011-12-31 GMT
#or
Sys.setenv(TZ=UTC)
c(d)
#[1] 2011-12-31 UTC
Hope it helps.
A.K.
- Original Message -
From: Andre Zege az...@yahoo.com
To: r-help@r-project.org r-help@r-project.org
Cc:
Sent: Thursday, November 15, 2012 9:36 PM
Subject: [R]
The short answer is sort of.
Medium is that survreg implements the model framework found in Kalbfeisch and Prentice,
The statistical analysis of failure time data, chapter, chapter 2.2. The ime variable T has
f(time) ~ X' beta + sigma * W
where W is an error term from some distribution and
Sally_roman sroman at umassd.edu writes:
Hi - I am using R version 2.13.0. I have run several GLMMs using
the glmmPQL function to model the proportion of fish caught in one
net to the total caught in both nets by length. I started with a
polynomial regression full model with three length
On 11/9/2012 7:06 AM, Jose Iparraguirre wrote:
Wrap Up Run 10k next March to raise vital funds for Age UK
Six exciting new 10k races are taking place throughout the country and we want
you to join in the fun! Whether you're a runner or not, these are
events are for everyone ~ from walking
Hello,
Try the following.
attr(eval(dx2x), gradient)
Hope this helps,
Rui Barradas
Em 16-11-2012 11:02, Marc Girondot escreveu:
I just discover the deriv function but I have a minor problem at the
end when using its result:
For example:
dx2x - deriv(~ A*x^2, x) ; dx2x
# it works fine:
#
Dear all,
maybe a simple problem but I found no solution for my problem.
I have a matrix Y with 23 000 rows and 220 colums. The entries are A, B or
C.
I want to extract all rows (as a matrix ) of the matrix Y where all entries of
a row are (for example) A.
Is there any solution? I tried the
Hi Peter,
On Fri, Nov 16, 2012 at 9:04 AM, Peter Kupfer peter.kup...@me.com wrote:
Dear all,
maybe a simple problem but I found no solution for my problem.
I have a matrix Y with 23 000 rows and 220 colums. The entries are A, B
or C.
A reproducible example with sample data is helpful.
I
Hey Sara,
first: Thanks for the fast reply! I checked the apply function and I found my
error.
For sure: I forgot to send an sample data. After sending the mail I recognized
it. Sorry about this!
Once again: Thanks for the fast reply and your help.
Best
Peter
Am 16.11.2012 um 15:26 schrieb
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
John Kane
Kingston ON Canada
-Original Message-
From: peter.kup...@me.com
Sent: Fri, 16 Nov 2012 15:04:31 +0100
To: r-help@r-project.org
Subject: [R] Deleting rows with special character
Dear
thanks
John Kane
Kingston ON Canada
-Original Message-
From: peter.kup...@me.com
Sent: Fri, 16 Nov 2012 15:32:23 +0100
To: sarah.gos...@gmail.com
Subject: Re: [R] Deleting rows with special character
Hey Sara,
first: Thanks for the fast reply! I checked the apply function and I
HI,
Not sure how your dataset looks like:
If it is like this:
set.seed(18)
mat1-matrix(sample(LETTERS[1:3],54,replace=TRUE),ncol=3)
mat1[apply(mat1,1,function(x) all(x==A)),]
#[1] A A A
which(apply(mat1,1,function(x) all(x==A)) )
#[1] 16
A.K.
- Original Message -
From: Peter
Readers,
If a vector consists of:
10
20
30
how to create a new vector based upon the results of calculations to
the elements, e.g. addition of successive elements, so that a new
vector would be:
30
50
i.e. 10+20, then 20+30, etc.?
__
All,
1. I will try and make this clear and concise. Please let me know any
information that would be helpful in figuring out this problem (I don't
know the relevant information to post). I am on linux- see below for
session information.
2. Problem:
working directory: home
an old version
Try this:
x - c(10, 20, 30)
head(x, -1) + tail(x, -1)
I hope it helps.
Best,
Dimitris
On 11/16/2012 3:50 PM, e-letter wrote:
Readers,
If a vector consists of:
10
20
30
how to create a new vector based upon the results of calculations to
the elements, e.g. addition of successive
Hello,
Try the following.
x - 1:3*10
idx - seq_along(x)[-1]
rowSums(cbind(x[idx], x[idx-1]))
Hope this helps,
Rui Barradas
Em 16-11-2012 14:50, e-letter escreveu:
Readers,
If a vector consists of:
10
20
30
how to create a new vector based upon the results of calculations to
the elements,
Hello,
I am using the mda package and in particular the fda routine to classify in
term of gear a set of 20 trips.
I preformed a flexible discriminant analysis (FDA) using a set of 151
trips.
FDAT1 - fda(as.factor(gear) ~ . , data =matrizR)
A total of 22 predictors were considered. 20 of
On Fri, Nov 16, 2012 at 2:52 PM, Stephen Sefick sas0...@auburn.edu wrote:
All,
1. I will try and make this clear and concise. Please let me know any
information that would be helpful in figuring out this problem (I don't know
the relevant information to post). I am on linux- see below for
On 16-11-2012, at 15:50, e-letter wrote:
Readers,
If a vector consists of:
10
20
30
how to create a new vector based upon the results of calculations to
the elements, e.g. addition of successive elements, so that a new
vector would be:
30
50
i.e. 10+20, then 20+30, etc.?
Nothing like posting to the list to figure it out yourself. defined the
little sourcing function in .Rprofile as the .First file. Now all is
well. Sorry for clutering everyones email boxes.
kind regards,
Stephen Sefick
On Fri 16 Nov 2012 08:52:00 AM CST, Stephen Sefick wrote:
All,
1. I
Hello,
Is it possible to import an Excel 2000 file (32-bit version) into R 2.15.1
64-bit version?
Thanks.
Best regards,
Cyril Hervy
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi again!
This might be more of a statistical question, but anyway:
If i train several support vector machines with different degrees of
polynomials, and as result, get that higher degrees not only have a higher test
error, but also a higher in-sample error, why is that?
I would assume i
Fixed it by using e1071 instead..
On 15.11.2012, at 17:43, Jessica Streicher wrote:
Guess it has something t do with the values in prob.model. The classifiers
with bad predicitons have positive values and the ones with good predictions
have negative values.
On 15.11.2012, at 17:18,
Actually i think i found the problem, its something about the probability model
again as it seems, if you just take the normal predictions everythings good.
Man does that probability stuff absolutely not work properly. Any suggestions
how to do ROC curves without it?
Or am i just generally
Hello,
I believe it is, but see package XLConnect. The vignette is very
helpfull, with lots of examples.
Hope this helps,
Rui Barradas
Em 16-11-2012 15:27, Hervy Cyril escreveu:
Hello,
Is it possible to import an Excel 2000 file (32-bit version) into R 2.15.1
64-bit version?
Thanks.
Best
On 2012-11-16 04:39, arun wrote:
HI,
Sys.setenv(TZ=GMT)
c(d)
#[1] 2011-12-31 GMT
#or
Sys.setenv(TZ=UTC)
c(d)
#[1] 2011-12-31 UTC
Hope it helps.
A.K.
Yes, but c(d) will still not have a 'tzone' attribute.
So the OP is okay as long as his further operations
do not depend on the presence
On 2012-11-16 01:30, Berend Hasselman wrote:
On 16-11-2012, at 09:43, e-letter wrote:
On 16/11/2012, Rolf Turner rolf.tur...@xtra.co.nz wrote:
Your question makes little sense. Functions have derivatives --- at
least some of them do. Data sets do not have derivatives. The
functions D(),
HI,
Thanks for the information.
d1-ymd_hms(2011123105)
Sys.setenv(TZ=UTC)
e2-c(d1)
e2
#[1] 2011-12-31 05:00:00 UTC
Sys.setenv(TZ=EST)
e1-c(d1)
attr(e1,tzone)-UTC
e1
#[1] 2011-12-31 05:00:00 UTC
#Looks similar,
attr(e1,tzone)
#[1] UTC
attr(e2,tzone)
#NULL
?c.POSIXct()
Using ‘c’ on
Thank you again all responders. Dan your solution was both easy miraculous.
--
View this message in context:
http://r.789695.n4.nabble.com/Can-you-have-a-by-variable-in-Lag-function-as-in-SAS-tp4649647p4649773.html
Sent from the R help mailing list archive at Nabble.com.
Hi Vijayan,
I just discovered http://www.rstudio.com/shiny/ which I think would
make it pretty easy to do this.
Best,
Ista
On Thu, Nov 15, 2012 at 1:31 AM, Vijayan Padmanabhan
v.padmanab...@itc.in wrote:
Dear R Group
I have a character vector from which I want to select a few elements and
On Nov 16, 2012, at 8:26 AM, Sarah Goslee sarah.gos...@gmail.com wrote:
Hi Peter,
On Fri, Nov 16, 2012 at 9:04 AM, Peter Kupfer peter.kup...@me.com wrote:
Dear all,
maybe a simple problem but I found no solution for my problem.
I have a matrix Y with 23 000 rows and 220 colums. The entries
On 12.11.2012 22:35, Worik R wrote:
When I say:
Sys.time()
[1] 2012-11-12 21:30:14 NZDT
But that is not what my clock on the wall and my system say. Cannot show
you my clock but...
worik@lemy:/tmp$ date
Tue Nov 13 10:32:20 NZDT 2012
Sys.time() is returning GMT
Confusing:
Above you
Hi
Can someone show me an easy way to multiple a weighted vector with an
matrix?
example below
mat1-matrix(sample(1:100,80,replace=TRUE),ncol=8)
w - 1/1:10
I want the first element in w to be multiplied by the first row of mat1 and
2nd element in w to be multiplied with the 2nd row and so on.
HI,
How about this?
library(zoo)
Period-c(201206,201207,201208)
as.yearmon(Period,format=%Y%m)
#[1] Jun 2012 Jul 2012 Aug 2012
A.K.
- Original Message -
From: PtitBleu ptit_b...@yahoo.fr
To: r-help@r-project.org
Cc:
Sent: Friday, November 16, 2012 2:56 AM
Subject: Re: [R] Creating
Thanks for the quick reply.
Sorry the attached file was not sent.
But you were right, it was a BOM, now the package compiles without
warning :-)
Matthieu Decorde
On 16/11/2012 13:32, Duncan Murdoch wrote:
On 12-11-16 5:14 AM, Matthieu Decorde wrote:
Hi,
I'm trying to build a modified
I am currently working with R's polycor package and I have encountered a
problem. I tried to follow the steps as outlined in the sem.pdf file where
a CFA model is run using polychoric correlations. Every time I run the
command sem(model, data, N=.), I get the following warning message:
Warning
Dear All,
I have a dataframe made up of individual beetles consisting of individual
number, family number, mother's family number, father's family number, and
sex of the beetle. I would like to pair up the individuals for breeding. I
would, however, like to avoid breeding beetles of the same
HI,
Try this:
dat1-read.csv(matrixGinger.csv,sep=\t)
res-apply(dat1,1,function(x) names(x)[x0.0165])
res1-t(sapply(res,`[`,seq(max(sapply(res,length)
res2-data.frame(id=rownames(res1),res1)
dat2-read.csv(Gingerdat.csv,sep=\t)
res3-merge(dat2,res2,by=id,sort=FALSE))
res3[1:5,1:5]
# id
I'm having trouble to do an infinite sum in R
I want to do the infinite sum of 1/(1+n)
how would I do this in R?
Thank You
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-do-an-infinite-sum-in-R-tp4649770.html
Sent from the R help mailing list archive at Nabble.com.
Hello together,
I have a data.frame, which I would like to export to excel.
This works without problems. My problem is, that i can't sum one of the
colums.
If i try this, i get the sum of this column.
sum_PT_PROG-sum(data_export_final_sort$PT_PROG,na.rm=TRUE)
sum_PT_PROG
[1] 130
But how can i
Hi R Users.
I have a simple question on a loop.
The following loop works fine:
r_t=list()
for(i in 1:500)
{
r_t[[i]]=h_t_half[[i]]%*%matrix(*z_t_m*[i,])
}
But indeed I need also that *z_t_m* varies. Let us suppose that *z_t_m* has
1000 replicates,
I have written the following loop that
Also, remember when you use it to connect to a DB: Teradata or Orcale or
MySQLmake sure you have JRE running and relevant DB jars are loaded in
class path.
.jclassPath() = to check class paths
.jinit() = to run JRE
--
View this message in context:
Hi A.K
Here is the error I get when I use %*%
dim(X)
[1] 71142 219
length(Weights)
[1] 71142
Wx-diag(Weights)%*%X
Error in array(0, c(n, p)) : 'dim' specifies too large an array
That is why I asked for different and faster method
--
View this message in context:
Hi
No i didn't get error it executed too fast, I had question about the Sum
squares in the weighted least square if u can help me I would I appreciated
Thanks
Date: Fri, 16 Nov 2012 09:46:59 -0800
From: ml-node+s789695n4649775...@n4.nabble.com
To: frespi...@hotmail.com
Subject: Re:
You can use the `gWidgets` package to do this kind of thing easily enough:
require(gWidgets)
my_vec - character(0)
items - state.name ## some values
handler - function(h,...) my_vec - svalue(h$obj)
w - gwindow()
g - ggroup(cont=w, horizontal=FALSE)
## one way to select one or more from many
Hi,
I search online and find rpar argument to specify random effect for
independent variables.
Is there a way to specify that for intercepts too?
Thanks,
Aaron
Management, PhD
Stanford University
[[alternative HTML version deleted]]
__
Le 16/11/12 11:49, Meli Massimiliano a écrit :
Hello All,
i would fit my data with a function like this :
y = a0 + a1 * exp(-x/a2) + a3 * exp(-x/a4) + a5 * exp(-x/a6) + a7 *
exp(-x/a8) + a9 * exp(-x/a10)
plus i have to impose that
a1 + a3 + a5 + a7 + a9 = 1
a1 , a3 , a5 , a7 , a9 = 0
Dear friends,
I have a csv file entitled ven.csv located in C:\\, this file contains only
two columns:ve and su I have written the following lines:
data=read.csv(c:\\ven.csv,header=TRUE,sep=;);
lm(ve~ su)
I have obtained the following message:
Error in eval(expr, envir, enclos) : object 've'
I'm trying to use Sweave to create a dynamic report of a variety of financial
data checks. I have an .R code file to pull the data from a database,
manipulate and filter it, and create individual data frames for each test. My
Sweave .RNW document then calls that file with source() to generate
Perhaps it would help to think before you compute.
albyn
On Fri, Nov 16, 2012 at 09:30:32AM -0800, Simon Bolivar wrote:
I'm having trouble to do an infinite sum in R
I want to do the infinite sum of 1/(1+n)
how would I do this in R?
Thank You
--
View this message in context:
Hello,
1. Don't call your dataset 'data', it's the name of an R function.
2. Imagine it's called 'dat'. Then you must use the lm() argument data =
dat. Like this:
lm(ve~ su, data = dat)
Hope this helps,
Rui Barradas
Em 16-11-2012 19:42, Sonia Amin escreveu:
Dear friends,
I have a csv file
Hello,
Try the following.
t(sapply(seq_along(w), function(i) mat1[i,]*w[i]))
Hope this helps,
Rui Barradas
Em 16-11-2012 16:34, frespider escreveu:
Hi
Can someone show me an easy way to multiple a weighted vector with an
matrix?
example below
On 16/11/2012 2:26 PM, Bush, Daniel P. DPI wrote:
I'm trying to use Sweave to create a dynamic report of a variety of financial
data checks. I have an .R code file to pull the data from a database,
manipulate and filter it, and create individual data frames for each test. My
Sweave .RNW
On 16-11-2012, at 20:42, Sonia Amin wrote:
Dear friends,
I have a csv file entitled ven.csv located in C:\\, this file contains only
two columns:ve and su I have written the following lines:
data=read.csv(c:\\ven.csv,header=TRUE,sep=;);
lm(ve~ su)
I have obtained the following message:
I quote Rolf Turner:
Learn something about R; don't just hammer and hope. Read the
introductory manuals and scan the FAQ.
The answer is that your data are in data, but until you make a
greater effort to learn R, I'm not sure this will be helpful to you.
Cheers,
Bert
On Fri, Nov 16, 2012 at
How to calculate the boxplots R? This question arises because we are building
manually boxplots, we consulted various literature sources for calculations
of the boxplot but our results differ from those generated by R, especially
when calculating the whiskers.
What is the procedure used by R to
My data:
I have raw data points that form a logit style curve as if they were a time
series. Which is to say they form 3 distinct lines with 3 distinct slopes
in backwards z pattern. A certain class of my data looks essentially flat
to the eye with marginal oscillation. What is important to me is
You would need an infinite amount of time and an infinite amount of numerical
precision, all to arrive at the conclusion that the answer is infinite. Or you
could take a short cut:
ans - Inf
---
Jeff Newmiller
On Nov 16, 2012, at 6:58 AM, bjmjarrett wrote:
Dear All,
I have a dataframe made up of individual beetles consisting of individual
number, family number, mother's family number, father's family number, and
sex of the beetle. I would like to pair up the individuals for breeding. I
would,
On Nov 16, 2012, at 2:04 PM, Elli ellilti_...@hotmail.com wrote:
How to calculate the boxplots R? This question arises because we are building
manually boxplots, we consulted various literature sources for calculations
of the boxplot but our results differ from those generated by R, especially
On 17/11/12 09:04, Elli wrote:
How to calculate the boxplots R? This question arises because we are building
manually boxplots, we consulted various literature sources for calculations
of the boxplot but our results differ from those generated by R, especially
when calculating the whiskers.
What
I haven't heard anything on this question. Is there something fundamentally
wrong with my question? Any feedback is appreciated.
Mark
On Nov 15, 2012, at 8:13 AM, Mark T. W. Ebbert wrote:
Dear Gurus,
Thank you in advance for your assistance. I'm trying to understand scope
better when
Dear R users,
i want to check matrices if they are identical when i change the rows or the
columns or the signs of the one or more columns
isomorphic - function (m1, m2) {
combs.c - combn(ncol(m1), 2)
nc - ncol(combs.c)
ind.c - vector(logical, nc)
for (i in 1:nc) {
m
Use mapply instead
On Fri, Nov 16, 2012 at 5:01 PM, billycorg candi...@gmail.com wrote:
Hi R Users.
I have a simple question on a loop.
The following loop works fine:
r_t=list()
for(i in 1:500)
{
r_t[[i]]=h_t_half[[i]]%*%matrix(*z_t_m*[i,])
}
But indeed I need also that *z_t_m*
Do you (OP) mean the partial sum of an infinite series? As your
question stands you don't need R.
On 16-Nov-12, at 12:39 PM, Rolf Turner wrote:
On 17/11/12 08:35, Albyn Jones wrote:
Perhaps it would help to think before you compute.
Fortune nomination!
cheers,
Rolf Turner
On Nov 16, 2012, at 12:16 PM, Mark Ebbert wrote:
I haven't heard anything on this question. Is there something fundamentally
wrong with my question? Any feedback is appreciated.
Perhaps failure to read this sig at the bottom of every posted message to rhelp?
PLEASE do read the posting
On Nov 16, 2012, at 9:30 AM, Simon Bolivar wrote:
I'm having trouble to do an infinite sum in R
I want to do the infinite sum of 1/(1+n)
how would I do this in R?
You could try submitting this job:
sum(1/(1+1:(2^31-1) ) ) # 2^31-1 being the highest integer in R at the moment
After I
Dear Laura,
As I explained to you when you wrote to me directly, you're not having trouble
with the polycor package, since you have AFAICS successfully computed
polychoric correlation among your variables. The error is produced when you
call sem(), apparently in the lavaan package (though you
HI,
set.seed(15)
mat1-matrix(sample(1:100,800,replace=TRUE),nrow=8000)
w - 1/1:8000
system.time(diag(w)%*%mat1)
# user system elapsed
# 54.235 0.444 54.792
system.time(sweep(mat1,MARGIN=1,w,`*`) )
# user system elapsed
# 0.220 0.044 0.265
Dear List,
I have a data matrix with 570 columns containing 95 (samples) with 6 replicates
each.
How can I calculate the mean of the replicates for 95 samples?
Thank you.
The information contained in this electronic e-mail transmission and any
attachments are intended only for the use of the
Dear list:
Can I use simple linear regression when I have proportion data for
both dependent and independent variables? Or, should I use beta
regression analysis? Or any suggestion?
Thanks!
SH
__
R-help@r-project.org mailing list
?aggregate will do it.
x - data.frame( height= c(50, 174, 145, 200, 210, 140, 175),
age_group=c(1,2,2,1,1,2,1),
ville= c(1,2,3,1,2,3,1))
aggregate(x$height,list(x$age_group, x$ville), mean)
or have a look at the plyr or datatable packages.
John Kane
Kingston ON Canada
On Nov 16, 2012, at 8:34 AM, frespider wrote:
Hi
Can someone show me an easy way to multiple a weighted vector with an
matrix?
example below
mat1-matrix(sample(1:100,80,replace=TRUE),ncol=8)
w - 1/1:10
I want the first element in w to be multiplied by the first row of mat1 and
2nd
Thanks. But aggregate will work on rows or columns. I need to calculate mean
for subsets of rows in a matrix
I.E.
Indx x1 x2 x3 x4 x5 x6 x7 x8 x9
1 25 30 15 8 12 9 18 21 89
2 52 35 42
Rui and Berend thank you for your help
before posting this mail, I change the name of my data and it becomes mat and
I tried with this line:
lm (ve~ su, data = mat)
I got this message:
Lm.fit error in (x, y, offset = offset = singular.ok singular.ok ...)
[[alternative HTML version
Hello,
The error message is not at all clear. Have you copied and pasted it?
Can you post a data example? Using ?dput, for instance.
dput(head(mat, 30)) # paste the output of this in a post
Rui Barradas
Em 16-11-2012 21:44, Sonia Amin escreveu:
Rui and Berend thank you for your help
before
1 - 100 of 120 matches
Mail list logo
