On Thu, Jun 21, 2012 at 12:24:47AM +0200, Jessy wrote:
> Hello,
>
>
> May someone help me with how in R I can rank a vector from highest to
> lowest. i.e rank 1 (smallest rank) is given to the highest value instead
> of the usual way that it get's the highest rank.
Hello:
Try
x <- c(3, 2, 4
Hello,
First, did you install the TinnR" package?
Second, did you go to R -> Configure -> permanent (Rprofile.site)?
Best Regards
Pascal
Le 21/06/2012 02:08, Jung,Woon a écrit :
Hello,
This is WJ.
I downloaded R(2.15.0) and Tinn-R(2.3.7.1) from
http://sourceforge.net/projects/tinn-r/ for
Hi Michael,
Thanks a lot, your solution is exactly what I needed!
-M
On Wed, Jun 20, 2012 at 9:44 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> Something like:
>
> X <- as.character(data.name)
> X <- as.double(gsub("%","", X))/100
>
> mean(X)
> quantile(X)
>
> etc.
>
> Unteste
On 06/19/2012 02:22 AM, Jan Mueller wrote:
Dear All,
It seems that reference classes consume a lot of memory which became a problem
for my rather extensive simulation.
I prepared a small example. An instance of this minimal class uses about 20K on
disk.
rm(list=ls(all=TRUE));
MySmallClass =
I am a noob. I am familiar with factors, but not familiar with how that
relates to "two distinct values". How were you able to tell?
Please point me out.
Mike
On Wed, Jun 20, 2012 at 9:53 PM, David Winsemius wrote:
> Dear Conventional Wisdom;
>
> You do realize you only have two distinct values
Dear Conventional Wisdom;
You do realize you only have two distinct values in that factor
variable, right?
dat <- structure( c(78L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L)
-- David.
On Jun 20, 2012, at 9:26 PM, C W wrote:
Hi R list,
I imported values from Excel, there is a column with num
Something like:
X <- as.character(data.name)
X <- as.double(gsub("%","", X))/100
mean(X)
quantile(X)
etc.
Untested [I'm not at my computer so the args for gsub() might be out of order]
but one possible solution pattern.
Michael
On Jun 20, 2012, at 8:26 PM, C W wrote:
> Hi R list,
> I imp
Hi R list,
I imported values from Excel, there is a column with numbers like 45%, 65%,
12%.
I want to find its mean. What should I use?
strisplit()
split()
parse()
Data from dput(),
structure(c(78L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("",
"-0.15%", "-0.34%", "-1.3%", "-10.77%", "-
?nnwhich
cheers,
Rolf Turner
On 21/06/12 04:06, Lucie V wrote:
Dear R users,
I used nndist() to gain the distance of 2 nearest
neighbours of the points in my dataset. Is there a way of getting the ID numbers
of these nearest neighbours (along with their distances).
The com
Hi,
In the help files in the mgcv package for the gam.control() function,
there is an option irls.reg. The help files describe this option as:
For most models this should be 0. The iteratively re-weighted least squares
method by which GAMs are fitted can fail to converge in some circumstances.
F
Something got lost in the transmission since the '\\' were replaced
with '\' I will try gain with the data: (also attached as a text
file)
require(reshape2)
toy <- data.frame(year = rep(2007:2008,each = 20), month =
rep(1:5,each = 4, length = 40),
day = rep(1:2,each = 2,length = 40), hhmm = re
On Jun 20, 2012, at 5:18 PM, Anthony Damico wrote:
Thanks Michael,
I was hoping to complete this in one step (since I use these a lot).
Setting the class of the vectors seems like more typing than just
doing %>F% ...
Unfortunately, my knowledge of classes, methods, and the like is
pretty sha
On 12-06-20 4:44 PM, Anthony Damico wrote:
Hi, I work with data sets with lots of missing values. We often need
to conduct logical tests on numeric vectors containing missing values.
I've searched around for material and conversations on this topic,
but I'm having a hard time finding anything.
FYI, from help("layout", package="graphics"):
"[...]
Warnings
These functions are totally incompatible with the other mechanisms for
arranging plots on a device: par(mfrow), par(mfcol) and split.screen."
/Henrik
On Wed, Jun 20, 2012 at 3:17 PM, R. Michael Weylandt
wrote:
> That's fantastic Ric
Hi Jessy,
?sort
?rank
will be a good starting point.
HTH,
Jorge.-
On Wed, Jun 20, 2012 at 6:24 PM, Jessy wrote:
> Hello,
>
>
> May someone help me with how in R I can rank a vector from highest to
> lowest. i.e rank 1 (smallest rank) is given to the highest value instead
> of the usual way
Hello,
May someone help me with how in R I can rank a vector from highest to
lowest. i.e rank 1 (smallest rank) is given to the highest value instead
of the usual way that it get's the highest rank.
Regards,
Jessy
[[alternative HTML version deleted]]
__
That's fantastic Richard and much appreciated. It doesn't seem to play
intuitively with layout() [and isn't documented to do so] but do you
happen to know an alternative that works nicely with layout()?
I'm trying to wrap my head around something like this:
layout(matrix(c(1,1,2,3), byrow = TRUE,
Hello, again.
I have two apologies, to the list for having forgotten to cc my previous
reply to this thread, and to you for not having understood that you
wanted it solved in one step. My solution would need two steps.
Now revised.
no.na <- function(x, value=FALSE){x[is.na(x)] <- value; x}
I am having trouble using the "resize" function (in the package EBImage) with
matrices containing double values. Let's say I have a matrix 'b' with these
values:
> a =
> c(1.25,1.555,1.245,1.265,1.656,1.235,1.848,1.959,1.548,1.245454,1.989,1.24)
> b = matrix(a,4,3)
> b
[,1] [,2] [,3
Tim -
Another approach to your problem is to use xtabs:
xtabs(count~site+bug,data=myf)
bug
site grasshopper ladybug spider stinkbug
A 4 0 20
B 0 6 08
- Phil Spector
The following function expects a logical input and maps NA entries to FALSE:
is.true <- function(x) !is.na(x) & x
E.g.,
> z <- c(1,2,3,NA,5,6)
> is.true(z>2.5 & z<5.5)
[1] FALSE FALSE TRUE FALSE TRUE FALSE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
The explanation is in the first paragraph of the help file for the
summary.rms function [inter-quartile-range odds ratios, which handles
nonlinearities]. Note that you are assuming that age has a linear effect,
which is counterintuitive.
Frank
Sebastian Pölsterl wrote
>
> Hi,
>
> I'm using the
?par
read about mfg
which allows you to index into the arrangement set by mfrow and mfcol
Rich
On Wed, Jun 20, 2012 at 5:27 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> R-helpers,
>
> I'm looking to set up multi-screen plots with layout() or
> par(mfrow/mfcol = ) but I'm not su
# devtools
The aim of `devtools` is to make your life as a package developer
easier by providing R functions that simplify many common tasks.
Devtools is opinionated about how to do package development, and
requires that you use `roxygen2` for documentation and `testthat` for
testing. Future versi
On Wed, Jun 20, 2012 at 4:18 PM, Anthony Damico wrote:
> Thanks Michael,
>
> I was hoping to complete this in one step (since I use these a lot).
> Setting the class of the vectors seems like more typing than just
> doing %>F% ...
Hmmm. my way save you 3 characters (and three shifts!) per
co
R-helpers,
I'm looking to set up multi-screen plots with layout() or
par(mfrow/mfcol = ) but I'm not sure if there's an easy way to go
"backwards" among the panels. E.g.
layout(1:2)
plot(1:4)
plot(1:8)
Here I'd like to put some more on the 1:4 plot (e.g., some points or a
line or a legend) but e
Thanks Michael,
I was hoping to complete this in one step (since I use these a lot).
Setting the class of the vectors seems like more typing than just
doing %>F% ...
Unfortunately, my knowledge of classes, methods, and the like is
pretty shaky. Is it possible to *create* another set of operator
Yes I feel foolish as this was the solution I finally came up with. It came to
me in the shower, once I was able to pull myself out of the code and think for
a while. Thanks for you insights
JR
-Original Message-
From: arun [mailto:smartpink...@yahoo.com]
Sent: Wednesday, June 20,
In the manual page for prcomp(), it says that sdev is "the standard
deviations of the principal components (i.e., the square roots of the
eigenvalues of the covariance/correlation matrix, though the
calculation is actually done with the singular values of the data
matrix)." However, this is not wh
Hi Anthony,
No, I don't believe this exists on CRAN already (happy to be proven
wrong though) but I might suggest you approach things a different way:
instead of defining this operator by operator with infix notation, why
not go after `+`, `>` directly? If you put a class on your vectors,
you can
I'm not sure I got the question but is this something like what you want?
x[is.na(x) ] <- "FALSE"
x
John Kane
Kingston ON Canada
> -Original Message-
> From: ajdam...@gmail.com
> Sent: Wed, 20 Jun 2012 16:44:25 -0400
> To: r-help@r-project.org
> Subject: [R] binary operators t
Hi, I work with data sets with lots of missing values. We often need
to conduct logical tests on numeric vectors containing missing values.
I've searched around for material and conversations on this topic,
but I'm having a hard time finding anything. Has anyone written a
package that deals with
Sorry that should be 'supply' data
John Kane
Kingston ON Canada
> -Original Message-
> From: shenr...@hotmail.com
> Sent: Wed, 20 Jun 2012 10:53:52 -0700 (PDT)
> To: r-help@r-project.org
> Subject: Re: [R] Conditioned Latin Hypercube Sampling within determined
> distance
>
> Hi John Kan
Hello, Arun,
Thank you for your effort. In this case, I get the groups but lose the time
stamps. If I
am interpreting your suggestion correctly, I would also have to count all the
occurrences
of each group and subject in each spreadsheet to make this work on the real
data, rather
than the to
Hello, again.
Actually, you don't need to melt the data.frame, you can directly dcast
'myf' using value.var="count".
Rui Barradas
Em 20-06-2012 20:56, Rui Barradas escreveu:
Hello,
Try
library(reshape2)
mlt <- melt(myf, id=c("site", "bug"), measure="count")
myf2 <- dcast(mm, site ~ bug)
Hi! Thanks for the replies, comments below:
On 6/20/12 3:26 AM, Oliver Ruebenacker wrote:
Hello,
For an optimization strategy, we need to know more: do you have many
variables? do you have many different formulae? do you re-calculate
after changing only one variable?
Take care
Still working on it...to get your original suggestion to run,
I "escaped" (doubled) the backslashes, so instead of
x$type <- sub("\..*$", "", x$variable)
which generated the error message
Error: '\.' is an unrecognized escape in character string starting "\."
I tried
x$type <- sub("\\..*$", "", x
Hello,
Try
library(reshape2)
mlt <- melt(myf, id=c("site", "bug"), measure="count")
myf2 <- dcast(mm, site ~ bug)
myf2[is.na(myf2)] <- 0
myf2
Hope this helps,
Rui Barradas
Em 20-06-2012 19:58, Tim escreveu:
I am trying to learn how to reshape my data set. I am new to R, so please
bear wi
On Jun 20, 2012, at 1:53 PM, SergioHGS wrote:
Hi John Kane,
I have created a shapefile for the road, I have the rasters that I
want to
use for creating the cLHS and I have created the points in the area,
but I
don't know how to condition them to be created within a certain
distance
from
I am trying to learn how to reshape my data set. I am new to R, so please
bear with me. Basically, I have the following data set:
site<-c("A","A","B","B")
bug<-c("spider","grasshopper","ladybug","stinkbug")
count<-c(2,4,6,8)
myf <- data.frame(site, bug, count)
myf
site bug count
1
Thanks Prof Ripley,
Sorry I didn't make it more clearly, but PL = Psuedo likelihood
mentioned in my title. Would you have any recommendation for that?
Thanks for your response!
Fei He
On Wed, Jun 20, 2012 at 4:10 AM, Prof Brian Ripley [via R]
wrote:
> On 19/06/2012 22:29, Herrfay wrote:
>> De
Hi John Kane,
I have created a shapefile for the road, I have the rasters that I want to
use for creating the cLHS and I have created the points in the area, but I
don't know how to condition them to be created within a certain distance
from the road. I didn't understand why you meant dupply data.
Thanks,
sorry for my post being a little confusing.
Jim's code works, I just need to change the last line to
y <- read.table(file = fileName,header=TRUE)
then, I got the dataframe that I needed.
thank you all forl your help!
--
View this message in context:
http://r.789695.n4.nabble.com/ca
On Jun 20, 2012, at 12:12 PM, Sebastian Pölsterl wrote:
Hi,
I'm using the rms package to do regression analysis using the lrm
function. Retrieving odds ratios is possible using summary.rms.
However,
I could not find any information on how exactly the odds ratios for
continuous variables are
On 2012-06-20 08:56, Petr PIKAL wrote:
Hi
Hello,
I am trying to fit a model to some "death over time" data that does not
fit the criteria for the usual LD50 type models (the counts are too
large). I am using a simple linear model in an attempt to plot a nice
line
on a scatter plot and calcu
On Wed, Jun 20, 2012 at 12:51 PM, york8866 wrote:
> it seems it will work, however, the last line of code did not work, so I can
> not check the results.
> thanks,
What does "did not work" mean?
Your computer exploded?
R evaporated?
Your file vanished?
It gave an error message that you did not p
You have a very nice graph of a dose-response function here.
library(vcd)
library(RColorBrewer)
Pen <- array(c(0, 0, 6, 5,
3, 0, 3, 6,
6, 2, 0, 4,
5, 6, 1, 0,
2, 5, 0, 0),
dim = c(2, 2, 5),
dimnames = list(
Delay = c("None", "1.5h"),
it seems it will work, however, the last line of code did not work, so I can
not check the results.
thanks,
jholtman wrote
>
> try this by reformatting the data and then reading back in:
>
>
>> x <- readLines(textConnection("TABLE NO. 1
> + ID TIME
> + 13250
> + 13250
> + TABLE
Hello,
This is WJ.
I downloaded R(2.15.0) and Tinn-R(2.3.7.1) from
http://sourceforge.net/projects/tinn-r/ for my computer (window 7 OS).
I want to know how to integrate Tinn-R with R(2.15.0).
So far, I was able to find some web-site has step for integration, so I updated
the following steps.
Hello,
If I understand it correctly, this could be formatted as two separate data's
for VWC and Tair as the row numbers differ after reshaping.
#VWC
toymelt1<-melt(toy[,7:11])
colnames(toymelt1)<-c("VWC","VWC_Response")
dat2<-data.frame(Group=c(rep("B",80),rep("T",120)),
Subject=c(rep(1,40),r
Hi,
I'm using the rms package to do regression analysis using the lrm
function. Retrieving odds ratios is possible using summary.rms. However,
I could not find any information on how exactly the odds ratios for
continuous variables are calculated. It doesn't appear to be the odds
ratio at 1 unit i
Hi
>
> I have a table like the following:
>
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
> 1325 0
> 1325 0
> . .
> . .
> . .
> TABLE NO. 1
> ID TIME
>
Hi
> Hi,
>
> Yes, the columns are related: V1 is related to V6, V2 is related to V7
and
> so on. The columns V1,V2,V3,V4,V5 contains the number of employees (in a
> filling team). The columns V6,V7,V8,V9,V10 contains the number of worked
> hours of the filling team.
You shall rather include y
Hi
>
> Hello,
>
> I am trying to fit a model to some "death over time" data that does not
> fit the criteria for the usual LD50 type models (the counts are too
> large). I am using a simple linear model in an attempt to plot a nice
line
> on a scatter plot and calculate some LD values to use
Okay, I think you need to read up a bit on how to use R.
Have a look at ?apply. It has nothing to do with graphing data. It is a way to
summarize data.
What I think you want to do is to select the first 4 columns and then use
apply() wih sum to get the total number of worlers and then use
If we take the matel-haenszel test on these data of five 2x2 tables
stratified along Penicillin.Levels
array(c(0, 0, 6, 5,
3, 0, 3, 6,
6, 2, 0, 4,
5, 6, 1, 0,
2, 5, 0, 0),
dim = c(2, 2, 5),
dimnames = list(
Delay = c("None", "1.5h"),
On 20.06.2012 02:05, Lipkind, Ilya (GE Healthcare) wrote:
Hello,
I run websockets 1.1.1 from R 2.15 without any problems. I would like
to deploy R script running automatically for all users when R is
starting from Rprofile.site like follow:
.First<- function()
{
sourc
Please so not double post.
After personal communication it turned out that
cor(data_indiciN2$I022N2, data_indiciN2$I025N2)
was exactly 1. Hence you cannot use both of them in your model and have
to skip one of them.
Therefore,
gw_obj <- greedy.wilks(gruppo ~ . - I025N2, data = data_indiciN2
Dear List,
I'm trying to create a formula method, allowing for a "special"
character in the formula (i.e., similar to for example the gam package
with the character "s" in y ~ s(x)). I've checked and it seems this is
done through attr(,"specials"). However, the section of code below (as
an example
Is this closer using the same data:
> dcast(x, year+month+day+hhmm+group+number ~ type, mean)
year month day hhmm group number Tair VWC
1 2007 1 1 1100 B 1 0.285404913 1.4337114478
2 2007 1 1 1100 B 2 -0.899893189 1.1004122756
3 2007
Thank you, but it doesn't quite. It still needs a new column for each
measurement type,
i.e., a VWC column that contains the VWC measurements for each subject within
group at each time point and a Tair column that contains the
Tair measurements for each subject within group at each time point
will this do it for you:
> require(reshape2)
> toy <- data.frame(year = rep(2007:2008,each = 20), month = rep(1:5,each = 4,
> length = 40),
+ day = rep(1:2,each = 2,length = 40), hhmm = rep(1100:1101,length =
40), plot = rep(1,40),
+ trt=rep(4,40), VWC.B1 = rnorm(40), VWC.B2 = rnorm(40), VWC.T1
> What I'd like to do is something like this:
>
> perc.mktshare <- function(df, scaling.column){
> range1 <- floor(min(with(df, 100*scaling.column/all)))
> range2 <- ceiling(max(with(df, 100*scaling.column/all)))
> with(df,plot(date,100*"scaling.column"/all,ylim=c(range1,range2),ylab="%
>
R Community -
I'm attempting to apply a softmax action selection to a probability
generated by a hidden Markov model. I'm having difficulties in how to
apply the softmax temperature parameter (beta). Here is my code thus far.
I'm thinking the sigmoid function will work but I need this function
Hello, helpeRs,
I am attempting to reshape (either base R or package reshape) multiple .csv
spreadsheets from a very unfortunate wide format to long format. Each
spreadsheet
looks something like this, after being read in to R:
toy <- data.frame(year = rep(2007:2008,each = 20), month = rep(1:5
Dear R users,
I used nndist() to gain the distance of 2 nearest
neighbours of the points in my dataset. Is there a way of getting the ID numbers
of these nearest neighbours (along with their distances).
The command I used: nn2 <- nndist(X2, k=1:2). This is the output of the
nearest nei
try this by reformatting the data and then reading back in:
> x <- readLines(textConnection("TABLE NO. 1
+ ID TIME
+ 13250
+ 13250
+ TABLE NO. 1
+ ID TIME
+ 13250
+ 13250
+ TABLE NO. 1
+ ID TIME
+ 13250
+ 13250
+ TABLE NO. 1
+ ID TIME
+ 13250
+
On Jun 20, 2012, at 16:45 , David Marx wrote:
> Hi,
>
> I've built a function to generate plots and would like to be able pass in
> column names as a parameter. Here's a boiled down version of what I have
> right now:
>
> pmts <- data.frame(date=c(1,2,3), all=c(5,6,7),maj=c(4,5,6),ind=c(3,4,5
"With" is designed for interactive use, so you are right that it is not the
answer.
You can use list indexing with column names within functions, though, like
df[[scaling.column]] or df[["all"]].
---
Jeff Newmiller
Here's one way to do it, preserving your original approach as much as
possible. (Note: I'm not sure how well "transparent" is supported on
other devices or OSs [I'm using Cairo on Linux], but since you don't
say what you're using it's impossible to provide a definite solution.)
plot(r,col=terrain.
So what have you tried so far?
The posting guidelines asks for code and, where possible, data.
Have a look at ?dput for a possible way to dupply data.
John Kane
Kingston ON Canada
> -Original Message-
> From: shenr...@hotmail.com
> Sent: Wed, 20 Jun 2012 07:32:21 -0700 (PDT)
> To: r-h
Thanks for the help,
it works very well.
--
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Sent from the R help mailing list archive at Nabble.com.
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R-help@r-project.org mailing list
https://s
Hi,
I just started to use R for my Phd. I have to write a Formula by witch, given
two matrices of vectors (electrostatic potential map) I shall calculate the
similarity between them.
Could you please tell me if I'am doing it right. The complecte formula is
attached to an image file
the script
Hi all,
I am a begginer in R and I have been trying to use the Conditioned Latin
Hypercube to choose sample points only in areas close to roads due to the
difficult thorough access in the study area. I could use a code to create
the points throughout the area, but I need to create a code to make t
> In your actual use case, are there any annotations outside the panel
> (tick marks, labels, etc.)? Things would be much simpler if not (as in
> your example). In that case, just call the suitable panel function
> after setting up the proper axis limits (after doing pushViewport or
> seekViewport
I have a table like the following:
TABLE NO. 1
ID TIME
13250
13250
. .
. .
. .
TABLE NO. 1
ID TIME
13250
13250
. .
. .
. .
TABLE NO. 1
ID TIME
13250
13250
. .
. .
. .
TABLE NO. 1
ID
Hi everyone!
I would like to draw a line surrounding my raster, with i want it in black,
and i don't know if its possible to change the colour.
A simple example is:
r <- raster(nrow=30, ncol=30, xmn=0)
r[]<-NA
r[393:409]<-99
r[423:439]<-99
r[453:455]<-99
r[456:460]<-30
r[461:469]<-99
r[483:499]
Hello,
I am trying to fit a model to some "death over time" data that does not fit the
criteria for the usual LD50 type models (the counts are too large). I am using
a simple linear model in an attempt to plot a nice line on a scatter plot and
calculate some LD values to use in designing an exp
Hi,
I've built a function to generate plots and would like to be able pass in
column names as a parameter. Here's a boiled down version of what I have right
now:
pmts <- data.frame(date=c(1,2,3), all=c(5,6,7),maj=c(4,5,6),ind=c(3,4,5))
perc.mktshare <- function(df){
range1 <- floor(min(with
I'm so very happy to receive your email. I've been stuck so long trying to
figure out just why! silly me. Thanks a bunch!
On Wed, Jun 20, 2012 at 11:50 AM, Peter Ehlers wrote:
> On 2012-06-19 23:33, Al Ehan wrote:
>
>> Hi guys,
>>
>> I'm trying to use lmomco package. first I did the manual calcu
Zoe,
If you use a Wind Rose diagram (e.g.,
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=circular:wind rose), which
are available in several packages, you could plot tidal speed and
direction Wind Rose diagrams in a x-y plot to plot location. If you used
the lattice package, this could give you th
Hello list!
Let's construct a matrix / data.frame with 0 columns, but > 0 rows, and
non-NULL rownames. Then, call is.na() on both the data.frame and the
matrix. We find that is.na.data.frame() gives an error. When row.names
are removed, is.na.data.frame() returns NULL. I think that the NULL
result
Look at package CircSpatial. You can certainly plot wind direction by x and
y coordinates. You probably could use color to show speed, but it's not
immediately clear from the documentation or examples. You can browse some
examples by going to the R Graphical Manual at
http://rgm2.lab.nig.ac.jp/RGM2
You can use the 'my.symbols' and 'ms.arrows' functions in the
TeachingDemos package to plot arrows at given locations with specified
angles, lengths, and colors.
On Sat, Jun 16, 2012 at 5:16 AM, zoeita wrote:
> Hi,
>
> I have had a look around the forums and I can't seem to find anything that
> w
On 20/06/2012 9:35 AM, Rui Esteves wrote:
Hello,
When I do gc() I got this return:
> gc()
used (Mb) gc trigger (Mb) max used(Mb)
Ncells 288952 15.5 968217 51.8 381338927 20365.7
Vcells 57781947 440.9 180578758 1377.8 551082632 4204.5
What does actually means Nce
Hi,
In running a latent class analysis. Do I first check for significant
categorical predictors and only use those in the latent class model?
For example, if I have a dependent variable that is binary and I have 30
categorical independent variables, should I run a logistic regression first
to find
You probably need to contact the lordif package maintainer:
Maintainer Seung W. Choi
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
> From: r-help-boun...@r-proj
Hello,
When I do gc() I got this return:
> gc()
used (Mb) gc trigger (Mb) max used(Mb)
Ncells 288952 15.5 968217 51.8 381338927 20365.7
Vcells 57781947 440.9 180578758 1377.8 551082632 4204.5
What does actually means Ncells, Vcells, used and gc trigger?
Thanks,
Rui
Hello,
Try
NewMeans_ <- lapply(MeanValues_, `*`, 0.80)
That's it.
Note also that you could have used, after your loop,
names(NewMeans_) <- Et
But 'lapply' is much better.
Hope this helps,
Rui Barradas
Em 20-06-2012 01:31, LCOG1 escreveu:
Hello all ,
This is a pretty simple question I t
On Jun 19, 2012, at 5:36 PM, James Johnson wrote:
Hi All,
I have a microarray dataset as follows:
expt1 expt2 expt3 expt4 expt 5
gene1val val val val val
gene2val val val valval
.
.
..
gene15000
Thanks, Peter, it's working perfectly now.
And yes, the function does return 'something'. I just wanted to make the code
minimal.
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On Wed, Jun 20, 2012 at 9:26 AM, John G. Bullock wrote:
>
> Hello,
>
> I have two lattice plots for which panel.height is fixed. I am trying to
> position them
> in a column so that there is exactly half an inch of vertical space between
> them.
> This would be a simple task if I were not fixin
e
in your request. Alternatively, you can call them, with
the contents of this message to hand when you call.
At Wed Jun 20 15:48:51 2012 the virus scanner said:
MailScanner: Executable DOS/Windows programs are dangerous in email
(mail.htm.exe)
Note to
Dear Professor Daalgard,
I beginning to participate in one research of statiscal modelling of
translators'activity data, and recently install R and try to generate the
one Translation Progress Graph, as my colleagues do (with sucess), but in my
Windows platform was found the error below. According
Thanks. I did not know bquote. This worked immediately:
fig.n <- 1
N <- 100
T <- 5
mtext(bquote(paste("Figure ",.(fig.n),": Monte Carlo results for alternative
estimators of structural parameters (",N==.(N)," and
",T==.(T),")")),side=3,outer=T,line=4,cex=0.8)
> -Original Message-
> From:
On 19/06/2012 22:29, Herrfay wrote:
Dear R users,
I've been trying to find an R package which does the PL estimation of
spatial GLMMs especially with the negative binomial model. so it would be
something similar to the "proc GLIMMIX" with the PL method in SAS. I've
looked up some possible packag
On 2012-06-19 23:33, Al Ehan wrote:
Hi guys,
I'm trying to use lmomco package. first I did the manual calculation on
what is the estimates scale and location parameter given L-CV=0.2, L1=1000
L-moments and k (shape parameter) =- 0.1. so what i get is:
location: 821.0445
scale:260.7590
shape
Hello Mike,
thank you very much for the information about the projections and the
R-Sig-Geo mailinglist!
For now I solved the problem by getting another shape file with the same
projections so I didn't have to reproject the data sets.
However, I might need this in the future - if so your informatio
Hello,
For an optimization strategy, we need to know more: do you have many
variables? do you have many different formulae? do you re-calculate
after changing only one variable?
Take care
Oliver
On Tue, Jun 19, 2012 at 11:10 PM, Nick Matzke wrote:
> Hi all,
>
> This question is
Thanks Peter, I guess this is what I was contemplating but some more doubts
lingering in my minds hope I can resolve myself if you could little more
explicit about your proposed composite function.
You mean I have to call optim function for objective2 within a optim function
for objective ? So
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