Spencer:
Just for fun, may I hazard a guess: It would be messy to retain time zones
if your were concatenating objects with more than one time zone among them.
Normalizing everything to a single zone probably also makes subsequent
operations on the results much easier.
Not that it couldn't be
On Jul 8, 2012, at 23:39 , ycyi121 wrote:
Hi,
I have great difficulty in simulation the a dataset based in a loading
matrix [c(1,1,1,2,3,3,3,4,4,3,2,2,1,1), 7, 2) and an error covariance matrix
is 2*I. I have to simulate a dataset with 7 variables and 50 rows. I search
a lot and did find
Good Morning,
I forward your email to the R-help list.
Regards.
Message original
Sujet: Re: [R] Problem with Apriori
Date : Fri, 6 Jul 2012 12:45:47 +0200
De : Jolinda Bartlett joli...@eighty20.co.za
Pour : Pascal Oettli kri...@ymail.com
Good morning
I am using Windows
Hi everyone,
I'm sure this is pretty basic but I couldn't find a clear example of how to
do this. I'm running a regression, say:
reg - lm(Y ~ x1 + year)
where x1 is a continuous variable and year is a factor with various year
levels. Individually, each year factor variable is not significant,
Hi there,
I am trying to make a VECM model which does a loop to pull of long run
impact coefficients. The problem is that to calculate these for a,b,c I use
the irf() function and they are stored in irf$a, irf$b, irf$c. What I would
really like is to be able to call irf$[variablename(x)] where I
I've been looking into the effects package and it seems to be a great tool
for plotting the probabilities of the
response variable by the predictors. However, I'm wonder if I can use the
effects package to plot the probabilities
on the y axis and one predictor on the x axis, with the curve having
Hello everyone,
I need some help regarding calling WinBUGS from R. I
have a model for WinBUGS and an R code which calls WinBUGS. On running the
code I am being shown the error message :
Error in file(con, wb) : cannot open the connection
In addition: Warning messages:
1: In
Hello,
I'm not completely sure I understand your problem, you should say what
is the package you are using.
If it's just a doubt on how to access the elements of what seems to be a
list, here is an example:
x - list(a=1:5, b=rnorm(4))
x$a
x[[ a ]]
x[[ 1 ]]
If it's from package vars, the
Dear All,
when I try to call blpConnect() in order to open a connection to the
Bloomberg on my machine, I receive following error message:
R version 2.15.1 (2012-06-22)
rJava Version 0.9-3
RBloomberg Version 0.4-150
Java environment initialized successfully.
Looking for most
-Original Message-
From: Martin Ivanov
I have the longitudes (lon) and latitudes (lat), and I have a
resolution (r), for example r = 0.004. The bounding box must
have the same number of digits as resolution.
Surely the issue is not the particular numeric resolution of the
-Original Message-
Error in file(con, wb) : cannot open the connection In
addition: Warning messages:
1: In file.create(to[okay]) :
cannot create file 'c:/Program
Files/WinBUGS14//System/Rsrc/Registry_Rsave.odc', reason
'Permission denied'
This tells you that you do not have
Alexander,
I agree, this feels like a java version issue. You could start by
telling us what version of Java you're running: open a command prompt
and type java -verision, and reply with the output. I think any
version 1.5 and up should work, but let's see what you have.
-John
On Mon, Jul 9,
I have a text file that has semi-colon separated values. The table is nearly
10,000 by 585. The files looks as follows:
***
First line: Skip this line
Second line: skip this line
Third line: skip this line
variable1 Variable2 Variable3 Variable4
Dear David,
Thanks for the tip about naming the levels first with levels before
assigning them.
However, your answer does not solve the original question that I requested
help on. I will post a simplified version as a separate query, partly
because the subject line on the present one seems to
Hie
I am trying to install Package DAAG this is the error I get
library(DAAG)
Loading required package: randomForest
Error: package 'randomForest' could not be loaded
In addition: Warning message:
In library(pkg, character.only = TRUE, logical.return = TRUE, lib.loc =
lib.loc) :
there is no
appears when I run the summary command.
The data file is large (585000 rows) and has no NA, - or blank values.
My script (in brief) is as follows, with results:
library(MASS)
## ADD DATA
Jdata- read.delim(/Analysis/20120709 JLittle data file.txt, header=T)
attach(Jdata)
names
Dear Mr. Holtman,
thank you for your reply.
I think I did say which mean I needed: all of these reaction times per test
subject. , which means that I need a file with the mean of reaction times
of each file / of each test subject (because file XYZ_34.txt is identical
with subject 34's data).
Hi all,
I have a usecase defined as follows. I have a list of popular products in a
file and a file specifying users and products used by them.
Now using map-reduce for each user I want to sort out the popular products
that the user is already using and recommend top 100 popular products for
Hello,
Your data example has dots in the column of interess. If those values
are ntegers, this might do it.
fun - function(x){
dat - read.table(x, skip=14)
H - as.numeric(gsub(\\., , dat[, 8]))
mean(H)
}
sapply(list.files(pattern=XYZ.*\\.txt), fun)
Now do what you
On 12-07-09 6:42 AM, Chiruka, Raymond wrote:
Hie
I am trying to install Package DAAG this is the error I get
library(DAAG)
Loading required package: randomForest
Error: package 'randomForest' could not be loaded
In addition: Warning message:
In library(pkg, character.only = TRUE,
On 09/07/12 01:43, Mikkel Grum wrote:
Dear useRs
I need to do graphs with dates in different languages on Ubuntu.
In Windows the following will plot the date axis labels in Spanish:
random.dates - as.Date(2001/1/1) + 70*sort(stats::runif(100))
language - Spanish
Sys.setlocale(LC_TIME,
Hello,
Try the following.
head - readLines(test.txt, n=4)[4]
dat - read.table(test.txt, skip=5)
names(dat) - unlist(strsplit(head, ))
dat
hope this helps,
Rui Barradas
Em 09-07-2012 11:23, vioravis escreveu:
I have a text file that has semi-colon separated values. The table is nearly
John,
thanks for your quick response. Here is what you requested:
java version 1.7.0_05
Java(TM) SE Runtime Environment (build 1.7.0_05-b05)
Java HotSpot(TM) Client VM (build 23.1-b03, mixed mode, sharing)
Thanks,
Alex
-Ursprüngliche Nachricht-
Von: John Laing
OK. Are you running 32 bit or 64 bit R? And 32 bit or 64 bit Java?
This can help shed light on the settings for the JVM being accessed by R:
require(rJava)
.jinit()
jvm - .jnew(java.lang.System)
jvm.props - jvm$getProperties()$toString()
jvm.props - strsplit(gsub(\\{(.*)}, \\1, jvm.props), ,
Hello,
Why does anova.lm sometimes return a p-value and at other times not ? Is
it because it recognizes nested models from non-nested ones ?
x-seq(1,100,1)
y-3*x+rnorm(100)
anova(lm(y~x),lm(y~x+I(x^2)),test=F)
Analysis of Variance Table
Model 1: y ~ x
Model 2: y ~ x + I(x^2)
Res.Df
Also: require(rms); ?plot.Predict
Frank
Greg Snow wrote
Try the following:
library(TeachingDemos)
?TkPredict
fit.glm1 - glm( Species=='virginica' ~ Sepal.Width+Sepal.Length,
data=iris, family=binomial)
TkPredict(fit.glm1)
(you may need to install the
32-Bit both.
-Ursprüngliche Nachricht-
Von: John Laing [mailto:john.la...@gmail.com]
Gesendet: Montag, 9. Juli 2012 14:39
An: Alexander Erbse
Cc: r-help@r-project.org
Betreff: Re: [R] Problem to establish Bloomberg connection / Package RBloomberg
/ function blpConnect()
OK. Are you
Dear Alex,
As explained in ?linearHypothesis, you can use matchCoefs() in the second
argument:
linearHypothesis(reg, matchCoefs(reg, year) , vcov=vcovHC(reg, HC1))
In addition, you can set the argument white.adjust=hc1 as an alternative to
using the vcov argument. Finally, in this case it
If you provide a simple example, with actual data and code so that we can
reproduce your issue, it would make it easier for readers of this list to
help.
Jean
MarkBeauchene markbeauch...@hotmail.com wrote on 07/06/2012 04:38:32 PM:
I have a class with 732 members, so using rpart.plot is
On 09/07/2012 07:12, Bert Gunter wrote:
Spencer:
Just for fun, may I hazard a guess: It would be messy to retain time zones
if your were concatenating objects with more than one time zone among them.
Actually, impossible as the design allows for only one timezone for each
object.
Dear Abraham,
I must admit that I don't really follow what you want to do. Disregarding the
fact that the example you provide doesn't converge to a proper solution, the
plot that you've requested will range over all values of bid at the median
home, which is 0. You may have intended home to be
ooohhh!
John Kane
Kingston ON Canada
-Original Message-
From: ted.hard...@wlandres.net
Sent: Sun, 08 Jul 2012 21:07:09 +0100 (BST)
To: r-help@r-project.org
Subject: Re: [R] a fortune?
Should this not be attributed to DescaRtes?
Ted.
On 08-Jul-2012 18:41:17 Achim
Right, I see it now. Thanks.
Who knows in another 100 years I may understand regex.
John Kane
Kingston ON Canada
-Original Message-
From: smartpink...@yahoo.com
Sent: Sat, 7 Jul 2012 16:19:54 -0700 (PDT)
To: jrkrid...@inbox.com
Subject: Re: [R] Splitting a character vector.
Without more information, we can only guess what you did, or what you
are seeing on the page that is different.
I'll make a random guess though. There are about 5 ways to paramaterize
the Weibull distribution. The standard packages that I know, however,
tend to use the one found in the
Indeed!! I made a spelling mistake! Correction:
Should this not be attributed to DescRtes?
Ted.
On 09-Jul-2012 13:31:03 John Kane wrote:
ooohhh!
John Kane
Kingston ON Canada
-Original Message-
From: ted.hard...@wlandres.net
Sent: Sun, 08 Jul 2012 21:07:09 +0100 (BST)
Please.
find an example here. With exactly the same data set, I run two hazard
models following the instructions for each function.
aftreg(formula = Surv(sta, sto, S) ~ a + b + c + d + e + f + g
, factor(F), data = data.frame(SURV),
dist = weibull, id = ID)
streg f1 f2 f3 f4 f5 a b c d g
On Jul 9, 2012, at 15:40 , Suresh Krishna wrote:
Hello,
Why does anova.lm sometimes return a p-value and at other times not ? Is it
because it recognizes nested models from non-nested ones ?
x-seq(1,100,1)
y-3*x+rnorm(100)
anova(lm(y~x),lm(y~x+I(x^2)),test=F)
Analysis of Variance
What is the R code for Ljung-Box Test in Statistics?
Sajeeka Nanayakkara
[[alternative HTML version deleted]]
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PLEASE do read the posting guide
Hi John,
I did some further checks and noticed that the java version, which is indicated
by the settings (I used the code you sent) is version 1.5.0_12, although I
installed version 1.7.0_05 shortly. The paths as shown by parameter
java.endorsed.dirs are wrong. They are pointing to an older
John,
the problem is now resolved. I changed the PATH environment variable. I
replaced the path leading to the old java installation by the new one and now
it works.
Thanks for your help.
Regards,
Alex
-Ursprüngliche Nachricht-
Von: Alexander Erbse
Gesendet: Montag, 9. Juli 2012
On 7/9/2012 9:17 AM, Javier Palacios Fenech wrote:
Please.
After, Terry's response I guess I was expecting to hear how your
comparison between R and STATA went when you used the R function,
survreg() for your analysis.
We still don't know what your data look like. The posting guide asks
Hi all-
I fit a zero-inflated Poisson model to model bycatch rates using an offset term
for effort. I need to apply the fitted model to a datasets of varying levels of
effort to predict the associated levels of bycatch. I am seeking assistance as
to the correct way to code this.
Thanks in
Kai Ying yingk at iastate.edu writes:
Hi,
I want using zero-inflated negative binomial regression model to
classify data(a vector of data), that is I want know each observed value is
more likely belong to the zero or count distribution(better with
relative probability). My data is some
I ran the same program in a different computer where it had run proper a
week ago. This time around in the log file in WinBUGS the program stopped
at a line which said :
save(C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/RtmpU3u46p/log.odc)
save(C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/RtmpU3u46p/log.txt)
and did
:
library(MASS)
## ADD DATA
Jdata- read.delim(/Analysis/20120709 JLittle data file.txt, header=T)
attach(Jdata)
names(Jdata)
[1] POINTID Lat_Y_pos JVeg5 Subregion Rock_U_Nam
Rock_Name Elevation Slope Aspect Hillshade
Stream_dist Coast_dist Coast_SE
[14
Hi,
Let say I have a numeric vector: x - c(1, 2, 3, 3).
I want on one hand numbers which are not duplicated ie 1,2 and duplicated
3.
so I did:
duplicated(x)
FALSE FALSE FALSE TRUE
unique(x)
1 2 3
which is not what I want. Is there a function in R to have the following
result:
Here is one way of doing it -- you can create your own functions:
x - c(1, 2, 3, 3)
allDup -
+ function (value)
+ {
+ duplicated(value) | duplicated(value, fromLast = TRUE)
+ }
duped - unique(x[allDup(x)])
duped
[1] 3
setdiff(unique(x), duped)
[1] 1 2
On Mon, Jul 9, 2012 at 12:42
Hello,
Maybe this function.
fun - function(x) x %in% x[duplicated(x)]
x - c(1, 2, 3, 3)
fun(x)
Hope this helps,
Rui Barradas
Em 09-07-2012 17:42, Nico902 escreveu:
Hi,
Let say I have a numeric vector: x - c(1, 2, 3, 3).
I want on one hand numbers which are not duplicated ie 1,2 and
Hello,
The function is
?Box.test
It has two types, Box-Pierce and Ljung-Box.
Hope this helps,
Rui Barradas
Em 09-07-2012 15:59, Sajeeka Nanayakkara escreveu:
What is the R code for Ljung-Box Test in Statistics?
Sajeeka Nanayakkara
[[alternative HTML version deleted]]
excellent!!! thanks a lot!!
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On 09.07.2012 18:19, PRAGYA SUR wrote:
I ran the same program in a different computer where it had run proper a
week ago. This time around in the log file in WinBUGS the program stopped
at a line which said :
save(C:/DOCUME~1/ADMINI~1/LOCALS~1/Temp/RtmpU3u46p/log.odc)
Yes that was the problem. Thank you very much. Can anyone tell me the
meaning of
The following object(s) are masked _by_ '.GlobalEnv':
beta
i was shown this notification after the results were printed.
On Mon, Jul 9, 2012 at 1:15 PM, Uwe Ligges
lig...@statistik.tu-dortmund.dewrote:
On
I am running a program which has an output containing four vectors named
meanfevs, meanfevns, pfevs, pfevns. I wish to return all four and be able
to access them later. I used the command
return(list(a=meanfevs,b=meanfevns,c=pfevs,d=pfevns))
it did give me the ouput. However the values did not get
Hi,
No, you won't be able to simply call a and have that work.
R returns these in a single object with components (elements) named a,b,c,d
Here's a concrete example:
func - function(x, y) return(list(a = x+1, b = y + 2))
out - func(3, 5)
out[[a]] # or out$a
out[[b]] # or out$b
give the
Indeed. You do not understand lists. The behavior you expect is not how R
works. Have you read An Introduction to R where this is explained
(section 6.1). Also chapter 10 and 10.7 in particular for scoping in R.
See also
?with
?within
?eval
and e.g. ?lm or ?xyplot for the ubiquitous use of data
I have four files that map populations to appropriate keys: statepop.csv,
countypop.csv, zip3.csv, and zip5.csv. I'm using the TIGER/Lines shape files:
tl_2009_12_state, tl_2009_12_county tl_2009_12_zcta3, and tl_2009_12_zcta5. I
have carefully followed a number of tutorials and can reproduce
I don't know how significant this is, but WolframAlpha's value of pi disagrees
with R's at about the 16th decimal:
Wpi-3.141592653589793238462643383279502884197169399375105
..^
Rpi-3.14159265358979311599796346854418516
Probably not of interest to anyone but astronomers,
Hello R Community,
I am using the Lavaan package in R 2.15.0 to analyze data collected from
1200 lakes across North America. My dataset includes 3 continuous
independent variables (LOG_NTL, LOG_PTL, and LOG_SR_A_D) and 1 continuous
dependent variable (BIOVOL) . I have successfully constructed
I am trying to make the lines thicker in a graph (for a ppt presentation).
Here is what I currently have:
plot(x,y,type=l, ylab=Number of OTUs, xlab=Number of Samples
Collected, col=Black, pch=1, ylim=c(0,6000))
points(x, Sobs$Chao_1_Mean, type=l, col=Gray, pch=1) (this is one of the
added
Dear all,
For my research I want to test additive interaction for a dichotomous
dependent variable. Can anyone help me to estimate this in R?
Wacholder describes this procedure in the American Journal of Epidemiology
in 1986 (Binomial regression in GLIM: estimating risk ratios and risk
Hallo Rob,
have you figured out a solution to the error? I've got the same problem.
regards
N.
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Sent from the R help
bbolker wrote
PtitBleu ptit_bleu at yahoo.fr writes:
Hello,
I'm trying to fit experimental data with a model and nls.
For some experiments, I have data with x from 0 to 1.2 and the fit is
quite
good.
But it can happen that I have data only the [0,0.8] range (see the
example
Dear Mr. Barradas,
your solution comes very close to what I want.
But I have two questions left:
First question: If R computes the mean for the reaction times of test
subject 34 (the example I provided above), it says 310112.0, but if I use
the mean-function in Excel it says 345.210. Apart
Dear all
I'm trying to calculate the number of shared species between sites.
I have a dataframe with in the first column the names of the sites; the
names of the other columns are the species names. To use the shared function
from the package RICH one needs to input different matrices. So
UCLA's Advanced Technical Services' Statistical Computing website often has
very good resources for comparing analyses between R, Stata, and SAS (
http://www.ats.ucla.edu/stat ). For accelerated failure time models, I
believe that it has some examples for Stata (
Hi,
I guess you should have fill=TRUE in the read.table.
dat1-read.table(text=
First line: Skip this line
Second line: skip this line
Third line: skip this line
variable1 Variable2 Variable3 Variable4
Unit1 Unit2 Unit3
10 0.1 0.01
Dear Peter,
Thank you very much for that excellent answer to a rather stupid question :)
I did not notice that the RSS actually increased for the model with more
parameters and so in this case the F-statistic is negative and therefore a
p-value from the F-distribution is meaningless. But I guess
In addition to the helpful guidance suggested already, you might investigate
the sas7bdat package, by Matt Shotwell.
I described it here:
http://sas-and-r.blogspot.com/2011/07/really-useful-r-package-sas7bdat.html
Ken
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If you look at the help for install.packages:
help(install.packages)
You will see that the first argument to install.packages takes a vector of
the package names. In your case, you are passing three arguments to the
function, so R is trying to install RExcelInstaller package into a library
Hello,
Just now I checked reading directly from .txt file instead of the one showed in
my earlier reply,
#Use skip=3 instead of 4.
dat1-read.table(dat1.txt,sep=,skip=3,fill=TRUE,header=TRUE)
dat1-dat1[-1,]
row.names(dat1)-1:nrow(dat1)
dat1
variable1 Variable2 Variable3 Variable4
1
Hi community. I'm fairly new to R and have a basic question. When I try to
install RExcel with the code:
install.packages(RExcelInstaller, rcom, rsproxy)
I get the message below:
Warning in install.packages(RExcelInstaller, rcom, rsproxy) :
'lib = rcom' is not writable
Error in
I'd suggest using lines() rather than points() to add lines to a plot.
The parameter to change line width is lwd rather than cex.
e.g.
lines(x, Sobs$Chao_1_Mean, col=Gray, lwd=2)
Reading ?par can be both an enlightening and confusing experience.
Sarah
On Mon, Jul 9, 2012 at 1:11 PM, peziza
Hi,
Try this:
#Duplicated:
x-c(1:3,3)
x==x[duplicated(x)]
#[1] FALSE FALSE TRUE TRUE
#Unique:
x[!x==x[duplicated(x)]]
#[1] 1 2
A.K.
- Original Message -
From: Nico902 descos...@ciml.univ-mrs.fr
To: r-help@r-project.org
Cc:
Sent: Monday, July 9, 2012 12:42 PM
Subject: [R]
?:::
-- Bert
On Mon, Jul 9, 2012 at 10:45 AM, Niels2 nielssch...@gmx.net wrote:
Hallo Rob,
have you figured out a solution to the error? I've got the same problem.
regards
N.
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Hello,
I am trying to determine LD50 and LD95 using dose.p in MASS however some of the
Residual variance is larger than the degrees of freedom. Please can anyone help
with any advice as to how i can correct for this?
Here is the model as inputted into R
y-cbind(dead,n-dead)
?par and have a look at lwd
John Kane
Kingston ON Canada
-Original Message-
From: jenkere...@gmail.com
Sent: Mon, 9 Jul 2012 10:11:32 -0700 (PDT)
To: r-help@r-project.org
Subject: [R] how to make plot lines thicker
I am trying to make the lines thicker in a graph (for a ppt
Hello,
There must be a difference in the file you are processing and in the one
excel and I are:
fun - function(x){
+ dat - read.table(x, skip=14)
+ dat[ , 8] - as.numeric(gsub(\\., , dat[, 8]))
+ mean(dat[, 8])
+ }
sapply(list.files(pattern=XYZ.*\\.txt), fun)
XYZ_34.txt
345210.4
This
Hi All,
I have two questions regarding install.packages().
Q1: may I run it in non-interactive mode, which means just install the packages
I want rather than letting me to choose which mirror etc?
Q2: I ran a R code in batch mode, for example, R CMD BATCH a.r
In a.r, I have some statements
I think the real problem is the first data line:
2 1 1 3 27 0 6 1.200.995
Notice the two periods in the value. The previous solution was
getting rid of all the periods. If you leave out this value, you get
339.5. if you change it to 1200.995, you get
Lee, Laura wrote
Hi all-
I fit a zero-inflated Poisson model to model bycatch rates using an offset
term for effort. I need to apply the fitted model to a datasets of varying
levels of effort to predict the associated levels of bycatch. I am seeking
assistance as to the correct way to
These sorts of parameters are all documented under ?par, but it's a
bit of a beast to read. The one you are looking for is lwd=
To wit
layout(1:2)
plot(1:5, type = l)
plot(1:5, type = l, lwd = 3)
Best,
Michael
On Mon, Jul 9, 2012 at 12:11 PM, peziza jenkere...@gmail.com wrote:
I am trying to
On Jul 9, 2012, at 20:23 , Lawrence, Adaku wrote:
Hello,
I am trying to determine LD50 and LD95 using dose.p in MASS however some of
the Residual variance is larger than the degrees of freedom. Please can
anyone help with any advice as to how i can correct for this?
Er, in what sense is
Hi,
On Mon, Jul 9, 2012 at 2:30 PM, Hui Du hui...@dataventures.com wrote:
Hi All,
I have two questions regarding install.packages().
Q1: may I run it in non-interactive mode, which means just install the
packages I want rather than letting me to choose which mirror etc?
Sure. If you read
Thanks Sarah.
I realized it as well and tried to run install.packages(plyr, lib = .Library)
as root. It works now.
Thanks again.
HXD
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: Monday, July 09, 2012 12:06 PM
To: Hui Du
Cc: r-help@r-project.org
Subject:
Hello,
So you want to split data.frame 'dat' by the values of the first column,
named 'site', and return the other columns.
lapply(split(dat, dat[[ 1 ]]), function(x) x[-1])
Hope this helps,
Rui Barradas
Em 09-07-2012 13:27, elpape escreveu:
Dear all
I'm trying to calculate the number
I'm using R to create a heatmap from a matrix using heatmap.2 - and i want
to group these images into one big image - What i usually use to achieve
this is layout() - but this doesn't work, as heatmap.2 uses layout, and
apparently layout does not work recursively.
Does anyone have any suggestions
Thanks Prof, I needed one more step on my system:
sudo locale-gen es_ES.UTF-8
Mikkel
From: Prof Brian Ripley rip...@stats.ox.ac.uk
Cc: R Help r-help@r-project.org
Sent: Monday, July 9, 2012 7:01 AM
Subject: Re: [R] axis.Date language
On 09/07/12 01:43,
Dear UseRs,
I'm making box plots from a data set that looks like this:
Chr Start End GeneDensity ReadCount_Explant ReadCount_Callus ReadCount_Regen
1 1 1 1 107.82 1.2431.047 1.496
2 1 10001 2 202.50 0.835
With base graphics, you could use par(mfrow) or par(mfcol) to define a simple
layout.
I don't know anything about heatmap.2, but I have used image() to make heatmaps
and have laid them out with par(mfrow).
Two ways to get a 2 row by 3 column layout, for example:
par(mfrow = c(3,2)) # 3
On 2012-07-09 11:07, arun wrote:
Hi,
Try this:
#Duplicated:
x-c(1:3,3)
x==x[duplicated(x)]
#[1] FALSE FALSE TRUE TRUE
#Unique:
x[!x==x[duplicated(x)]]
#[1] 1 2
A.K.
Try the above approach with
x - c(1,2,3,3,3,4,4,5)
I think Rui's solution is preferable.
Peter Ehlers
-
On Mon, Jul 09, 2012 at 07:52:18AM -0500, Jim Plante wrote:
I don't know how significant this is, but WolframAlpha's value of pi
disagrees with R's at about the 16th decimal:
Wpi-3.141592653589793238462643383279502884197169399375105
..^
On Jul 9, 2012, at 21:08 , Lawrence, Adaku wrote:
Hello,
Thanks for getting back to me. I was of the impression that once the res.
var. is larger than the df then the data was overdispersed and as such the
model was not a best fit. Is this true?
Not without qualification. There are
On Jul 9, 2012, at 21:26 , Rui Barradas wrote:
Hello,
So you want to split data.frame 'dat' by the values of the first column,
named 'site', and return the other columns.
lapply(split(dat, dat[[ 1 ]]), function(x) x[-1])
Got to be easier to split(dat[-1], dat[[1]]), no?
-pd
Thanks for your reply. I do have a copy of Zero Inflated Models and
Generalized Linear Mixed Models with R and have been using that as a guide.
I applied the predict function (type=count) to the dataset for which I
built the model to compare the predicted bycatch numbers to the observed to
ensure
Ouch!
Rui Barradas
Em 09-07-2012 21:39, peter dalgaard escreveu:
On Jul 9, 2012, at 21:26 , Rui Barradas wrote:
Hello,
So you want to split data.frame 'dat' by the values of the first column, named
'site', and return the other columns.
lapply(split(dat, dat[[ 1 ]]), function(x) x[-1])
Hi all,
I am using avPlots... and it has the following:
Hit Return to see next plot:
But I just wanted to everything to be saved into a pdf file...
How to skip these Returns?
Thanks a lot!
[[alternative HTML version deleted]]
__
Laura Lee laura.lee at ncdenr.gov
Mon Jul 9 22:51:40 CEST 2012
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Thanks for your
Maybe read the help for avPlots, especially the part about the ask argument:
ask If TRUE, ask the user before drawing the next plot; if FALSE don't ask.
Sarah
On Mon, Jul 9, 2012 at 5:16 PM, Michael comtech@gmail.com wrote:
Hi all,
I am using avPlots... and it has the following:
hi all,
does anyone know if R can do exact logistic regression with correlated
binary data?
Thanks!!
Yue
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View this message in context:
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hi all,
I have a binary data set and am now confronted with a separation issue. I
have two predictors, mood (neutral and sad) and game type (fair and
non-fair). By separation, I mean that in the non-fair game, whereas 20%
(4/20) of sad-mood participants presented a positive response (coded as 1)
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