# Perhaps I misunderstand your original need, but
## I added a few lines to your data and used dput() to get the below data (I
named "df")
df<- structure(list(age = c(15L, 20L, 15L, 10L, 10L, 12L, 17L, 17L,
11L, 12L, 16L, 20L, 23L, 14L, 22L, 16L, 10L, 11L, 21L, 10L, 13L,
17L), sex = struct
system is one option, as it turns out, this is common enough that there are
special functions for it, see ?file.copy
More generally, this is a great use of apropos:
apropos("file")
Shows a lot of options. An alternative work flow would be a shell script that
runs r scripts and also interacts
Natalia Vizcaíno Palomar gmail.com> writes:
> I've been reading for many days trying to predict with lmer but I haven't
> managed to do it.
> I've fitted an allometric model for trees where I have included climatic
> variables and diameter in the fixed part and
> in the random part I've included
I'm somewhat confused on how you intend to use browser() after
rerouting output to a sink
Beyond that, how are you running your script? source() has some
arguments that encourage it to "say" alot more to the sink() command.
Michael
On Tue, Nov 1, 2011 at 6:15 PM, stat.kk wrote:
> Hi,
>
> I
On 02/11/11 15:45, Erin Hodgess wrote:
Never mind...I used paste and all is well.
sorry for the trouble.
You *might* want to consider using file.rename() rather than
system(). Dunno what the pro's and con's might be.
Still have to use paste(), but! :-)
cheers,
Rolf
___
2011/11/1 Ernest Adrogué :
> Hi,
>
> On ocasion, you need to subscript an array that has an arbitrary
> (ie. not known in advance) number of dimensions. How do you deal with
> these situations?
> It appears that it is not possible use a list as an index, for
> instance this fails:
>
>> x <- array(N
Never mind...I used paste and all is well.
sorry for the trouble.
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
R-help@r-project.org mailing li
Dear R People:
I have a variable named file1 which contains the name of a file. I
would like to copy that file to a different directory. Can I do that
via the system command or is there a better way, please?
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathemat
On 02/11/11 13:43, Ernest Adrogué wrote:
Sorry for not stating my problem in a more clear way. What I want is,
given an array of n dimensions, overwrite it by iteratating over its
"outermost" dimension... OK, in the previous example, I would like
to do
x<- array(NA, c(2,2,2))
for (i in 1:2) {
2/11/11 @ 13:10 (+1300), Rolf Turner escriu:
> On 02/11/11 11:14, Ernest Adrogué wrote:
> >Hi,
> >
> >On ocasion, you need to subscript an array that has an arbitrary
> >(ie. not known in advance) number of dimensions. How do you deal with
> >these situations?
> >It appears that it is not possible
Hi,
I use the epicalc package which provides the function aggregate.numeric.
Unfortunately aggregate.numeric produces warnings when aggregate is used by
functions not under my control on a numeric value. If I don't load epicalc,
aggregate.default is used instead by these functions and does not
1/11/11 @ 20:22 (-0400), Comcast escriu:
> Leaving the indices empty should give you what I'm guessing you want/expect.
>
> x[,,2]#. TRUE would also work, just not in a list.
Exactly, but this only works if x has three dimensions. What I want is
x[,,2] if x has three dimensions, x[,,,2] if
Hello, I need help getting two data sets to merge. The structure of
my
two data sets are:
> str(bcusip)
'data.frame': 1391 obs. of 3 variables:
$ bond_id : Factor w/ 1391 levels "AAGH","AAGI",..: 1 2 3 4 5 6
$ Freq: num 41361 4126 5206 10125 45536 ...
$ CUSIP_ID: Factor w/ 1391 levels "
Here's a hack, but perhaps you might want to rethink what type of
output you want.
# Function:
g <- function(arr, lastSubscript = 1) {
n <- length(dim(arr))
commas <- paste(rep(',', n - 1), collapse = '')
.call <- paste('arr[', commas, lastSubscript, ']', sep = '')
eval(parse(text = .c
will this do it:
> x <- read.table(textConnection("arrive depart intercept
+ 11 1
+ 12 1
+ 13 1
+ 12 2
+ 13 2
+ 13 3
+ 22 2
+ 23 2
+ 33
The print method is the issue:
> t.out <- t.test(b1,b2,var.equal=T)
> t.out$p.value
[1] 4.108001e-38
> t.out$statistic
t
-15.93656
albyn
On Tue, Nov 01, 2011 at 02:40:15PM -0500, Jonathan Thayn wrote:
> Sometimes the p.value returned by t.test() is the same that I calculate using
> p
Posted to another thread a response to this posting ( and to all those who
wanted R on an iPad, I say "forget it"
-
The if function only takes an argument of length 1 (as the warning says):
?"if"
Many such confusions are resolved by looking at :
?ifelse
--
David
On Nov 1, 2011, at 4
Plot doesn't create an object to be plotted later like in ggplot: rather it, by
default, makes a new plot and erases the histogram.
> hist(z)
> lines(z, y, lwd = 1)
should work, though you may need to add freq=FALSE to hist()
MW
On Nov 1, 2011, at 6:42 PM, Jim Silverton wrote:
> I need a pl
The if function only takes an argument of length 1 (as the warning says):
?"if"
Many such confusions are resolved by looking at :
?ifelse
--
David
On Nov 1, 2011, at 7:22 PM, Rich Shepard wrote:
> On Tue, 1 Nov 2011, stat.kk wrote:
>
>> I would like to export all my workspace (even with th
Yes,Ii did fail to read your post carefully and agree do.call seems
roundabout, but alternatives look even more tortured.
(You might want to include more context in the future.)
On Nov 1, 2011, at 8:30 PM, Ernest Adrogué wrote:
> 1/11/11 @ 20:22 (-0400), Comcast escriu:
>> Leaving the indices
Hi:
When I call str() on an mlogit object, I seem to get warnings. This
code is from an example provided in the mlogit documentation:
library(mlogit)
data("Train", package="mlogit")
tr<-mlogit.data(Train, shape="wide", choice="choice", varying=4:11,
sep="", alt.levels=c(1,2), id="id")
ml.train<-m
Shahab;
You would be well advised not to seek private tutoring from someone on the
Internet who tells you that a p-value of 0.008736 is "not significant".
On Nov 1, 2011, at 8:09 PM, Raphael Saldanha wrote:
> Hi Shahab,
>
> This test shows that there is some positive statistical correlation
Hi:
Here's what I got when I ran your code:
library('glmnet')
> x=matrix(rnorm(100*20),100,20)
> y=rnorm(100)
> cv.fit=cv.glmnet(x,y)
> predict(cv.fit,newx=x[1:5,])
1
[1,] 0.1213114
[2,] 0.1213114
[3,] 0.1213114
[4,] 0.1213114
[5,] 0.1213114
> coef(cv.fit)
21 x 1 sparse Matrix of cla
Leaving the indices empty should give you what I'm guessing you want/expect.
x[,,2]#. TRUE would also work, just not in a list.
David.
On Nov 1, 2011, at 6:14 PM, Ernest Adrogué wrote:
> Hi,
>
> On ocasion, you need to subscript an array that has an arbitrary
> (ie. not known in advance)
Hi Shahab,
This test shows that there is some positive statistical correlation, BUT
the p-value of the test - this is, the level of significance - shows that
the correlation is not statistically significant at 95% confidence level.
So, the correlation may be equal to zero.
To understand this conc
On 02/11/11 11:14, Ernest Adrogué wrote:
Hi,
On ocasion, you need to subscript an array that has an arbitrary
(ie. not known in advance) number of dimensions. How do you deal with
these situations?
It appears that it is not possible use a list as an index, for
instance this fails:
x<- array(NA
Hi:
After cleaning up your data, here's one way using the plyr and
reshape packages:
d <- read.csv(textConnection("
Individual, A, B, C, D
Day1, 1,1,1,1
Day2, 1,3,4,2
Day3, 3,,6,4"), header = TRUE)
closeAllConnections()
d
library('plyr')
library('reshape')
# Stack the variables
dm <- melt(d, id
Does this work?
library('plyr')
# Function to return a data frame if it has one row, else return NULL:
f <- function(d) if(nrow(d) == 1L) d else NULL
> ddply(RV09, .(set, month), f)
record.t trip set month stratum NAFO unit.area time dur.set distance
15 913 110 351 3O
On Tue, Nov 1, 2011 at 12:40 PM, Jonathan Thayn wrote:
> Sometimes the p.value returned by t.test() is the same that I calculate using
> pt() and sometimes it's not. I don't understand the difference. I'm sure
> there is a simple explanation but I haven't been able to find it, even after
> look
On Tue, 1 Nov 2011, stat.kk wrote:
I would like to export all my workspace (even with the evaluation of
commands) to the text file.
Have you looked at .Rhistory? If you save your workspace when you quit a
session with R it's put in that file. You can always read it anywhere you
have a text e
Works in the newly released 2.14.0:
> X = c(60593.23, 71631.17, 75320.1)
> round(X, -4)
[1] 6 7 8
Dennis
On Tue, Nov 1, 2011 at 10:07 AM, Wendy wrote:
> Hi all,
>
> I have a list of numbers, e.g., X = c(60593.23, 71631.17, 75320.1), and want
> to round them so the output is Y = c(
Hi,
I would like to export all my workspace (even with the evaluation of
commands) to the text file. I know about the sink() function but it doesnt
work as I would like. My R-function looks like this: there are instructions
for user displayed by cat() command and browser() commands for fulfilling
Hi there,
I am trying to use predict() with an object returned by cv.glmnet(), and get
the following error:
no applicable method for 'predict' applied to an object of class "cv.glmnet"
What's wrong?
my code:
x=matrix(rnorm(100*20),100,20)
y=rnorm(100)
cv.fit=cv.glmnet(x,y)
predict(cv.fit,newx=
Hi,
On ocasion, you need to subscript an array that has an arbitrary
(ie. not known in advance) number of dimensions. How do you deal with
these situations?
It appears that it is not possible use a list as an index, for
instance this fails:
> x <- array(NA, c(2,2,2))
> x[list(TRUE,TRUE,2)]
Error
I have a dataset called "results" that looks like this:
arrive depart intercept
11 1
12 1
13 1
12 2
13 2
13 3
22 2
23 2
33 3
Sometimes the p.value returned by t.test() is the same that I calculate using
pt() and sometimes it's not. I don't understand the difference. I'm sure there
is a simple explanation but I haven't been able to find it, even after looking
at the code for t.test.default. I apologize if this is a bas
Hi,
I am an R novice and I am trying to do something that it seems should be fairly
simple, but I can't quite figure it out and I must not be using the right words
when I search for answers.
I have a dataset with a number of individuals and observations for each day (7
possible codes plus miss
If no one has a better solution, split it, take a sample of size X from both
and put it back together.
hgwelec wrote:
>
> Dear members,
>
> Consider the following data frame (first 4 rows shown)
>
>
> age sex class
> 15 m low
> 20 f high
> 15 f low
> 10 m low
>
> in
Hi,
I am not really familiar with Correlation foundations, although I read
a lot. So maybe if someone kindly help me to interpret the following
results.
I had the following R commands:
correlation <-cor( vector_CitationProximity , vector_Impact, method =
"spearman", use="na.or.complete")
cor_tes
I need a plot with a histogram z and a plot of z versus y superimposed.
This is not working...
z = rnorm(1000)
y = rnorm(1000,1,2)
hist(z)
dx = plot(z, y, lwd = 1)
lines(dx, add=TRUE)
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R
I perhaps made that a little too complicated:
this will also work:
v2 = xts(ct3[,3], as.POSIXct(paste(ct3[,1], ct3[,2]), format =
"%m/%d/%Y %H:%M:%S"))
identical(v, v2)
TRUE
On Tue, Nov 1, 2011 at 6:06 PM, R. Michael Weylandt
wrote:
> Your first problem is that you aren't using paste() properl
Your first problem is that you aren't using paste() properly: print
out paste(ct3[, 1:2]) and take a look at it.
This works:
apply(head(ct3[,1:2]),1,paste,collapse = " ")
You also have the format argument to POSIXct wrong. See ?strptime for details.
So the whole line (if you want to do it in on
Thanks for the fast response and your comments!
That works perfect!
Another little mystery solved ;)
Stefan
From: Felipe Carrillo [mailto:mazatlanmex...@yahoo.com]
Sent: Tuesday, November 01, 2011 3:54 PM
To: Schreiber, Stefan; r-help@r-project.org
Subject: Re: [R] factor level iss
Stefan:
Use the droplevels function...
dat <- read.table(textConnection("
treat yield
1 cont 98.7
2 cont 97.2
3 cont 96.1
4 cont 98.1
5 10 103.0
6 10 101.3
7 10 102.1
8 10 101.9
9 30 121.1
10 30 123.1
11 30 119.7
12 30 118.9
13 60 109.9
14 60 110.1
15 60 1
first of all, the subsetting line is overly complicated.
dat.sub<-dat[dat$treat!='cont',]
will work just fine. R does exactly what you're describing. It knows
the levels of the factor. Once you remove 'cont' from the data, that
doesn't mean that the level is removed from the factor:
> df<-dat
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Schreiber, Stefan
> Sent: Tuesday, November 01, 2011 2:29 PM
> To: r-help@r-project.org
> Subject: [R] factor level issue after subsetting
>
> Dear list,
>
> I cannot figure out
Hope this helps:
http://rss.acs.unt.edu/Rdoc/library/randomForest/html/rfImpute.html
Ken Hutchison
On Nov 1, 2554 BE, at 5:29 PM, Sascha Vieweg wrote:
> Hello
>
> I am working on my first attempt to impute missing data of a data set with
> systematically incomplete answers (school performance
To all Windows users:
The binary build of 2.14.0 that was uploaded yesterday was missing Cairo
support. I have rebuilt it, and uploaded a new copy.
You can tell which one you have by running "svg()", which works on the
new one, but not the old one.
You can tell which one is on your CRAN mi
Hello
I am working on my first attempt to impute missing data of a data
set with systematically incomplete answers (school performance
tests). I was googling around for some information and found
Amelia (Honaker et al., 2010) and the mi package (Yu-Sung et al.,
n.d.). However, since I am new
Dear list,
I cannot figure out why, after sub-setting my data, that particular item
which I don't want to plot is still in the newly created subset (please
see example below). R somehow remembers what was in the original data
set. A work around is exporting and importing the new subset. Then it's
Perhaps use tapply() to split by the survey unit and write a little
identity function that returns only those rows you want, then patch
them all back together with something like simplify2array().
Michael
On Tue, Nov 1, 2011 at 1:16 PM, Aurelie Cosandey Godin wrote:
> Dear list,
>
> After readin
Bert,
How do you define "elementary"? And, do we play this like "Name That Tune"?
"Bert, I can do that calculation in X operations." Or, maybe like Jeopardy,
"What is the number X?" Or maybe we could play "Are You Smarter Than a
Seventh-Grader?" I'm just asking.
Dan
Daniel J. Nordlund
Also be aware of the IEEE standard of rounding to even:
> round(61000, -4)
[1] 6
> round(62000, -4)
[1] 6
> round(65000, -4)
[1] 6
> round(66000, -4)
[1] 7
> round(76000, -4)
[1] 8
> round(75000, -4)
[1] 8
>
notice what 65000 rounds to and what 75000 rounds to.
On Tue, Nov
Yes, I agree with David that this looks like an error.
However, for fun, one might ask: what is the fewest number of R elementary
math operations that would produce such a result -- this might be good for
clever 6th or 7th graders, for example.
For here, I leave this as an exercise for the reader
Or does the middle number have two digits switched? 76131.17 would
round up to 8 very nicely.
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
Could you divide by your desired order of magnitude, use ceiling and then
re-multiply?
Michael
On Nov 1, 2011, at 3:42 PM, David Winsemius wrote:
>
> On Nov 1, 2011, at 1:07 PM, Wendy wrote:
>
>> Hi all,
>>
>> I have a list of numbers, e.g., X = c(60593.23, 71631.17, 75320.1), and want
>> t
On Nov 1, 2011, at 1:07 PM, Wendy wrote:
Hi all,
I have a list of numbers, e.g., X = c(60593.23, 71631.17, 75320.1),
and want
to round them so the output is Y = c(6, 8, 8).
Under what notion of "rounding" would that be the result?
I tried
Y<-round(X,-4), but it gives me Y =
the following is a more appropriate forum for your question, seeing as this
has nothing to do with R (per se).
http://stats.stackexchange.com/questions
good luck.
peak99 wrote:
>
> Hello,
>
> I have some patient data for my masters thesis with three groups (n=16, 19
> & 20)
>
> I have com
Hello,
I have some patient data for my masters thesis with three groups (n=16, 19 &
20)
I have completed compiling the results of 7 tests, for which one of these
tests the variances are unequal.
I wish to perform an ANOVA between the three groups but for the one test
with unequal variance (<0.0
Dear list,
After reading different mails, blogs, and tried a few different codes without
any success, I am asking your help!
I have the following data frame where each row represent a survey unit with the
following variables:
> names(RV09)
[1] "record.t" "trip" "set" "month" "
Hi all,
I have a list of numbers, e.g., X = c(60593.23, 71631.17, 75320.1), and want
to round them so the output is Y = c(6, 8, 8). I tried
Y<-round(X,-4), but it gives me Y = c(6, 7, 8). Do anybody know
how to round up a number to 10^4?
Thank you in advance.
Wendy
--
Dear David,
Thank you for your follow-up and extra-suggestion.
Your additional stab at my problem indeed looks like it'd work too; as I
previously wrote, I had already devised a work-around but was nonetheless left
wondering whether a more elegant and compact solution was still escaping my
kno
I think a lag() command will work for you -- but a more helpful piece
of advise is probably to use xts or zoo packages rather than the base
ts. Most people find ts very hard to work with and find that all their
difficulties magically go away when they use those classes.
In xts: it would be somethi
On Nov 1, 2011, at 5:05 AM, Martin Ivanov wrote:
Dear R users,
I want to plot not only triangles point up and triangles point down,
which is easy using the "pch" argument to "points". I want to plot
left and right pointing triangles as well. They must be fillable
with colour.
I browsed a
On 11/01/2011 12:32 PM, loubna181 wrote:
Hi,
Thanks all for your responses, but as I m a new user of R while trying to
apply what David suggests I dont know what *"dorm" *refers to.
dfrm[c(rownames(dfrm[*dorm*$Y==1,]), sample(rownames(dfrm[dfrm$Y==0]),
0.10)) , ]
I suspect that dorm was a typo
Hi,
Thanks all for your responses, but as I m a new user of R while trying to
apply what David suggests I dont know what *"dorm" *refers to.
dfrm[c(rownames(dfrm[*dorm*$Y==1,]), sample(rownames(dfrm[dfrm$Y==0]),
0.10)) , ]
But to give you more details , I'm working on a table calles balance from
Hello,
I am working a marketing mix model with different specification
(that is different lags , degree of polynomial and different end point
specification ) for each of the independent variables and will appreciate any
help on the package I can used to this exercise and also a sa
I'm trying to compute sample size requirements for a binomial exact test.
we want to show that the proportion is at least 90% assuming that it is
95%, with 80% power so any asymptotic approximations are out of the
questions. I was planning on using binom.test to perform the simple test
against a
See rug() and use col=2 to get red. So, as an example ...
x <- rchisq(100, df=2)
hist(x)
abline(v=median(x), lty=2)
rug(x, col=2)
TF
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wendy
Sent: Tuesday, November 01, 2011 10:22 AM
T
On Nov 1, 2011, at 5:18 AM, Stefano Conti wrote:
Dear David,
My ultimate purpose is to generate a text file encoding a LaTeX
table for later inclusion in a report; while I'm aware of, and
familiar with, Sweave, such table would feature _some_ whole or
partial crossing horizontal lines (\
Hi,
You can use function segments() to draw them.
Regards,
Carlos Ortega
www.qualityexcellence.es
2011/11/1 Wendy
> Hi all,
>
> I want to make a histogram like the one show
> http://nar.oxfordjournals.org/content/39/suppl_1/D1011/F1.expansion.html
> here , but I did not figure out how to add
Hi all,
I want to make a histogram like the one show
http://nar.oxfordjournals.org/content/39/suppl_1/D1011/F1.expansion.html
here , but I did not figure out how to add the red marks at the bottom of
the bars. Could anybody help? Thank you very much
--
View this message in context:
http://r.789
Perhaps something like this:
X = jitter(1:10)
Y = jitter(3*X-5, factor = 3)
X[3] = NA
m = lm(Y~X)$fitted.values
fits <- rep(NA, length(X)); fits[as.numeric(names(m))] <- m;
cbind(X,Y,fits)
Michael
On Tue, Nov 1, 2011 at 8:52 AM, Johannes Radinger wrote:
> Hi,
>
> I have got a dataset with
> t
hello,
over a month ago i asked this question and then wrote to the list that i
had to postpone the problem.
anyway, now i tried it again with the XLConnect package and was successfull!
i used these commands:
wb <- loadWorkbook("MyExcelFile.xlsx", create=TRUE)
createSheet(wb,name="MySheetname")
Alexandre,
The output from corr.test is a list of matrices. To export one of those
matrices, simply specify which one you want:
Using the example from my previous note:
> library(psych)
> examp <- corr.test(sat.act)
> mat.c.p <- lower.tri(examp$r)*examp$r + t(lower.tri(examp$p)*examp$p)
> mat
On Nov 1, 2011, at 13:20 , Filoche wrote:
> Hi everyone.
>
> I'm trying to use small letter phi in a graph produced in R. However, the
> small letter phi does not look as it should.
>
> In fact, it looks like this:
>
> http://r.789695.n4.nabble.com/file/n3963311/Untitled.png
>
> instead of w
Try changing phi to varphi.
Michael
On Tue, Nov 1, 2011 at 8:20 AM, Filoche wrote:
> Hi everyone.
>
> I'm trying to use small letter phi in a graph produced in R. However, the
> small letter phi does not look as it should.
>
> In fact, it looks like this:
>
> http://r.789695.n4.nabble.com/file/n
On 01.11.2011 13:20, Filoche wrote:
Hi everyone.
I'm trying to use small letter phi in a graph produced in R. However, the
small letter phi does not look as it should.
In fact, it looks like this:
http://r.789695.n4.nabble.com/file/n3963311/Untitled.png
Yes, an excellent phi.
instead of
Hi everyone.
I'm trying to use small letter phi in a graph produced in R. However, the
small letter phi does not look as it should.
In fact, it looks like this:
http://r.789695.n4.nabble.com/file/n3963311/Untitled.png
instead of what is here http://en.wikipedia.org/wiki/Phi
Here's the code I
That worked, thanks!
On Fri, Oct 28, 2011 at 12:27 PM, bbolker [via R] <
ml-node+s789695n3948463...@n4.nabble.com> wrote:
> ethan.shepherd gmail.com> writes:
>
> >
> > I'm getting errors when running what seems to be a simple Weibull
> > distribution function:
> >
>
> [snip]
>
> > If I change
Just to close my issue: When changing the data types from xts to numeric,
the problem went away and the speed of processing increased dramatically.
(seems obvious in hindsight)
--
View this message in context:
http://r.789695.n4.nabble.com/Interesting-Memory-Management-Problem-Windows-tp3944243
Dear all,
I've been reading for many days trying to predict with lmer but I haven't
managed to do it.
I've fitted an allometric model for trees where I have included climatic
variables and diameter in the fixed part and
in the random part I've included the experimental sites where trees are and
al
hello,
using the package "descr" i created a crosstab e.g.:
Table <- crosstab(age,question,weight=pond)
"age": being a variable that consists of the surveyed data of the ages of
children: the values are "6 years", "7 years" and "8 years".
"question": is a variable that consists of the answers of
Hi,
I have got a dataset with
the variables Y,X1,X2,X3.
Some of these variables contain NAs. Therefore
incomplete datasets aren't recognized when
I am doing a regression like:
model <- lm(Y~X1+X2+X3)
so the resulting vector of resid(model) is
obviousely shorter then the original variables.
How
For short:
1. I wonder if you cannot make the two steps one and hence only need one
lapply anway.
2. Why not just try out?
3. If it does not work, ordinary loops may be able to.
best,
Uwe Ligges
On 01.11.2011 09:24, Alaios wrote:
Dear all,
I want for a given data set to call a funciton man
On 01.11.2011 09:47, Muhammad Rahiz wrote:
Thanks for those who replied.
I know what a p-value is and the links given reaffirm my understanding.
The code below gives p=0.26. This is more than 0.05 - hence null
hypothesis i.e. attributed by chance.
But I did not specify the significance level
On Mon, Oct 31, 2011 at 11:27 PM, Vinny Moriarty wrote:
> Thanks for everyone's input so far, it is greatly appreciated. But I've got
> one last task I could use some advice on
>
>
> Here are the first few lines of my data set:
>
> site,time_local,time_utc,reef_type_code,sensor_type,sensor_depth_m
You can use mvtnorm package
http://cran.r-project.org/web/packages/mvtnorm/index.html
>-Original Message-
>From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]
>On Behalf Of Anera Salucci
>Sent: 01 November 2011 11:22
>To: r-help@r-project.org
>Subject: [R] multivariat
Hi,
one solution is to use sink. Check ?sink to see explanation and following
example.
sink("sink-examp.txt")
i <- 1:10
outer(i, i, "*")
sink()
Andrija
On Tue, Nov 1, 2011 at 10:43 AM, AlexC wrote:
> Hello,
>
> Thank you for your replies. I cannot run the function rcor.test even when
> having
Dear All,
How can I generate multivariate random variable (not multivariate normal )
I am in urgent
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Hello,
Thank you for your replies. I cannot run the function rcor.test even when
having loaded package ltm. Perhaps it has to do with the fact that I am
using the latest version of R and this package wasn't created under that
version
The function corr.test in package psych works fine. Is ther
Dear David,
My ultimate purpose is to generate a text file encoding a LaTeX table for later
inclusion in a report; while I'm aware of, and familiar with, Sweave, such
table would feature _some_ whole or partial crossing horizontal lines (\hline
or \cline with varying arguments, which need to be
Many thanks for your comments. The median of the r_i is something around
1000. And for the time being there are no covariates, though this might
change in the future. We are only starting to exploit a very nice data set.
Regarding the probability of being in the data, p, I would say it is indeed
c
I am writing a GUI for my R script. It is a very basic form consisting of
textboxes and buttons.
I tried to run the following example to learn how to read value currently
entered into the textbox:
( it requires tcltk/tcltk2 packages )
# Create the widgets
base <- tktoplevel()
list <- t
On 11/01/2011 09:50 AM, Prof Brian Ripley wrote:
> On Tue, 1 Nov 2011, Paul Hiemstra wrote:
>
>> On 10/31/2011 06:30 PM, Cable, Sam B Civ USAF AFMC AFRL/RVBXI wrote:
>>> I am trying to install R on a pretty up-to-date CentOS system. I have
>>> tried installing 2.14.0 and 2.13.2. In both cases, th
On Tue, 1 Nov 2011, Paul Hiemstra wrote:
On 10/31/2011 06:30 PM, Cable, Sam B Civ USAF AFMC AFRL/RVBXI wrote:
I am trying to install R on a pretty up-to-date CentOS system. I have
tried installing 2.14.0 and 2.13.2. In both cases, the configure step
fails, and does not produce a Makefile. So
Dear R users,
I want to plot not only triangles point up and triangles point down,
which is easy using the "pch" argument to "points". I want to plot left and
right pointing triangles as well. They must be fillable with colour.
I browsed a little in the documentation, tried rotating the up and
Thanks for those who replied.
I know what a p-value is and the links given reaffirm my understanding.
The code below gives p=0.26. This is more than 0.05 - hence null
hypothesis i.e. attributed by chance.
But I did not specify the significance level at 0.05. So I'm wondering if
any part of t
On Tue, Nov 1, 2011 at 4:49 AM, wrote:
> Hello,
>
> I have been following the thread dated Monday, October 9, 2006 when Kamila
> Naxerova asked a question about plotting elliptical shapes. Can you explain
> the equations for X and Y. I believe they used the parametric form of x and y
> (x=r co
Dear all,
I want for a given data set to call a funciton many time. That bring us all to
the notion of lapply (or any other variance). My problem is that this data set
should also be created from another lapply
As for example
DataToAnalyse <- return_selected_time_interval(TimeStamps,Time[
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