Ken Ervin wrote:
for (i in 1:2858) {
...
write.table(Data_frame_name,file=~/file/goes/here.csv, append
= FALSE, sep = \t,row.names = FALSE, col.names = c(Calc_1, Calc_2,
Calc_3),qmethod = double)
}
use something like
filename = paste(~/file/myfile,i,.csv,sep=)
On 10/08/2009 01:15 PM, darrenh65 wrote:
Hi,
I am attempting to create a polar chart (using plotrix polar.plt) and I need
to colour a segment of the grid to indicate an area of interest, say from
90-120 degrees, for arguments sake. Is this possible? I can colour the whole
grid by passing the
What kind of filter are you using? Since your models are expressed in state
space form I suggest that you fit your models by maximizing the log likelihood
function of the Kalman filter output (see e.g. FKF-package). Using the obtained
log likelihood values you might perform a likelihood ratio
Hello!
I'm still working on my problem, which also occurs with the predict.lm()
function. - Providing newdata, which is a data.frame with all variables
being numeric, as str() shows, R tells me the following:
ar1.xpred.test.pred - predict(ar1.xpred.fitted, regdata.test, se.fit =
FALSE)
Wolfgang Waser wrote:
I subjected each of about 50 critters (about 10 each in 5 distinct
populations) to 4 consecutive treatments (exposure to increasing
concentrations), with one measurement per treatment and individual.
...
Since I'm dealing with 'paired' data, and also with a
Dear all,
Is there any R package which would help in analysing election results between
two elections? Does anybody know any good papers which are related to this
field? I am a statistician and my main research area so far has been regression
and classification modelling. The analysis of two
What are you trying to find out from your analysis and what techniques do you
have in mind for doing it?
It might be as well to sort out this first (and tell us if you have) rather
than look round for a package first and then decide on what you want to do in
the analysis.
Having said all
LinZhongjun wrote:
Dear R-help lists:
Hello!
I ran WinBUGS under R using the function bugs.
But I kept having the following error message:
Error in FUN(X[[3L]], ...) :
.C(..): 'type' must be real for this format
Why did I have the error message?What's the problem?How to get rid of the
Whatever your final choice is, I think you should really try to use the
distribution's package management system. I don't use the distributions you
cited so I can't give you precise hints, but I'm sure you can easily find
tutorials on how to install/remove software. It will make things a lot
hi everyone!
i want to check the autocorrelation function for a univariate time series
(streamflow) in a data frame as below:
DF - read.table(D:/file path)
DF
year jan feb mar apr .. dec
1966 0.504 0.406 0.740 0.241 0.429
1967 0.683 0.529
*additional:
the lags i am expecting is in months.. since i am trying to predict monthly
streamflow.
thanks again
--
View this message in context:
http://www.nabble.com/acf-for-a-univariate-time-series-in-a-data-frame-tp25799751p25799784.html
Sent from the R help mailing list archive at
I use R on my Ubuntu 9.04 laptop, which was installed by aptitude install
way
Now i'm learning a book of R which needs MASS and DAAG installed.
So i followed the instructions:
install.packages(MASS)
library(MASS)
MASS works fine.
But DAAG doesn't. Anyone who could help would be appreciated a
Cheers guys that's helpful. Doug, you're right, my code for ff should have
been
for (i in 1:length(y))
{if (f1[i]==after f3[i]==1) ff[i]-1, after
else if(f1[i]==after f3[i]==2) ff[i]-2, after
else if(f1[i]==before f3[i]==1) ff[i]-1, before
else if(f1[i]==before f3[i]==2) ff[i]-2, before}
On Thu, Oct 8, 2009 at 1:57 AM, Paul Chatfield p.s.chatfi...@rdg.ac.uk wrote:
Cheers guys that's helpful. Doug, you're right, my code for ff should have
been
for (i in 1:length(y))
{if (f1[i]==after f3[i]==1) ff[i]-1, after
else if(f1[i]==after f3[i]==2) ff[i]-2, after
else
Dear All,
Let 499 piece-wise lines were buit up by 500 pair of observations, via R code
below.
x - 1:500
y - rnorm(500)
plot(x, y, type = 'b')
I was trying to compute all the slopes for the lines which were connected
between two adjacent points. For instance, slopes of lines between first
That's solved it. Superb!
All you probably need is to make f2 a factor (e.g., y ~ factor(f2) |
f1). Otherwise dotplot() doesn't know which one to treat as
categorical.
-Deepayan
--
View this message in context:
Hi,
Like this perhaps?
slope = diff(y) / diff(x)
str(slope)
num [1:499] 1.5068 -1.8406 2.1745 0.0676 -2.6088 ...
HTH,
baptiste
2009/10/8 FMH kagba2...@yahoo.com:
Dear All,
Let 499 piece-wise lines were buit up by 500 pair of observations, via R
code below.
x - 1:500
y - rnorm(500)
Dear all
when I try to install distr, the following error appears, I am using R in
windows. can u suggest me?
Error in normalizePath(path) :
path[1]=C:\Program Files\R\R-2.9.1\library/distr: The system cannot find
the file specified
--
This message was sent on behalf of msig...@yahoo.com
Do you really need to keep all 2858 dataframes around, or are you just
using them to create the object to write out? If you really have to
keep that many, I would suggest that you use a list since you will
really polute your workspace with that many objects from the
standpoint if you ever wanted
Your function is referencing 'cand2' but where is it defined? There
are other objects referenced in the function that do not appear to be
passed in as parameters. Are they defined globally? What do you
think the following statement in your function is doing:
Here is one way of doing it:
n - 20
set.seed(2)
# create test dataframe
x - as.data.frame(matrix(sample(1:2,n*6, TRUE), nrow=n))
x
V1 V2 V3 V4 V5 V6
1 1 2 2 2 1 1
2 2 1 1 2 2 1
3 2 2 1 2 1 2
4 1 1 1 1 1 2
5 2 1 2 2 1 1
6 2 1 2 1 2 2
7 1 1 2 1
First, you should define your function as :
test - function(cand2,phi,lambda,
whatever-arguments-you-want-to-use-further){... insert code here ...}
All variables you use inside a function only exist within that
function. Your parameters/arguments is the interface between the
function and the
Quite a strange error. The function installed.packages should be in
the utils package.
Try ?installed.packages and see if you get a help file. If it doesn't
find that function, there is a problem with your R installation.
It might be something went wrong with downloading the package itself.
DAAG
I answered the wrong question. Here is the code to find all the
matches for each row:
n - 20
set.seed(2)
# create test dataframe
x - as.data.frame(matrix(sample(1:2,n*6, TRUE), nrow=n))
x
x.col - c(1,3,5)
# match against all the other rows
x.match1 - apply(x[, x.col], 1, function(a){
.mat -
What you need to do is to specify that you don't want labels on the second plot:
## (2) Does plot points where (x-axis) they should be
xlim - range(x1)
plot(x1, y1, type=l, xlim=xlim)
par(new=TRUE)
plot(x2, y2, type=p, xlim=xlim, xlab='', ylab='', axes=FALSE)
axis(4)
On Wed, Oct 7, 2009 at 4:29
On Oct 8, 2009, at 4:27 AM, Yanyuan Zhu wrote:
I use R on my Ubuntu 9.04 laptop, which was installed by aptitude
install
way
The is an R-SIG-Debian which is going to have more people with
knowledge of your particular OS.
Now i'm learning a book of R which needs MASS and DAAG
How to determine restricted variable in SVAR and SVEC? There are some values
which set to be zero and others set to be NA.. How to determine values that set
to be 0? Thanks
Regards,
Arif
Hi,
I'm using the sm.regression function from the package sm. Everything works
fine, but I'd like to have the the RGL display only with the smoothed surface
and without the data points displayed.
Is there an option to do it? I can reduce their size, using size parameter,
but I haven't
I have the following data set, representing the the estimated number of
some event (est), when the actual number
was 3, 4, ..., 15. The numbers in the cells are the observed
*frequencies* of each combination of (actual, estimated),
so each column (a3 -- a15) gives a single discrete frequency
Another approach is:
n - 20
set.seed(2)
x - as.data.frame(matrix(sample(1:2, n*6, TRUE), nrow = n))
x.col - c(1, 3, 5)
values - do.call(paste, c(x[x.col], sep = \r))
out - lapply(seq_along(ind), function (i) {
ind - which(values == values[i])
ind[!ind %in% i]
})
out
Best,
Dimitris
Dear all,
I want to split the strips in xyplot and push them into the margins ...
Tried to find this in common documentation (such as Deepayan's book) on
lattice ... but so far without success ...
Here is the situation:
xyplot(Speed~Count|Lane*Day,...)
where Speed and Count are numeric, Lane
Hi,
I would like to find out the coordinates of the intersection points of 2
density curves. I did a search but i didn't get any significant results. I
really hope some of you have some ideas. here it is an example:
set.seed(123)
x1 - rnorm(100, 1, 1)
x2 - rnorm(100, 0, 1)
d1 - density(x1)
Hi all,
I am a beginar user of R and try to do some CART analysis with it.
With rpart we can get several terminals and draw it in the TREE plot. Now
I am trying to draw a plot like this: x-axis is each terminal's value, and
y-axis is those observe values. Does anyone has idea what gramma should
Hi,
Try the useOuterStrips function in the latticeExtra package.
HTH,
baptiste
2009/10/8 Christian Ritter crit...@ridaco.be:
Dear all,
I want to split the strips in xyplot and push them into the margins ...
Tried to find this in common documentation (such as Deepayan's book) on
lattice
The standard asymptotic theory of likelihood ratio tests assumes that
you are testing a submodel, which is not the case here. Moreover, even
when testing submodels, there are other assumptions that often are not
met in the case of DLMs - the typical example being hypothesised
values on the
Hello everybody,
I want to run the following R-code, but it does'nt work.
res.vgm = variogram(ertrag ~ rep, # !!!
+ loc = ~ xpos + ypos, # !!!
+ width = start.range / 10,
+ data = d)
Error in vector(double, length) : element 2 is
On Oct 8, 2009, at 4:33 AM, msig...@yahoo.com wrote:
Dear all
when I try to install distr, the following error appears, I am
using R in windows. can u suggest me?
I'm not a Windows user, but the obvious questions would be How?.
What commands or method? and Which? (version of Windows
fb7c7e870910080701o1683a41q7d4295d97af8c...@mail.gmail.com
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
Hi=2C
=20
Thanks so much for your answer. It works beautifully for my example and of =
course i will test it on thereal data as
Try this (possibly after scaling the rows or columns to 1):
library(gplots)
with(as.data.frame.table(as.matrix(jevons[-1])), balloonplot(Var1, Var2, Freq))
On Thu, Oct 8, 2009 at 9:24 AM, Michael Friendly frien...@yorku.ca wrote:
I have the following data set, representing the the estimated
On Thu, Oct 8, 2009 at 4:26 PM, Monica Pisica pisican...@hotmail.comwrote:
fb7c7e870910080701o1683a41q7d4295d97af8c...@mail.gmail.com
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
Hi=2C
=20
Thanks so much for your answer. It
Dear All,
I am a researcher, working at the Department of Software Engineering at
University of Minho, Portugal. I am working on in the area of Distributed
data mining and attempting to develop distributed data mining system in R
Environment.
I have an enquiry.
Suppose I have 3 different nodes
It should also be remembered that there will often be multiple
intersections of such density estimates.
I hope this is not being done in support of a data-driven
discretization of two group comparisons. Such practices are to be
eschewed (as they are even worse than obfuscation).
--
David
Hi,
On Oct 8, 2009, at 11:06 AM, wesley mathew wrote:
Dear All,
I am a researcher, working at the Department of Software Engineering
at
University of Minho, Portugal. I am working on in the area of
Distributed
data mining and attempting to develop distributed data mining system
in „R‰
57d3c6e9-402b-4991-a53b-71e25af78...@comcast.net
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
No worries there =3B-)) But thanks for the reminder all the same!
=20
Monica
CC:
Dear All,
I have temperature series for 10 different sea levels, where there are 3000
observations for each level. I am planning to use multiple time
series modelling on these series, with/ without correlation structure for the
residuals.
Is there any package that can be used to do this
Gabor Grothendieck wrote:
Try this (possibly after scaling the rows or columns to 1):
library(gplots)
with(as.data.frame.table(as.matrix(jevons[-1])), balloonplot(Var1, Var2, Freq))
Thanks, Gabor --
That's not bad as a first cut; I can work with that.
--
Michael Friendly Email:
I am attempted to examine the temporal independence of my data set and think
I need an unordered multinomial logistic regression (or logit model) with
repeated measures to do so. The data in question is location of chickens.
Chickens could be in any one of 5 locations when a snapshot sample was
Hi ,
I want to convert a my list:
myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like c(1,2,3,4,5,6,7,8,9,10,11)
I realized that this is possible by
c(do.call(cbind,myList))
But only when list[[1]] through list[[3] have equal length of vectors within
them
On Oct 8, 2009, at 11:13 AM, delnatan wrote:
Hi ,
I want to convert a my list:
myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like c(1,2,3,4,5,6,7,8,9,10,11)
?unlist
I realized that this is possible by
c(do.call(cbind,myList))
But only when list[[1]]
Dear useRs and developeRs,
In the last few months we have had an increasing number of requests
about deSolve and dynamic modelling. We have also received several
requests to establish a SIG of our own about such things. We think
this is a good idea -- here it is!
Hi Peng,
On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the current working directory is always '~' no
matter where my current directory is in the terminal. I'm wondering if
there is a way to make the
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always quit unexpectedly.
What could cause the problem? Is it because that I have a 32 bit R
installed already installed in the same machine.
Regards,
Hi,
On Oct 8, 2009, at 11:59 AM, Peng Yu wrote:
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always quit unexpectedly.
What could cause the problem? Is it because that I have a 32 bit R
drlucyasher wrote:
I am attempted to examine the temporal independence of my data set and
think I need an unordered multinomial logistic regression (or logit model)
with repeated measures to do so.
What I have thought of so far is
glm(Location~ Pen| Bird*AMPM*Time,
If you want do.call, you can use c indeed of cbind:
do.call(c, list(1:5, 6:8, 9:11))
On Thu, Oct 8, 2009 at 12:13 PM, delnatan delna...@gmail.com wrote:
Hi ,
I want to convert a my list:
myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like
Ok, nevermind.
all I had to do was unlist(myList).
delnatan wrote:
Hi ,
I want to convert a my list:
myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like c(1,2,3,4,5,6,7,8,9,10,11)
I realized that this is possible by
Apologies for my overhasty and thus inaccurate reply. I of course assumed that
model2 is a submodel of model1. But thanks to Giovanni: a closer look at your
equations shows that this is only the case if at == bt (and of course same
innovations/ disturbances et, etat) and xt e.g. something like
If I understand your intent, I believe you can get what you want much faster
(no interpreted loops and linear times) by looking at this slightly
differently.
First of all, the choice of columns is unimportant, as indexing can be used
to create a data frame containing only the columns of
Maithili
I find it really hard to believe that a beta distribution would be a
reasonable probability model for loss data. There would have to be an
upper bound on the size of losses. What is the process that
generates the data. Is there any natural upper bound? Why is there a
lower
On Oct 8, 2009, at 11:59 AM, Peng Yu wrote:
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always quit unexpectedly.
It is hard to say. If you were really trying to run R from the image,
then
On Oct 8, 2009, at 12:53 PM, Albyn Jones wrote:
Maithili
I find it really hard to believe that a beta distribution would be a
reasonable probability model for loss data. There would have to be
an upper bound on the size of losses. What is the process that
generates the data. Is
Dear all,
In mucking around with ggplot2, I've hit the following snag,
library(ggplot2)
# this returns a grob, OK
GeomAbline$icon()
# lines[GRID.lines.9]
# this returns the function icon, OK
GeomAbline$icon
# proto method (instantiated with ): function (.)
# linesGrob(c(0, 1), c(0.2, 0.8))
#
Quoting David Winsemius dwinsem...@comcast.net:
In insurance situation there is typically a cap on the covered
losses and there is also typically an amount below which it would
not make sense to offer a policy. So a minimum and a maximum are
sensible assumptions about loss distributions
I am trying to use the adapt function in the package adapt. To make sure
I am using it correctly, I am trying a toy example that should yield a
result of 2/3.
Suppose the function is f(x,y) = x*y^2 and I want to integrate over f as
Int_0^1 Int_0^2 x*y^2 dxdy
Where the limits of integration for
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
function-internal operators. For example
var=1
foo-function(var){var=var+1}
foo(var)
var
[1] 1
but the goal is to get the var equal to 2 in this
On Thu, Oct 8, 2009 at 10:57 AM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
Hi Peng,
On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the current working directory is always '~' no
matter
... without answering my previous question, I have just found a
fortune-hate workaround,
getIcon - function(geom){
st - paste(Geom, firstUpper(geom),$icon, sep=)
eval(parse(text=st))
}
getIcon(abline)
I'm still curious about the get() behaviour though.
Best,
baptiste
See assign, you can use '-' assignment:
foo - function(var) var - var + 1
On Thu, Oct 8, 2009 at 2:14 PM, devol sund...@gmail.com wrote:
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
Is it possible to do what you mentioned somehow outside of the
function. I mean that there's some function and the only thing I can
do is to point on the variable to be modified inside the function
without any possibilities to modify the very function.
2009/10/8 Henrique Dallazuanna
Hi,
with assign,
foo - function(var){
assign(var, var+1, envir = .GlobalEnv)
}
var =1
foo(2)
var
# [1] 3
HTH,
baptiste
2009/10/8 devol sund...@gmail.com:
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view)
On Oct 8, 2009, at 1:13 PM, Albyn Jones wrote:
Quoting David Winsemius dwinsem...@comcast.net:
In insurance situation there is typically a cap on the covered
losses and there is also typically an amount below which it would
not make sense to offer a policy. So a minimum and a maximum are
On Oct 8, 2009, at 1:14 PM, Peng Yu wrote:
On Thu, Oct 8, 2009 at 10:57 AM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
Hi Peng,
On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the
On Oct 8, 2009, at 1:14 PM, devol wrote:
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
function-internal operators. For example
var=1
foo-function(var){var=var+1}
foo(var)
var
[1] 1
Use
I am uncertain what you mean. Possibly...
x - 1
f - function()x-2
f()
x
[1] 2
HOWEVER, this is very dangerous and most unwise as it depends on scoping
rules and where f is called to determine exactly which x is being assigned
the value of 2. So I think you would do well to reconsider what you
Bert, Jim, Dimitris and Joris,
Thank you all very much for your prompt help and suggestions.
After trying the ideas out, I have decided to go with Bert's approach
since it is by far the fastest of the lot.
Thanks again!
Rama Ramakrishnan
On Oct 8, 2009, at 12:49 PM, Bert Gunter wrote:
Thanks all of you for your help!!!
2009/10/8 Bert Gunter gunter.ber...@gene.com:
I am uncertain what you mean. Possibly...
x - 1
f - function()x-2
f()
x
[1] 2
HOWEVER, this is very dangerous and most unwise as it depends on scoping
rules and where f is called to determine exactly which
hi all,
It's not clear to me how (or if) I can pass multiple values for lty to a key in
xyplot?
I've tried: lines=list(lty=1:3), to no avail.
Do I need to use something other than auto.key?
(Deepayan, if you're out there, I have your book and must admit the answer
isn't jumping out at me.)
Hello all,
In order to get the desired order of panels within lattice I realized I needed
to define my factor levels.
I now see how to properly do it (attached code), but:
1. I don't understand why I have to use factor() on a variable that is already
a factor, and why levels() alone doesn't work
Hi Ben,
It looks like you're getting into trouble because you've
converted the body of a for-loop into a function.
You don't need to do that with foreach, and some of
your problems will be solved by reverting to something
more like your original for-loop, I suspect.
However, it also looks like
Bert Gunter wrote:
If I understand your intent, I believe you can get what you want much faster
(no interpreted loops and linear times) by looking at this slightly
differently.
First of all, the choice of columns is unimportant, as indexing can be used
to create a data frame containing only the
devol wrote:
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
function-internal operators. For example
var=1
foo-function(var){var=var+1}
foo(var)
var
[1] 1
but the goal is to get
Dear R-Help Team,
I have been trying to sort (all columns of) a matrix:
a-matrix(a-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
[,1] [,2]
[1,]11
[2,]33
[3,]44
[4,]66
[5,]66
[6,]44
[7,]66
[8,] 56 56
[9,]44
Harold,
Try this:
ff - function(x) x[1]*x[2]^2
adapt(2, lo = c(0,0), up = c(1,2), fun = ff)
The answer should be 4/3, since the first variable is integrated from 0 to 1
and the second variable (second-degree) is integrated from 0 to 2.
The answer is 1/3, if you do,
adapt(2, lo = c(0,0),
Untested, but
a.sorted - apply(a,sort,2)
?
Kajan Saied wrote:
Dear R-Help Team,
I have been trying to sort (all columns of) a matrix:
a-matrix(a-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
[,1] [,2]
[1,]11
[2,]33
[3,]44
[4,]66
[5,]6
You need to define sorted in the context of a matrix. By each row
individually, by each column individually, or by column using the whole matrix,
or by row using the whole matrix?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf
Try this:
a[order(a[,1], a[,2]),]
On Thu, Oct 8, 2009 at 1:50 PM, Kajan Saied kajan.sa...@gmail.com wrote:
Dear R-Help Team,
I have been trying to sort (all columns of) a matrix:
a-matrix(a-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
[,1] [,2]
[1,] 1 1
[2,] 3 3
I assume you want to sort the whole dataset (matrix) by ordering one column
in ascending order (and order all other columns appropriately).
a-matrix(a-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
#sort by first column
a[order(a[,1]),]
#sort by second column
a[order(a[,2]),]
#both give
Right. My guess is that Kajan wants:
a[do.call(order,data.frame(a)),]
## this generalizes to an arbitrary number of columns
## do.call() is a very powerful and useful R feature worth learning about
Yet another reason why the posting guide asks for a simple, proper,
reproducible example.
--
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
Sent: Thursday, October 08, 2009 11:41 AM
To: 'Erik Iverson'; 'Kajan Saied'; r-help@r-project.org
Subject: Re: [R] sort (all columns of) a matrix
Right. My guess
Dear R help lists,
Hello!
I ran WinBUGS under R using the function bugs, but I kept getting the following
error message.
Error in FUN(X[[3L]], ...) :
.C(..): 'type' must be real for this format
I can not find the reason.
I am freaking out.
Could any one have a look at the model codes
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of William Dunlap
Sent: Thursday, October 08, 2009 12:01 PM
To: Bert Gunter; Erik Iverson; Kajan Saied; r-help@r-project.org
Subject: Re: [R] sort (all columns of) a matrix
...
I have been using R the past couple of years to run models on data we
are collecting. I recently got a new computer and updated to a new
version of R (2.60 - 2.90). Since the update, I cannot get my syntax to
run. I have tried copying the file it is looking for into many different
directories to
Hello,
I have a non-linear function (exponential) that I am trying to display
the line with the data in a plot, is there a command similar to abline()
for the function I created, if not what is the best way to display the
fitted line. Example code below.
### Example
z -
Hi all,
I have a question about hist()
1)
t1 - hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And should the counts be '1 1 1 1 1' ?
Is there any other function to count frequency of discrete data?
Thanks.
-k
[[alternative HTML
For simply counting frequency of discrete data, table , xtabs, and ftable will
work.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Khanh Nguyen
Sent: Thursday, October 08, 2009 2:47 PM
To: r-help@r-project.org
Subject:
On 9/10/2009, at 6:52 AM, Folkes, Michael wrote:
Hello all,
In order to get the desired order of panels within lattice I
realized I needed to define my factor levels.
I now see how to properly do it (attached code), but:
1. I don't understand why I have to use factor() on a variable that
Change the breaks argument:
t1 - hist(1:5, 0:5)
t1$counts
On Thu, Oct 8, 2009 at 4:47 PM, Khanh Nguyen kngu...@cs.umb.edu wrote:
Hi all,
I have a question about hist()
1)
t1 - hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And should
On Oct 8, 2009, at 3:39 PM, Douglas M. Hultstrand wrote:
Hello,
I have a non-linear function (exponential) that I am trying to
display the line with the data in a plot, is there a command similar
to abline() for the function I created, if not what is the best way
to display the fitted
On Oct 8, 2009, at 3:47 PM, Khanh Nguyen wrote:
Hi all,
I have a question about hist()
1)
t1 - hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And should the counts be '1 1 1 1 1' ?
It depends on the way the bins are defined.
Had you forced
On 9/10/2009, at 8:47 AM, Khanh Nguyen wrote:
Hi all,
I have a question about hist()
1)
t1 - hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1?
RTFM --- include.lowest = TRUE and right = TRUE by default.
And should the counts be '1 1
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