On 15 February 2012 19:00, Jeff Newmiller wrote:
> What part of "read the posting guide" did you not understand? The "provide
> commented, minimal, self-contained, reproducible code" part?
Off topic: why is there so much unfriendliness on this thread ? Both
the above and DW's post seem unnecessa
Brilliant Sarah ! I love seeing such unexpected and creative applications.
I'm not a weaver but am a knitter (and a knotter actually) and have
mused about using R to help design elements of textured knitting
patterns e.g. as seen in single-colour, traditional fisherman's
jumpers from England and
Another search term is "geometric mean regression".
For simple models you can try the lmodel2 package.
Michael
On 2 February 2011 04:31, David Winsemius wrote:
>
> On Feb 1, 2011, at 10:41 AM, misil wrote:
>
>>
>> I wanna to do a Regression type 2 or Regression with X measured with
>> error
Hello,
I've been waiting to see if anyone else would answer this.
I've previously used random reallocation of objects to groups
(clusters) as a monte-carlo test of the informativeness of groups, as
described here:
http://lastresortsoftware.blogspot.com/2010/09/monte-carlo-testing-of-classificati
Hi Amelia,
You statement...
dbGetPreparedQuery(con, "INSERT INTO output(df) VALUES (?)", data.frame(output))
...is the problem.
To insert an entire data.frame into the database use dbWriteTable.
To insert with debSendPreparedQuery or dbGetPreparedQuery, the number
of "?" in the VALUES specifie
Hi Ed,
The sn package can generate random values from skewed Normal and t
distributions.
Also see here: http://azzalini.stat.unipd.it/SN/faq-r.html
Michael
On 11 January 2011 23:37, Ed Keith wrote:
> This is not exactly an R specific question, but I think the people on this
> list can probabl
Hello,
The answer to this one drops out of the answer to your previous question...
m <- matrix(1:16, nrow=4)
end <- c(2,3,1,3)
ii <- cbind(sequence(end), rep(1:length(end), end))
x <- m[ ii ]
Hope this helps,
Michael
2011/1/11 zhaoxing731 :
> Hello
>
> Suppose I have a matrix mat=(1:16,2)
>
Sagga K
>
> --- On *Tue, 11/1/11, Michael Bedward * wrote:
>
>
> From: Michael Bedward
> Subject: Re: [R] Generation of Normal Random Numbers
> To: "saggak"
> Cc: r-help@r-project.org
> Received: Tuesday, 11 January, 2011, 7:24 AM
>
> m <- c(1004
Hello,
Here is one way...
m <- matrix(1:16, nrow=4)
end <- c(2, 3, 1, 3)
ii <- cbind(sequence(end), rep(1:length(end), end))
sums <- tapply(m[ ii ], ii[ , 2], sum)
And here is another way...
sums <- mapply(function(col, lastrow) sum(m[1:lastrow, col]), 1:ncol(m), end)
Hope this helps,
Michael
m <- c(1004.1, 1028.3, 1044.3, 861.4)
s <- c(194.5899, 158.7052, 123.3000, 285.8695)
x <- mapply(function(mi, si) rnorm(25, mi, si), m, s)
Hope this helps,
Michael
On 11 January 2011 17:44, saggak wrote:
> Dear R helpers
>
> I have a data frame as given below
>
> df = data.frame(A = c(776,827,
Hi Sebastian,
You might also find the proto package useful as a way of restricting
the scope of variables. It provides a more intuitive (at least to me)
way of packaging variables and functions up into environments that can
be related in a hierarchy.
Michael
On 10 January 2011 23:48, Sebastien B
Just to add to the silly solutions, here's how I would have done it...
mu <- 40
sdev <- 10
days <- 100:120 # range to explore
p <- 0.8
days[ match(TRUE, qnorm(0.2, mu*days, sqrt(sdev * sdev * days)) >= 4000) ]
Michael
On 9 January 2011 08:48, Bert Gunter wrote:
> If I understand what you have
Hi Diego,
It depends on your what your research questions are. You haven't told us :)
For example, if you wanted to know whether (a) the environmental
distance between lakes is correlated with spatial distance and (b) if
the relationship changes over time you might do a series of Mantel
tests. T
Hello Diego,
This might not be relevant, but on reading your question the first
idea that struck me was that ordination trajectories of your lakes
over time might be more informative than clustering.
Michael
On 5 January 2011 01:31, Diego Pujoni wrote:
> Dear R-help,
>
> In my Master thesis I m
Hi Amy,
I'm not sure if I understand your question correctly so let me know if
the following is off track.
Starting with your example, here is how to create a data.frame and
write it to a new table in a new database file...
my.data = data.frame(X = c("US", "UK", "Canada", "Australia",
"Newzealan
Hello Walt,
Have a look at the bnlearn and deal packages.
Michael
On 24 December 2010 01:29, Data Analytics Corp.
wrote:
> Hi,
>
> Does anyone know of a package for or any implementation of a Bayesian Belief
> Network in R?
>
> Thanks,
>
> Walt
>
>
>
> Walter R. Paczkow
Hello Ufuk,
Here is one way to do it...
# make up some data for this example
x <- matrix(runif(7 * 20), ncol=7)
# a data.frame is most convenient for lm
x <- as.data.frame(x)
colnames(x) <- c("x1", "x2", "x3", "x4", "x5", "x6", "y")
# a list to hold lm results
x.lm <- list()
# run regressions
Hello Mark,
This is how I do it but it's longer than your code :)
unique.rows <- function (df1, df2) {
# Returns any rows of df1 that are not in df2
out <- NULL
for (i in 1:nrow(df1)) {
found <- FALSE
for (j in 1:nrow(df2)) {
if (all(df1[i,] == df2[j,])) {
found <- T
Hi Christiaan,
That looks like the join function in the plyr package.
Michael
On 16 December 2010 22:06, christiaan pauw wrote:
> Hi everybody
>
> Im on R version 2.11.1 on Mac OS X
>
> I am working through David Kahle's example of using ggplot2 with Rgooglemaps
> (found here:
> https://github
Hi Anirban,
You can do it like this...
lm.foobar <- lm(foo ~ bar)
some.data <- rnorm(200)
predict(lm.foobar, newdata=list(bar=some.data))
Hope that helps.
Michael
On 16 December 2010 17:05, Anirban Mukherjee wrote:
> Hi all,
>
> Suppose:
>
> y<-rnorm(100)
> x1<-rnorm(100)
> lm.yx<-lm(y~x1)
>
Hello Samaire,
I don't know much about eyesight measurements (other than that my own
would probably be 0.1, 0.2) but I'll attempt some suggestions...
For objective (1) you will have to define what you mean by
significantly different. If you had data on left/right values from a
larger population t
Hi Joe,
Just for info, I've done this in the past by applying lowess followed
by diff to a vector, then identifying points with change of sign in
the diffs.
Michael
On 14 December 2010 14:22, Joe Stuart wrote:
> Never mind. I did find this package, which seems to do the trick. Thanks
>
> http:/
Hello Erin,
Try this...
plot(x, y, type="b", pch=16)
Michael
On 13 December 2010 18:11, Erin Hodgess wrote:
> Dear R People:
>
> When I plot using type="b", I have circles and lines, which is as it should
> be.
>
> Is there a way to have filled in circles using the type argument,
> please? O
Hi Haoda,
I couldn't find a package that implements this, although I'm not
familiar with the field so there could be something but using
different terminology.
However, looking at the the Google preview of Nelson (2003) which is
cited by the page that you linked to, the calculations seem very
sim
Just to follow up on Robert's comment,
If you do an ls() you'll see that you've created objects V1, V2 in
your global environment.
A very similar question was discussed last week (I think... it's all a
blur) in the context of using "<-" instead of "=" with named function
arguments.
Michael
On
Hi Michael,
Sorry if I'm being slow, but I've read your post three times and still
can't quite work out what you're trying to do (the changing variables
names are a bit confusing).
I use RSQLite a lot and might be able to help if you could explain
your inputs and desired output in simple terms.
Hello Martin,
You were almost there :)
T1 <- subset(daten1, Geschlecht=="M" & GG=="A")
Hope this helps.
Michael
On 10 December 2010 22:25, Martin Spindler wrote:
> Dear all,
>
> I have a dataframe of the following strucutre
>
> numacc_b coverage_b Geschlecht GG
> 1 0 1
Just to add to Michael F's comments: I've looked for that elbow many a
time but never found it :) Admittedly, I typically deal with fairly
noisy, ecological data, but I think it's a mistake to try to identify
the "optimal" number of dimensions. Better instead to concentrate on a
"useful" number, i
Below is a toy function with one way of doing it. There are bound to
be better ways :)
function(niter = 10, time.out = 3) {
pretend.task <- function() {
Sys.sleep(0.5)
}
start <- proc.time()
for (iter in 1:niter) {
pretend.task()
cur <- proc.time() - start
if (c
The offset arg is your friend...
x <- 1:10
y <- 42 + 2*x + rnorm(length(x), 0, 0.5)
# we suspect the intercept might be 42 !
lm( y ~ 0 + x, offset=rep(42, length(x)))
Michael
On 4 December 2010 13:42, cborley87
wrote:
>
> Hi,
>
> Im fitting some simple linear models to data from the olympic r
Hello Bill,
Have a look at the example at the bottom of the help page for ?qqplot
Michael
On 4 December 2010 11:19, <5...@queensu.ca> wrote:
> Hi there,
>
>
> I am doing a test to see the the residual is distributed in the form of
> t-distribution and trying to plot the residuals and the t-di
Hello,
Please post a sample of your code so people here can understand what
you are trying to do.
Michael
On 4 December 2010 11:00, rushabhbm wrote:
>
> Guys,
> I am new to R so please excuse if I am not very clear.
>
> My problem is: I have a 'for' loop in which I am defining a Dataframe df
>
It's only obvious when someone points it out :)
fubar is not created because, in the test x > 3 returned FALSE, which
means the cat function doesn't get used, which means the y arg (fubar
<- 6) is never required and therefore not evaluated.
Evil isn't it ?
Michael
On 3 December 2010 20:18, Ivan
Sounds just like the subset function (?)
x <- as.data.frame(matrix(sample(5, 100, rep=TRUE), ncol=10))
subset(x, V1 > 3 & V2 < 5)
Michael
On 3 December 2010 19:05, Santosh Srinivas wrote:
> Hello Group,
>
> Is there an easy way to query a data.frame or data.table (this is
> fast!) for multiple
Hi Aaron,
Following up on Ivan's suggestion, if you want the column order to
mirror the row order...
mo <- order(rowSums(MAT), decreasing=TRUE)
MAT2 <- MAT[mo, mo]
Also, you don't need all those extra c() calls when creating
inputData, just the outermost one.
Regarding your second question, you
And just to add to Ivan's comment, if you are using the rowSums or
colSums functions with a matrix or data.frame they also have the na.rm
argument.
Michael
On 1 December 2010 20:16, Ivan Calandra wrote:
> Hi,
>
> (a) sum() and mean() have a na.rm argument that should be set to TRUE.
>
> (b) let'
Here is one way...
f <- function(a, b, op="==") {
call <- call(op, a, b)
result <- eval(call)
# possibly do other stuff
result
}
> f(1, 2)
[1] FALSE
> f(1, 2, "<")
[1] TRUE
Michael
On 1 December 2010 13:54, randomcz wrote:
>
> Hi guys,
>
> How to pass an operator to a function. For examp
Hi Bill,
Have a look at the influence.measures function...
my.lm <- lm( ... )
influence.measures( my.lm )
Hope this helps,
Michael
On 30 November 2010 00:13, Schwab,Wilhelm K wrote:
> Hello all,
>
> I fit a linear model to some data and used plot() to create diagnostic plots
> for the fit;
Hi Marianne,
How about this...
ac.ad <- unstack(dat1, choice ~ cond1:cond2)[, c("A.C", "A.D")]
acad.xtab <- with(ac.ad, table(A.C, A.D))
Michael
On 29 November 2010 20:18, Marianne Promberger
wrote:
> Dear list,
>
> I have data like this:
>
> dat1 <- data.frame(subject=rep(1:10,2),
>
On 29 November 2010 15:09, Joshua Wiley wrote:
> (I hope I'm like wine and get better with age or)
>
Sigh, me too - but I suspect I'm heading more towards vinegar
Michael
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
Hello again Wendy,
Actually, the simex package is probably a more useful suggestion...
http://www.stat.uni-muenchen.de/~helmut/Texte/Simex_Rnews.pdf
Michael
On 29 November 2010 13:55, Michael Bedward wrote:
>>> In case you haven't see it, the glm function accepts an option
>> In case you haven't see it, the glm function accepts an optional
>> weights argument.
>>
>
> Thanks for the reply. But the philosopy behind weighting is the assumption
> of unequal variance in the y values. In normal regression one assumes that
> the x values are known without error
>
> Wendy
S
Hi Wendy,
In case you haven't see it, the glm function accepts an optional
weights argument.
Michael
On 29 November 2010 09:42, Wendy Anderson wrote:
> I have a glm regression (quasi-poisson) of log(mu) on x but I have varying
> degrees of confidence in the x values, and can attach a numerical
Hello Lorezo,
Try this...
order(sapply(mylist, min))[1]
Michael
On 26 November 2010 11:23, Lorenzo Cattarino wrote:
> Hi R-users,
>
>
>
> I have a list
>
>
>
> mylist <- list(c(0.79, 0.92, 0.91, 0.86, 0.96, 0.96, 0.95, 0.94, 0.99),
> c(0.28, 0.45, 0.59, 0.69, 0.80, 0.87, 0.95, 0.94, 0.98), c(
Is this what you want ?
printCount <- function(N) {
for (i in 1:N) {
cat(i, "\n")
}
}
Depending on your platform, output from the print function may not
appear until after your function has finished. You can try using
flush.console() to give it a nudge...
printCount <- function(N) {
fo
Hello Kayce,
My (very basic) understanding is that you can't directly compare the
coefficients across models that have different response variables, nor
could you use AIC and similar metrics of model goodness of fit.
Instead, I think you have to carefully define what you mean by "reveal
similar po
Hi Hana,
Use the paste function to create your file names.
for ( i in 1:dim(M)[1] ) save( M[i,,], file=paste("M_", i, ".img", sep="") )
Alternatively, use the sprintf function to get names with leading
zeroes for easier sorting of files:
for (i in 1:dim(M)[1] ) save( M[i,,], file=sprintf("M_%03
/1:
http://cran.r-project.org/doc/Rnews/Rnews_2008-1.pdf
Hope this helps.
Michael
On 22 November 2010 15:03, Ahmed Attia wrote:
> Hi Dr Michael,
>
> Attached is an example for the linear plus plateua model but in SAS, this
> exactly what I need to do in R.
>
> Ahmed
>
>
Sorry, "chage-point" should be "change-point"
On 22 November 2010 14:44, Michael Bedward wrote:
> Hi Ahmed,
>
> Does 'quadratic plateau' model refer to a chage-point or bent-cable
> regression with quadratic on one side and an asymptote on the othe
Hi Ahmed,
Does 'quadratic plateau' model refer to a chage-point or bent-cable
regression with quadratic on one side and an asymptote on the other ?
If so, you might like to look at the bentcableAR package. There is a
background article here:
http://faculty.washington.edu/gchiu/Articles/bentcable-
Ah, this looks like Australian data :)
One simple way would be to use the lowess function and fiddle with the
f parameter (like bandwidth).
Michael
On 22 November 2010 14:18, Roslina Zakaria wrote:
> Hi,
>
> I would like to overlap the cdf curve for observed and generated data Here is
> my cod
On 20 November 2010 20:57, Stephen Liu wrote:
> Hi Michael,
>
> Thanks for your advice.
>
> data() only displays a list of files. Is there an easy to show the brief
> summary of files rather than calling each file.
What sort of summary do you want ? The data() command should display
names and a
Type data() to list the numerous example data sets included with the
standard R distribution.
Michael
On 20 November 2010 20:42, Stephen Liu wrote:
> Hi folks,
>
> Please advise where can I down free data files for learning R? Google search
> brought me many, not easy for to screen. TIA
>
> B.
e execution at the point it was and let your write commands in console
> to check what was going on.
>
> Is that possible?
> Regards
> Alex
>
> --- On Thu, 11/18/10, Michael Bedward wrote:
>
> From: Michael Bedward
> Subject: Re: [R] How to catch warnings
> To: "
made things easier. One
> more question what If I want to halt or pause the program when a warning
> happens? Right now I get only a message printed but it would be nicer if the
> execution is paused so to try to print more values.
>
> Best REgards
>
> Alex
>
> ---
Hi Alex,
Something like this ?
x <- 1:4
y <- list(good=2:5, bad=3:5)
for (yy in y) {
tryCatch( x <- cbind(x, yy),
warning=function(w) cat("problem values: ", yy, "\n")
)
}
Michael
On 18 November 2010 03:19, Alaios wrote:
> Hello when my code executes I receive the message tha
Hi Nick,
I've used MCMC to fit change point regressions to a variety of
ecological data and prefer this approach to strucchange and similar
because I feel I have more control over the model, ie. I find it
easier to tailor the form of the model to biological / demographic
processes. I also find the
t; If not, do not spend to much time, since I can solve the problem by
> implementing list with all partitions of set {1,2} and {1}. (only two
> partitions for {1,2} ...)
>
> Best wishes
> Diana
>
> Am 12.11.2010 12:06, schrieb Michael Bedward:
>>
>> You're wel
Hi Kere,
Step 1 is choose an appropriate distribution :) Do you have one in
mind ? Or are you interested in examining the effects of survival
times having the same mean but generated with alternative
distributions ?
One ready-rolled alternative is the SimSurv method in package prodlim.
To find
On 16 November 2010 16:10, wangwallace wrote:
>
> Michael, I really appreciate your help.
>
> but I got the following error message when I wan trying to run the function
> written by you:
>
> Error in out[i, ] <- apply(help[, c(grp1 + 1, grp2 + 5)], 2, sample, 1) :
> number of items to replace is
Hello,
Is this what you want ?
sampleX <- function(X, nGrp1, nsamples)
# X is matrix or data.frame with cols for two groups of variables
# with grp1 in cols 2:5 and grp2 in cols 6:9
#
# nGrp1 <- number of variables to sample from group 1
#
# nsamples <- number of rows in output matrix
if (nGrp
("f", x)
eval(fcall)
}
On 13 November 2010 20:24, Michael Bedward wrote:
> Hello Marius,
>
> NULL is not the same as missing. You could something like this in
> various ways. Here are a couple...
>
> g <- function(x) {
> if (missing(x)) {
> f()
> }
Hello Marius,
NULL is not the same as missing. You could something like this in
various ways. Here are a couple...
g <- function(x) {
if (missing(x)) {
f()
} else {
f(x)
}
}
or change f to detect null args
g <- function(x) {
if (missing(x)) {
x <- NULL
}
f(x)
}
f <- fu
Fancy that... vector spam :)
Michael
On 12 November 2010 20:30, Jeff Musgrave wrote:
> Now you and your Vector Team can make more money.
> Offer your current client base a chance to buy and sell
> a product in high demand. An item that increases in value every day.
> Are you ready for this no-i
ere the first element contains the one-elemnt partion, the
> second to fourth the three 2-element-partions and the fifth the 3-element
> partition?
> Do you know or is it my job to implement this by myself (o.k. but
> time-consuming ..)
>
> Best wishes
> Diana
>
> Am 12.11.201
Hi Ivan,
> I had already seen your solution (that does work).
> If you're right about the issue in "my" function, then the error message is
> confusing ('could not find the function "get<-" '). Moreover, I assign()ed
> and get() "x" from the .GlobalEnv, so there shouldn't be a problem with
> scopi
Hi Diana,
Have a look at the setparts function in the partitions package.
Michael
On 12 November 2010 20:03, Diana wrote:
>
> Hi
>
> I am new on this forum. I am searching for a function in R which provides
> all partitions of a set, say for the set
> {1,2,3}
> you get
> {{1,2,3}}
> {1,{2,3}}
>
}
>
> assign2("y", 1:4)
> Error in attr(get(x), "creation.time") <- Sys.time() :
> could not find function "get<-"
>
> Why doesn't it work?
> If I remove the attr() part,
> identical(y, get("y")) returns TRUE, so why attr() canno
On 12 November 2010 02:21, David Winsemius wrote:
>
>> The fastest and easiest solution is
>>
>> t(A) %*% A
>
> That is really elegant. (Wish I could remember my linear algebra lessons as
> well from forty years ago.) I checked it against the specified output and
> found that with one exception t
richt-
>> Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
>> Auftrag von Michael Bedward
>> Gesendet: Donnerstag, 11. November 2010 10:56
>> An: friedericksen.h...@gmail.com
>> Cc: r-h...@stat.math.ethz.ch
>> Betreff: Re: [R] How
As a hack you could do this...
assign("=", assign2)
Michael
On 11 November 2010 21:30, Barry Rowlingson
wrote:
> On Thu, Nov 11, 2010 at 9:37 AM, Tal Galili wrote:
>
>> 4) My real intention is to somehow change the "<-" operator (not simply the
>> assign). I am unsure as to how to do that.
>
Hi Tal,
Here's a way of doing the first bit...
assign2 <- function(x, ...) {
xname <- deparse(substitute(x))
assign(xname, ...)
x <- get(xname)
attr(x, "creation.time") <- Sys.time()
assign(xname, x, pos=.GlobalEnv)
}
Michael
On 11 November 2010 20:37, Tal Galili wrote:
> My objecti
uentially as I iterate through
> the data, how would you recommend it??
>
> Thanks,
>
> -N
>
> On 11/11/10 12:03 AM, Michael Bedward wrote:
>>
>> All values in a matrix are the same type, so if you've set up a matrix
>> with a character column then your nume
Hi Friedericksen,
This function will do it. No doubt there are more elegant ways :)
rmatch <- function(x, name) {
pos <- match(name, names(x))
if (!is.na(pos))
return(x[[pos]])
for (el in x) {
if (class(el) == "list") {
out <- getEl(el, name)
if (!is.null(out)) return(o
factors.
>
> -N
>
> On 11/10/10 11:16 PM, Michael Bedward wrote:
>>
>> Hello Noah,
>>
>> If you set these names...
>>>
>>> names(results)<- c("one", "two", "three")
>>
>> this won't work...
>>&g
Hello Noah,
If you set these names...
> names(results) <- c("one", "two", "three")
this won't work...
> results[results$c < 100,]
because you don't have a column called "c" (unless that's just a typo
in your post).
> I tried making it a data.frame with
> foo <- data.frame(results)
>
> But that
Hello Judit,
The code below is a toy simulation function that takes as arguments a
matrix representing the initial population, a dispersal kernel, global
survival probability and max number of iterations to run. It doesn't
implement exactly what you described in your post but if you study the
cod
If you want to assign to a variable in your workspace, rather than a
local variable in your function, you can use the <<- operator (double
headed arrow) like this...
mat <- function(i) {
for (k in i:10) {
y[k] <<- k+1
f[k] <<- y[k-1] / 2
}
}
Type ?"<<-" for the help page.
Michael
O
Hi James,
The following is probably more an expression of empathy than direct help :)
I've used rjags quite a bit but have often found it to be very picky
about the formulation of the model. Scripts that run without problem
under WinBUGS (for instance) provoke errors with jags and more often
than
Hi Sachin,
That's OK - you don't need to know the dimensions up front and you can
add new vectors, or elements to an existing vector, as required.
# empty list to start with
X <- list()
# we get a vector
v1 <- c(1, 2, 3, 4, 5)
# add it to the ragged array
X <- c(X, list(v1))
# get another coup
You want the paste command (cat is for printing to the console)...
A$period <- paste(A$period, 0, sep="")
Michael
On 10 November 2010 16:09, wrote:
>
> Hi All,
>
> Suppose I want to concatenate a zero to all the values to a column called
> period in data frame A. I want to do the following bu
Hello Sachin,
You have a "ragged array" and you can easily store this as a list of vectors...
x <- list(c(0,0,1,1), c(1,3,5), 4, c(7, -1, 8, 9, 10, 6))
The only gotcha with this is that you will then need to use double
brackets for the first index when retrieving values (single brackets
will ret
Don't know if this is "efficient" but I think it works...
yn <- c("Y", "N")
X <- expand.grid(x1=yn, x2=yn, x3=yn, x4=yn)
Yp <- c(0.6, 0.5, 0.8, 0.9)
X$prob <- apply(X, 1, function(x) cumprod(ifelse(x == "Y", Yp,
1-Yp))[length(x)])
Michael
On 9 November 2010 17:05, Kate Hsu wrote:
> Dear r user
Perhaps just use the ftable function to generate a flat contingency
table and look for counts below some threshold.
Michael
On 9 November 2010 09:25, Alan Chalk wrote:
> Regarding unusual combinations of factors in categorical data.
> Are there any R packages that can be used to identify the ou
Hello Jumlong,
For Normal distribution see the help page for pnorm.
For dealing with unknown (empirical) distributions, look at ecdf.
Hope this helps
Michael
On 8 November 2010 16:29, Jumlong Vongprasert wrote:
> Dear all
> I have problem with calculate probability, I have data x1,...
Hello,
One approach would be to fit your distribution using MCMC with, for
example, the rjags package. Then you can use the "zeroes trick" or
"ones trick" to implement your new distribution as described here...
http://mathstat.helsinki.fi/openbugs/data/Docu/Tricks.html
You will find a summary of
Hi Andreas,
Try this...
# forget to assign result set
dbSendQuery(con, "select * from df")
# retrieve the result set just created
rs <- dbListResults(con)[[1]]
Then you can do dbClearResult or whatever.
Michael
On 4 November 2010 19:56, Andreas Borg wrote:
> Hello R-help members,
>
> I have
rained in (0,1).
>
> OK so some addition info. I know each of the X2 is in (0,1). Is there any
> method available?
> Jim
>
> On Sat, Oct 23, 2010 at 8:31 AM, Michael Bedward
> wrote:
>>
>> Hi Jim,
>>
>> You don't mention whether you have any prior
Hi Jim,
You don't mention whether you have any prior information regarding X2
that can be used to constrain values imputed for it. I think you will
need some because without it values sampled for b and X2 respectively
will just "see-saw" against each other.
Michael
On 22 October 2010 18:37, Jim
Hi Muhammad,
Have a look at the biclust package...
http://cran.r-project.org/web/packages/biclust/index.html
Michael
On 21 October 2010 18:00, Muhammad Yaseen wrote:
> *Hi Folks,*
> *
> *
> *I want to do two-way joining or clustering as described in STATISTICA
> website *http://www.statsoft.com
Hello Steve,
> I've been asked to help evaluate a vegetation data set, specifically to
> examine it for community similarity. The initial problem I see is that the
> data is ordinal. At best this only captures a relative ranking of
> abundance and ordinal ranks are assigned after data collection
nally posted as your
objective (?)
Michael
On 15 October 2010 22:49, Michael Bedward wrote:
> Hi John,
>
> The word "species" attracted my attention :)
>
> Like Dennis, I'm not sure I understand your idea properly. In
> particular, I don't see what you ne
ilies it will be the same as doing the
> simulation experiment outline in the method above?
>
> Thanks
>
> John
>
>
>
>
> On 15 Oct 2010, at 12:49, Michael Bedward wrote:
>
> Hi John,
>
> The word "species" attracted my attention :)
>
> Like Den
Hi John,
The word "species" attracted my attention :)
Like Dennis, I'm not sure I understand your idea properly. In
particular, I don't see what you need the simulation for.
If family F has Fn species, your random expectation is that p * Fn of
them will be at risk (p = 0.0748). The variance on t
stalled I look forward to trying it out shortly.
Thanks again.
Michael
On 15 October 2010 03:17, Rainer M Krug wrote:
>
>
> On Thu, Oct 14, 2010 at 3:15 AM, Michael Bedward
> wrote:
>>
>> Hi Juan,
>>
>> Yes, you can use EMD to quantify the difference bet
Some quick ideas...
One very easy way would be to round them all to integer degrees and
remove the duplicates - or even just let the duplicates overwrite each
other in the plot.
A step up from that would be to create a matrix at some resolution
(e.g. 180 x 360 for a 1 degree global grid) and coun
Hello Julia,
I'm afraid your code had multiple problems: variables declared but not
used, incorrect or unnecessary use of the "c" function, out-of-bounds
subscripts and overwriting of result objects.
Rather than point them all out in detail I've modified your code so
that it works (see below). Pl
; Juan
>
>
> On Wed, Oct 13, 2010 at 4:39 AM, Michael Bedward
> wrote:
>>
>> Just to add to Greg's comments: I've previously used 'Earth Movers
>> Distance' to compare histograms. Note, this is a distance metric
>> rather than a parametric
is available:
>
> http://r.789695.n4.nabble.com/Measure-Difference-Between-Two-Distributions-td2712281.html#a2713505
>
> HTH,
> Dennis
>
> On Tue, Oct 12, 2010 at 7:39 PM, Michael Bedward
> wrote:
>>
>> Just to add to Greg's comments: I've previous
Super ! An option for vertical plotting would be very nice.
Michael
On 13 October 2010 22:19, Jim Lemon wrote:
> On 10/13/2010 07:11 PM, elpape wrote:
>>
>> Dear everyone,
>>
>> I would like to create a kite chart in which I plot densities (width of
>> the
>> vertical kites) in relation to sedi
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