Deepayan Sarkar writes:
> You can specify a fixed-width fontfamily if that helps:
>
> levelplot(matrix(seq(4,120,l=9),3,3),
> colorkey = list(at = seq(0, 120, 20),
> labels = list(labels = c(' 0',' 20',' 40','
> 60',' 80','100','120'),
>
On 26/06/12 00:39, Kathie wrote:
Dear R users,
I'd like to compute X like below.
X_{i,t} = 0.1*t + 2*X_{i,t-1} + W_{i,t}
where W_{i,t} are from Uniform(0,2) and X_{i,0} = 1+5*W_{i,0}
Of course, I can do this with "for" statement, but I don't think it's good
idea because "i" and "t" are too
I have an epidemiologic dataset and want to do a weighted logistic
regression. I have a continuous exposure variable and I want to see
if it is associated with a binary outcome. The accuracy of the
exposure value is highly variable and I want the logistic regression
to assign greater weight/impor
Hello,
The following worked with the supplied dataset, named 'Sunday'.
dat <- Sunday
dat$SunDate <- as.POSIXct(Sunday$SunDate, format="%m/%d/%Y %H:%M")
dat <- na.exclude(dat)
h <- hours(dat$SunDate)
# 1. Two different displays
aggregate(dat$SunScore, by=list(h), mean)
aggregate(SunScore ~ h, da
Hi,
Try this:
example1<-data.frame(V1=c("rs4685","rs4685","rs788018","rs788023"),V2=c("2:198257795","2:198257795","2:198265526","2:198283305"))
example2<-data.frame(V1=c("rs4685","rs4675","rs788018","rs788023"),V2=c("2:198257795","2:198258795","2:198265526","2:198283305"))
subset(example2,!(V1 %in
Hey everyone,
I've been reading an old scientific paper (well, not that old, about 15
years) and I want to verify the authors' statistical results. The paper is
fairly unclear about what exactly they did, and none of the relatively
simple commands I'm familiar with are producing results similar to
Hello everyone,
I would like to have an Excel SpreadSheet with a Macro that allows me to
select between the following Models (in the form of a combobox): AR, MA,
ARMA and ARIMA. And depending on the model chosen, give Excel the parameters
(for p, q, d and P, D, Q).
I want to find a way to connect
rrdf is incredibly helpful, but I've notice that the rrdf package for mac
hasn't been working for some time: http://goo.gl/5Ukpn . wondering if there
is still a plan to maintain that in the long run, or if there is some other
alternative to read RDF files.
[[alternative HTML version delete
Hi Elliot
This works on Win 7 ver 2.15
useOuterStrips(combineLimits(
xyplot(x + y ~ d | g, groups = h, data = dat, type = 'l',
scales = list(y = list(relation = "free"),
x = list( at =seq(from =
as.Date("2011-01-01"), to = as.Date("2011-10-01"), by = "3 month"),
On Jun 25, 2012, at 12:30 , Jim Holtman wrote:
> you have to load the 'foriegn' package first.
>
'foreign' works better
> Sent from my iPad
...with spellcheck disabled, it seems ;-)
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Freder
xx <- structure(list(X1000s = c(47L, 37L, 17L), X600s = c(63L, 64L,
62L), X500s = c(75L, 45L, 25L), X250s = c(116L, 11L, 66L), X100s = c(125L,
25L, 12L), X50s = c(129L, 19L, 29L), X10s = c(131L, 61L, 91L),
X5s = c(131L, 131L, 171L), X3s = c(131L, 186L, 186L), X1s = c(131L,
186L, 186L))
Hi,
I'm sorry for not explaining well. I'm trying to do my best, but it's
alittle complicated for me. I will send what I've tried but I think I should
also provide some files to make possible for you to run the codes I'm
running, but i couldn't find a way to do it.
I have 2 rasters to represent en
Oh, whoops. When I responded the first time, I had done dput(head(Sunday,
100)) too, but it didn't look useful. It was only just now that I saw that
it basically prints out the vectors. Sorry about that.
Anyways, here's dput(head(Sunday, 100)):
> dput(head(Sunday, 100))
structure(list(SunDate
thanks
it works brilliantly
wrote:
> Hello,
>
> It is recomended to post data using dput(). Something like
>
> dput(head(DF, 20)) # or 30
>
> Then, paste the output of this command in a post.
>
> Untested, but I think it should work:
>
>
> # Create a logical index into 'example2'
> ix <
I want to fit a bi-Weibull distribution, as defined by Evans et al. in their
book Statistical distributions:
http://www.scribd.com/doc/57260630/164/FIVE-PARAMETER-BI-WEIBULL-DISTRIBUTION
I would like to know if there is any package in R that already does it, or any
quick procedure to accomplish
There's no way we can tell you the "best way" to display your
information, because we don't know anything about it. The best display
method has a lot to do with what the data are, and what you're trying
to illustrate. That said, here are two possibilities, one using the
bar graph you requested, and
Good Afternoon, I'm trying to create a graph that displays the best way the
following information.
For instance organized by bar graph, A, B, C
Source X1000s X600s X500s X250s X100s X50s X10s X5s X3s X1s
1 A 476375 116 125 129 131 131 131 131
2 B 3764
When I try your code I'm told that 'mongo' is not a defined class. Removing
that field from the definition, I do not get a print on definition.
> mongoDbUser = setRefClass("mongoDbUser",
+ fields = list(
+ auth = "logical",
+ host = "character",
+ username = "character",
+ pa
Hello,
It is recomended to post data using dput(). Something like
dput(head(DF, 20)) # or 30
Then, paste the output of this command in a post.
Untested, but I think it should work:
# Create a logical index into 'example2'
ix <- example2$V1 %in% setdiff(example2$V1,example1$V1)
example2[ix,
hi,
I have 2 files example 1 and example 2 and would like to know what is in
example2 and not in example1 (attached)
V1 contain data which could be in duplicated which I am using as identifiers
I used setdiff(example2$V1,example1$V1) to find the identifiers which
are specific to example2:
[
I will try again.
1. If you have 5 groups, and split each in half, then you have 10 groups.
If MH is applicable to 5 groups, then it should be applicable to the 10
groups as
long as they are disjoint groups.
2. I don't think MH applies to your example because the groups do not have
similar behavi
I'm having some trouble using the latticeExtra 'combineLimits' function
with a Date x-variable:
require(lattice)
set.seed(12345)
dates <- seq(as.Date("2011-01-01"), as.Date("2011-12-31"), "days")
dat <- data.frame(d = rep(dates, 4),
g = factor(rep(rep(c(1,2), each = length(date
Oh I was wrong. R package 'GridR' seems to be what one needs to run R scripts
on Condor directly ... interesting. thanks again.
-
http://censix.com
--
View this message in context:
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Sent from
How do I specify a "tight" y-axis, where the plot completely fills the
y-axis range, inside the prepanel function? For example, consider the
following code:
require(lattice)
set.seed(12345)
x <- 1:1000
y <- cumsum(rnorm(length(x)))
prepanel.test <- function(x, y, groups = NULL, subscripts = NUL
Thanks Drew
a first glance at Condor tells me that this will very likely do what I need
in terms of flexible allocation of (cpu) resources. Unfortunately there is
not much R integration so I will need to see if setting up Condor and
submitting R jobs to it is worth the effort.
Cheers
Soren
Hi all!
Is there any way to compare mixdist models with a different number of
components using AIC or BIC?
I'm looking for some similar functionality as in flexmix, but not sure
which to stick with: mixdist or flexmix..
Yakir Gagnon
cell+1 919 886 3877
office +1 919 684 7188
Johnsen Lab
Biolog
Martin,
I had a nagging feeling that there must be more to this than my previous
reply suggested, and have been digging a bit further. Basically I would
expect these p-values to not be great if you had nested random effects
(such as main effects + their interaction), but your case looked rathe
and finally...
thingy <- function(x) {
x <- C(x, poly, 1)
tmp <- contrasts(x)
contrasts(x, 1) <- 2 * tmp / sum(abs(tmp))
x
}
dat2 <- with(data.catapult,
data.frame(
Distance,
h=thingy(h),
s=thingy(s),
l=thingy(l),
e=thingy(e),
b=
... but this is tantalisingly close:
dat1 <- with(data.catapult,
data.frame(
Distance,
h=C(h, poly, 1),
s=C(s, poly, 1),
l=C(l, poly, 1),
e=C(e, poly, 1),
b=C(b, poly, 1)
)
)
lm4 <- lm(Distance ~ .^2, data = dat1)
summary(lm4)
... wish I kne
While lm() is a linear modeling, the constraints make it easier to solve with a
nonlinear
tool. Both my packages Rvmmin and Rcgmin (I recommend the R-forge versions as
more
up-to-date) have bounds constraints and "masks" i.e., fixed parameters.
I am actually looking for example problems of this
They are coding the variables as factors and using orthogonal
polynomial contrasts. This:
data.catapult <- data.frame(data.catapult$Distance,
do.call(data.frame, lapply(data.catapult[-1], factor, ordered=T)))
contrasts(data.catapult$h) <-
contrasts(data.catapult$s) <-
contrasts(data.catapult$l) <-
Staleno:
As always, you need to read the Help file carefully. From ?aov:
"The main difference from lm is in the way print, summary and so on handle
the fit: this is expressed in the traditional language of the analysis of
variance rather than that of linear models. "
summary.aov() computes seque
thanks peter. I was thinking more about t but you're right in that there's
an i there also. my
bad ( twice ).
On Mon, Jun 25, 2012 at 9:37 AM, Petr Savicky wrote:
> On Mon, Jun 25, 2012 at 05:39:45AM -0700, Kathie wrote:
> > Dear R users,
> >
> > I'd like to compute X like below.
> >
> > X_{i
On Mon, Jun 25, 2012 at 05:39:45AM -0700, Kathie wrote:
> Dear R users,
>
> I'd like to compute X like below.
>
> X_{i,t} = 0.1*t + 2*X_{i,t-1} + W_{i,t}
>
> where W_{i,t} are from Uniform(0,2) and X_{i,0} = 1+5*W_{i,0}
>
> Of course, I can do this with "for" statement, but I don't think it's
On Fri, Jun 22, 2012 at 3:45 PM, APOCooter wrote:
>
> [snip and rearrange]
>
>> try to put your data in a proper time series class (zoo/xts if I
>>might give a personal-ish plug) which will make all these calculations
>>much easier.
>
> I thought that's what I was doing with as.POSIXlt?
as.POSIXl
Arrgh yes I did mean dput(head(mydata, 100)). Thanks for catching it.
John Kane
Kingston ON Canada
> -Original Message-
> From: michael.weyla...@gmail.com
> Sent: Fri, 22 Jun 2012 14:25:30 -0500
> To: jrkrid...@inbox.com
> Subject: Re: [R] Questions about doing analysis based on time
>
On 06/20/2012 05:36 PM, Mikko Korpela wrote:
> Hello list!
>
> Let's construct a matrix / data.frame with 0 columns, but > 0 rows, and
> non-NULL rownames. Then, call is.na() on both the data.frame and the
> matrix. We find that is.na.data.frame() gives an error. When row.names
> are removed, is.n
Hi Kathryn: I'm sorry because I didn't read your question carefully enough.
arima.sim won't help because you don't have a normal error term. I think
you have to loop unless someone else knows of a way that I'm not aware of.
good luck.
On Mon, Jun 25, 2012 at 8:39 AM, Kathie wrote:
> Dear R us
Hi: Use arima.sim because when you have recursive relationships like that,
there's no way to
not loop. If arima.sim doesn't allow for the trend piece (0.1*t ), then you
can do that part seperately using cumsum(0.1*(1:t)) and then add it back in
to what arima.sim gives you. I don't remember if arim
Hello.
Thank you for the help. However, I'm not sure your reply answers my question.
Let me rephrase:
I'm trying to reproduce the values in the second table in the example on
http://www.itl.nist.gov/div898/handbook/pri/section4/pri472.htm. The table
shows the summary of the linear model, which
Help anyone? Is this impossible? I tried to use the initFields method
but it did not change anything.
Le 25/06/2012 10:55, Julien Cochennec a écrit :
Hi,
New to this list, and glad I found R5 which si a great improvement to
R for my work.
Something seems odd anyway, when I do a setRefClass, th
Hi,
I am in the process of migrating my machine learning scripts from bash +
libsvm to R and for the past weeks I have been battling a substantial
problem when using the tune.svm function from the e1071 package. The
tune.svm function that performs the grid-search for the best cost and gamma
parame
Dear R users,
I'd like to compute X like below.
X_{i,t} = 0.1*t + 2*X_{i,t-1} + W_{i,t}
where W_{i,t} are from Uniform(0,2) and X_{i,0} = 1+5*W_{i,0}
Of course, I can do this with "for" statement, but I don't think it's good
idea because "i" and "t" are too big.
So, my question is that
Is
Hi Soren
Have you looked into using the condor scheduler
(http://research.cs.wisc.edu/condor/)? I'm not aware of it linking up
with multicore or other parallel processing code inside R, but I've
used it to run multiple R processes on a variable number of processors
where N can both increase and de
Hi,
I'm doing a Market Basket Analysis in which I have a list of transaction
id's in column 2 and transactions(product names) in column 1. I read this
into a transaction file using a
txn<-read.transaction(file="data.csv",format='single', rm.duplicates=F,
cols=c(1,2))
If I want to use the apriori
Hello, again.
Sorry, I've used 'sum' when it should be 'mean'.
Rui Barradas
Em 25-06-2012 12:49, Rui Barradas escreveu:
Hello,
It's not that complicated.
f <- function(index, dat, offset=5){
x <- data.frame(Emp=dat[, index], Hours=dat[, index + offset])
aggregate(Hours~Emp, data=
Hello,
It's not that complicated.
f <- function(index, dat, offset=5){
x <- data.frame(Emp=dat[, index], Hours=dat[, index + offset])
aggregate(Hours~Emp, data=x, sum)
}
sp <- read.table("supermarkt.txt")
lapply(1:5, f, sp)
Also, please provide context. (Quote the post you
>
> I do now know how to navigate through the table.
> But my question is, what kind of graphical and numerical summaries can I
> make with keeping in mind that I am interested in the average working
hour
> per employee?
Rather vague question without data or code, so rather vague answer.
For nu
The posting guide asked for 'at a minimum' information, which you have
not provided (nor do we know what graphics device this is).
But on many OSes and devices simply use the Unicode character "\u00b1".
Or use plotmath (see ?plotmath).
On 25/06/2012 12:03, gianni lavaredo wrote:
dear resera
I do now know how to navigate through the table.
But my question is, what kind of graphical and numerical summaries can I
make with keeping in mind that I am interested in the average working hour
per employee?
--
View this message in context:
http://r.789695.n4.nabble.com/Apply-on-columns-tp4633
dear reserachers,
I am looking for a expression about a special symbol (±) in a Legend and
add on front of "Mean ± SD" the symbol of bracket
sorry if the example is not great
disptest <- matrix(rnorm(10),nrow=10)
disptest.means<- rowMeans(disptest)
plot(1:10,disptest.means)
dispersion(1:10,disp
you have to load the 'foriegn' package first.
Sent from my iPad
On Jun 25, 2012, at 5:12, niloo javan wrote:
> Hi dear all,
> I am trying to import a dataset from SPSS into R (R x64 2.15.0).
> I used both
> read.spss("D:/Untitled2.sav", use.value.labels = TRUE, to.data.frame = TRUE,
> max.va
Hi All
In the past I have worked with parallel processing in R where a function F
is applied to the elements of a list L. The more cpu cores one has, the
faster the process will run. At the time of launching the process for (F,L)
I will have a certain fixed number of cpu's that I can use. I have t
Hi dear all,
I am trying to import a dataset from SPSS into R (R x64 2.15.0).
I used both
read.spss("D:/Untitled2.sav", use.value.labels = TRUE, to.data.frame = TRUE,
max.value.labels = Inf,
trim.factor.names = FALSE,trim_values = TRUE, reencode = NA,
use.missings = to.data.frame)
&
read.spss("D
Hello.
I'm a new user of R, and I have a question regarding the use of aov and
lm-functions. I'm doing a fractional factorial experiment at our production
site, and I need to familiarize myself with the analysis before I conduct
the experiment. I've been working my way through the examples provide
Hi,
New to this list, and glad I found R5 which si a great improvement to R
for my work.
Something seems odd anyway, when I do a setRefClass, the initialize
method execute at the class definition.
I would like it to execute only when an instance is created.
How can I do that? I found no way to
Dear Rui,
Thanks. I check the spell in the code, no wrong (just spell wrong in
the error message). After I download the newest 2.15 version, the
code work fine with me, too. I guess it resulted from the old version
(2.7.1) I used.
2012/6/25 Rui Barradas :
> Hello,
>
> Your code works with me
Hello,
Instead of replacing "record_start" with newlines, you can split the
string by it. And use 'gsub' to make it prettier.
x <- readLines("test.txt")
x
y <- gsub("\"", "", x) # remove the double quotes
y <- unlist(strsplit(y, "record_start,")) # do the split, with comma
# remove leading
Dear Mr Dunlap,
Your solution works really fine. Thank you for your time,
Best wishes,
Luigi
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: 23 June 2012 18:48
To: Luigi
Cc: 'Martin Maechler'; r-help@r-project.org
Subject: RE: [R] Boxplot with Log10 and base-expo
On 12-06-24 4:50 PM, Søren Højsgaard wrote:
Dear all,
Indexing matrices from the Matrix package with [i,j] seems to be very slow. For
example:
library(rbenchmark)
library(Matrix)
mm<- matrix(c(1,0,0,0,0,0,0,0), nr=20, nc=20)
MM<- as(mm, "Matrix")
lookup<- function(mat){
for (i in
I have a comma separated data file with no carriage returns and what
I'd like to do is
1. read the data as a block text
2. search for the string that starts each record "record_start", and
replace this with a carriage return. Replace will do, or just add a
carriage return before it. The str
Hi
>
> I have the graph plotted with x axis(-50 to 250) and y axis (-50 to
500).I
> need the x axis values(-50 to 250) with spacing between two tick marks
as
> 1or 0.1.The graph should be wider to get enough resolution.
For the first case you shall have some special display as you will need a
On 25/06/2012 09:32, Rune Haubo wrote:
According to standard likelihood theory these are actually not
t-values, but z-values, i.e., they asymptotically follow a standard
normal distribution under the null hypothesis. This means that you
Whose 'standard'?
It is conventional to call a value of t
Hello,
Your code works with me, unchanged:
> boot<-boot(c,table,100, sim="ordinary", stype="i",simple=TRUE)
> boot
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = c, statistic = table, R = 100, sim = "ordinary",
stype = "i", simple = TRUE)
Bootstrap Statistics :
original bias
According to standard likelihood theory these are actually not
t-values, but z-values, i.e., they asymptotically follow a standard
normal distribution under the null hypothesis. This means that you
could use pnorm instead of pt to get the p-values, but an easier
solution is probably to use the clm-
I have the graph plotted with x axis(-50 to 250) and y axis (-50 to 500).I
need the x axis values(-50 to 250) with spacing between two tick marks as
1or 0.1.The graph should be wider to get enough resolution.
--
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I am doing boot with a large database, thus want to use simple=TRUE to
reduce the memory used.
I alreday set up sim="ordinary", stype="i" , but I don't know how to
set "n=0". In fact, I don't know what does "n=0" mean?
For example,
a<-c(1:1000)
b<-c(2:1001)
c<-cbind(a,b)
library(boot)
table<-func
Hi,
Here's one approach:
plot_one <- function(d){
with(d, plot(Xvar, Yvar, t="n")) # set limits
with(d[d$param1 == 0,], lines(Xvar, Yvar, lty=1)) # first line
with(d[d$param1 == 1,], lines(Xvar, Yvar, lty=2)) # second line
}
par(mfrow=c(2,2))
plyr::d_ply(data, "Subject", plot_one)
HTH,
Hello,
Just as a complement, if you want to see a solution, and why.
x <- 3/5
y <- 2/5
z <- 1/5
#
x - (y + z) # not zero
x == y + z
all.equal(x, y + z)
#
x - y - z # not zero
x - y == z
all.equal(x - y, z)
# To see why it works: abs diff less than tolerance
equal <- function(x, y, tol=.Machi
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