Displaying 20 results from an estimated 11000 matches similar to: "New User"
2012 Nov 18
2
[lattice] format and rotation of strip text
Thanks to the lattice gurus on this list, and having reference to the
excellent open-access Sarkar 2008
ISBN 978-0-387-75968-5
e-ISBN 978-0-387-75969-2
http://dx.doi.org/10.1007/978-0-387-75969-2
I now know how to label lattice panels by variable value: see thread
starting @
https://stat.ethz.ch/pipermail/r-help/2012-November/329450.html
(and demonstrated below). This allows me to use
2011 Nov 28
1
code problem with the optim() function
Please forgive me if someone has seen this duplicated email, because I sent
the email to the wrong address just now.
Hi, all,
I'm estimating an inter-rater coefficient, Aickin (1990)'s alpha. It's been
a long time for me to figure out how to make it work, but it's still of no
avail. The problem seems to be my code with the optim() function. Can
anyone help me take a look at the
2011 Jan 20
4
LMTP & home, chroot, mail userdb fields.
Hello, Dovecot.
I'm using postfix + dovecot with pure virtual users. postfix uses
standard virtual transport, and dovecot fetches such fields from
userdb:
chroot: "/usr/home/hosted/v-mail/%d/%n"
home: "/"
mail: "maildir:."
Everything works Ok -- dovecot founds users' mail.
Now, after upgrade to dovecot2, I want to use it LMTP server as
2002 Dec 01
1
generating contrast names
Dear R-devel list members,
I'd like to suggest a more flexible procedure for generating contrast
names. I apologise for a relatively long message -- I want my proposal to
be clear.
I've never liked the current approach. For example, the names generated by
contr.treatment paste factor to level names with no separation between the
two; contr.sum simply numbers contrasts (I recall an
2008 Jan 02
3
Find missing days
Hi,
I have a data.frame like this:
y <- rnorm(60)
lev <- gl(3,20, labels=paste("lev", 1:3, sep=""))
date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5),
by=60*60*24))
date1 <- date1[-c(3,4,15,34,38,40)]
df <- data.frame(lev=lev, date1=date1, y=y)
I would like to produce a new data.frame with missing days in df$date1
in each df$lev, like this:
lev
2020 Oct 20
2
How to assign the same levels to a dataframe column?
Hello,
I have a dataframe df with a column x that has these unique values:
"L" "M" "V" "N" "H". I can assign a factor to it:
```
df$x = as.factor(df$x)
> [1] L M V N H
Levels: H L M N V
```
I now need to get a subset of this dataframe. I could do the same
thing as before on the subset sf, but I would like to avoid the
possibility that not all
2008 Jul 23
1
R2WinBUGS problem
Dear friends - I'm on winXP, R 2.71 - I have with some help dveloped
this multivariate normal model, which gives very plausible results in
WinBUGS even without any
initial values specified. However, when I then try to run the same model
via the bugs function in R2WinBUGS with inits specified as inits=NULL
the program stops in a dead end. So I have tried to make inits for the
bugs function
2012 Nov 18
2
[newbie] convert 3D spatial array to dataframe
summary: how to convert a 3D array (as obtained from ncdf4::ncvar_get)
to a dataframe suitable for use by lattice::levelplot(), e.g.,
levelplot(conc ~ lon * lat | lev, data = data.frame)
details:
I have atmospheric data in netCDF files specifying gas concentrations
over a 3D space with dimensions longitude, latitude, and (vertical)
level. I need to plot the data by level. Since there are
2003 Jan 10
1
plot() and lines() multivariate problem
Dear list
I'm trying to solve the following problem since 2 days with no success.
Could someone help a newbie, please ?
I have a dependant variable which is diameter of shell, and two factors
which are level on the beach (3 levels 'B', 'E' and 'H') and a hydrodynamics
indice (3 levels 1, 2 and 3)
To study diameter variations according to these factors, I may use
2006 Mar 11
2
Draw level lines on the surface of a bivariate function
Hello,
Is it possible to draw level lines on the surface of a bivariate function?
In the following example, to draw surface and levels lines for a
multivariate normal law,
I use persp, trans3d, contourLines and lines,
but if the lines are correctly drawn, some parts of them are, of
course, visible
even if they are drawn on a non visible "face".
Any suggestion to avoid this problem
2009 Feb 11
2
Question about apply()
Hello, everyone!
Assume you have this data:
data <- structure(c(66.609375, 67.09375, 66.40625, 66.734375, 67.109375,
66.875, 66.09375, 65.921875, 66.546875, 66.140625, 66.140625,
65.65625, 65.875, 65.59375, 65.515625, 66.09375, 66.015625, 66.140625,
66.109375, 66.421875, 1702.7, 1647.7, 1649.4, 1639.9, 1696.4,
1710.9, 1690.2, 1677.9, 1694.4, 1713.9, 1713.9, 1705.4, 1708.4,
1692.9, 1689.6,
2020 Oct 20
0
How to assign the same levels to a dataframe column?
Stop using as.factor() and start using factor()? Be explicit about what levels you want and in what order.
As for extracting subsets, use split() and do your subsets on each set of rows with the same level. It will be on you to decide what statistical properties (proportions?) you want to maintain between the full data set and your subset.
On October 20, 2020 6:42:27 AM PDT, Luigi Marongiu
2007 Oct 15
1
avoiding a loop?
Hi All:
I feel like there must be a slick R-native no-loop way to get the counts
for the entries in a factor, but I'm unable to see how. Right now I'm
doing this:
hn.ent<-data.frame(rep(0,length(levels(hnf))), row.names=levels(hnf))
colnames(hn.ent)<-c("count")
for (lev in levels(hnf)) {
hn.ent[lev,"count"] <- sum(lev == hnf)
}
I'm sure there's a
2010 Aug 10
1
one (small) sample wilcox.test confidence intervals
Dear R people,
I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example:
> sample <- c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06)
> summary(sample)
Min. 1st Qu. Median Mean 3rd Qu. Max.
2009 Apr 17
1
cast function in package reshape
Hello R useRs,
I have a function which returns a list of functions :
freq1 <- function(x) {
lev <- unique(x[!is.na(x)])
nlev <- length(lev)
args <- alist(x=)
if (nlev == 1) {
body <- c("{", "sum(!is.na(x))", "}")
f <- function() {}
formals(f) <- as.pairlist(args)
body(f) <- parse(text = body)
namef <-
2010 Jul 15
2
taking daily means from hourly data
I have a data frame (morgan) of hourly river flow, river levels and wind direction and speed thus:
Time hour lev.morgan lev.lock2 lev.lock1 flow direction velocity
1 2009-07-06 15:00:00 15 3.266 3.274 3.240 1710.6 180.282 4.352
2 2009-07-06 16:00:00 16 3.268 3.272 3.240 1441.8 192.338 5.496
3 2009-07-06 17:00:00 17 3.268
2012 Nov 16
2
Question about contour3d and writeWebGL: rgl and misc3d package
I saw that in rgl:::writeWebGL that "Polygons will only be rendered as
filled; there is no support in WebGL for wireframe or point rendering.". I
found that you can easily use contour3d to make reproducible contour web
figures, such as (taken from contour3d help)
library(AnalyzeFMRI)
a <- f.read.analyze.volume(system.file("example.img", package=
2009 Feb 03
1
7.1-stable (righ after release) locks up on soekris net5501 every day
Hello, Freebsd-stable.
I installed 7.1-STABLE on my new Soekris net5501. Kernel config is
in attach.
This unit lock up in strange way every day. It is pingable, but no
access to host on any network protocol (sshd, named, etc are not
answering), and serial console (only one this unit has) DOESN'T ANSWER
too!
Only way to un-freeze it is cold reboot.
I've thought, it is
2018 Mar 22
1
adjusted values
Hi all,
I am fitting a linear mixed model with lme4 in R. The model has a single
factor (des_days) with 4 levels (-1,1,14,48), and I am using random
intercept and slopes.
Fixed effects: data ~ des_days
Value Std.Error DF t-value p-value
(Intercept) 0.8274313 0.007937938 962 104.23757 0.0000
des_days1 -0.0026322 0.007443294 962 -0.35363 0.7237
des_days14 -0.0011319
2008 Aug 20
1
Improvements to write.arff (PR#12574)
Full_Name: Martin C. Martin
Version: 2.7.1
OS: Ubuntu
Submission from: (NULL) (75.150.115.86)
The function write.arff, in the foreign library:
- Can produce relation names with invalid characters
- Doesn't use colnames() for attribute names when writing a matrix.
Here's a better version:
write.arff <- function (x, file, eol = "\n")
{
if (file == "")