This behavior is typical for "fast" Fourier Transform algorithms. You need to
decide what theory you want to use in dealing with your missing data, and apply
that yourself. There are some packages that can help you (e.g. Lomb-Scargle or
discrete slow Fourier transform). Search with package sos o
Hello
I am using package multicore for parallel computing in a Altix UltraViolet
1000 server with 64 CPUs and 960 GB of RAM memory. Access is managed by
means of a SGE queue system. This is the first time I am using parallel
computing and my experience with supercomputers is quite l
Dears Tinn-R users,
Below a link to the new user list on Google groups:
https://groups.google.com/forum/#!forum/tinn-r
All the best,
--
///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\
Jose Claudio Faria
Estatistica
UESC/DCET/Brasil
joseclaudio.faria at gmail.com
Telefones:
55(73)3680.5545 - UES
Hi,
Please show a reproducible example.
countrydiff <- c("Albania", "Algeria", "Belarus", "Canada", "Germany")
long_df <- data.frame(country_name = c("Algeria", "Guyana", "Hungary",
"Algeria", "Canada", "Iran", "Iran", "Norway","Uruguay", "Zimbabwe") )
ifelse(long_df$country_name %in% countrydif
Dear list-members,
I have the following problem: I have a vector (countrydiff) with length 72
and another vector (long_df$country_name) which is about 12000 long.
Basically what I want to do is to if the factor level (or string name) in
long_df$country_name appears on the countrydiff, then long_d
Hello,
i know there are several functions on r to calculate the power of a test.
f.e the functions of pwr packet. In this packet the use the power function
(1-\beta) to calculate. My question is how we get this function for anova
(pwr.anova.test)?
I know that
1-\beta=P(H1|H1) = P(F>F_[1-\alpha,k-
Hi everyone,
I am a very new user of r (doing most of my previous stuff in vba). I am now
mandated to draw a 3-d surface of a mine pit hole. I have all the location
points (around 5000 points) of the pit in a CSV file under 3 column X, Y & Z.
However, going from page to page on the web, I could
Hi,
Suggestion 1: read
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and bookmark it for future reference.
Suggestion 2:
set.seed(123)
countrydiff <- letters[1:5]
long_df <- data.frame(country_name = sample(letters[1:8], 20, replace=TRUE))
long_df$povdat <
Hi to everyone, I am tring to make a Anova with repeated measures,my data set
looks like:
participantes <- c("1", "2", "3", "4", "5", "6", "7", "8", "1", "2", "3", "4",
"5", "6", "7", "8", "1", "2", "3", "4", "5", "6", "7", "8")
grupo <- factor(c(rep("A", 8), rep("B", 8), rep("C", 8)))valor <-
Hi
Extension does not specify file format. You can rename any file with any
extension without changing its nature. However slk stays for symbolic link and
therefore it just brings actual file to Excel.
Maybe you could start to play with
http://stat.ethz.ch/R-manual/R-devel/library/base/html/f
Dear Arun
Thanks for your reply, it made me realize that the problem was not in the
code but in the levels() of the factors. Some countries had some extra
spacing which made the ifelse() function not work. So if I modify your code
(added space to countrydiff), it will then look something like thi
Hi Adel,
If the problem is the spacing, then
library(stringr)
1*(long_df$country_name %in% str_trim(countrydiff))
# [1] 1 0 0 1 1 0 0 0 0 0
A.K.
Dear Arun
Thanks for your reply, it made me realize that the problem was
not in the code but in the levels() of the factors. Some countries had
som
Hi everyone,I am a very new user of r. I am now mandated to draw a 3-d surface
(and possibly rotating) of a mine pit hole. I have all the location points
(around 5000 points) of the pit in a CSV file under 3 column X, Y & Z. The
fllowing gives a 3d scatter but i would like to have a surface.
at
On Wed, Dec 18, 2013 at 2:52 PM, Simon Delay-Fortier
wrote:
> Hi everyone,
>
> I am a very new user of r (doing most of my previous stuff in vba). I am now
> mandated to draw a 3-d surface of a mine pit hole. I have all the location
> points (around 5000 points) of the pit in a CSV file under 3
Thanks! that works, more or less. Although the wonky behaviour of mapply that
David pointed out is irritating. I tried deleting the $call item from the
models produced and passing them to stargazer for reporting the results, but
stargazer won't recognize the results even though the class is expl
Dennis, how would your function be modified to allow it to be more flexible in
future.
I'm thinking like:
> f <- function(x='Dependent variable', y='List of Independent Variables',
> z='Data Frame')
> {
>form <- as.formula(paste(x, y, sep = " ~ "))
>glm(form, data =z)
> }
I tried that t
Folks:
1. Haven't closely followed the thread. I'm responding only to Simon's post.
2. ?formula ## Especially note the use of "."
So just make an appropriately constructed data frame for the data
argument of glm:
## example
> df <- data.frame(y=rnorm(9),x1=runif(9), x2=1:9)
> glm(y~.,data=df)
Hi,
I want my output to be like this:
Value
BXR
abc
DHH abc
DHK abc def
DSL def ghi
DSM abc def ghi DSS def ghi
DST ghi
DIW abc
DIL abc ghi
My input dataset is this with colnames name and Value:
name Value abc BXR abc DHH abc DHK def DHK def DSL ghi DSL abc DSM
def DSM ghi
The standard error of the curve cannot be extracted from the summary
information you have.
The variance is made up of two terms, one of which is a sum over all the death times, of a
quadratic term per death time. That term involves the variance matrix of the Cox model
coefficients, the target
Hi,
Try:
dat1<- read.table(text="name Value
abc BXR
abc DHH
abc DHK
def DHK
def DSL
ghi DSL
abc DSM
def DSM
ghi DSM
def DSS
ghi DSS
ghi DST
abc DIW
abc DIL
ghi DIL",sep="",header=TRUE,stringsAsFactors=FALSE)
aggregate(name~Value,data=dat1,paste,collapse=" ")
#or
li
I would like to predict a new response from a fitted linear model where the
new data is a single case with a missing value. My reading of the help on
predict() is inconclusive on whether this is possible.
Leaving out the missing value or setting it to NA both fail but differently,
see example code
I found all the color transparency was defined with character color, or rgb
color. What if I have number code and still try to modify the transparency?
For example:
>x=c(1:5)
> color=c(2,2,3,4,5)
> plot(x, col=color)
> plot(x, col=color,pch=20)
here I defined color by numbers, how can I modify
I'll re-enter the fray.
The data set is an example where coxph is incorrect; due to round off error it
is treating
a 5 column rank 3 matrix as if it were rank 4. This of course results in 0 digits of
precision.
Immediate fix, for the user, is to add "toler.chol= 1e-10" to their coxph
call.
On 13-12-18 2:08 PM, capricy gao wrote:
I found all the color transparency was defined with character color, or rgb
color. What if I have number code and still try to modify the transparency?
For example:
x=c(1:5)
color=c(2,2,3,4,5)
plot(x, col=color)
plot(x, col=color,pch=20)
here I define
As far as I can discern, your question makes no sense at all.
Suppose you *know* that y = 2 + 3*x1 + 4*x2.
Now what should you predict when x1 = 6 (with x2 "missing"/unknown)?
See fortune("magic").
On 19/12/13 07:18, Chris Wilkinson wrote:
I would like to predict a new response from a fitte
The take home message that you should be learning from your struggles
is to "Not Use The 'assign' Function!" and "Do Not Use Global
Variables Like This".
R has lists (and environments) that make working with objects that are
associated with each other much simpler and fits better with the
function
Try something like the following. Because lm() evaluates many
of its arguments in nonstandard ways, f() manipulates the call
and then evaluates it in the frame from which f() was called.
It also puts that environment on the formula that it creates so
it can refer to variables in that environment.
Another option to consider if your goal is to create a word file with
1 or more plots in it (possibly intermingled with text and other
output) is to use the knitr or pander packages (or odfWeave or sweave
or ...). This way you can create a script (or template file) that
sets a couple of options up
Â
Hello,
Â
Iâm trying to install R 3.02 for windows on a Windows 7 machine (we were just
upgraded from Windows XP).
Itâs a networked machine in a work environment. I have administrator rights.
Program files go on C; data is stored on D (it's a partitioned drive).
Â
I get the message 'fail
I checked as you suggested. However, I found that the number in those functions
are the number of colors. In contrast, my number here means a specific color,
for example, 2 in my code means "red", 3 in my code means "green"
On Wednesday, December 18, 2013 1:18 PM, Duncan Murdoch
wrote:
you will need to specify colours as RGB values and then set transparency
via the "alpha" argument.
e.g.: color=rgb(0,0,0,alpha=0.3)
# will give black (0,0,0) and a transparency of 30%.
Best wishes
Christoph
On 18/12/2013 23:23, capricy gao wrote:
I checked as you suggested. However, I foun
Another nice thing about your solution is that circles look like circles,
and not like diamonds (when viewed on screen).
Thanks.
Kevin Wright
On Mon, Dec 16, 2013 at 8:02 PM, Steve Taylor wrote:
> Unfortunately the win.metafile() device does not support semi-transparent
> colours, which I lik
As Duncan suggested, this will probably get you what you want.
You can set the transparency using alpha.f in adjustcolor().
x <- c(1:5)
color <- c(2,2,3,4,5)
color_transparent <- adjustcolor(color, alpha.f = 0.3)
plot(x, col = color, pch = 20, cex = 4)
plot(x, col = color_transparent, pch = 20,
On 13-12-18 5:23 PM, capricy gao wrote:
I checked as you suggested. However, I found that the number in those
functions are the number of colors. In contrast, my number here means a
specific color, for example, 2 in my code means "red", 3 in my code
means "green"
You didn't read very caref
Hi,
You could try:
Either:
dat1 <- read.table("Test.txt",header=TRUE)
dim(dat1)
#[1] 4735 4735
dat2 <- read.table("1991res.txt",header=TRUE)
dim(dat2)
#[1] 574 574
m1 <- as.matrix(dat1)
m2 <- as.matrix(dat2)
library(data.table)
d1 <-
data.table(Name1=as.vector(outer(rownames(m1),colnames(m1),past
dear all,
i would like to plot the value of different response groups. when i simply
use plot(y ~ x) i obtain a series of boxplots. i would rather use dots. i
also tried with stripchart(y ~ x), which gives better results but does not
place properly the labels since place them alphabetically.
in ad
Hello,
I've built a PLSR model to predict the concentrations of mixture components
from experimental data using the 'pls' library. I calculated the Q residual
(or lack of fit) and T squared value for each of the samples used to build
the model in order to assess how well each sample is described b
I suspect this could be SYLK format, a very old spreadsheet exchange format.
The only hit that I get from RSiteSearch("sylk") is read.gnumeric.sheet, which
depends on an external program "ssconvert" to extract CSV. In the plus
department, it is a text-based format that has been implemented numer
On 12/19/2013 11:18 AM, Luigi Marongiu wrote:
dear all,
i would like to plot the value of different response groups. when i simply
use plot(y ~ x) i obtain a series of boxplots. i would rather use dots. i
also tried with stripchart(y ~ x), which gives better results but does not
place properly t
Hi there,
I am new to R and programming in general and am looking for help with
writing a function with dates and times. I have checked around but am still
a bit stuck.
Basically, I have data in the format "dd/mm/ HH:MM" and I have to
calculate how much time has passed between various events.
use POSIXct instead of Date:
> x <- c("2013-12-12 12:00:00", "2013-12-15 03:15:23")
> # convert using as.POSIXct
>
> times <- as.POSIXct(x, format = "%Y-%m-%d %H:%M:%S")
> times
[1] "2013-12-12 12:00:00 EST" "2013-12-15 03:15:23 EST"
>
> difftime(times[2], times[1], units = 'secs')
Time differenc
I apologize if this is a simple question.
I am running GLMM's using glmmML and model averaging with MuMIn. One of the
parameter estimates for a parameter (firefreq) in the best model is giving
a positive number, where in reality I know this to be a negative
correlation.
I have checked and double c
You didn't say what gave that message (and there is no 'R 3.02': did you
mean 3.0.2?).
If this came from the installer, that is not part of R: look for info on
'Inno Setup'.
On 18/12/2013 21:43, Barry DeCicco wrote:
Â
Hello,
Â
I’m trying to install R 3.02 for windows on a Windows 7 machin
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