(1) Please do not hijack another thread to ask an off-topic question -
start a new one instead.
(2) Your question refers to a topic called 'restricted least squares'.
Search the R-help archives on that subject and you should get a number
of answers. Someone answered a similar question here within t
Any time you use ``par(new=TRUE)'', you're probably doing something
wrong. Avoid it like the plague.
Following up on Michael Weylandt's post: plotCI() indeed has an xlim
argument (i.e. the "..." arguments which get passed to plot.default()
can include xlim).
Note that plotCI() also has an ***
Please advice on the package I should use to run a linear regression model
(weighted least squared) with linear equality constraint. I initially tried
"constrOptim" but it turned out that it only supported
http://solve-graph-linear-inequalities.blogspot.com/2011/11/blog-post.html
linear inequality
log(x1+1) just like your formula
best,
Josh
On Tue, Nov 15, 2011 at 9:54 PM, arunkumar wrote:
> Hi
>
> i have few clarification regarding the output from the fitted(object).
>
> I'm creating a object using the following formula
> dataset <- read.csv("~/data.csv")
>
> obj = lm (formula="lo
Hi there I need some
[URL=http://calculate-conditional-and-distributio.blogspot.com/2011/11/blog-post.html]Statistics
Help[/URL]!! What statistical tests should i use/consider (pref in
SPSS) in regards to an invertebrate survey i have carried out comparing
urban street trees to urban park t
Hi
i have few clarification regarding the output from the fitted(object).
I'm creating a object using the following formula
dataset <- read.csv("~/data.csv")
obj = lm (formula="log(y+1)~log(x1+1)+(log(x2+1)+log(x3+1)", data=dataset)
fitted(obj)
here the output of the fitted(obj) considers th
Hi,
I am looking for a function which takes as input
the spatial coordinate in longitute/latitude and
the distance in metric and returns the second coordinate
(of course depending on an angle as well).
So far I have only found functions which calculate
the distance between two coordinates.
Thank
I came across the package 'ineq' that computes a variety of inequality measures
(e.g. gini, theil etc). I want to compute the Theil index (racial segregation)
and decompose the total into sub-components (by geog levels). I think the
package doesn't report the decomposition (correct me if I'm wro
Hi:
(1) Here is why your e-mails look mangled on this list:
[[alternative HTML version deleted]]
R-help is a text-only list, so please change your mailer's
settings to send ASCII text rather than HTML.
(2) The print method you see displayed in dd1 is equivalent to the following:
ddm * low
Does plotCI take an xlim argument?
Something like plotCI(..., xlim=range(Data1$Year, Data2$Year))
Michael
On Nov 15, 2011, at 11:51 PM, Vinny Moriarty wrote:
> Hello All,
>
> Many thanks to the help I have received so far.
>
>
> Here is an example data set I hope to plot
>
> Data1
> Year
Hi:
Does this work?
# library('lattice')
levs <- as.vector(quantile(volcano,c(0,0.1,0.5,0.9,0.99,1)))
levelplot(volcano, at = levs,
colorkey = list(labels = list(at = levs,
labels = levs) ))
HTH,
Dennis
On Tue, Nov 15, 2011 at 1:12
Hello All,
Many thanks to the help I have received so far.
Here is an example data set I hope to plot
Data1
Year Data SE
1 20052 0.01
2 20064 0.01
3 20075 0.01
4 20082 0.01
5 20093 0.01
6 20106 0.01
Data2
Year Data SE
1 2006 32 1
2 2007 100 2
3 2008 60 4
Yes, you probably need some sort of C compiler, but why can't you just
download the appropriate binary directly? I just did on OS X 10.5.8
(admittedly for R 2.13.2, not 2.14) with no problems. The output of
sessionInfo()
install.packages("RCurl")
if you don't mind please.
Thanks,
Michael
On Tu
Can you post working examples of your data using the dput() function?
There are so many types of time series in R and so many different
things you could mean that it's just easier to work with real data.
Michael
On Tue, Nov 15, 2011 at 4:28 PM, Sarwarul Chy wrote:
> Hello Michael,
> Thanks for y
Try something like this:
X = matrix(rnorm(25), 5); colnames(X) <- rownames(X) <- letters[1:5]
Y <- dist(X)
library(reshape)
melt(as.matrix(Y))
Michael
On Tue, Nov 15, 2011 at 7:03 PM, ram basnet wrote:
> Dear R users,
>
> I am not good in R-language programming. So, i need your help.
>
> I wa
How to do the nearest neighbor density estimation in R?
Is there any function/package to perform k-Nearest Neighbor based density
estimation in R
Thank you very much!
Thanks to a dozen of volunteer moderators!
[[alternative HTML version deleted]]
_
PRI gmail.com> writes:
>
> Write an R program to draw 10,000 random samples using the Polya Urn Scheme
> where the initial contents of the urn is one white ball and one black ball.
> Accumulate the proportion of white balls after each draw into one vector.
> Upload your R script file.
> HINT: Yo
Copying David W from another thread:
We don't do homework here. In fact, your instructor is probably watching.
That said, try googling or, even better, using the rseek search. (Or even
better-er RSiteSearch() in your terminal)
Michael
On Nov 15, 2011, at 4:16 PM, Jan Kraner wrote:
> Dear Si
Jan Kraner gmx.ch> writes:
>
> Dear Sir/Madam:
>
> Could you please help me with this exercise and how I could solve it in R?:
>
> Attachment (PastedGraphic-1.pdf): application/pdf, 84 KiB
>
> I especially have problems with the selection of the 40 datasets,
> once taken the 110 (because the
Steffen,
Did you ever have luck getting rid of the tick marks?...
I like your idea and have modified it, but yes, the tick marks need to go.
Steffen Fleischer wrote:
>
> Dear all,
>
> I want to draw a graph that contains the scatterplot matrix in the lower
> panel and coefficients in the u
On 16/11/11 10:10, anaraster wrote:
I've seen some questions regarding the output of multiple objects from a
function, however the suggestions all end up suggesting the use of
return(list(result1=result1, result2=result2 , result3=result3)).
How can I return multiple objects that are 2 big to b
Hi,
On Tue, Nov 15, 2011 at 4:31 PM, sophy1987 wrote:
> Hi, everyone
> When I ran this cript, There is Error in substring(tmp.subject, tmp.end[ex]
> + 1, tmp.start[ex + 1] - 1) :
> invalid substring argument(s)
>
> Could someone figure out what the problem is?
Not without a reproducible exampl
Hi,
The obvious answer is don't use attach() and you'll never have
that problem. And see further comments inline.
On Tue, Nov 15, 2011 at 6:05 PM, Steven Yen wrote:
> Can someone help me with this variable/data reading issue?
> I read a csv file and transform/create an additional variable (calle
Hi,
On Tue, Nov 15, 2011 at 4:10 PM, anaraster wrote:
> I've seen some questions regarding the output of multiple objects from a
> function, however the suggestions all end up suggesting the use of
> return(list(result1=result1, result2=result2 , result3=result3)).
>
> How can I return multiple
Thank you so much for the help! It is really appreciated!
Dongli
On Nov 15, 2011, at 9:19 AM, Marc Schwartz wrote:
> Hi Dongli,
>
> Sorry for the delay in following up.
>
> You might want to read the dsDesignManual.pdf document, which is available in
> the 'inst/doc' folder in the package so
Can someone help me with this variable/data reading issue?
I read a csv file and transform/create an additional variable (called y).
The first set of commands below produced different sample statistics
for hw11$y and y
In the second set of command I renameuse the variable name yy, and
sample st
Hi, everyone
When I ran this cript, There is Error in substring(tmp.subject, tmp.end[ex]
+ 1, tmp.start[ex + 1] - 1) :
invalid substring argument(s)
Could someone figure out what the problem is?
for(i in 1:length(genebody[,1])){
tmp.id<-as.vector(genebody[i,1])
On 11/15/2011 06:46 PM, Kevin Burton wrote:
What is wrong with the following?
x<- 1:2
if(x[1]> 0)
{
if(x[2]> 0)
{
print("1& 2> 0")
}
else
{
Hello Michael,
Thanks for your reply. What I want to do is something like this? For
example, I have a continuous time series y=x(t), and another discrete time
series z=w(t).
Xdiff(i)=Max. difference between x(t) and w(t) in interval i
Ndiff(i)=Min. difference between x(t) and w(t) in interval
On Nov 15, 2011, at 3:57 PM, PRI wrote:
Can someone help me out with this problem?
Write an R program to draw 10,000 random samples using the Polya Urn
Scheme
where the initial contents of the urn is one white ball and one
black ball.
Accumulate the proportion of white balls after each dra
Write an R program to draw 10,000 random samples using the Polya Urn Scheme
where the initial contents of the urn is one white ball and one black ball.
Accumulate the proportion of white balls after each draw into one vector.
Upload your R script file.
HINT: You can simulate sampling one ball from
I've seen some questions regarding the output of multiple objects from a
function, however the suggestions all end up suggesting the use of
return(list(result1=result1, result2=result2 , result3=result3)).
How can I return multiple objects that are 2 big to be added to a list?
--
View this mess
A follow-up on the data/variable issue I posted earlier:
Here was what I did, which might was obviously causing the problem:
I inserted the following line in my file "Rprofile.site":
setwd("z:/R")
Then, as soon as I run R (before I read any data) I issue
summary(mydata)
I get summary statis
Can someone help me out with this problem?
Write an R program to draw 10,000 random samples using the Polya Urn Scheme
where the initial contents of the urn is one white ball and one black ball.
Accumulate the proportion of white balls after each draw into one vector.
Upload your R script file.
HI
Dear R users,
I am not good in R-language programming. So, i need your help.
I want to convert my lower-triangle value of symmetry matrix into a vector with
their row and column name.
I found a function called "sm2vec" in "corpcor" package but it give only a
vector of values but not row and col
Hi,
Please don't cross post.
It seems that ggplotGrob has been replaced by new functions. You can
define it as
ggplotGrob <- function(x) ggplot2:::gtable_gTree(ggplot2:::ggplot_gtable(x))
and it seems to work as before with grid.arrange().
HTH,
baptiste
On Wed, Nov 16, 2011 at 3:26 AM, Joha
Forescasting using predict() in an object of class arimax when there is an
outlier IO in the model.
Hi R users
I have a problem when a use the predict() method in an object of class
arimax ( These objects are the results of the implementation of the function
arimax() from the TSA library) . The obj
Hi,
Check out this R-help thread from 2007.
http://tolstoy.newcastle.edu.au/R/e2/help/07/06/19513.html
Cheers,
Ben
On Nov 15, 2011, at 6:46 PM, Kevin Burton wrote:
> What is wrong with the following?
>
>
>
> x <- 1:2
>
> if(x[1] > 0)
>
> {
>
>if(x[2] > 0)
>
>
On Tue, Nov 15, 2011 at 3:46 PM, Kevin Burton wrote:
> What is wrong with the following?
AFAIK the else needs to follow the end brace of if {} on the same
line, at least at the main level.
Peter
>
>
>
> x <- 1:2
>
> if(x[1] > 0)
>
> {
>
> if(x[2] > 0)
>
> {
>
>
What is wrong with the following?
x <- 1:2
if(x[1] > 0)
{
if(x[2] > 0)
{
print("1 & 2 > 0")
}
else
{
print("1 > 0")
}
}
els
1. Please post in plain text, not HTML (as the posting guide asks!)
2. This might actually be an R question -- is
?lower.tri
what you want?
-- Bert
On Tue, Nov 15, 2011 at 2:50 PM, Jeff Newmiller
wrote:
> This is R-help, not the linear algebra hotline. Please stay on topic.
> --
On Tue, Nov 15, 2011 at 2:38 PM, Juliet Ndukum wrote:
> Given a vector;> ab = seq(0.5,1, by=0.1)> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0
> The euclidean distance between the vector elements is given by the lower
> triangular matrix > dd1 = dist(ab,"euclidean")> dd1 1 2 3 4 52 0.1
>
On Nov 15, 2011, at 5:38 PM, Juliet Ndukum wrote:
Given a vector;> ab = seq(0.5,1, by=0.1)
> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0
The euclidean distance between the vector elements is given by the
lower triangular matrix
> dd1 = dist(ab,"euclidean")
> dd11 2 3 4 5
2 0.1
3 0.2
Given a vector;> ab = seq(0.5,1, by=0.1)> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0
The euclidean distance between the vector elements is given by the lower
triangular matrix > dd1 =
dist(ab,"euclidean")> dd1 1 2 3 4 52 0.1 3 0.2 0.1
4 0.3 0.2 0.1 5 0.4 0.3 0.2 0.
This is R-help, not the linear algebra hotline. Please stay on topic.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
Given a vector;> ab = seq(0.5,1, by=0.1)> ab[1] 0.5 0.6 0.7 0.8 0.9 1.0
The euclidean distance between the vector elements is given by the lower
triangular matrix > dd1 = dist(ab,"euclidean")> dd1 1 2 3 4 52 0.1
3 0.2 0.1 4 0.3 0.2 0.1 5 0.4 0.3 0.2 0.1
Hmm... you know I'd love to run a study on how software and other
information displays affect the speed, accuracy and reliability with which
people make insights about data.
Tom
On Tue, Nov 15, 2011 at 4:44 PM, Stavros Macrakis wrote:
> > Last time, I was told that I couldn't list my R package a
Dear R users,
I am happy to announce the initial release of the "mvProbit" package
on CRAN (version 0.1-0). This package provides tools for econometric
analysis with Multivariate Probit Models. While these models can be
estimated also by several other statistical software packages (e.g.
LIMDEP/NLO
> Last time, I was told that I couldn't list my R package and associated
papers as a research activity with
> substantial impact because it was outside my official scope of work.
(Even though I wrote it so I could
> *do* my work.)
That seems wrong. My impression is that "method" papers were frequ
Thanks Michael,
Op 11/14/2011 3:30 PM, R. Michael Weylandt schreef:
You need to explicitly pass th to your function with the ... argument
of integrate.
That was a point I was missing!
Thanks again,
This solved my problems for this time.
Gerrit.
E<- function(th){
integrate(function(x,th)
Generally I have several versions of R installed on any PC running
Windows 7 and have never had any problem choosing which version of R
was to be the default one. In the past the updated version did not
always register as the default but there was and is a utility
RSetReg.exe in the bin directory
Given the example:
R> (levs <- quantile(volcano,c(0,0.1,0.5,0.9,0.99,1)))
0% 10% 50% 90% 99% 100%
94 100 124 170 189 195
R> levelplot(volcano,at=levs)
How can I make the key categorical with the size of the divisions equally
spaced in the key? E.g., five equal size rectangles wit
David:
You need to re-read ?tapply _carefully_. Note that:
FUN the function to be applied, or NULL. In the case of functions
like +, %*%, etc., the function name must be backquoted or quoted.
Now note that in tapply(whatever, byfactor, mean), "mean" _is_ a function.
However in tapply (what
There's a slight variant that might be even more helpful if you need
to line the data up with how you started: ave(). I'll let you work out
the details, but the key difference is that it returns a vector that
has the 90th percentile for each group, each time that group appears,
instead of the summa
1) tapply will work for quantile, but the syntax was a little off: try this
tapply(Cape $ ResponseTime, Cape $ Grid_ID, quantile, c(0.05, 0.95))
The fourth argument is additional parameters passed to the function
FUN which here is quantile. You could also do this
tapply(Cape $ ResponseTime, Cape
Hi:
Summary:
I am trying to determine the 90th percentile of ambulance response times for
groups of data.
Background:
A fire chief would like to look at emergency response times at the 90th
percentile for 1 kilometer grids in Cape Coral, Florida. I have mapped out
ambulance response times on a
It's not clear what it means for the differences to be "of increasing
order" but if you simply mean the differences are increasing, perhaps
something like this will work:
library(caTools)
X = cumsum( 2*(runif(5e4) > 0.5) - 1) # Create a random Walk
Y = runmean(X, 30, endrule = "mean", align = "rig
Excellent, as.matrix() didn't work but switched it to as.numeric() around
the definition of both variables in the function and it did work:
rF <- function(x, a, b) cor(as.numeric(x[a]), as.numeric(x[b]), use =
"complete.obs")
maindata$rFcor<-apply(maindata,1,FUN=rF,a=174:213,b=214:253)
Thanks
OK, I think I see the problem. Rather than setting method="nAGQ" I need
nAGQ=1. Doing so throws the following error:
"Warning messages:
1: glm.fit: algorithm did not converge
2: In mer_finalize(ans) : gr cannot be computed at initial par (65)
Error in diag(vcov(fitmodel)) :
error in evaluat
Is the whole thing a data frame? Then any multi-column subset is also a data
frame. Try adding a as.matrix() wrapper in the definition of rF.
Michael
On Nov 15, 2011, at 3:14 PM, "Rob Griffin" wrote:
> Error in cor(x[a], x[b], use = "complete.obs") : 'x' must be numeric
>
> This is strange,
I hadn't considered altering the font. Thank you I will try that.
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: Tuesday, November 15, 2011 1:53 PM
To: Kevin Burton
Cc: r-help@r-project.org
Subject: Re: [R] Plot alignment with mtext
Hi Kevin,
On Tue, Nov 15,
Wow, that is a very clever way to do it.
Thank you very much!
Cheers,
Syrvn
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_
Error in cor(x[a], x[b], use = "complete.obs") : 'x' must be numeric
This is strange, it works on your example (and you've understood what I'm
trying to do perfectly), but then when I use it on the original data it
comes up with the error above
I've checked str() and the columns are all numeri
Never mind-I fixed it.
My script is throwing the following error:
"Error in glmer(formula = modelformula, data = data, family = binomial(link =
logit), :
Argument ‘method’ is deprecated.
Use ‘nAGQ’ to choose AGQ. PQL is not available."
I remember hearing somewhere that PQL is no longer av
Yet another solution. This time using the LaF package:
library(LaF)
d<-c(1,4,7,8)
P1 <- laf_open_csv("M1.csv", column_types=rep("double", 10), skip=1)
P2 <- laf_open_csv("M2.csv", column_types=rep("double", 10), skip=1)
for (i in d) {
M<-data.frame(P1[, i],P2[, i])
}
(The skip=1 is needed as l
On Nov 15, 2011, at 2:43 PM, Sverre Stausland wrote:
Thanks David - this is pretty close to what I am looking for. However,
the output of vec[2] now includes the row number [1] and quotations
marks at the endpoints of the row. Is there an easy way to exclude
those?
The usual method is to use
Hi Kevin,
On Tue, Nov 15, 2011 at 2:36 PM, Kevin Burton wrote:
> I would like the text plotted with 'mtext' to be alighned like it is for
> printing on the console. Here is what I have:
You don't provide any of the info in the posting guide (OS may be
important here), or a reproducible example,
I'm getting the following error in a script: "Error: could not find function
"lmer." I'm wondering of my lme4 package is installed incorrectly. Can
someone tell me the installation procedure? I looked at the support docs but
couldn't translate that into anything that would work.
__
This problem is not for the faint of heart. Doug Bates, author of
nls(...) has said that a general purpose implementaion of R code for
multiresponse nonlinear regression is unlikely in the near future. You
have a large set of issues to deal with here. First, you have a system
of differential eq
Thanks David - this is pretty close to what I am looking for. However,
the output of vec[2] now includes the row number [1] and quotations
marks at the endpoints of the row. Is there an easy way to exclude
those?
Thanks
Sverre
On Tue, Nov 15, 2011 at 8:11 AM, David Winsemius wrote:
>
> On Nov 14
I would like the text plotted with 'mtext' to be alighned like it is for
printing on the console. Here is what I have:
> print(emt)
ME RMSE MAE
MPE MAPE MASE
original -1.034568e+07 1.097695e+08 2.433160e+07 -31.30554 3
Hello,
Can you please help me with this? I am also stack in the same problem.
Sam
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Take advantage of a 20% discount on the most recent R books from Chapman &
Hall/CRC!
We are pleased to offer our latest R books at a 20% discount through our
website. To take advantage of this offer, simply visit www.crcpress.com, choose
your titles and insert code AZL02 in the 'Promotion C
Dear all,
I am trying to install a package from bioconductor (biomaRt) for which I
need the RCurl package. I get the following main error message when I try
to install RCurl (and its dependencies).
configure: error: no acceptable C compiler found in $PATH
See `config.log' for more details.
ERROR:
Hi Chris,
On Tue, Nov 15, 2011 at 1:47 PM, Chris82 wrote:
> Hi R users,
>
> I want to colored points by their value
>
> for example:
>
> x <- c(1,2,3,4)
> y <- c(1,2,3,4)
> z <- c(2,3,4,9)
>
> y and x are coordinates
>
> z is the value of the coordinates
>
> points(x,y,col= rainbow(z))
In the ge
Try either col=z or col=rainbow(max(z))[z] depending on what color scheme you
want.
Michael
On Nov 15, 2011, at 1:47 PM, Chris82 wrote:
> Hi R users,
>
> I want to colored points by their value
>
> for example:
>
> x <- c(1,2,3,4)
> y <- c(1,2,3,4)
> z <- c(2,3,4,9)
>
> y and x are coord
I'm not sure your request completely makes sense: marginal means and variances
are not sufficient to give the joint distribution; even if you can be assured
it is bivariate normal, you still need a correlation. Just a heads up
Michael.
PS - next time would you please use a slightly more nuance
Dear Louise
On 15 November 2011 19:03, lstevenson wrote:
> Hi all,
>
> I'm trying to estimate model parameters in R for a pretty simple system of
> equations, but I'm having trouble. Here is the system of equations (all
> derivatives):
> eqAlgae <- (u_Amax * C_A) * (1 - (Q_Amin / Q_A))
> eqQuota
Hi all,
I'm trying to estimate model parameters in R for a pretty simple system of
equations, but I'm having trouble. Here is the system of equations (all
derivatives):
eqAlgae <- (u_Amax * C_A) * (1 - (Q_Amin / Q_A))
eqQuota <- (p_max * R_V) / (K_p + R_V) - ((Q_A-Q_Amin)*u_Amax)
eqResource <- -C
I was able to solve this problem by going back to nls and obtaining the
initial parameter estimates through optim. When I used nlsList with my
dataset, it took 2 minutes to solve and was not limited by the bounds. Now I
have the bounds working and it takes 45 seconds to solve. Here is the new
code:
Hi R users,
I want to colored points by their value
for example:
x <- c(1,2,3,4)
y <- c(1,2,3,4)
z <- c(2,3,4,9)
y and x are coordinates
z is the value of the coordinates
points(x,y,col= rainbow(z))
something like that
But haven't found any solution at the moment.
Thanks.
Chris
--
View
file.path() is much better for this than paste(), e.g.
dir <- "C:/Users/Desktop"
pathname <- file.path(dir, "bs_dev_segment_file.csv")
temp_data <- read.csv(pathname)
/Henrik
On Tue, Nov 15, 2011 at 10:08 AM, R. Michael Weylandt
wrote:
> Try pasting them together like
>
> paste(dir, "...")
>
>
On Nov 15, 2011, at 11:21 AM, Gyanendra Pokharel wrote:
Hi all,
I have the mean vector mu<- c(0,0) and variance sigma <- c(10,10),
now how
to sample from the bivariate normal density in R?
Can some one suggest me?
I did not fine the function "mvdnorm" in R.
But when you typed ?mvdnorm R sh
Hi Rob,
Here is one approach:
## define a function that does the calculations
## (the covariance of two vectors divided by the square root of
## the products of their variances is just a correlation)
rF <- function(x, a, b) cor(x[a], x[b], use = "complete.obs")
set.seed(1)
bigdata <- matrix(rno
Try pasting them together like
paste(dir, "...")
You may need to use the collapse argument. Alternatively, change your working
directory to dir with setwd().
M
On Nov 15, 2011, at 9:12 AM, aajit75 wrote:
> Hi List,
>
> I am new to R, this may be simple.
>
> I want to store directory path
Hi Aajit,
try using the ?paste function to combine the variable with your
directly and the filename into one string, and then pass that to
read.csv() or whatever
paste(dir, "/bs_dev_segment_file.csv", sep = '')
HTH,
Josh
On Tue, Nov 15, 2011 at 6:12 AM, aajit75 wrote:
> Hi List,
>
> I am new
Just as an update on this problem:
I have managed to get the variance for the selected columns
Now all I need is the covariance between these 2 selections -
the two target columns are and the aim is that a new column contain a
covariance value between these on each row:
maindata[,c(174:213)] an
Hi all,
I have the mean vector mu<- c(0,0) and variance sigma <- c(10,10), now how
to sample from the bivariate normal density in R?
Can some one suggest me?
I did not fine the function "mvdnorm" in R.
Best
Gyan
[[alternative HTML version deleted]]
Hi List,
I am new to R, this may be simple.
I want to store directory path as parameter which in turn to be used while
reading and writing data from csv files.
How I can use dir defined in the below mentioned example while reading the
csv file.
Example:
dir <- "C:/Users/Desktop" #location of
take a look at the structure of what Sys.time returns.
str(Sys.time)
and now at ?strptime!
> format(Sys.time(),format='%d-%H-%M-%S')
[1] "15-09-55-55"
> format(Sys.time(),format='%Y')
[1] "2011"
> format(Sys.time(),format='%m')
[1] "11"
>
Hope that helps,
Justin
On Tue, Nov 15, 2011 at 9:48
Hello,
with Sys.time() you get the following string:
"2011-11-15 16:25:55 GMT"
How can I extract the following substrings:
year <- 2011
month <- 11
day_time <- 15_16_25_55
Cheers,
Syrvn
--
View this message in context:
http://r.789695.n4.nabble.com/Extract-pattern-from-string-tp4073432p4
Hi Michael,
Your strings were long so I made a bit smaller example. Sarah made
one good point, you want to be using gsub() not sub(), but when I use
your code, I do not think it even works precisely for one instance.
Try this on for size, you were 99% there:
## simplified cases
form1 <- c('produ
I'm using Bill Browne's MLPowSim to do some sample size estimation for a
multilevel model. It creates an R program to carry out the estimation using
lmer in the lme4 library. When there are predictors with more than two
categories one has to modify the code generated to account for the multino
Hi Michael,
You need to take another look at the examples you were given, and at
the help for ?sub():
The two ‘*sub’ functions differ only in that ‘sub’ replaces only
the first occurrence of a ‘pattern’ whereas ‘gsub’ replaces all
occurrences. If ‘replacement’ contains backreferen
Good afternoon list,
I have the following character strings; one with spaces between the maths
operators and variable names, and one without said spaces.
form<-c('~ Sentence + LEGAL + Intro + Intro / Intro1 + Intro * LEGAL +
benefit + benefit / benefit1 + product + action * mean + CTA + help + me
Thank you. I mainly didn't know about the vector/matrix printing rules.
Kevin
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Tuesday, November 15, 2011 10:43 AM
To: Kevin Burton; r-help@r-project.org
Subject: RE: [R] Controlling the precision of the digits prin
On Tue, Nov 15, 2011 at 9:00 AM, Catarina Miranda
wrote:
> Hello;
>
> I am having a problems with the interpretation of models using ordered or
> unordered predictors.
> I am running models in lmer but I will try to give a simplified example
> data set using lm.
> Both in the example and in my rea
Thanks. Yes, I meant nrow(dataset)+1 (typo...)
Sammy
On Mon, Nov 14, 2011 at 1:29 AM, Petr PIKAL wrote:
> >
> > dataset[ nrow(dataset), ] <- c ("Male", 5, "bad")
> >
> > The above seems to have worked to append a row in place of a rbind().
> This
>
> No. It overwrites your last row. You maybe m
When you print a vector R uses a single
format for the whole vector and tries to
come up with one format that displays all the values
accurately enough. For a matrix (or data.frame)
it uses a different format for each column, so perhaps
you would like the output of:
> matrix(a, nrow=1, dimnames
... In addition, the following may also be informative.
> f <- paste("day", 1:3)
> contrasts(ordered(f))
.L .Q
[1,] -7.071068e-01 0.4082483
[2,] -7.850462e-17 -0.8164966
[3,] 7.071068e-01 0.4082483
> contrasts(factor(f))
day 2 day 3
day 1 0 0
day 2 1
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