This seems a little impractical to me. People respond so much at random
and most only tackle questions with which they feel comfortable. As
it's not a competition in any sense, it's going to be hard to rank
people in any effective way. But suppose you succeed in doing so, then
what?
To me a muc
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of alfieim29
Sent: Saturday, 1 December 2007 9:26 AM
To: r-help@r-project.org
Subject: [R] Quantiles and QQ plots
I have 20 variables:
5,9,6,1,5,9,7,4,5,6,3,2,4,8,9,6,1,8,4,8
[WNV] I think you have 20 val
Well, you can do it:
assign(paste("myname", name[1], sep = "."), seq(1:10))
but this looks clumsy to me, as if you were trying to make R do
something the same way that you do things in some other system. This is
nearly always a bad idea. My advice is to learn how to do things the R
way.
Bill V
Here's how to generate a million
psample <- rpois(100, lambda = 20)
so guess how you would generate one.
The main point is that if you really want to generate more than one, do
it all at once like the above, do not do it in a loop.
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4
Does this do it for you?
> x <- read.table(textConnection(" id dayk
+ 56 -1 566
+ 63 -1 680
+ 73 -1 773
+ 56 2 298
+ 63 2 273"), header=TRUE)
> x$percent <- ave(x$k, list(x$id), FUN=function(.data){
+ (.data - .data[1]) / .data[1] * 100
+ })
>
> x
id day k percent
1
>From: Bernd Jagla <[EMAIL PROTECTED]>
>Date: 2007/11/30 Fri PM 09:24:34 CST
>To: r-help@r-project.org
>Subject: [R] compare strings
below assumes that there are only 2 columns in
the dataframe or that the string columns are the
first two columns ?
which(df[,1] == df[,2])
>Sorry for the questi
It helps writing down these question, you are then getting much closer to an
answer...
summary(as.integer(t3[,2]) == as.integer(t3[,4]) & as.integer(t3[,3]) ==
as.integer(t3[,5]))
will compare two pairs of column pairs and give a count of flase and true
rows...
-B
|-Original Message-
|F
Just found out that I am actually dealing with factors:
> t3[1,5] == t3[1,3]
Error in Ops.factor(t3[1, 5], t3[1, 3]) : level sets of factors are
different
But the problem remains... how can I compare them?
Thanks,
Bernd
|-Original Message-
|From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECT
Sorry for the question, but I really cannot find the right search terms to
find an answer..
I have a data frame with strings in some of the columns.
I want to know all the rows where the strings in both columns are equal.
How do I do this?
Thanks,
Bernd
[[alternative HTML
Your vector:
x=c(5,9,6,1,5,9,7,4,5,6,3,2,4,8,9,6,1,8,4,8)
Check descriptives:
mean(x)
sd(x)
get quantiles for a normal with mean and standard deviation of x:
qnorm(c(0.25,0.5,0.75),mean=mean(x),sd=sd(x))
Output:
[1] 3.76997 5.5 7.23003
to get the quantile plot install library "car" and
Assuming that there is a -1 day for every day
in the data frame and all the -1 days are at
the beginning, try this where k[match(id, id)]
is the vector of corresponding day -1 k-values:
DF <- data.frame(id = c(56, 63, 73, 56, 63),
day = c(-1, -1, -1, 2, 2),
k = c(566, 680, 773, 298
Sorry about that, it was sent by accident.
I have a data frame that looks something like this:
id dayk
56 -1 566
63 -1 680
73 -1 773
56 2 298
63 2 273
Of course, it is a very simplified version of the
real data frame I am working on. I need to add another
colum
I was just using colorRampPalette to generate a range of color. You might try:
Lab.palette <- colorRampPalette(c("dark red", "yellow", "dark green"))
to see if this is closer. You can use the palette in other plot
functions; this was just the easiest to show what was happening.
On Nov 30, 200
Jim,
This is not right as I think. The color code generated did not
show a grading pattern. For example, it should go from
very red---red---less red---darkgreen---very green coinciding
with the descending order of values, just like the very left panel
shown in
http://www.bme.unc.edu/research/B
Hi,
I have a data frame that looks something like this:
id dayk
656 -1 566
663 -1 680
673 -1 773
675 -1 761
704 -1 685
714 -1 636
717 -1 421
719 -1 645
727 -1 761
731 -1 663
11 735 -1 603
12 738 -1 865
13 742 -1 594
14 744 -1
Try this. This will plot a similar scale for data.
require(lattice)
# generate some data
x <- matrix(sample(0:8,100,replace=TRUE),10) # create 9 value
# create a color palette for outpu
Lab.palette <- colorRampPalette(c("red", 'yellow','green','blue'))
levelplot(x,col.regions=Lab.palette(10))
O
Hi Jim,
Thanks but I imagine the transformed data would
not actually represent the pattern from the original
data. Is there other alternatives? Ideally, I would
like to have the color bar looking like the one
in http://www.bme.unc.edu/research/Bioinformatics.FunctionalGenomics.html
Best,
All
If you are just going to have 3 values for coloring, then you could
just convert your data into 1, 2, 3:
new.data <- ifelse(old.data > 2, 3, ifelse(old.data < 2, 1, 2))
dim(new.data) <- dim(old.data)
and then plot it.
On Nov 30, 2007 5:09 PM, affy snp <[EMAIL PROTECTED]> wrote:
> Hi list,
> My d
It was not my intention to start any kind of war.
I do want to thank Katharine Mullen for sharing to link to the review in R
News. It helps me understand why there might be some controversy over my
positive experience with The R Book.
As I mentioned in an earlier post, I'm working my through th
I have 20 variables:
5,9,6,1,5,9,7,4,5,6,3,2,4,8,9,6,1,8,4,8
How do I calculate the corresponding quantiles from a normal distribution
with the same mean and variance as the sample?
Also, how do I draw a QQ plot of the data?
Thanks for any help!
--
View this message in context:
http://www.na
gutierrez.m05 wrote:
>
> I am interested in using the nonlinear forecasting techniques developed
> by Sugihara et al. In particular the simplex projection and the S-Map
> (see this website for details and reprints: http://iod.ucsd.edu/simplex/
> ).
> I've looked through CRAN but could not fin
Hi list,
My data set is comprised of 47 columns and about 700 rows.
Most of the values would be around 2, while some will go beyond
in either direction, higher or lower. Is there a way to specify the
parameter of col or others if necessary to have the range of representing
colors be 08, and mak
Also just MHO, but I think it is a good idea, if for no other reason
than it creates an additional incentive for people to respond and to
do so in a thoughtful way. I think the rating systems on some other
boards, and those used at commercial sites such as Amazon, testify to
the fact that people of
Maybe I haven't been on the list long enough, but how bad is the R
advice on this list? Is a rating system even necessary?
As Mark Leeds pointed out, there is the issue that what a beginner
thinks is good advice isn't what an expert thinks in good advice. There
is also the converse, expert advi
Pedro Emmanuel Alvarenga Americano do Brasil a écrit :
> Hello friends of R list,
>
> Im a physician and Im not that good in statistics. I have posted similar
> email in the epi-sig list but no one aswered so far. Im cunducting a
> systematic review on two diagnostic test for a particular tropical
>From: "Doran, Harold" <[EMAIL PROTECTED]>
>Date: 2007/11/30 Fri PM 02:12:36 CST
>To: R Help <[EMAIL PROTECTED]>
>Subject: [R] Rating R Helpers
Of course, it's just MHO, but I think your suggestion is more work than
neccesary because it becomes pretty obvious who the R-experts are once you've
be
alfieim29 wrote:
> How do I generate a value in R from a poisson distribution with mean 20?
Just type ?Poisson ...
Best,
Uwe Ligges
> Thanks!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the post
A nice model might be http://www.workingwithrails.com/.
It also suggests other possible measures of authority:
* member of R core
* packages written by/contributed to
* R conference attended/presented at
* contributer to wiki
Hadley
On 11/30/07, Doran, Harold <[EMAIL PROTECTED]> wrote:
> Si
On 11/30/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Thanks,
>
> I'll finaly figure it out. It was not very intuitive though, even looking
> at the help file. The final code, for whom it might interest is:
>
> xyplot(Volume ~ Age|Farm,
> groups = Treatment,
>
On 11/30/07, Uwe Ligges <[EMAIL PROTECTED]> wrote:
>
>
> Tom La Bone wrote:
> > IMHO "The R Book" is far better than indicated in that review and should be
> > near the top of the list for beginners looking for a "manual" for R.
>
> We can start a big war about that book now, but we should discuss
Doran, Harold wrote:
> Since R is open source and help may come from varied levels of
> experience on R-Help, I wonder if it might be helpful to construct a
> method that can be used to "rate" those who provide help on this list.
>
> This is something that is done on other comp lists, like
> http:
Since R is open source and help may come from varied levels of
experience on R-Help, I wonder if it might be helpful to construct a
method that can be used to "rate" those who provide help on this list.
This is something that is done on other comp lists, like
http://www.experts-exchange.com/.
I t
Ettinger, Nicholas wrote:
> Hello all,
>
> For the not-so-deeply-mathematically/statistically-literate, what is the
> simplest command with which to choose three "random" numbers from the
> integer list 1 up to 9?
?sample
as in
> sample(1:9, 3) # your result may vary!
[1] 4 7 1
or, if you nee
See ?sample
Uwe Ligges
Ettinger, Nicholas wrote:
> Hello all,
>
> For the not-so-deeply-mathematically/statistically-literate, what is the
> simplest command with which to choose three "random" numbers from the
> integer list 1 up to 9?
>
> After reading the ".Random.seed" help page along with
x <- 1:9
sample(x,3, replace=TRUE)
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Ettinger, Nicholas
> Sent: Friday, November 30, 2007 2:48 PM
> To: r-help@r-project.org
> Subject: [R] How to choose 3 random numbers from 1 up to 9?
>
> Hello all,
Thanks,
I'll finaly figure it out. It was not very intuitive though, even looking
at the help file. The final code, for whom it might interest is:
xyplot(Volume ~ Age|Farm,
groups = Treatment,
data = my_data,
panel= panel
Tom La Bone wrote:
> IMHO "The R Book" is far better than indicated in that review and should be
> near the top of the list for beginners looking for a "manual" for R.
We can start a big war about that book now, but we should discuss useful
things instead. I am currently reviewing this book for
Hello all,
For the not-so-deeply-mathematically/statistically-literate, what is the
simplest command with which to choose three "random" numbers from the
integer list 1 up to 9?
After reading the ".Random.seed" help page along with several others, it
seems like the one simple choice might be:
ro
help.search('poisson')
On Nov 30, 2007 12:46 PM, alfieim29 <[EMAIL PROTECTED]> wrote:
>
> How do I generate a value in R from a poisson distribution with mean 20?
>
> Thanks!
> --
> View this message in context:
> http://www.nabble.com/Generating-a-value-tf4922234.html#a14086120
> Sent from the R
Coming to this late, but hopefully not too late...
You may want to try mixture of regression models:
install.packages("flexmix")
require("flexmix")
## simulate some data
x1 <- rnorm(100, sd=5)
y1 <- rnorm(100, mean=x1)
x2 <- rnorm(50, sd=5)
y2 <- rnorm(50, mean=-5 + 0.5 * x2)
x <- c(x1, x2)
y <-
On Thu, 29 Nov 2007, stathelp wrote:
>
> I have this maximum liklihood estimate problem
>
> i need to find the roots of the following:
Dear 'stathelp',
Did you follow the Posting Guide?
It would suggest that you issue something like
help.search("roots")
or
help.search("root")
Thanks Peter!
On 11/30/07, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
>
> tom soyer wrote:
> > Hi,
> >
> > I am trying to find a function in R that would calculate the critical
> value
> > of t for a given probability and df. For example, if p=5% and df=20,
> then
> > does R have a function to calc
tom soyer wrote:
> Hi,
>
> I am trying to find a function in R that would calculate the critical value
> of t for a given probability and df. For example, if p=5% and df=20, then
> does R have a function to calculate the t value? I tried:
>
>
>> dt(0.05,df=20)
>>
> [1] 0.3934718
>
> That do
Hi,
I am trying to find a function in R that would calculate the critical value
of t for a given probability and df. For example, if p=5% and df=20, then
does R have a function to calculate the t value? I tried:
> dt(0.05,df=20)
[1] 0.3934718
That doesn't look right to me. Could someone tell me
That worked! Many thanks (to all of you)
Take care
Jon
On Nov 30, 2007 10:50 AM, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> Jonathan Weeks wrote:
> > The file is attached.
> >
> > Also, to restate what I said earlier. The package passes the checks
> > without any errors or warnings under R 2.5
Jonathan Weeks wrote:
> The file is attached.
>
> Also, to restate what I said earlier. The package passes the checks
> without any errors or warnings under R 2.5, but now that it is running
> under 2.6.1 I get a warning (no errors) about having .Rd files without
> associated .R files (the datase
How do I generate a value in R from a poisson distribution with mean 20?
Thanks!
--
View this message in context:
http://www.nabble.com/Generating-a-value-tf4922234.html#a14086120
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-p
As with many things, I suspect that someone has created plots of this
sort, but no-one seems to have contributed a function to do so. If
you find/create one, send it to me and I'll be glad to include it in
either gmodes or gplots as appropriate.
One place to check before writing one is Fra
The file is attached.
Also, to restate what I said earlier. The package passes the checks without
any errors or warnings under R 2.5, but now that it is running under 2.6.1 I
get a warning (no errors) about having .Rd files without associated .R files
(the datasets are in an .RData file in the dat
Read ?panel.superpose carefully and pay attention to the comments about the
group.number argument passed to panel.groups. You can test the value of this
argument and adjust the plot accordingly.
Bert Gunter
Genentech Nonclinical Statistics
7-7374
-Original Message-
From: [EMAIL PROTECTE
> "JW" == Jonathan Weeks <[EMAIL PROTECTED]>
> on Fri, 30 Nov 2007 08:46:50 -0700 writes:
JW> Thank you both for your help. I still don't know what is happening. I
JW> uploaded my package (plink_0.0-3) to ftp://cran.r-project.org/incoming
That's a place NOBODY can get it - but
I am not sure I understand what you mean by
programatically but does this example help at all?
aa <- 1:10
plot(aa)
mmean <- 5
sdd <- .25
text(c(2,3),6, labels=c( "mean", "sd"))
text(c(2,3),5.5, labels=c(mmean, sdd))
--- Felipe Carrillo <[EMAIL PROTECTED]> wrote:
> Hello all:
> Is it possible t
There probably is a better way to do this but I'd
suggest getting an overall total and then splitting
the data.frame into it's 3 component parts then apply
the equations below to get the sums and the number of
missing values for each subject over all four
data.frames.
===
Dear readers,
I need to prevent the output of very large
integers in scientific format.
So I use format, but now the integers in the
resulting tab delimited output file are padded.
I tried justify="none" or "left", width=0 but nothing
helps.
1\tA\t1\t1
2\ta\t1\t1
^^^
thank y
Hi everyone,
I am trying to change some of the default colors in a
multipanel-multigroup lattice graphic, but when I change the panel color
to white, the only group that is plotted is the last one, as panel.fill
overwrites all the previous ones. Is there any way to set the panel
background color
Luis Orlindo Tedeschi wrote:
> Folks; I have looked for this subject extensively using all the resources I
> have available and could not find a solution; therefore I am sending you this
> email seeking for help. I need to solve a discontinuous nonlinear function,
> but R 2.6 keeps giving me an
Hi
I am unable to impute using the MICE command in R when imputing
a binary variable using linear discriminant analysis. To illustrate my
problem I have created a dataset, which consists of 1 continuous and 1
binary variable. The continuous variable is complete and the binary
variable is partia
I have this maximum liklihood estimate problem
i need to find the roots of the following:
[sum (from i=1 to n) ] ((2(x[i]-parameter)/(1+(x[i]-parameter)^2))=0
given to me is the x vector which has length 100
how would I find the roots using R?
I have 2 thoughts.. 1 is using a grid search
Hi everybody,
I've been looking for a function that combines all variables from a
dataset because I need to do multivariate regression.
If we have linear regression with an expression like f(x) = a0 +
sum(ai*xi) what I want to do is something like f(x) = a0 + sum(ai*xi) +
sum(sum(bij * xi * xj
Hello all,
I'm attempting to fit a generalized linear mixed-effects model using lmer
(R v 2.6.0, lmer 0.99875-9, Mac OS X 10.4.10) using the call:
vidusLMER1 <- lmer(jail ~ visit + gender + house + cokefreq + cracfreq +
herofreq + borcur + comc + (1 | code), data = vidusGD, family = binomial,
co
Not exactly sure what name you want, but try:
assign(paste("myname", name[1]), seq(1:10))
On Nov 30, 2007 11:12 AM, livia <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I have got a vector, for example seq(1:10), and I have an array "name"
> defined before which contain the vector name I would like t
Hello,
I have got a vector, for example seq(1:10), and I have an array "name"
defined before which contain the vector name I would like to use. So
name[1]= "price". I would like to name the vector in the following codes:
paste("myname", name[1]) <- seq(1:10)
But it does not work. Could anyone g
?get
get(code[1])
On Nov 30, 2007 7:26 AM, livia <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I define a dataframe named "Price" first and there are other dataframes as
> well. Then I built an array named "code" containing all the dataframe name.
> So If I call code[1], it is "Price".
>
> Basicly, I
Try this:
> a <- 1
> b <- 2
> mat <- matrix(c("a", "a*b", "a+b", "b"), 2)
> mat
[,1] [,2]
[1,] "a" "a+b"
[2,] "a*b" "b"
> matrix(sapply(mat, function(x) eval(parse(text = x))), nrow(mat))
[,1] [,2]
[1,]13
[2,]22
On Nov 30, 2007 10:06 AM, BOISSON, Pascal <[EMAIL PROTE
Thank you both for your help. I still don't know what is happening. I
uploaded my package (plink_0.0-3) to ftp://cran.r-project.org/incoming
If you have any other suggestions they would be greatly appreciated
Take care
Jon
On Nov 29, 2007 10:30 AM, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> P
Try the reshape package
This should do what you want.
library(reshape)
df1 <- melt(mydata, id=c("ID"),
measured=c("a1","a2","a3")) ; df1
stats <- cast(df1, ID ~ variable , mean) ; stats
=
--- affy
Try:
plot(Sepal.Length ~ Sepal.Width, iris)
legend("topleft", capture.output(summary(iris[1:2])), cex = 0.7)
On Nov 30, 2007 10:16 AM, Felipe Carrillo <[EMAIL PROTECTED]> wrote:
> Hello all:
> Is it possible to make annotations such as the
> summary(sd,mean,min,max)programmatically on the graph
Hello all:
Is it possible to make annotations such as the
summary(sd,mean,min,max)programmatically on the graph window?
Felipe D. Carrillo
Fishery Biologist
US Fish & Wildlife Service
California, USA
_
Hello all,
I posted a similar question recently but I suspect that it was not well
enough formulated to trigger any answers. So I try again:
Is there a way to produce boxplots (or something similar) that uses the
estimated fixed effects of an lme{nlme} object? When I want to know the
mean, I use
Dear R_Users,
I am confronted to a technical issue that I wouldn't even know how to start to
handle and where to look at. It is about data manipulation in a dataframe and
value substitution
Imagine I have a matrix describing a set of chemical equations.
Colnames are Components names and Rowname
Vectorize f using ifelse in place of if:
> f <- function (t,k,l,u) {
+ r0 <- 346.1
+ x<- t-l
+ ifelse(x < 0, r0, (r0-u)*exp(-k*x)+u)
+ }
>
> prof1 = data.frame(
+ t = c(0, 3, 6, 12, 18, 24, 36, 48, 72, 96),
+ y1 = c(346.12, 343.13, 327.44, 280.95, 242.76, 228.67,
+193.48, 174.29, 164.81, 1
Hello,
I define a dataframe named "Price" first and there are other dataframes as
well. Then I built an array named "code" containing all the dataframe name.
So If I call code[1], it is "Price".
Basicly, I would like to call the datafram with reference to the
array"code".e,g, If I want to call
Folks; I have looked for this subject extensively using all the resources I
have available and could not find a solution; therefore I am sending you this
email seeking for help. I need to solve a discontinuous nonlinear function, but
R 2.6 keeps giving me an error. Below is a simple example:
f
Thanks, Katharine!
Your suggestion is probably simple enough. However,
my modification below is needed to do what is required
(yours labels the tick-marks at 200,400,600,800,1000;
mine does it at 0,200,400,600,800,1000).
Thanks, and best wishes,
Ted.
On 30-Nov-07 13:19:32, Katharine Mullen wrote:
On Sun, 25 Nov 2007, Prof Brian Ripley wrote:
> There is a reproducible example in the MASS survival chapter. I've
> reported this to the survival maintainer more than once: it's a partial
> matching problem.
Some diging shows that this was fixed in survival 2.34 (but not in the SVN
for surviva
something like this sounds like what you want:
labs<-NA
x<-200
for(i in 1:5){ labs<-append(labs, c(rep(NA,3),x)); x<-x+200 }
plot(1:900, xaxt="n", xlim=c(0, 1000))
axis(side = 1, at = seq(0,1000,by=50), labels=labs)
On Fri, 30 Nov 2007 [EMAIL PROTECTED] wrote:
> Hi Folks,
> I'm advising someon
Hello friends of R list,
Im a physician and Im not that good in statistics. I have posted similar
email in the epi-sig list but no one aswered so far. Im cunducting a
systematic review on two diagnostic test for a particular tropical disease.
I found a free software that make most of the analysis
Ted,
I don't think you can include the axis ticks in plot(), but you can
specify them in axis() using
plot(x,y,xlim=c(0,1000), axes=F)
box()
axis(1,at=seq(0,1000,50),labels=F)
axis(1,at=seq(0,1000,200),labels=T)
Obviously, if you want this to be variable for different xlim etc, your
simplest ans
Hi everybody!
I have a problem when use the StatET plug-in of Eclipse.
When I use run plot() , it only output a ps file in working directory
below is the code(between #)
#
> x <- rbinom(2000,1000,0.25)
x <- rbinom(2000,1000,0.25)
mean(x)
>
Hi Folks,
I'm advising someone who's a beginner with R,
and therefore wants the simplest answer possible.
The issue is to produce a plot using
plot(x,y,...)
where, on the X-axis, the tick-marks should be
on the lines of:
-- Range of X-axis: 0:1000
-- tick-marks labelled "0","200",...,"800","1
1. rollapply is for creating zoo objects from zoo objects --
not for creating lists from zoo objects. 2. The
subscripting is wrong. 3. Its possible that you need a different
representation for out but that depends on what you want to
do with it. At any rate try this and note that there is no
zoo
[EMAIL PROTECTED] wrote:
>
> Hello R Fundi,
>
> The poetic title of this post is a request for help in regard to a
> "simple" plotting question.
> I have displayed the mean observations of a series variables using
> barplot(). On the same figure I have colored the bars to represent the
>
You should RODBC library. It has some functions to fetch data from excel.
You can use the following code
>channel<-odbcConnectExcel("filename.xls") # replace filename with the name
of ur file
>xldata<-sqlFetch(channel,"excel sheet name", colnames = T)
hope this helps
Regards,
Rana Shoaaib Mehm
Dear R users.
I have zoo object "size_june" containing market-capital values:
> dim(size_june) # market-cap data of 625 firms for 20 years
[1] 20 625
> class(size_june)
[1] "zoo"
> size_june # colnames = "size.firmcode"
size.34020 size.4710 size.11050 size.10660 size.9540 size.8060
hi there!
i am searching for a possibility to plot the cluster fusion number against the
distance in hierarchical clustering. i used the following code:
distances = dist(my data[, c(1:11)],
method="euclidean")
eward = hclust(distances, method="ward")
plot(eward, labels=my data[, 12], hang=0,cex
On Tue, 2007-11-20 at 16:39 +0100, Irene Mantzouni wrote:
> Hi all!
>
> Is it possible to model a multiple regression in which the response becomes
> zero when one of the two covariates is zero?
> lm(y~ x1+x2) and y=0 if x1=0.
> However, when x1=0, y=x2+1(intercept).
> Does this mean I cann
I am interested in using the nonlinear forecasting techniques developed
by Sugihara et al. In particular the simplex projection and the S-Map
(see this website for details and reprints: http://iod.ucsd.edu/simplex/ ).
I've looked through CRAN but could not find any package with functions
that al
Howdy Gurus
I am try to compute envelope of Khat from simulations of complete spatial
randomness using R package of "Splancs."
I got the following error:
> UL.khat <- Kenv.csr(length(X_coord), bnd, nsim=100, s)
Doing simulation 1
Error in runif(n, min, max) : invalid arguments
In addition: War
Hello,
I am hoping for some advice as to how I can prorate a number of scale
items that comprise a score. At least 69 of 159 cases have at least 1
value missing (65 cases have H7 missing). The maximum number of missing is
5.
I want to compute a total score, a score for the H items, the R items an
Dear all
I'm sorry, I provide us more information. One example of dataset is:
classage DeathsAtRisk
1 1 20-34 1 100
2 2 20-34 2 98
3 3 20-34 2 250
4 4 20-34 1 90
> > Are you interested in equality constraints or inequality constraints?
>
> No, I am interested in 2 kinds of inequality constraints:
> 1) monotonic splines
> 2) positive coefficients of the variables, which are not splines.
> It seems that pcls should be able to deal with both of them, if smoot
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