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CCd to r-devel as suggested by Peter.
On 09/02/12 19:28, Hadley Wickham wrote:
>>> I'm exploring using a version control system to keep better
>>> track of changes to the packages I maintain. I'm leaning
>>> towards git (although mercurial also looks
On Feb 10, 2012, at 04:20 , sezgin ozcan wrote:
> I use mac.
> I tried this command also
> a<-read.table("clipboard",sep=”\t”,row.names=1,header=T)
Something is wrong with those quotes around \t...
Also, last I needed this, using file="clipboard" to read from the clipboard
didn't work on Mac,
Actually, is there any way to get at additional information beyond the
classProbs? In particular, is there any way to find out the
associated weights, or otherwise the row indices into the original
model matrix corresponding to the tested instances?
On Thu, Feb 9, 2012 at 4:37 PM, Yang Zhang wro
On Feb 10, 2012, at 06:08 , R. Michael Weylandt wrote:
> Add -1 to your glm to remove the intercept term -- that will force
> Friday to have its own coefficient.
That's one answer.
Another one is that the coefficient (in the model with an intercept term) for
Friday is zero with an s.e. of ze
Is it possible to use ROCR to plot a simple recall@p plot? I.e., a
plot where the x-axis is the position into the ranked test set, and
the y-axis is the recall, so you can see what's the recall in the top
10% of the ranked results.
I searched through the performance() manual but found nothing.
(
Hi Aidan,
str is your friend:
> str(g)
'data.frame': 9 obs. of 3 variables:
$ Date: chr "2011-12-23" "2011-12-30" "2012-01-06" "2011-12-23" ...
$ variable: Factor w/ 3 levels "Price","Yield",..: 1 1 1 2 2 2 3 3 3
$ value : num 86.78 86.04 86.44 9.74 9.54 ...
You haven't turned the
On Thu, Jan 12, 2012 at 5:27 PM, Darran King wrote:
> Hi all
>
> New to R and GGplot2 but loving the potential. I am trying to plot four
> separate point plots by looping over the data and plotting a different
> subset each time.
>
> When I plot the data as a point plot, the size of the points is
Hi Mario,
If you're still having problems, I'd suggest sending a small
reproducible example
(https://github.com/hadley/devtools/wiki/Reproducibility) to the
ggplot2 mailing list.
Hadley
On Tue, Jan 17, 2012 at 12:32 PM, Mario Giesel wrote:
> Hello, R-List,
> I'm getting error messages when addi
Hi Simon,
You might want to try sending a small reproducible example
(https://github.com/hadley/devtools/wiki/Reproducibility) to the
ggplot2 mailing list.
Hadley
On Tue, Jan 31, 2012 at 10:53 PM, sjlabrie wrote:
> Hi,
>
> I am looking for a way to plot bar on a map instead of the standard poin
Hi Raimund,
To increase your chances of getting help, I'd recommend using the
ggplot2 mailing list, and reducing your example down to the essence of
the problem. For example, the theme components don't affect the
problem, but make the code longer, and so harder to understand.
Hadley
On Mon, Feb
Dear list,
Please excuse my ignorance, but I'm trying to model some data using the lme
package. vot is a numeric response, and condition, location and obs are all
categories.
This works:
> anova(vot.lme <- lme(vot ~ condition * location *
obs,data=mergedCodesL,random= ~1 |patient))
I think you need to read the man pages and the four vignettes. A lot
of your questions have answers there.
If you don't specify the resampling indices, they ones generated for
you are saved in the train object:
> data(iris)
> TrainData <- iris[,1:4]
> TrainClasses <- iris[,5]
>
> knnFit1 <- train
Dear All
This is the first time I use this help forum. My apologies in case I did
not follow a predefined format.
My question:
I use the package "hwriter".
How can I produce a matrix (html) that allows scrolling (up and down,
left and right) or, alternatively, prduces a fixed row for the hea
Melissa,
par(new=T) works as many times as you use it. You don't provide data,
but (assuming it is not NULL) more likely your n=500 qqplot was just
obscuring the points of the n=50 plot.
Reverse the order (i.e. qqplot 500 first, 50, 5 last) and see if all
three are there (as there are more 500 you
Thanks Hadley, of course I should have spotted that.
On Fri, Feb 10, 2012 at 12:46 PM, Hadley Wickham wrote:
> Hi Aidan,
>
> str is your friend:
>
>> str(g)
> 'data.frame': 9 obs. of 3 variables:
> $ Date : chr "2011-12-23" "2011-12-30" "2012-01-06" "2011-12-23" ...
> $ variable: Factor
A couple of clarifications for you.
1. I write mixed effects Cox models as exp(X beta + Z b), beta = fixed
effects coefficients and b = random effects coefficients. I'm using
notation that is common in linear mixed effects models (on purpose).
About 2/3 of the papers use exp(X beta)* c, i.e.,
hi,
I have a vector as follow:
myVec = c(1,4,10,30)
How can I get a vector that show the interval of the numbers.
I need to get this result:
distance
>3,6,20
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I have coded a time series from simulated data:
simtimeseries <- arima.sim(n=1024,list(order=c(4,0,0),ar=c(2.7607, -3.8106,
2.6535, -0.9258),sd=sqrt(1)))
#show roots are outside unit circle
plot.ts(simtimeseries, xlab="", ylab="", main="Time Series of Simulated Data")
# Yule
--
Have a look at function diff(), e.g.,
diff(c(1,4,10,30))
I hope it helps.
Best,
Dimitris
On 2/10/2012 1:33 PM, alend wrote:
hi,
I have a vector as follow:
myVec = c(1,4,10,30)
How can I get a vector that show the interval of the numbers.
I need to get this result:
distance
I was able to make a scatterplot but ...
1) what does the "86" mean? The "86" shows up on the graph as well.
> scatterplot (Shells/TotalEggs ~ Sector, data = data.to.analyze)
[1] "86"
2) Also how do you change the Y axis title? I don't want it to read
Shells/TotalEggs, instead I would like it t
Hi everybody,
I'm looking for an optimal way to split a big matrix (e.g. ncol = 8,
nrow=8) into small square submatrices (e.g. ncol=2, nrow=2)
For example
If I have
> h
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]19 17 25 33 41 49 57
[2,]2 10 18 26 34
Dear All,
I have questions about the function "rpart" to construct a regression tree
in
R code.
My problem is how to change the splitting criteria.
In the "rpart" we have : parms=list(split="..") , I ask you if in this
command is it possible to use an another splitting criterion to substitute
i made jDialog through JGR package(Java gui 4 R),
i want to export it with imported external package e.g.bio3d,RCOR , into
.exe
or if not possible convert code to executable jar file
how should i go, give me package name,example codes (if possible)
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Hi all,
I want to use graph.maxflow function for my work. Given a network, I could
get a result, but I could not figure out how the function get the result.
Did anyone have the same problem before? How did you resolve it? Does anyone
know any other functions/programs for max-flow analysis.
Thank
Dear R users,
I'm having trouble with calculating pvalues for my 2d dataset. First I
performed clustering and I would like to get some info about the strength
of cluster membership for each point. I've calculated (thanks to nice
people help) the multivariate normal densities (mnd) using dmvnorm fu
Hello to everyone.
I'm fitting a glm.nb model to a count data. I'm using about 8 predictive
variables.
Once a run the script I do get a result but it tells me that the iteration
limit has been reached.
So, can i trust the results given by the model?
Could it be a multicollinearity problem?
Thank
Hi all,
I have a large dataset with ~8600 observations that I want to compress to
weekly means. There are 9 variables (columns), and I have already added a
"week" column with 51 weeks. I have been looking at the functions:
aggregate, tapply, apply, etc. and I am just not savvy enough with R to
fi
Hi all,
I am trying to fill a 904x904x904 array, but at some point of the loop R
states that the 5.5Gb sized vector is too big to allocate. I have looked at
packages such as "bigmemory", but I need help to decide which is the best
way to store such an object. It would be perfect to store it in thi
When I tried to run svm on the same data frame, memory usage as reported
by top(1) doubled to 4GB almost right away and the function never
returned (has been running for ~15 hours now). ^C does not stop it.
This is most unusual, libsvm has always seemed very fast.
This is R version 2.13.1 (2011-07
such as a=[1,2,3 ],b =[2,4] I want to get a new one [1 2 3 4 5]. Thank you.
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How could you combine a and b as given to get your "new one"? It
doesn't have the same elements...
?union
?intersect
?rbind
?c
Michael
On Fri, Feb 10, 2012 at 9:55 AM, summer wrote:
> such as a=[1,2,3 ],b =[2,4] I want to get a new one [1 2 3 4 5]. Thank you.
>
> --
> View this message in conte
On Fri, Feb 10, 2012 at 12:02:25AM -0800, xiddw wrote:
> Hi everybody,
>
> I'm looking for an optimal way to split a big matrix (e.g. ncol = 8,
> nrow=8) into small square submatrices (e.g. ncol=2, nrow=2)
>
> For example
>
> If I have
>
> > h
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,
Hi guys,
AFAICT, this conversation now discusses some domain-specific issues.
However, in reply to OP and OQ, I'll shamelessly advertise the package that
Gergely Daróczi and I have released recently. It's named "rapport", and you
can grab it from CRAN or GitHub (https://github.com/aL3xa/rapport/).
On Fri, Feb 10, 2012 at 3:02 AM, xiddw wrote:
> Hi everybody,
>
> I'm looking for an optimal way to split a big matrix (e.g. ncol = 8,
> nrow=8) into small square submatrices (e.g. ncol=2, nrow=2)
>
> For example
>
> If I have
>
>> h
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,] 1
>> To be strictly correct, shouldn't that be:
>>
>> formula<- eval(substitute( value*v*LEFT ~ RIGHT, list(LEFT=LEFT,
>> RIGHT=RIGHT)))
>>
>> ?
>
>
> I think it probably doesn't matter. The difference is that mine gives a
> pure language object, whereas yours gives a formula object. The formula
>
On Feb 10, 2012, at 7:14 AM, Lucas wrote:
Hello to everyone.
I'm fitting a glm.nb model to a count data. I'm using about 8
predictive
variables.
Once a run the script I do get a result but it tells me that the
iteration
limit has been reached.
Most regression functions have a control pa
Dear J.,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jhope
> Sent: February-10-12 3:44 AM
> To: r-help@r-project.org
> Subject: [R] Q - scatterplots
>
> I was able to make a scatterplot but ...
>
> 1) what does the "86"
> * Sam Steingold [2012-02-10 10:01:54 -0500]:
>
> When I tried to run svm on the same data frame, memory usage as reported
> by top(1) doubled to 4GB almost right away and the function never
> returned (has been running for ~15 hours now). ^C does not stop it.
> This is most unusual, libsvm has a
First, make sure the file name is correct.
Second, try
list.files("Users:/sezginozcan/Downloads/")
to see if the file is listed and you are looking in the correct folder.
Third, you may be able to locate and open the file using
A <- read.table(file.choose(), sep="\t", row.names=1, header=T)
-- Forwarded message --
From: Gyanendra Pokharel
Date: Fri, Feb 10, 2012 at 11:13 AM
Subject: Re: [R] Importing a CSV file
To: sezgin ozcan
You save your file in your own folder either in desktop or in document
where ever you want and serach the file using the command
mydatafil
A little more on topic is that there are tools for R to nicely format output.
The simplest is probably the R2wd package which will transfer objects
and text to word, you can send a matrix or data frame to word as a
nicely formatted table or wdVerbatim will send output to word but make
sure that th
Hello,
I've encountered a very weird issue with the method subset(), or maybe this
is something I don't know about said method that when you're subsetting
based on the columns of a data frame you can only use constants (0.1, 2.3,
2.2) instead of variables?
Here's a look at my data frame called 'e
On Fri, Feb 10, 2012 at 08:15:39AM -0800, vaneet wrote:
> Hello,
>
> I've encountered a very weird issue with the method subset(), or maybe this
> is something I don't know about said method that when you're subsetting
> based on the columns of a data frame you can only use constants (0.1, 2.3,
>
This is likely a representation issue, as in R FAQ 7.31.
?"==" suggests that using identical and all.equal is a better strategy:
x1 <- 0.5 - 0.3
x2 <- 0.3 - 0.1
x1 == x2 # FALSE on most machines
identical(all.equal(x1, x2), TRUE) # TRUE everywhere
Sa
If you want to pause for the person to look at a plot before going on
to the next plot then just do:
> par(ask=TRUE)
This will actually allow your loop to continue with calculations while
the user looks at the plot but will pause before drawing the next plot
(hitting enter in the command window o
Thanks guys, both those solutions work. I really appreciate the help!
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__
In your second example the boot function will still generate its
random indicies before your internal function calls sample, so the
seed will be different when you call sample from what you originally
set it to. If you really want to know what the boot function does,
look at its code (it does some
I have started using the circular package but it is not recognizing the
function arrows.circular. I attempted to use the example provided in the
circular manual. Here is the example code using the circular package:
plot(rvonmises(10, circular(0), kappa=1))
arrows.circular(rvonmises(10, cir
Your sessionInfo() would be helpful. Also, just to check: you did do
library(circular)
right?
Sarah
On Fri, Feb 10, 2012 at 11:55 AM, Chosid, David (MISC)
wrote:
> I have started using the circular package but it is not recognizing the
> function arrows.circular. I attempted to use the example
Hi David,
You need to load the package before running your code:
# install.packages('circular')
require(circular)
plot(rvonmises(10, circular(0), kappa=1))
arrows.circular(rvonmises(10, circular(0), kappa=1))
arrows.circular(rvonmises(10, circular(0), kappa=1), y=runif(10), col=2)
arrows.circ
Your script is rather inefficient with spurious cbind calls. Any
particular reason not to use
?ar directly ?
Call:
ar.yw.default(x = simtimeseries, order.max = 4)
Coefficients:
1234
1.9440 -1.9529 0.8450 -0.2154
Order selected 4 sigma^2 estimated as 15.29
To
Yes, sorry to not be clear. library(circular) was run.
From: Jorge I Velez [mailto:jorgeivanve...@gmail.com]
Sent: Friday, February 10, 2012 12:03 PM
To: Chosid, David (FWE)
Cc: r-help@r-project.org
Subject: Re: [R] function arrows.circular not working
Hi David,
> -Original Message-
> I wonder why it is still standard practice in some circles to
> search for "outliers" as opposed to using robust/resistent methods.
At the risk of extending an old debate and driving us off list topic, here are
three possible reasons:
i) Identifying outliers i
I think one of your problems (the others have been addressed by
others) is that you want the expression x$y to represent a column of x
whose name is stored in y (not the name y itself). The problem here
is that the $ notation is a magical shortcut and like any other magic
if used incorrectly is li
Hi,
On Fri, Feb 10, 2012 at 12:15 PM, Chosid, David (MISC)
wrote:
> Yes, sorry for not being more clear. Here is the sessionInfo():
>
>> sessionInfo()
> R version 2.9.1 (2009-06-26)
> i386-pc-mingw32
That is definitely the place to start. Your version of R is 2.5 years old,
and circular is up t
Hi,
I'd like to debug in a loop (using debug() and browser() etc but not print()
). I'am looking for the first occurence of NA.
For instance:
tab = c(1:300)
tab[250] = NA
len = length(tab)
for (i in 1:len){
if(i != len){
tab[i] = tab[i]+tab[i+1]
}
}
I do not want to do "Browse[2]> n"
I've looked around and I just can't find anything that will work for my
needs. This is a bit of a 2 part question but pertaining to the same topic
so bare with me.
The first is with my qq plot. On the Y axis of my qq plot it'll have my
sample quantities but because my data is log-normal it'll sho
Hello,
Does anyone know if it is possible to find out whether a calling function
exited with an error or not inside of an on.exit() call? I'd like to use the
same error cleanup function inside of several functions, and for aesthetic
reasons would prefer not to surround the body of all of these f
Thanks to Peter and Gabriel for their comments,
The first way which Peter commented was my first approach, however as I use
it for a 'large' matrix (in my case nrow = ncol = 512) and I want to split
it into very small matrices (e.g. nrow = ncol = {4, 8, 16}) both nested
loops I think are a quite e
You can add
if(is.na(tab[i])) browser()
or
if(is.na(tab[i])) break
see inline
On Fri, Feb 10, 2012 at 7:22 AM, ikuzar wrote:
> Hi,
>
> I'd like to debug in a loop (using debug() and browser() etc but not
> print()
> ). I'am looking for the first occurence of NA.
> For instance:
>
> tab = c(1
It's a good idea to acknowledge that you've found a
solution on the R-help list, rather than just to me.
That way the answer appears in the list archives, and
other people will know you no longer need help with
this problem.
Sarah
On Fri, Feb 10, 2012 at 12:57 PM, Chosid, David (MISC)
wrote:
>
Dear all,
I have some questions regarding the validity an implementation of
statistical tests based on linear models and loess. I've searched the
R-help arhives and found several informative threads that related to my
questions, but there are still a few issues I'm not clear about. I'd be
grateful
All,
I am looking for an optimization library that does well on something as chaotic
as the Schwefel function:
schwefel <- function(x) sum(-x * sin(sqrt(abs(x
With these guys, not much luck:
> optim(c(1,1), schwefel)$value
[1] -7.890603
> optim(c(1,1), schwefel, method="SANN", control=list
Hello,
I am presently trying to get a feel for the various packages out there that
allow me to both analyze and simulate random fields. The package
RandomFields is nice, but there are still a few aspects of its
implementation that are confusing to me and I was hoping someone could help
clarify th
Hi All,
I am attempting to simulation an inventory model on R and I am having some
problems. I believe the problem is when I define my demand rate is stays
constant throughout so when I do need to reorder the model does not
recognise it as I have the initial supply arrival time set to infinity at
I think you must have given the path to the file wrong.
Assuming you are using R.app, try
infile <- file.choose()
And then give infile to the read.csv() function.
Then print the value of infile to find out what the path really is.
Probably, you wanted
'/Users/sezginozcan/Downloads/beer.dat
Sorry for not being more clear - I'm interested in accessing these
indices from within the trainControl summaryFunction, not afterward
(from the train object).
As for the weights, I'm referring to the weights argument passed into
train.
On Fri, Feb 10, 2012 at 5:50 AM, Max Kuhn wrote:
> I think
(I couldn't find answers to this question in the documentation)
On Fri, Feb 10, 2012 at 11:59 AM, Yang Zhang wrote:
> Sorry for not being more clear - I'm interested in accessing these
> indices from within the trainControl summaryFunction, not afterward
> (from the train object).
>
> As for the
On 2/9/2012 6:24 PM, ilai wrote:
You do not provide mlm.influence() so your code can't be reproduced.
Or did you mean to put lm.influence() in the formals to your hatvalues.mlm ?
If yes, then 1) you have a typo 2) lm.influence doesn't allow you to
pass on arguments, maybe try influence.lm inste
yaxt='n' in ?par and ?axis are your friends.
# A plot on log scale labeled with original:
plot(x,log(y),yaxt='n')
axis(2,at=pretty(log(y)),labels=round(exp(pretty(log(y)
Works for qqnorm and boxplots, as well as other top level fun.
By the way this is a FAQ.
On Fri, Feb 10, 2012 at 9:43 AM
For these type of setups, I typically turn to "default" values, e.g.
hatvalues.mlm <- function(model, m=1, infl=NULL, ...)
{
if (is.null(infl)) {
infl <- mlm.influence(model, m=m, do.coef=FALSE);
}
hat <- infl$H
m <- infl$m
names(hat) <- if(m==1) infl$subsets else apply(infl$s
I am very new to R and programming and thank you in advance for your patience
and help with a complete novice!
I am working with a large multivariate data set that has 10 explanatory
environmental variables (e.g. temp, depth) and over 60 response variables
(each is a separate species). My data fr
On 2/10/2012 4:09 PM, Henrik Bengtsson wrote:
So people may prefer to do the following:
hatvalues.mlm<- function(model, m=1, infl, ...)
{
if (missing(infl)) {
infl<- mlm.influence(model, m=m, do.coef=FALSE);
}
hat<- infl$H
m<- infl$m
names(hat)<- if(m==1) infl$subsets
A good way to solve problems like this is to write a function that
will work with one variable, and then use one of the apply family
of functions (in this case sapply) to do it for each of your
variables. In this case, such a function would be something like
this:
make1plot = function(var){
Hi David,
Totally forgot line 3! So "513 red" should be ok but "420 red" and "917
yellow" aren't.
Jeffrey
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: johjeff...@hotmail.com
> Subject: Re: [R] Table rearranging
> Date: Tue, 7 Feb 2012
Good Day,
I fit a multivariate linear regression model with 3 dependent variables and
several predictors using the lm function. I would like to use stepwise
variable selection to produce a set of candidate models. However, when I pass
the fitted lm object to step() I get the following error:
On Feb 10, 2012, at 4:47 PM, Jeffrey Joh wrote:
Hi David,
Totally forgot line 3! So "513 red" should be ok but "420 red" and
"917 yellow" aren't.
I don't have a plyr solution but this is a base solution:
dat[ as.logical( ave(dat$door, dat$date,
FUN=function(x) {"open" %i
Yes, I know how to statically specify lambda choices. I was asking
about whether there's a way to let glmnet guide that, instead of
specifying it in advance. Again, sorry if I could've been more clear.
On Thu, Feb 9, 2012 at 7:15 PM, Max Kuhn wrote:
> You can adjust the candidate set of tuning
"Anova.mlm" would be one way to do model selection.
On Fri, Feb 10, 2012 at 4:29 PM, Fugate, Michael L wrote:
> Good Day,
>
> I fit a multivariate linear regression model with 3 dependent variables and
> several predictors using the lm function. I would like to use stepwise
> variable select
My data:
> dput(mydata)
structure(list(V1 = c(1328565067, 1328565067.05, 1328565067.1,
1328565067.15, 1328565067.2, 1328565067.25), V2 = c(0.0963890795246276,
0.227296347215609, 0.240972698811569, 0.221208948983498, 0.230898231782485,
0.203282153087549), V3 = c(0.0245045248243853, 0.083567941170357
damn, that's pretty clever. +1
thanks
On Thu, Feb 9, 2012 at 8:11 PM, ilai wrote:
> Your attempt was just overly complicated. All you needed was
>
> threshold <- c( .2 , .4 , .5 )[ df$track ]
> df$value <- pmax(threshold, df$value)
> df # desired outcome
>
> Cheers
>
> On Thu, Feb 9, 2012 at 3:
Hi list
I need some help for drawing some histograms
I have a dataframe , say,
X Y Z T
I want to draw a histogram Z-T for each value of the couple (X-Y).
When I use thus syntax
> library(lattice)
> histogram(law[,3] ~ law[,66] | law[,1] )
it draws multiple histograms but by selecting distinc
On Feb 10, 2012, at 7:05 PM, Adel ESSAFI wrote:
Hi list
I need some help for drawing some histograms
I have a dataframe , say,
X Y Z T
I want to draw a histogram Z-T for each value of the couple (X-Y).
When I use thus syntax
library(lattice)
histogram(law[,3] ~ law[,66] | law[,1] )
Per
Hello
i need to simulate data for and arch and garch process, n=1000.
I've been generating arch and garch data but i can't find the way to set an
starting error, epsilon(t=0) and sigma(t=0) to run the command garchSim(). I
generated the process on Excel and i know that for the Arch process has a
me
I'm not quite sure what you mean, but perhaps this will help:
library(splines)
mydata <- structure(list(V1 = c(1328565067, 1328565067.05, 1328565067.1,
1328565067.15, 1328565067.2, 1328565067.25), V2 = c(0.0963890795246276,
0.227296347215609, 0.240972698811569, 0.221208948983498, 0.230898231782485
Someone supplied me with an SPSS datafile that caused a buffer
overflow and then a crash when reading it in R. Unfortunately I can't
supply the dataset at hand and I have a hard time reproducing it with
a toy example. But I found at least 2 issues that might be related. I
would like to know which o
Michael,
On 10 February 2012 18:11, R. Michael Weylandt
wrote:
> I'm not quite sure what you mean, but perhaps this will help:
> spf <- splinefun(mydata$V2)
> splInt <- function(low, up) integrate(spf, low, up)$value
Sort of, I'm looking to get the nth order integral, where the only
thing I know
You might find the pairs2 function in the TeachingDemos package useful.
On Fri, Feb 10, 2012 at 1:13 PM, jarvisma wrote:
> I am very new to R and programming and thank you in advance for your patience
> and help with a complete novice!
>
> I am working with a large multivariate data set that has
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