On Wed, 17 Dec 2014, Francisco M. da Rocha wrote:
Hallo there,
I would like to work with CHAID, but the newest version of R does have
it. So I thought I could use an older version of R which accepts or has
the library CHAID. Could you tell me which version it is and where to
download it?
A
# I have a dataframe that contains 2 columns:
CaseID <- c('1015285',
'1005317',
'1012281',
'1015285',
'1015285',
'1007183',
'1008833',
'1015315',
'1015322',
'1015285')
Primary.Viol.Type <- c('AS.Age',
'HS.Hours',
'HS.Hours',
'HS.Hours',
'RK.Records_CL',
'OT.Overtime',
'OT.Overtime',
'OT.Overtime'
Hi,
when I am trying to deug a package, I am getting the error.
> library(quantstrat)
> setBreakpoint("strategy.R#3",envir=environment(strategy))
No source refs found.
In R studio, I get a strange message, "Breakpoints will be activated when an
updated version of the quantstrat package is loa
One of the reasons lists are useful is that you can put various things in them
and then you have an object name that you can hard code into your program, yet
still use variables to find objects in that list. That is you do not need to
directly use the get function at all.
foo[[var]]
If you eve
Thank you. I was flummoxed by late night tiredness.
getElement via ?"[" helped me. I am flubbergusted with your speedy answer.
ce
-Original Message-
From: "Bert Gunter" [gunter.ber...@gene.com]
Date: 12/17/2014 10:35 PM
To: "ce"
CC: "r-help@r-project.org"
Subject: Re: [R] how to make th
?"["
Read the docs! Go thru an R tutorial.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Wed, Dec 17, 2014 at 7:24 PM, ce wrote:
> Dear all,
>
> If I h
Dear all,
If I have a list like this how I can get an object of it with a variable :
foo<-list(A = c(1,3), B =c(1, 2), C = c(3, 1))
var <- "A"
get(paste("foo$",'A',sep=''))
Error in get(paste("foo$", "A", sep = "")) : object 'foo$A' not found
__
R-he
Dear all,
I read my a character matrix from a text file. Some of them have greek
characters. To reserve the special characters, I used stringsAsFactors=F using
read.table. I notice that I can’t print these character string using print(),
but I can use cat():
> print("LC\246\302")
[1] "LC\246\302
I can answer, but I think you'll get a better one if you tell us the
environment in which you're working -- RStudio, R's GUI, Rterm,... and
the platform (Windows, MAC, _nix).
Very briefly, but perhaps inadequately, the script is what you write
and send to R to execute or perhaps store in its works
Ian,
ls() is your friend. To learn more about ls(), type ?ls
Clint
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600
"out of your depth" does not serve as a legitimate excuse -- for me
anyway. There are many good tutorials on regular expressions out
there. Go through one. Ditto with R data handling. "An Introduction to
R" (ships with R) is one that's right at hand.
Although others may be more inclined than I am
Using lme4 how does one define a 2 factor factorial model with both factors
being random?
Specifically I am just trying to recreate the results from Montgomery's
Design of Experiments book (7th edition), example 13.2. In this example
there are 2 random factors and I want to include the interaction
I apologize that I am very new to R and programming in general. I do not
understand the difference between the script, the workspace, and the
history, and what saving each one means.
I seem to be doing fine writing commands and going through lessons and
examples (I'm using Learn R in a Day) but wh
Dear list,
I have an optimization problem that I would like to solve by Maximum
Likelihood.
I have analytical functions for the first and second derivatives of my
parameters.
In addition, some parameters are constrained between 0 and 1, while some
others can vary freely between -Inf and +Inf.
I a
Hallo there,
I would like to work with CHAID, but the newest version of R does have it.
So I thought I could use an older version of R which accepts or has the
library CHAID.
Could you tell me which version it is and where to download it?
Thanks a lot in advance.
Francisco
___
I have a large dataset (~50,000 rows, 96 columns), of hospital
administrative data.
many of the columns are clinical coding of inpatient event (using ICD-10).
A simplified example of the data is below
> dput(dat_unmatched)
structure(list(ID = structure(c(4L, 3L, 2L, 1L), .Label = c("BCM3455",
"BZD
If you want to use lattice then the following should give you some tips
Read ?lattice::xyplot
testmatrix<-matrix(c(1,2,3,4,3,6,12,24),nrow=4,ncol=2)
testylabels<-c('w1','x1','y1','z1')
dtf <- data.frame(testmatrix)
dtf
testmatrix
dotplot(X2 ~ X1, dtf)
dtf[,1] = factor(dtf[,1])
dotplot(X2 ~ X1, dtf
Dear Dennis, David, Jeff, and Denes,
Thanks for your helps and comments. The simple one seems good enough for
my works.
Best,
Steve
On Wed, Dec 17, 2014 at 5:46 AM, Dénes Tóth wrote:
>
> Dear Jeff,
>
> On 12/17/2014 01:46 AM, Jeff Newmiller wrote:
>
>> You are chasing ghosts of performance pa
Tena koe Sachin
The following might help you understand what is going on and how to rectify it.
cost <- function(x) {x[,1]*x[,2]}
ttMat <- matrix(1:4, ncol=2)
ttMat
cost(ttMat)
cost(ttMat[1,])
cost(as.matrix(ttMat[1,]))
cost(t(as.matrix(ttMat[1,])))
cost(matrix(ttMat[1,], ncol=2))
str(ttMat)
str
as.matrix(c(1,1)) gives a matrix with only one column, but your function
assumes you have at least two columns (you refer to x[,2]). Please make your
examples reproducible (run it yourself in a fresh instance of R) to obtain best
results with questions on this list.
However, you might just be
On 17/12/2014 12:24, r...@openmailbox.org wrote:
Subscribers,
For this example:
library(lattice)
testmatrix<-matrix(c(1,2,3,4,3,6,12,24),nrow=4,ncol=2)
testylabels<-c('w1','x1','y1','z1')
dotplot(testmatrix, scales=list(y=list(labels=testylabels), xlab=NULL))
legend('bottomright', 'legend', c
Subscribers,
For this example:
library(lattice)
testmatrix<-matrix(c(1,2,3,4,3,6,12,24),nrow=4,ncol=2)
testylabels<-c('w1','x1','y1','z1')
dotplot(testmatrix, scales=list(y=list(labels=testylabels), xlab=NULL))
legend('bottomright', 'legend', col=c('blue', 'pink'))
Error in strwidth(legend, unit
nlsLM and nls share a numerical gradient approximation and pop up the
"singular gradient" quite often at the start. Package nlmrt and a very
alpha nls14 (not on CRAN) try to use analytic derivatives for the
Jacobian (most optimization folk will say singular Jacobian rather than
singular gradien
Dear Jeff,
On 12/17/2014 01:46 AM, Jeff Newmiller wrote:
You are chasing ghosts of performance past, Denes.
In terms of memory efficiency, yes. In terms of CPU time, there can be
significant difference, see below.
The data.frame
function causes no problems, and if it is used then the OP w
Wolfgang Obermeier writes:
> how is it possible that the loadings of the second or even third component of
> a PLS-Analysis show higher values than the first component? Somebody got an
> idea??
The loadings of a PLS regression are simply the coefficients that are
multiplied with the X variables
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