Hi
I like to use logical values directly in computations if possible.
yourData[,10] <- yourData[,9]/(yourData[,8]+(yourData[,8]==0))
Logical values are automagicaly considered FALSE=0 and TRUE=1 and can be used
in computations. If you really want to change 0 to 1 in column 8 you can use
yourDa
Hello R users,
I’m hoping to run an exploratory and confirmatory factor analysis on a
psychology survey instrument. The data has been collected from
multiple groups, and it’s likely that the data is hierarchical/has 2nd
order factors.
It appears that the lavaan package allows me to run a multiple
Now THAT is a fortune:
My preferences are colored by my experience looking for problems in
other people's code. My own code is always easy to understand but
code other people write can be difficult. :-)
On Sun, Jul 27, 2014 at 8:38 PM, William Dunlap wrote:
> The problem with Don's
> if (cond
On Sun, 27 Jul 2014 02:08:25 AM Maede Nouri wrote:
> hello
> I am new to R language. I fitted a linear model and my output has
about 300
> coefficients. I need to definition a function that join my model &
these
> coefficients ! this is difficult because number of coefficients is many.
I
> thi
On Sun, 27 Jul 2014 01:41:51 AM kafi dano wrote:
> Hi. I have a problem in R. I want the command of least median
squares and
> the package of this estimator
>
>
Hi Kafi,
Try the MASS (lmsreg) and robust packages.
Jim
__
R-help@r-project.org mailing l
I now have two win7s in a local area network called A and B.
A now can run shiny app, and it shows that it listen 192.168.1.100:6271. And A
can visite the app in A's own chrome using 192.168.1.100:6271.
Then I want to B to visite A's shiny app. I type 192.168.1.100:6271 in chrome,
but it doesn
On 7/27/2014 8:03 PM, super wrote:
Tks for your opinions, let me describe the background for my question:
The orignal purpose is to call the function which defined in anothor function,
e.g.
f1<-function(){
x<-1
f2<-function(x)
{x}
f3<-function(){
1+1
}
a<-1
return(x+a)
}
The real function f1 m
Tks for your opinions, let me describe the background for my question:
The orignal purpose is to call the function which defined in anothor function,
e.g.
f1<-function(){
x<-1
f2<-function(x)
{x}
f3<-function(){
1+1
}
a<-1
return(x+a)
}
The real function f1 may be very long, And i know there is
The problem with Don's
if (condition) {
Results <- something
} else {
Results <- somethingElse
}
is that in a long sequence of if-then-else-if... you have
to check every branch to make sure Results got assigned to
(or that the remaining branches contained a return() or a stop()).
On 7/27/2014 4:29 PM, MacQueen, Don wrote:
As long as people are sharing their preferences . . .
I find return() useful in a scenario like the following:
Myfun <- function() {
{a few lines of code}
if (condition) return(whatever)
{Many, many lines of code}
Results
}
Which I find p
On 27/07/2014, 7:29 PM, MacQueen, Don wrote:
> As long as people are sharing their preferences . . .
>
>
> I find return() useful in a scenario like the following:
>
> Myfun <- function() {
> {a few lines of code}
> if (condition) return(whatever)
> {Many, many lines of code}
> Results
>
As long as people are sharing their preferences . . .
I find return() useful in a scenario like the following:
Myfun <- function() {
{a few lines of code}
if (condition) return(whatever)
{Many, many lines of code}
Results
}
Which I find preferable to
Myfun <- function() {
{ a few lin
Just an (ignorable) opinion
I'm not sure I would agree on the exception handling view, but maybe
it often boils down to:
Do you prefer:
a) function(...)
{
if(cond1) {do one} else{
if(cond2) {do two}} else {
if(cond3) {do three}}
results
}
## versus
b) function(...)
{
if(cond1) {do one; ret
Hi. I have a problem in R. I want the command of least median squares and the
package of this estimator
Kafi Dano Pati
Ph.D candidate ( mathematics/statistics)
Department of mathematical Science/ faculty of Science
University Technology Malaysia
81310 UTM, Johor Bahru, Johor, Malaysia
H
Well, he did say it was his opinion. Goto has been pretty effectively
eliminated from modern programming languages, while return has not.
IMHO the nature of the return statement resembles exception handling more than
normal control flow... so I avoid using it. Exceptions are exceptional, and
no
On 7/27/2014 10:34 AM, William Dunlap wrote:
This is a real hack, but you can redefine return in your function:
f <- function() {
+ return("early return")
+ "last value in function"
+ }
f()
[1] "early return"
f <- function() {
+ return <- function(x)x
+ return("early return"
This is a real hack, but you can redefine return in your function:
> f <- function() {
+ return("early return")
+ "last value in function"
+ }
> f()
[1] "early return"
> f <- function() {
+ return <- function(x)x
+ return("early return")
+ "last value in function"
+ }
> f()
[1]
i try this to remove the "return" statement in the body(f):
tmp<-as.character(body(f))
a<-sub("return","#return",tmp)
but i don't know how to transfer the character form back to call form or
language form or expression form ?
Is the method i posted could be implemented correctly, or Is there oth
If the argument to the function you
give in the 'by' call were 'z' instead
of 'x', then it would have been easier
to see what was wrong.
You are using 'xx' inside the function
rather than 'x'. I think you want
something like:
function(z) {cor(z$Data1.MEAN,
z$Data2.MEAN)}
Pat
On 27/07/2014
You should study the theory of regression before proceeding, since the chance
of obtaining a significant (useful) result with 400 inputs is negligible.
You should also read the help files for the functions you are using (e.g. ?lm),
which in this case mention the coef() function in the See Also s
I find bquote() handy for this sort of manipulation. E.g.,
> f <- function() { 1 }
> origBody <- body(f)
> newBody <- bquote({ .(origBody) ; .(addedStuff) }, list(origBody=origBody,
> addedStuff=quote(function(){})))
> body(f) <- newBody
> f
function ()
{
{
1
}
function() {
Hello all,
I strongly believe, that my issue is quite easy to resolve. However, I have
no idea how to find/debug what is wrong and I would blame myself for
insufficient knowledge about data tables and some useful functions in R! as
apply, sapply, lapply.
My question is as follows:
I would like t
On 27.07.2014 15:41, super wrote:
Suppose that I had a function as below:
f<-function() {
return(1)
}
i want to change the body of f to the form like this:
f<-function(){
1
function() {}
}
How can i do the task using body(f) or something else solutions?
See ?body:
body(f) <- as.call(c(as.n
hello
I am new to R language. I fitted a linear model and my output has about 300
coefficients. I need to definition a function that join my model & these
coefficients ! this is difficult because number of coefficients is many.
I think there is not a provided query for this. I also used "fitted
Suppose that I had a function as below:
f<-function() {
return(1)
}
i want to change the body of f to the form like this:
f<-function(){
1
function() {}
}
How can i do the task using body(f) or something else solutions?
[[alternative HTML version deleted]]
Hello,
If I understand it correctly the following should do it. Note that it
removes both columns and rows.
idx <- corData > 0.5
diag(idx) <- FALSE
idx2 <- which(apply(idx, 2, function(x) !any(x)))
corData[-idx2, -idx2]
Use corData[, -idx2] to remove only columns.
Hope this helps,
Rui Bar
Dear all,
I have written the following code for correlation calculations. I want to
create a new matrix from corData) with correlation more than 0.5 only and
the rest of the columns should be removed. How can I do it?
set.seed(1234)
data<-matrix(rnorm(100),nrow=10)
data[,1]<-100*(data[,2]+data[
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