Hi Patrick,
Thanks for posting. The wrapper function works as it should. If you do not
specify return the function returns the last object declared, and this is
exactly what I intended. Doing some research on my own I found out that the
problem was here:
x <- c(1: length(linH))
Hi,
I used this command to produce a table:
(tab1 = ftable(SEX, ESTCIV, Q1))
Q1 B L M N
SEXOESTCIV
F A 11 13 4 2
E 1 0 0 0
M A 5 0 3 1
E
Hi
I made a simple spredsheet for PSD using Rosin Rammler equation and I am lazy
to transform it to R. However for single purpose you can use nls.
Reverse your cumulative values
PSD$cum<-cumsum(PSD$ret)
plot(PSD$size, PSD$cum)
fit<-nls(cum~ exp(-((size/r)^gama))*100, data=PSD, start=c(r=80, g
I don't know if this is still relevant for you, but probably for someone
else.
I ran into the same problem an found two dirty workarounds that can help
in many cases (but not all). Both can also be combined:
1. repeat the call to "segmented" with different seeds and different
initial estima
hi,
excuse me
where did I mistake in these procedures?
(I want to fit these models to my data)
install.packages('drc',rep='http://cran.um.ac.ir')
install.packages('car',rep='http://cran.um.ac.ir')
install.packages('abind',rep='http://cran.um.ac.ir')
library(car)
library(abind)
library(alr3)
lib
Hi,
How to execute ms-Excel Macro(*.xlsm) using R function ? I tried but not
get. There are method to call from R function from Excel macro, but i need
Excel macro to execute from R. Is it possible ? I am using Excel-2010 macro.
Thanks in advance,
Antony.
--
View this message in context:
h
Dear list,
I'm posting in the R-help list due to:
- Not knowing a better place for it;
- I would like to know the opinion of more specialized people.
What is the best place to classify polynomial regressions (Y = bo +
b1X + b2X^2 + ... + bnX^n): single or multiple linear regression?
Regards,
--
What does your original data look like? I seems to me that it would be better
to give us that information since you may not want to use ftable() at all.
Ideally you should give us the original data or a sample of it. If it's
confidential just replace the actual values with fake data. The struct
Hi,
Try:
library(reshape2)
dcast(as.data.frame(tab1), SEX+ESTCIV~Q1,value.var="Freq") ##not tested.
A.K.
On Friday, December 20, 2013 8:03 AM, "silv...@uel.br" wrote:
Hi,
I used this command to produce a table:
(tab1 = ftable(SEX, ESTCIV, Q1))
Q1 B L M N
Dear Dr. José Faria,
I think that the best category to put polynomial regressions is single
regressions. Although, in polynomial regressions there are more then one
term as in multiple regressions this is an adjustment consequence, not a
design consequence. So, to me this is sufficient to justify
HI,
May be this helps:
set.seed(45)
dat1 <- as.data.frame(matrix(sample(0:1,100*5,replace=TRUE),ncol=5))
dat1$Newvar <- 1*(!!rowSums(dat1))
A.K.
Hello.
I have a problem combining a number of variables. I have five
columns with binary variables with the values 0 and 1. I would like to
comb
Trying to understand environments is not for the faint of heart. If
lists do what you want then I would stick with a list and not worry
about the environments. Most of the time that you deal with
environments everything happens automatically behind the scenes and
you don't need to worry about the
Thanks a lot. It works!
On Thursday, December 19, 2013 10:53 PM, Jim Lemon wrote:
On 12/20/2013 08:19 AM, capricy gao wrote:
> I have played around with it and found that the only color could be changed.
> But I really would like to change the width...
>
Hi Capricy,
Try this on the first e
My first thought was to use Reduce, but I think for this case that is
a bit of overkill. You can have a vector or list of functions and
just use sapply/lapply on the list of functions then sum the result.
A quick example:
> funs <- c(sin,cos,tan)
> sapply( funs, function(f) f(pi/6) )
[1] 0.50
Thanks William.
I was convinced by pmax, until I played with the following example today. I
tried to reproduce a computation from a paper. Here is the code:
P<- function(x) {
ab<-100*exp((0.0435-0.0269-0.5*0.3315^2)*4.3122+x*sqrt(4.3122)*0.3315)
return(ab)
}
integrand<- function(x){
cd<-
-1
You should try to match up values of integrand() instead of the value of its
integral
on a single integral if you want to see if the function is being computed
correctly.
integrate(f) calls f(x) where x is a vector of values (typically 21 values).
It expects
that f() is vectorized: that f(x[i])
Makes sense. Simple and works. Thank you Greg.
Sent from my iPhone
> On 20 Dec 2013, at 16:15, Greg Snow <538...@gmail.com> wrote:
>
> My first thought was to use Reduce, but I think for this case that is
> a bit of overkill. You can have a vector or list of functions and
> just use sapply/lapp
I used the by() function on a data.frame to get sums of the data grouped by 2
factors. The function worked however the output is in a class called 'by'. Not
familiar with this class. How can I turn the output into a nice table where
columns represent values of factor1, row represent values of fa
On Dec 20, 2013, at 5:01 AM, wrote:
> Hi,
>
> I used this command to produce a table:
It's not actually a 'table'.
>
> (tab1 = ftable(SEX, ESTCIV, Q1))
is.table(tab1) # will return FALSE
>
> Q1 B L M N
> SEXOESTCIV
>F A 11 13
On Dec 20, 2013, at 9:38 AM, Onur Uncu wrote:
> I used the by() function on a data.frame to get sums of the data grouped by 2
> factors. The function worked however the output is in a class called 'by'.
> Not familiar with this class. How can I turn the output into a nice table
> where columns
SAS uses words. R uses symbols.
See this:
http://xkcd.com/1306/
(Yes, I know IML uses plenty of symbols. It's just supposed to be funny.
And somewhat true.)
--
Kevin Wright
[[alternative HTML version deleted]]
__
R-help@r-project.org maili
Hi,
Try:
as.table(by(warpbreaks[,1],warpbreaks[,-1],sum))
#or to convert to data.frame
as.data.frame(as.table(by(warpbreaks[,1],warpbreaks[,-1],mean)))
A.K.
On Friday, December 20, 2013 12:39 PM, Onur Uncu wrote:
I used the by() function on a data.frame to get sums of the data grouped by 2
Hi,
You can also try:
library(reshape2)
dcast(as.data.frame(as.table(by(warpbreaks[,1],warpbreaks[,-1],sum))),wool~tension,
value.var="Freq")
A.K.
On , arun wrote:
Hi,
Try:
as.table(by(warpbreaks[,1],warpbreaks[,-1],sum))
#or to convert to data.frame
as.data.frame(as.table(by(warpbreaks[,1
On Dec 20, 2013, at 9:45 AM, David Winsemius wrote:
>
> On Dec 20, 2013, at 9:38 AM, Onur Uncu wrote:
>
>> I used the by() function on a data.frame to get sums of the data grouped by
>> 2 factors. The function worked however the output is in a class called 'by'.
>> Not familiar with this clas
dear all,
I have a dataset composed by different classes: I have two main classes,
"active (Act.)" and "latent (Lat.)", furher subdivided in the gene
subclasses "IP10" and "MIG" and then into 7 stimulus sub-subclasses "ESAT6"
to "PHA".
Would it be possible to test in a direct way the statistical di
You seem to be falling prey to a common misconception that "R" is some
monolithic tool, when in fact it is a herd of cats.
The "by" function, from the "base" package, returns a list of results returned
by your function. One approach to making a data frame out of that is to use the
simplify2arra
You've found an interesting corner of Circle 1
of 'The R Inferno'.
http://www.burns-stat.com/documents/books/the-r-inferno/
The issue is that your 'n' in the final case is
slightly less than 11. So:
> n
[1] 11
> as.integer(n)
[1] 10
> 1:n
[1] 1 2 3 4 5 6 7 8 9 10 11
The mystery to m
Hello!
I am using function multinom:
library(nnet)
library(MASS)
mnl<-multinom(myDV~. ,data=mydata,na.action="na.omit", MaxNWts = 2000,
maxit = 1000)
I can see the resulting coefficients:
print(mnl)
But how could I grab them? I am not finding them when I do:
str(mnl)
Thank you!
--
Dimitri
Hello,
Try
coef(mnl)
Hope this helps,
Rui Barradas
Em 20-12-2013 20:21, Dimitri Liakhovitski escreveu:
Hello!
I am using function multinom:
library(nnet)
library(MASS)
mnl<-multinom(myDV~. ,data=mydata,na.action="na.omit", MaxNWts = 2000,
maxit = 1000)
I can see the resulting coefficien
That's it! Thanks a lot!
On Fri, Dec 20, 2013 at 4:18 PM, Rui Barradas wrote:
> Hello,
>
> Try
>
> coef(mnl)
>
> Hope this helps,
>
> Rui Barradas
>
> Em 20-12-2013 20:21, Dimitri Liakhovitski escreveu:
>
> Hello!
>>
>> I am using function multinom:
>>
>> library(nnet)
>> library(MASS)
>>
>> m
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