Displaying 20 results from an estimated 1100 matches similar to: "solve nonlinear equation using BBsolve"
2010 Apr 29
1
BB package
Hi I would like to solve a system of nonlinear equations below using dfsane function
mn <- 142.36; vr <- 9335.69 ; sk <- 0.81; kur <- 0.25
test_fn <- function(p)
{ f <- rep(NA, length(p))
f[1] <- p[1]*(p [2]+p[3])- mn
f[2] <- - vr + 2*p[1]*p[2]*p[3]*(p[4]-1)+p[1]*(p[2]+p[3])^2
f[3] <- - sk + (p[1]*(p [2]+p[3])^3*(p[1]+1)*(p[1]+2)-6*p[1]*p[2]*p[3]*(p[2]+
2011 Aug 15
0
Stopping criterion in option "control" of BBsolve( )
Dear Dr. Gilbert, Dr. Varadhan and all R-help list members,
I'm using the function BBsolve( ) and I have some questions on the stopping
criterion "maxit" and "noimp" specified in the option "control". Here
are two such examples I'm having problem with. In these two examples, the
function BBsolve( ) always stops at iteration 100, overlooking the values
2012 Dec 11
1
Solving Simultaneous nonlinear equations
Dear:
I am having trouble solving simultaneous nonlinear equations by R. I have
been using BBsolve (BB) to do so. Though the function is very strong, still
the program doesn't converge. I have tried all (according to my small
knowledge) the options described in the help file.
Now I am trying to find something else than BBsolve for solving
simultaneous nonlinear equations by R. Any idea or
2011 Apr 28
1
DLSODA error
Dear R-users,
I'm running an MLE procedure where some ODEs are solved for each iteration
in the maximization process. I use mle2 for the Maximum likelihood and
deSolve for the ODEs.
The problem is that somewhere along the way the ODE solver crashes and I get
the following error message:
DLSODA- Warning..Internal T (=R1) and H (=R2)
are
such that in the machine, T + H = T on the next
2018 Jan 24
1
random sample set for regression
Hi,
I'm not a programmer, so I have a question about R functions,
I run the Random Forest regression models, but
I would like to run the random forest model 1000 times with different
random sample set. to check the uncertainty of the regression model
estimates.
exemple of data:
#################################
table= all
Y: all$AGB
X variables:
Variables=as.matrix(all[, c( "min",
2009 Apr 03
1
Discriminant Analysis - Obtaining Classification Functions
Hello!
I need some help with the linear discriminant analysis in R.
I have some plant samples (divided into several groups) on which I
measured a few quantitative characteristics. Now, I need to infer some
classification rules usable for identifying new samples.
I have used the function lda from the MASS library in a usual fashion:
lda.1 <- lda(groups~char1+char2+char3, data=xxx)
I'd
2012 Dec 20
1
Problem with simulation
Dear:
I am having problem with simulation. Sometimes for 100 iteration, my
program works but sometimes give error message regarding some "built-in
function" which I never used
in my code. I have no idea about the problem. Any suggestion, comments or
idea regarding this would be highly appreciable. Thanks.
I have used ---- "maxLik()" and "BBsolve()" in my code. Error
2011 Mar 31
0
dfsane arguments
Hi there,
I'm trying to solve 2 nonlinear equations in 2 unknowns using the BB
package.
The first part of my program solves 3 ODEs using the deSolve package. This
part works. The output is used as parameter values in the functions I need
to solve.
The second part is to solve 2 equations in 2 unknowns. This does not work. I
get the error message "unexpected end of input". So what
2006 Mar 11
1
Non-linear Regression : Error in eval(expr, envir, enclos)
Hi..
i have an expression of the form:
model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+
2012 Jun 14
0
fixed trimmed mean for j-group
Hello...i want to find the empirical rate for type 1 error using fixed
trimmed mean. To make it easy, i'm referring to journal given by this
website
http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf.
I already run the programme and there is no error in it but i got zero for
the empirical rate of type 1 error. The empirical rate for the type 1 error
given in the journal
2008 Feb 01
2
the "union" of several data frame rows
Hi,
I have a question about how to obtain the union of several data frame
rows. I'm trying to create a common key for several tests composed of
different items. Here is a small scale version of the problem. These
are keys for 4 different tests, not all mutually exclusive:
id q1 q2 q3 q4 q5 q6
1 A C
2 B D
3 A D B
4 C D B D
I would like
2012 Jul 07
0
fixed trimmed mean for group
Hello,
I haven't found errors in your code. I implemented the test in the paper
(the first, fixed symetric mean) and it also gives me zero Type I
errors, when alpha = 0.05. Try to see the value of min(pv) or to plot
the histogram of 'pv', hist(pv) and you'll see that there are no
significant p-values, at that level.
Anyway I'll continue to look at it, but my first
2004 Oct 04
3
Beginners problem
Hi,
I'm new to R and have a problem with a little test program (see below).
Why doesn't <<- in function rk4
assign the new value to y so that it is seen in rktest. I thought that
<<- does exactly this. But it seems that I
didn't get it right. I would be very appreciative for an explanation of
that behaviour of <<-. I know how to
write the whole thing so that it
2010 Sep 26
2
Finding Zeros of a Function
Dear All,
I need to find the (possible multiple) zeros of a function f within an
interval. I gave uniroot a try, but it just returns one zero and I need
to provide it with an interval [a,b] such that f(a)f(b)<0.
Is there any function to find the multiple zeros of f in (a,b) without
constraints on the sign of f(a) and f(b)?
Many thanks
Lorenzo
2012 May 15
2
Renaming names in R matrix
I have the following matrix:
> dat
[,1] [,2] [,3] [,4]
foo 0.7574657 0.2104075 0.02922241 0.002705617
foo 0.0000000 0.0000000 0.00000000 0.000000000
foo 0.0000000 0.0000000 0.00000000 0.000000000
foo 0.0000000 0.0000000 0.00000000 0.000000000
foo 0.0000000 0.0000000 0.00000000 0.000000000
foo 0.0000000 0.0000000 0.00000000 0.000000000
and given this:
2009 Oct 19
1
updating columns using other column as reference
Dear R-gurus,
Just supose I have a dara.frame that looks like
myDF<-read.table(stdin(),head=T,sep=",")
codID,namesp,k1,k2,k3,k4
1,spA,2,5,6,3
2,spB,4,5,4,6
3,spC,2,1,5,6
4,spC,5,4,3,2
5,spD,1,2,3,4
6,spE,2,4,3,1
I need to update the columns k1-k4 with the namesp, but
considering the math between Kx and codID.
My desired output must looks like:
codID,namesp,k1,k2,k3,k4
1997 Apr 30
2
R-alpha: New Incomplete Beta Function
Here is a drop-in replacement for the R incomplete beta function.
src/math/pbeta.c
It is a slightly modified version of the cephes library one from
Netlib. In the few cases I tried it seems to give at least 14
digit agreement with the one in S-PLUS (its hard to get more).
I'm not sure what performance is like. I'd like to know if it
helps with some of the problems which have been
2006 Nov 29
2
How to solve differential equations with a delay (time lag)?
Hi,
I would like to solve a system of coupled ordinary differential equations,
where there is a delay (time lag) term. I would like to use the "lsoda"
function "odesolve" package. However, I am not sure how to specify the
delay term using the syntax allowed by odesolve.
Here is an example of the kind of problem that I am trying to solve:
> library(odesolve)
2013 Apr 25
2
Vectorized code for generating the Kac (Clement) matrix
Hi,
I am generating large Kac matrices (also known as Clement matrix). This a tridiagonal matrix. I was wondering whether there is a vectorized solution that avoids the `for' loops to the following code:
n <- 1000
Kacmat <- matrix(0, n+1, n+1)
for (i in 1:n) Kacmat[i, i+1] <- n - i + 1
for (i in 2:(n+1)) Kacmat[i, i-1] <- i-1
The above code is fast, but I am curious about
2011 Jul 27
2
fitting sine wave
Dear R-helpers
?
I have 7 data points that I want to fit a continuous curve to, that should look similar to a sine wave
My data points would mark the local minima and maxima respectively.
This is what I?ve got so far. And I would keep doing so, but sadly nls() then says that it has reached the maximum number of Iterations?
?