Displaying 20 results from an estimated 600 matches similar to: "problems with nonlinear fits using nls"
2012 Oct 02
4
[LLVMdev] llvm-g++ does not work!
Hi,
I am using PinaVM which is a prototype of a SystemC front-end based on
"LLVM". The only version that it works with is 2.8. Also to test
PinaVM, we need llvm-g++ (I think clang does not work). However, when
I want to run an example, i get the following error, which i think is
related to llvm-g++:
reza at RezaUbuntu:~/pinavm-pinavm/systemc-examples/jerome-chain$ make promela
2009 Mar 09
3
Shorewall Rules and Configurations
Hi,
I need a help... I''m a beginner with shorewall.
I have two shorewall firewalls, each with a link.
FW (a) - w/ openVPN
eth0 = 192.168.150.5/24
eth1 = 192.168.200.5/24
eth2 = public IP
eth3 = 192.168.120.5/24
tun240 = 10.240.255.1
/etc/shorewall/zones
all zones declared as ipv4
/etc/shorewall/interfaces
#ZONE INTERFACE BROADCAST OPTIONS
tlm eth0
2012 Oct 02
0
[LLVMdev] llvm-g++ does not work!
The issue here (even if you get dragonegg working) is that the thing that most newer linuxes install when you apt-get llvm-gcc isn't actually llvm-gcc, it's gcc with the dragonegg plugin. Even if the plugin issues are sorted out, the "fake" llvm-gcc doesn't support -emit-llvm so this wouldn't work.
You'll probably need to pull a 2.8 of it from llvm.org or a
2007 Mar 19
4
matrix similarity comparison
Good morning to you all,
I have a problem with a set of matrices that I want to compare.
I want to see the similarity between them, and to be able to extract the
differences between them.
They have all the same number of columns and rows, and correspond
presence absence data:
for example:
m1 <- matrix(c(1,0,0,0,1,0,1,1,1,1,1,1), 3,4)
m2 <- matrix(c(1,0,1,0,1,0,0,1,0,1,0,1), 3,4)
I
2007 Jul 12
1
problems with memory in Mac
Dear friends,
I am having some doubts about the amount of memory that is being used by
R in my Mac (MacBook Pro, 2Gig). Is there a way to increase the amount
of memory used?
When I type:
> mem.limits()
the result is:
nsize vsize
NA NA
and I can't change it, tough my computing in R isn't using all the
memory at it's disposal.
Best regards,
Carlos
--
Carlos GUERRA
2009 Mar 13
0
Polices, Rules and Configurations - No Success (#/etc/shorewall/policy)
Hello,
I forgot to put my #/etc/shorewall/policy file:
# /etc/shorewall/policy
###############################################################################
#SOURCE DEST POLICY LOG LIMIT: CONNLIMIT:
# LEVEL BURST MASK
#
adm net DROP info
tlm net DROP info
#
net adm DROP
2004 Dec 15
3
adding perspectives to existing persp plots
I've created a perspective plot using 'persp' in the graphics package.
I'd like to add a second plane of z values to the existing plot, but I
cannot seem to do this using 'persp'. Is there an analogue to 'lines' or
'points' for perspectives?
Corey.
corey.bradshaw at cdu.edu.au
2003 Aug 28
3
(no subject)
Dear All,
A couple of questions about the nls package.
1. I'm trying to run a nonlinear least squares
regression but the routine gives me the following
error message:
step factor 0.000488281 reduced below `minFactor' of
0.000976563
even though I previously wrote the following command:
nls.control(minFactor = 1/4096), which should set the
minFactor to a lower level than the default
2007 Apr 15
1
nls.control( ) has no influence on nls( ) !
Dear Friends.
I tried to use nls.control() to change the 'minFactor' in nls( ), but it
does not seem to work.
I used nls( ) function and encountered error message "step factor
0.000488281 reduced below 'minFactor' of 0.000976563". I then tried the
following:
1) Put "nls.control(minFactor = 1/(4096*128))" inside the brackets of nls,
but the same error message
2012 Apr 17
3
error using nls with logistic derivative
Hi
I?m trying to fit a nonlinear model to a derivative of the logistic function
y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function with nls)
The derivative calculated with D function is:
> logis<- expression(a/(1+exp((b-x)/c)))
> D(logis, "x")
a * (exp((b - x)/c) * (1/c))/(1 + exp((b - x)/c))^2
So I enter this expression in the nls function:
2003 Sep 06
9
Scanner for 4.8
Dear All,
Can anyone recommend a scanner that works well on 4.8.
Thanks in advance,
Regards,
Dave
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2008 Sep 02
2
nls.control()
All -
I have data:
TL age
388 4
418 4
438 4
428 5
539 10
432 4
444 7
421 4
438 4
419 4
463 6
423 4
...
[truncated]
and I'm trying to fit a simple Von Bertalanffy growth curve with program:
#Creates a Von Bertalanffy growth model
VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4,
start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE)
#Scatterplot of the data
plot(TL~age, data=box5.4,
2009 Oct 02
1
nls not accepting control parameter?
Hi
I want to change a control parameter for an nls () as I am getting an error
message "step factor 0.000488281 reduced below 'minFactor' of 0.000976562".
Despite all tries, it seems that the control parameter of the nls, does not
seem to get handed down to the function itself, or the error message is
using a different one.
Below system info and an example highlighting the
2017 Jan 28
2
[PATCH] Add missing HAVE_CONFIG_H guards
Hi,
Attached patch adds some missing HAVE_CONFIG_H guards to some places in
celt/arm. Please review.
Thanks,
--Michael
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2017 Jan 28
2
[PATCH] Reduce the scope of Ne10 includes
Hi,
Attached patch modifies some of the Ne10 includes in celt/arm. The original
includes worked, but ended up pulling in extra Ne10 headers that were
unnecessary. This allows libopus to be built with just the Ne10 DSP module
installed on the user's machine. Please review.
Thanks,
--Michael
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2013 Oct 03
2
SSweibull() : problems with step factor and singular gradient
SSweibull() : problems with step factor and singular gradient
Hello
I am working with growth data of ~4000 tree seedlings and trying to fit non-linear Weibull growth curves through the data of each plant. Since they differ a lot in their shape, initial parameters cannot be set for all plants. That’s why I use the self-starting function SSweibull().
However, I often got two error messages:
2012 Jan 25
1
solving nls
Hi,
I have some data I want to fit with a non-linear function using nls, but it
won't solve.
> regresjon<-nls(lcfu~lN0+log10(1-(1-10^(k*t))^m), data=cfu_data,
> start=(list(lN0 = 7.6, k = -0.08, m = 2)))
Error in nls(lcfu ~ lN0 + log10(1 - (1 - 10^(k * t))^m), data = cfu_data, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
Tried to increase minFactor
2008 Apr 14
3
Logistic regression
Dear all,
I am trying to fit a non linear regression model to time series data.
If I do this:
reg.logis = nls(myVar~SSlogis(myTime,Asym,xmid,scal))
I get this error message (translated to English from French):
Erreur in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start =
list(xmid = aux[1], :
le pas 0.000488281 became inferior to 'minFactor' of 0.000976562
I then tried to set
2006 Aug 04
1
gnlsControl
When I run gnls I get the error:
Error in nls(y ~ cbind(1, 1/(1 + exp((xmid - x)/exp(lscal)))), data = xy, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976563
My first thought was to decrease minFactor but gnlsControl does not contain
minFactor nor nlsMinFactor (see below). It does however contain nlsMaxIter
and nlsTol which I assume are the analogs of
2007 Sep 05
3
'singular gradient matrix’ when using nls() and how to make the program skip nls( ) and run on
Dear friends.
I use nls() and encounter the following puzzling problem:
I have a function f(a,b,c,x), I have a data vector of x and a vectory y of
realized value of f.
Case1
I tried to estimate c with (a=0.3, b=0.5) fixed:
nls(y~f(a,b,c,x), control=list(maxiter = 100000, minFactor=0.5
^2048),start=list(c=0.5)).
The error message is: "number of iterations exceeded maximum of