Re: [R] lme4 Error Help: “maxstephalfit…pwrssUpdate”
Thanks. I used the most current version of lme4 that is why I was a bit concerned. My data seems appropriate and with lme4 working last week on a very similar data set, I was left a bit confused. Since I only starting implementing this technique, does anybody have some pointers on what I should look for that may potentially cause some issues? -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Craig O'Connell Verzonden: maandag 28 april 2014 3:20 Aan: r-help@r-project.org Onderwerp: [R] lme4 Error Help: maxstephalfit pwrssUpdate I am using a mixed model to assess the effects of various variables (i.e. treatment, density, visibility) on bee behavior (e.g., avoidance frequency - total avoidances per total visits; feeding frequency, and mating frequency). Bee individuals is my random factor (n=63 different bees), whereas treatment type, animal density, and air visibility are my fixed factors. However, when I run my models, I immediately get an error that I cannot fix. Here is a sample of my data: Bee TreatmentVisitsAvoid FeedingMatingDensity Visibility 1 C 5 0 5 0 54 2 C 4 0 3 0 54 3 C 3 0 3 0 54 ... 63 1 PC 2 0 1 1 54 2 PC 3 0 0 3 54 3 PC 1 0 0 0 54 ... 63 1 M 5 0 1 3 54 2 M 3 2 0 0 54 3 M 2 0 0 2 54 ... 63One I create my .txt file, I being my coding in R by first loading lme4. After that, my coding starts off as follows: barrierdat = read.table(GLMMROW.txt, header=TRUE) barrierdat barrierdat$Visibility = as.factor(barrierdat$Visibility); barrierdat$Density= as.factor(barrierdat$Density); p01.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee), family=poisson, data=egghead); # null model; p02.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Treatment, family=poisson, data=egghead); p03.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Visibility, family=poisson, data=egghead); p04.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Density, family=poisson, data=egghead);However, upon immediately running my models (e.g. p01.glmer), I receive the error: Error: (maxstephalfit) PIRLS step-halvings failed to reduce deviance in pwrssUpdate Does anybody know what the issue is? I ran similar data several weeks ago and had no issues. Any Suggestions on how to proceed? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. * * * * * * * * * * * * * D I S C L A I M E R * * * * * * * * * * * * * Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lme4 Error Help: “maxstephalfit…pwrssUpdate”
I am using a mixed model to assess the effects of various variables (i.e. treatment, density, visibility) on bee behavior (e.g., avoidance frequency - total avoidances per total visits; feeding frequency, and mating frequency). Bee individuals is my random factor (n=63 different bees), whereas treatment type, animal density, and air visibility are my fixed factors. However, when I run my models, I immediately get an error that I cannot fix. Here is a sample of my data: Bee TreatmentVisitsAvoid FeedingMatingDensity Visibility 1 C 5 0 5 0 54 2 C 4 0 3 0 54 3 C 3 0 3 0 54 ... 63 1 PC 2 0 1 1 54 2 PC 3 0 0 3 54 3 PC 1 0 0 0 54 ... 63 1 M 5 0 1 3 54 2 M 3 2 0 0 54 3 M 2 0 0 2 54 ... 63One I create my .txt file, I being my coding in R by first loading lme4. After that, my coding starts off as follows: barrierdat = read.table(GLMMROW.txt, header=TRUE) barrierdat barrierdat$Visibility = as.factor(barrierdat$Visibility); barrierdat$Density= as.factor(barrierdat$Density); p01.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee), family=poisson, data=egghead); # null model; p02.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Treatment, family=poisson, data=egghead); p03.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Visibility, family=poisson, data=egghead); p04.glmer = glmer(Avoidance~offset(log(Visits))+(1|Bee)+Density, family=poisson, data=egghead);However, upon immediately running my models (e.g. p01.glmer), I receive the error: Error: (maxstephalfit) PIRLS step-halvings failed to reduce deviance in pwrssUpdate Does anybody know what the issue is? I ran similar data several weeks ago and had no issues. Any Suggestions on how to proceed? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Need Help Plotting Line for multiple linear regression
Hello, My name is Craig and I need help plotting a line for a multiple linear regression in R. Here is my sample data (filename: convis.txt) Output of convis.txt is (vis and density being predictors of either avoidance or entrance): vis den avoid entrance 1 10 1 0. 0. 2 10 3 0.8750 0. 38 3 0.8180 0.0300 48 3 0.6670 0.0667 58 1 0.2110 0. 66 1 0.2500 0. 7 10 1 0.3000 0. 8 10 1 0.1050 0. 98 1 0.7000 0.1000 10 3 5 0.1176 0.0588 11 3 5 0.3077 0.1150 12 3 9 0.9090 0.0900 13 3 7 0.7778 0.1110 14 3 5 0.5560 0.1110 15 3 1 0.5710 0. 16 3 4 0.5710 0. In order to do the multiple regression, I used the following coding: double=read.table(convis.txt,header=TRUE) attach(double) double stem(vis) stem(den) stem(avoid) stem(entrance) plot(entrance,vis*den) *as means to see how the interaction between visibility and density may impact entrance behaviors model6=lm(entrance~vis*den) model6 summary(model6) *abline(model6) *Here is the issue as I used this for my simple linear regression technique, but do not know what to use for a multiple regression* If anybody can provide some feedback on this, it would be greatly appreciated. Kind Regards, Craig -- Craig O'Connell University of Massachusetts Dartmouth Marine Biologist www.youtube.com/craigpoconnell craigosea.blogspot.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Predator Prey Models
Dear R-users,
I am currently modifying a previously developed predator prey model and
was curious if there was a way to add in a disturbance to the model (let's
say at time t=100). The disturbance can be the introduction of 40 prey
(N=40) and 10 predators (Pred = 10). I would like to see my model go from a
state of equilibrium (up to t = 99), show this disturbance (at t = 100) and
then slowly work its way back to equilibrium. Does anybody know if this
could be done?
LVmod0D - function(Time, State, Pars) {
with(as.list(c(State, Pars)), {
IngestPred - rI * N * Pred
GrowthN - rG * N * (1 - N/K)
MortPred - rM * Pred
dN - GrowthN - IngestPred
dPred - IngestPred * AE - MortPred
return(list(c(dN, dPred)))
})
}
pars - c(rI = 0.1, rG = 0.9, rM = 0.8, AE = 0.9, K = 20)
yini - c(N = 20, Pred = 20)
times - seq(0, 200, by = 1)
y(time=100)-c(N=10, Pred=5)
print(system.time)
zout - as.data.frame(lsoda(yini, y(time=100), times, LVmod0D, pars))
plot(zout[,1],zout[,2], type=l,col=black, ylim=range(0,25),
xlab=time,ylab=population)
points(zout[,1],zout[,3],type=l,col=red)
title(main=prey (black) predator (red))
Thanks much for your help!
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Re: [R] MATLAB vrs. R
Thank you Peter. That is very much helpful. If you don't mind, I continued
running the code to attempt to get my answer and I continue to get inf inf
inf... (printed around 100 times).
Any assistance with this issue. Here is my code (including your corrections):
myquadrature-function(f,a,b){
npts=length(f)
nint=npts-1
if(npts=1)
error('need at least two points to integrate')
end;
if(b=a)
error('something wrong with the interval, b should be greater than a')
else
dx=b/real(nint)
end;
npts=length(f)
int=0
int - sum(f[-npts]+f[-1])/2*dx
}
#Call my quadrature
x=seq(0,2000,10)
h = 10.*(cos(((2*pi)/2000)*(x-mean(x)))+1)
u = 1.*(cos(((2*pi)/2000)*(x-mean(x)))+1)
a = x[1]
b = x[length(x)]
plot(x,-h)
a = x[1];
b = x[length(x)];
#call your quadrature function. Hint, the answer should be 3.
f=u*h;
val = myquadrature(f,a,b); ? ___This is where issue arises.
result=myquadrature(val,0,2000) ?
print(result) ?
Thanks again,
Phil
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Re: [R] MATLAB vrs. R
alain,
Perhaps i'm still entering the code wrong. I tried using your
result=myquadrature(f,0,2000)
print(result)
Instead of my:
val = myquadrature(f,a,b)
result=myquadrature(val,0,2000)
print(result)
...and I am still getting an inf inf inf inf inf...
Did you change any of the previous syntax in addition to changing the result
statement?
Thank you so much and I think my brain is fried! Happy Holiday.
Craig
Date: Mon, 11 Oct 2010 09:59:17 +0200
From: alain.guil...@uclouvain.be
To: craigpoconn...@hotmail.com
CC: pda...@gmail.com; r-help@r-project.org
Subject: Re: [R] MATLAB vrs. R
Hi,
The first argument of myquadrature in result shouldn't be val but f I
guess. At least it works for me
result=myquadrature(f,0,2000)
print(result)
[1] 3
Regards,
Alain
On 11-Oct-10 09:37, Craig O'Connell wrote:
Thank you Peter. That is very much helpful. If you don't mind, I continued
running the code to attempt to get my answer and I continue to get inf inf
inf... (printed around 100 times).
Any assistance with this issue. Here is my code (including your
corrections):
myquadrature-function(f,a,b){
npts=length(f)
nint=npts-1
if(npts=1)
error('need at least two points to integrate')
end;
if(b=a)
error('something wrong with the interval, b should be greater than a')
else
dx=b/real(nint)
end;
npts=length(f)
int=0
int- sum(f[-npts]+f[-1])/2*dx
}
#Call my quadrature
x=seq(0,2000,10)
h = 10.*(cos(((2*pi)/2000)*(x-mean(x)))+1)
u = 1.*(cos(((2*pi)/2000)*(x-mean(x)))+1)
a = x[1]
b = x[length(x)]
plot(x,-h)
a = x[1];
b = x[length(x)];
#call your quadrature function. Hint, the answer should be 3.
f=u*h;
val = myquadrature(f,a,b); ? ___This is where issue arises.
result=myquadrature(val,0,2000) ?
print(result) ?
Thanks again,
Phil
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--
Alain Guillet
Statistician and Computer Scientist
SMCS - IMMAQ - Université catholique de Louvain
Bureau c.316
Voie du Roman Pays, 20
B-1348 Louvain-la-Neuve
Belgium
tel: +32 10 47 30 50
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Re: [R] MATLAB vrs. R
Daniel,
That's it! Thanks. Your help is very much appreciated. I'm hoping to
nail down the code conversion from MATLAB to R, but it seems to be a bit more
difficult that I had anticipated.
Craig
From: djnordl...@frontier.com
To: djnordl...@frontier.com; r-help@r-project.org
Date: Mon, 11 Oct 2010 16:12:48 -0700
Subject: Re: [R] MATLAB vrs. R
I apologize for the noise. I didn't clean up the code enough. See below.
snip
Craig,
I haven't seen an answer to this yet, so let me jump in. You seem to have
some stuff still leftover from MATLAB. Here is some cleaned up code that
produces the result you expect. I don't think the value of dx was being
correctly computed in your code. I did not change the assignment operator
you used (=), but in R the preferred operator is - (without the
quotes).
myquadrature - function(f,a,b){
npts = length(f)
nint = npts-1
if(npts = 1) error('need at least two points to integrate')
if(b = a) error('something wrong with the interval, b should be greater
than a') else dx=b/nint
The 2 'if' statements above should have been
if(npts = 1) stop('need at least two points to integrate')
if(b = a) stop('something wrong with the interval, b should be greater than
a') else dx=b/nint
sum(f[-npts]+f[-1])/2*dx
}
#Call my quadrature
x = seq(0,2000,10)
h = 10*(cos(((2*pi)/2000)*(x-mean(x)))+1)
u = (cos(((2*pi)/2000)*(x-mean(x)))+1)
a = x[1]
b = x[length(x)]
plot(x,-h)
a = x[1];
b = x[length(x)]
#call your quadrature function. Hint, the answer should be 3.
f = u*h
result = myquadrature(f,a,b)
result
Hope this is helpful,
Dan
Daniel Nordlund
Bothell, WA USA
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[R] MATLAB vrs. R
I need to find the area under a trapezoid for a research-related project. I
was able to find the area under the trapezoid in MATLAB using the code:
function [int] = myquadrature(f,a,b)
% user-defined quadrature function
% integrate data f from x=a to x=b assuming f is equally spaced over the
interval
% use type
% determine number of data points
npts = prod(size(f));
nint = npts -1; %number of intervals
if(npts =1)
error('need at least two points to integrate')
end;
% set the grid spacing
if(b =a)
error('something wrong with the interval, b should be greater than a')
else
dx = b/real(nint);
end;
npts = prod(size(f));
% trapezoidal rule
% can code in line, hint: sum of f is sum(f)
% last value of f is f(end), first value is f(1)
% code below
int=0;
for i=1:(nint)
%F(i)=dx*((f(i)+f(i+1))/2);
int=int+dx*((f(i)+f(i+1))/2);
end
%int=sum(F);
Then to call myquadrature I did:
% example function call test the user-defined myquadrature function
% setup some data
% velocity profile across a channel
% remember to use ? for help, e.g. ?seq
x = 0:10:2000;
% you can access one element of a list of values using brackets
% x(1) is the first x value, x(2), the 2nd, etc.
% if you want the last value, a trick is x(end)
% the function cos is cosin and mean gives the mean value
% pi is 3.1415, or pi
% another hint, if you want to multiple two series of numbers together
% for example c = a*b where c(1) = a(1)*b(1), c(2) = a(2)*b(2), etc.
% you must tell Matlab you want element by element multiplication
% e.g.:c = a.*b
% note the .
%
h = 10.*(cos(((2*pi)/2000)*(x-mean(x)))+1); %bathymetry
u = 1.*(cos(((2*pi)/2000)*(x-mean(x)))+1); %vertically-averaged cross-transect
velocity
plot(x,-h)
% set begin and end points for the integration
a = x(1);
b = x(end);
% call your quadrature function. Hint, the answer should be 3.
f=u.*h;
val = myquadrature(f,a,b);
fprintf('the solution is %f\n',val);
This is great, I got the expected answer of 3.
NOW THE ISSUE IS, I HAVE NO IDEA HOW THIS CODE TRANSLATES TO R. Here is what I
attempted to do, and with error messages, I can tell i'm doing something wrong:
myquadrature-function(f,a,b){
npts=length(f)
nint=npts-1
if(npts=1)
error('need at least two points to integrate')
end;
if(b=a)
error('something wrong with the interval, b should be greater than a')
else
dx=b/real(nint)
end;
npts=length(f)
_(below this line, I cannot code)
int=0
for(i in 1:(npts-1))
sum(f)=((b-a)/(2*length(f)))*(0.5*f[i]+f[i+1]+f[length(f)])}
%F(i)=dx*((f(i)+f(i+1))/2);
int=int+dx*((f(i)+f(i+1))/2);
end
%int=sum(F);
Thank you and any potential suggestions would be greatly appreciated.
Dr. Argese.
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