'non-finite-function value' Error when call 'integrate' with 'maxLik' function - r
I want to use the maximum likelihood method to estimate the parameters of a function which involves integration. However, when i tried to run it, i got a message that "non-finite function value".
Here are the codes:
library("maxLik")
integrand <− function(x,para)
{
# print(para)
# dv, dv_T, dn, dn_T are the actual input data, here only gives one set of value for easy debugging
dv <- -1.4527280
dv_T <- 0.2038909
dn <- 17.5174383
dn_T <- 26.6993828
beta10 = para[1];
beta11 = para[2];
beta12 = para[3];
alpha1 = para[4];
beta20 = para[5];
beta21 = para[6];
beta22 = para[7];
alpha2 = para[8];
U_Cur = exp(beta10 + beta11*dv + beta12*dn + alpha1*x);
U_Tgt = exp(beta20 + beta21*dv_T + beta22*dn_T + alpha2*x);
# Update Start - 03/10/2021
if (is.infinite(U_Cur))
{
if (U_Cur<0)
{U_Cur=-2^1000}
else
{U_Cur=2^1000}
}
if (is.infinite(U_Tgt))
{
if (U_Tgt <0)
{U_Tgt =-2^1000}
else
{U_Tgt =2^1000}
}
# Update End - 03/10/2021
# print(U_Cur)
# print(U_Tgt)
P <- (U_Cur/(U_Cur+U_Tgt))
# print(P)
return(P)
}
integrand2 <− function(para1) {integrate( integrand, lower = -2 , upper = 2, para=para1)$value }
MLE<-maxLik(logLik=integrand2,start=c(0.5, 0.2, 0.1, 0.3, -0.5, 0.2, 0.1, 0.6))
After debug, i find the reason of the error is the value of U_Cur or U_Tgt sometimes may be 0 or Inf.
I have no idea about how to deal with that, I really appreciate any related proposal.
===================================
EDIT:
I have update the value of U_Cur, U_Tgt, if they are (-)infinite, set their value as (-)2^1000 (the code is added in the function body above, between “# Update - 03/10/2021”).
I could be misreading your code, but it looks like you are maximizing the following (log)likelihood:
LL = int^2_{-2} exp(5+x+theta) dx = exp(5+theta)int^2_{-2} exp(x) dx.
That likelihood will be maximized by sending theta to infinity. Is that the log likelihood you want?
Related
(R) Error in optim - attempt to apply non-function, when function is defined
not sure what I'm doing wrong here. I'm trying to get a cross-validation score for a mixture-of-two-gammas model.
llikGammaMix2 = function(param, x) {
if (any(param < 0) || param["p1"] > 1) {
return(-Inf)
} else {
return(sum(log(
dgamma(x, shape = param["k1"], scale = param["theta1"]) *
param["p1"] + dgamma(x, shape = param["k2"], scale = param["theta2"]) *
1
(1 - param["p1"])
)))
}
}
initialParams = list(
theta1 = 1,
k1 = 1.1,
p1 = 0.5,
theta2 = 10,
k2 = 2
)
for (i in 1:nrow(cichlids)) {
SWS1_training <- cichlids$SWS1 - cichlids$SWS1[i]
SWS1_test <- cichlids$SWS1[i]
MLE_training2 <-
optim(
par = initialParams,
fn = llikGammaMix2,
x = SWS1_training,
control = list(fnscale = -1)
)$par
LL_test2 <-
optim(
par = MLE_training2,
fn = llikGammaMix2,
x = SWS1_test,
control = list(fnscale = -1)
)$value
}
print(LL_test2)
This runs until it gets to the first optim(), then spits out Error in fn(par, ...) : attempt to apply non-function.
My first thought was a silly spelling error somewhere, but that doesn't seem to be the case. Any help is appreciated.
I believe the issue is in the return statement. It's unclear if you meant to multiply or add the last quantity (1 - param["p1"])))) to the return value. Based on being a mixture, I'm guessing you mean for it to be multiplied. Instead it just hangs at the end which throws issues for the function: return(sum(log(dgamma(x, shape = param["k1"], scale = param["theta1"]) * param["p1"] + dgamma(x, shape = param["k2"], scale = param["theta2"]) * (1 - param["p1"])))) ## ISSUE HERE: Is this what you meant? There could be other issues with the code. I would double check that the function you are optimizing is what you think it ought to be. It's also hard to tell unless you give a reproducible example we might be able to use. Try to clear up the above issue and let us know if there are still problems.
Storing information during optim()
I have a general function I have provided an example below if simple linear regression:
x = 1:30
y = 0.7 * x + 32
Data = rnorm(30, mean = y, sd = 2.5);
lin = function(pars = c(grad,cons)) {
expec = pars[1] * x + pars[2];
SSE = sum((Data - expec)^2)
return(SSE)
}
start_vals = c(0.2,10)
lin(start_vals)
estimates = optim(par = start_vals, fn = lin);
## plot the data
Fit = estimates$par[1] * x + estimates$par[2]
plot(x,Data)
lines(x, Fit, col = "red")
So that's straight forward. What I want is to store the expectation for the last set of parameters, so that once I have finished optimizing I can view them. I have tried using a global container and trying to populating it if the function is executed but it doesn't work, e.g
Expectation = c();
lin = function(pars = c(grad,cons)) {
expec = pars[1] * x + pars[2];
Expectation = expec;
SSE = sum((Data - expec)^2)
return(SSE)
}
start_vals = c(0.2,10)
estimates = optim(par = start_vals, fn = lin);
Expectation ## print the expectation that would relate to estimates$par
I know that this is trivial to do outside of the function, but my actual problem (which is analogous to this) is much more complex. Basically I need to return internal information that can't be retrospectively calculated. Any help is much appreciated.
you should use <<- instead of = in your lin function, Expectation <<- expec,The operators <<- and ->> are normally only used in functions, and cause a search to be made through parent environments for an existing definition of the variable being assigned.
R code: "Error in parse unexpected end of input" when using non-linear solver
I am using the non-linear solver (nls) in R, but cannot run my model due to a parser error that I have no idea how to debug. Could someone please offer some advice on how to fix this? Code: Bass.nls <- nls( Zt[which(!is.na(Zt))] ~ M * ( ((P+Q)^2 / P) * exp(-(P+Q) * days) ) / (1+(Q/P)*exp(-(P+Q)*days))^2, start = list(M=Z[tInt], P=0.03, Q=0.38), lower = list(Y[tInt], 0,0), upper = list(2e10, 1,1), algorithm = "port", trace = TRUE) Error: Error in parse(text = x, keep.source = FALSE) : :2:0: unexpected end of input 1: ~
The error Unexpected end of input occurs because of a missing paranthesis. You might have forgotten to finish the call to the function with a closing paranthesis - ")". Look at the section 6.4 of this link for a better explanation. You should write the script as described in Johannes' answer as it gets easy for you to debug the mistake then.
My advice would be to write a script that contains the call in a more readable form, like
Bass.f <- function(days, M, P, Q) {
M * (((P + Q)^2 / P) * exp(-(P + Q) * days)) /
(1 + (Q / P) * exp(-(P + Q) * days))^2
}
Then you can call the function to see if it is written correctly, like
Bass.f(1, 100, 0.03, 0.38)
and then try with the call
Bass.nls <- nls( Zt[which(!is.na(Zt))] ~ Bass.f(days, M, P, Q),
start = list(M = Z[tInt], P = 0.03, Q = 0.38),
lower = list(Y[tInt], 0, 0),
upper = list(2e10, 1, 1),
algorithm = "port", trace = TRUE)
Does this still give the same error? If yes, it would be useful to see the data you are using, i.e. the output of
dput(Zt)
Z[tInt]
days
R fit user defined distribution
I am trying to fit my own distribution to my data, find the optimum parameters of the distribution to match the data and ultimately find the FWHM of the peak in the distribution. From what I've read, the package fitdistrplus is the way to do this. I know the data takes the shape of a lorentzian peak on a quadratic background.
plot of the data:
plot of raw data
The raw data used:
data = c(0,2,5,4,5,4,3,3,2,2,0,4,4,2,5,5,3,3,4,4,4,3,3,5,5,6,6,8,4,0,6,5,7,5,6,3,2,1,7,0,7,9,5,7,5,3,5,5,4,1,4,8,10,2,5,8,7,14,7,5,8,4,2,2,6,5,4,6,5,7,5,4,8,5,4,8,11,9,4,8,11,7,8,6,9,5,8,9,10,8,4,5,8,10,9,12,10,10,5,5,9,9,11,19,17,9,17,10,17,18,11,14,15,12,11,14,12,10,10,8,7,13,14,17,18,16,13,16,14,17,20,15,12,15,16,18,24,23,20,17,21,20,20,23,20,15,20,28,27,26,20,17,19,27,21,28,32,29,20,19,24,19,19,22,27,28,23,37,41,42,34,37,29,28,28,27,38,32,37,33,23,29,55,51,41,50,44,46,53,63,49,50,47,54,54,43,45,58,54,55,67,52,57,67,69,62,62,65,56,72,75,88,87,77,70,71,84,85,81,84,75,78,80,82,107,102,98,82,93,98,90,94,118,107,113,103,99,103,96,108,114,136,126,126,124,130,126,113,120,107,107,106,107,136,143,135,151,132,117,118,108,120,145,140,122,135,153,157,133,130,128,109,106,122,133,132,150,156,158,150,137,147,150,146,144,144,149,171,185,200,194,204,211,229,225,235,228,246,249,238,214,228,250,275,311,323,327,341,368,381,395,449,474,505,529,585,638,720,794,896,919,1008,1053,1156,1134,1174,1191,1202,1178,1236,1200,1130,1094,1081,1009,949,890,810,760,690,631,592,561,515,501,489,467,439,388,377,348,345,310,298,279,253,257,259,247,237,223,227,217,210,213,197,197,192,195,198,201,202,211,193,203,198,202,174,164,162,173,170,184,170,168,175,170,170,168,162,149,139,145,151,144,152,155,170,156,149,147,158,171,163,146,151,150,147,137,123,127,136,149,147,124,137,133,129,130,128,139,137,147,141,123,112,136,147,126,117,116,100,110,120,105,91,100,100,105,92,88,78,95,75,75,82,82,80,83,83,66,73,80,76,69,81,93,79,71,80,90,72,72,63,57,53,62,65,49,51,57,73,54,56,78,65,52,58,49,47,56,46,43,50,43,40,39,36,45,28,35,36,43,48,37,36,35,39,31,24,29,37,26,22,36,33,24,31,31,20,30,28,23,21,27,26,29,21,20,22,18,19,19,20,21,20,25,18,12,18,20,20,13,14,21,20,16,18,12,17,20,24,21,20,18,11,17,12,5,11,13,16,13,13,12,12,9,15,13,15,11,12,11,8,13,16,16,16,14,8,8,10,11,11,17,15,15,9,9,13,12,3,11,14,11,14,13,8,7,7,15,12,8,12,14,9,5,2,10,8)
I have calculated the equations which define the distribution and cumulative distribution:
dFF <- function(x,a,b,c,A,gamma,pos) a + b*x + (c*x^2) + ((A/pi)*(gamma/(((x-pos)^2) + (gamma^2))))
pFF <- function(x,a,b,c,A,gamma,pos) a*x + (b/2)*(x^2) + (c/3)*(x^3) + A/2 + (A/pi)*(atan((x - pos)/gamma))
I believe these to be correct. From what I understand, a distribution fit should be possible using just these definitions using the fitdist (or mledist) method:
fitdist(data,'FF', start = list(0,0.3,-0.0004,70000,13,331))
mledist(data,'FF', start = list(0,0.3,-0.0004,70000,13,331))
This returns the statement 'function cannot be evaluated at initial parameters> Error in fitdist(data, "FF", start = list(0, 0.3, -4e-04, 70000, 13, 331)):the function mle failed to estimate the parameters, with the error code 100' in the first case and in the second I just get a list of 'NA' values for the estimates.
I then calculated a function to give the quantile distribution values to use the other fitting methods (qmefit):
qFF <- function(p,a,b,c,A,gamma,pos)
{
qList = c()
axis = seq(1,600,1)
aF = dFF(axis,a,b,c,A,gamma,pos)
arr = histogramCpp(aF) # change data to a histogram format
for(element in 1:length(p)){
q = quantile(arr,p[element], names=FALSE)
qList = c(qList,q)
}
return(qList)
}
Part of this code requires calling the c++ function (by using the library Rcpp):
#include <Rcpp.h>
#include <vector>
#include <math.h>
using namespace Rcpp;
// [[Rcpp::export]]
std::vector<int> histogramCpp(NumericVector x) {
std::vector<int> arr;
double number, fractpart, intpart;
for(int i = 0; i <= 600; i++){
number = (x[i]);
fractpart = modf(number , &intpart);
if(fractpart < 0.5){
number = (int) intpart;
}
if(fractpart >= 0.5){
number = (int) (intpart+1);
}
for(int j = 1; j <= number; j++){
arr.push_back(i);
}
}
return arr;
}
This c++ method just turns the data into a histogram format. If the first element of the vector describing the data is 4 then '1' is added 4 times to the returned vector etc. . This also seems to work as sensible values are returned. plot of the quantile function:
Plot of quantiles returned for probabilities from 0 to 1 in steps of 0.001
The 'qmefit' method can then be attempted through the fitdist function:
fitdist(data,'FF', start = list(0,0.3,-0.0004,70000,13,331), method = 'qme', probs = c(0,0.3,0.4,0.5,0.7,0.9))
I chose the 'probs' values randomly as I don't fully understand their meaning. This either straight-up crashes the R session or after a brief stuttering returns a list of 'NA' values as estimates and the line <std::bad_alloc : std::bad_alloc>
I am not sure if I am making a basic mistake here and any help or recommendations are appreciated.
In the end I managed to find a work-around for this using the rPython package and lmfit from python. It solved my issue and might be useful for others with the same issue. The R-code was as follows:
library(rPython)
python.load("pyFit.py")
python.assign("row",pos)
python.assign("vals",vals)
python.exec("FWHM,ERROR,FIT = fitDist(row,vals)")
FWHM = python.get("FWHM")
ERROR = python.get("ERROR")
cFIT = python.get("FIT")
and the called python code was:
from lmfit import Model, minimize, Parameters, fit_report
from sklearn import mixture
import numpy as np
import matplotlib.pyplot as plt
import math
def cauchyDist(x,a,b,c,d,e,f,g,A,gamma,pos):
return a + b*x + c*pow(x,2) + d*pow(x,3) + e*pow(x,4) + f*pow(x,5) + g*pow(x,6) + (A/np.pi)*(gamma/((pow((x-pos),2)) + (pow(gamma,2))))
def fitDist(row, vals):
gmod = Model(cauchyDist)
x = np.arange(0,600)
result = gmod.fit(vals, x=x, a = 0, b = 0.3, c = -0.0004, d = 0, e = 0, f= 0, g = 0, A = 70000, gamma = 13, pos = row)
newFile = open('fitData.txt', 'w')
newFile.write(result.fit_report())
newFile.close()
with open('fitData.txt', 'r') as inF:
for line in inF:
if 'gamma:' in line:
j = line.split()
inF.close()
FWHM = float(j[1])
error = float(j[3])
fit = result.best_fit
fit = fit.tolist()
return FWHM, error, fit
I increased the order of polynomial to obtain a better fit for the data and returned the FWHM, its error and the values for the fit. There are likely much better ways of achieving this but the final fit is as I needed.
Final fit. Red data points are raw data, the black line is the fitted distribution.
Catching the print of the function
I am using package fda in particular function fRegress. This function includes another function that is called eigchk and checks if coeffients matrix is singular.
Here is the function as the package owners (J. O. Ramsay, Giles Hooker, and Spencer Graves) wrote it.
eigchk <- function(Cmat) {
# check Cmat for singularity
eigval <- eigen(Cmat)$values
ncoef <- length(eigval)
if (eigval[ncoef] < 0) {
neig <- min(length(eigval),10)
cat("\nSmallest eigenvalues:\n")
print(eigval[(ncoef-neig+1):ncoef])
cat("\nLargest eigenvalues:\n")
print(eigval[1:neig])
stop("Negative eigenvalue of coefficient matrix.")
}
if (eigval[ncoef] == 0) stop("Zero eigenvalue of coefficient matrix.")
logcondition <- log10(eigval[1]) - log10(eigval[ncoef])
if (logcondition > 12) {
warning("Near singularity in coefficient matrix.")
cat(paste("\nLog10 Eigenvalues range from\n",
log10(eigval[ncoef])," to ",log10(eigval[1]),"\n"))
}
}
As you can see last if condition checks if logcondition is bigger than 12 and prints then the ranges of eigenvalues.
The following code implements the useage of regularization with roughness pennalty. The code is taken from the book "Functional data analysis with R and Matlab".
annualprec = log10(apply(daily$precav,2,sum))
tempbasis =create.fourier.basis(c(0,365),65)
tempSmooth=smooth.basis(day.5,daily$tempav,tempbasis)
tempfd =tempSmooth$fd
templist = vector("list",2)
templist[[1]] = rep(1,35)
templist[[2]] = tempfd
conbasis = create.constant.basis(c(0,365))
betalist = vector("list",2)
betalist[[1]] = conbasis
SSE = sum((annualprec - mean(annualprec))^2)
Lcoef = c(0,(2*pi/365)^2,0)
harmaccelLfd = vec2Lfd(Lcoef, c(0,365))
betabasis = create.fourier.basis(c(0, 365), 35)
lambda = 10^12.5
betafdPar = fdPar(betabasis, harmaccelLfd, lambda)
betalist[[2]] = betafdPar
annPrecTemp = fRegress(annualprec, templist, betalist)
betaestlist2 = annPrecTemp$betaestlist
annualprechat2 = annPrecTemp$yhatfdobj
SSE1.2 = sum((annualprec-annualprechat2)^2)
RSQ2 = (SSE - SSE1.2)/SSE
Fratio2 = ((SSE-SSE1.2)/3.7)/(SSE1/30.3)
resid = annualprec - annualprechat2
SigmaE. = sum(resid^2)/(35-annPrecTemp$df)
SigmaE = SigmaE.*diag(rep(1,35))
y2cMap = tempSmooth$y2cMap
stderrList = fRegress.stderr(annPrecTemp, y2cMap, SigmaE)
betafdPar = betaestlist2[[2]]
betafd = betafdPar$fd
betastderrList = stderrList$betastderrlist
betastderrfd = betastderrList[[2]]
As penalty factor the authors use certain lambda.
The following code implements the search for the appropriate `lambda.
loglam = seq(5,15,0.5)
nlam = length(loglam)
SSE.CV = matrix(0,nlam,1)
for (ilam in 1:nlam) {
lambda = 10ˆloglam[ilam]
betalisti = betalist
betafdPar2 = betalisti[[2]]
betafdPar2$lambda = lambda
betalisti[[2]] = betafdPar2
fRegi = fRegress.CV(annualprec, templist,
betalisti)
SSE.CV[ilam] = fRegi$SSE.CV
}
By changing the value of the loglam and cross validation I suppose to equaire the best lambda, yet if the length of the loglam is to big or its values lead the coefficient matrix to singulrity. I recieve the following message:
Log10 Eigenvalues range from
-5.44495317739048 to 6.78194912518214
Created by the function eigchk as I already have mentioned above.
Now my question is, are there any way to catch this so called warning? By catch I mean some function or method that warns me when this has happened and I could adjust the values of the loglam. Since there is no actual warning definition in the function beside this print of the message I ran out of ideas.
Thank you all a lot for your suggestions.
By "catch the warning", if you mean, will alert you that there is a potential problem with loglam, then you might want to look at try and tryCatch functions. Then you can define the behavior you want implemented if any warning condition is satisfied. If you just want to store the output of the warning (which might be assumed from the question title, but may not be what you want), then try looking into capture.output.