I'm using the future.callr that creates a new thread(?) every time a future is requested so it is calculated separately and the main R script can keep moving on.
I'm getting the following warning when my futures come back:
Warning message:
UNRELIABLE VALUE: Future (‘<none>’) unexpectedly generated random numbers without specifying argument 'seed'. There is a risk that those random numbers are not statistically sound and the overall results might be invalid. To fix this, specify 'seed=TRUE'. This ensures that proper, parallel-safe random numbers are produced via the L'Ecuyer-CMRG method. To disable this check, use 'seed=NULL', or set option 'future.rng.onMisuse' to "ignore".
In the actual code I'm running, it's just loading some data and I don't know why (or care EDIT: I do care, see comments below) that it's generating random numbers. How do I stop that warning from being shown (either fixing the rng generation or just ignoring it)?
I've got a lot of lines with futures so I am hoping to be able to just set the option at the beginning somehow and not have to add it to every line.
Here's an example and my attempt to ignore the warning.
library(future.callr)
set.seed(1234567)
future.seed = TRUE
#normal random number - no problem
a<-runif(1)
print(a)
#random number in future, using callr plan
plan(callr, future.rng.onMisuse = 'ignore')
b %<-% runif(1)
print(b)
See help("%seed%", package = "future").
You could either use %seed% like below:
b %<-% runif(1) %seed% TRUE # seed is set by future pkg
print(b)
# or
b %<-% runif(1) %seed% 1234567 # your seed
print(b)
Or to disable checking
options(future.rng.onMisuse = "ignore")
b %<-% runif(1)
print(b)
Related
How to reduce the number of iterations for PAM clustering algorithm in the cluster package?
I am trying to produce a couple of plots showing how pam works, so trying to reduce the number of iterations to 2. I have cloned the cluster repo to my working directory, where I have edited the pam.q file (directory ./cluster/R) for nMax to be equal to 2.
# original
nMax <- 65536 # 2^16 (as 1+ n(n-1)/2 must be < max_int = 2^31-1)
# modified
nMax <- 2
However, even with no changes applied to the original file, pam algorithm fails to run. If I load it by typing in library(cluster) instead, it works as supposed, but this way I have no ability to manipulate the number of iterations.
Sample code of what I'm trying to achieve is displayed below:
# -- Working code --
library(datasets)
data(iris)
library(cluster)
df <- data.frame(iris$Petal.Length, iris_modified$Petal.Width)
pam.res <- pam(df, k = 2)
pam.res
# -- Failing Code --
library(datasets)
data(iris)
source("./cluster/R/pam.q")
df <- data.frame(iris$Petal.Length, iris_modified$Petal.Width)
pam.res <- pam(df, k = 2)
pam.res
This is the error I'm getting, when running the "Failing Code" above:
Error in pam(clust_ex, k = 2) : object 'cl_Pam' not found
I expect the same output as for the working code, when I am linking the pam.q file directly instead of loading the library.
Is there something I'm not doing quite right in the way I import the q file? Or is there another way to change the number of iterations the pam algorithm performs?
Nmax is the maximum number of objects.
Its not the maximum number of iterations.
Is also not sufficient to just modify the .q file.
It's probably easier to do this with ELKI...
Issue 1
I have an objective function, gFun(modelOutput,l,u), which returns 0 if the simulated output is in interval [l,u], otherwise it returns a positive(!) number.
OFfun <- function(params) {
out <- simulate(params)
OF <- gFun(out,0,5)
return(OF)
}
The objective function is called from the optim function with some tolerance settings.
fitval=optim(par=parms,fn=OFfun,method="SANN",control = list(abstol = 1e-2))
summary(fitval)
My issue is that the optimization doesn't stop if the OFfun == 0.
I have tried with the condition below:
if (OF == 0){
opt <- options(show.error.messages=FALSE)
on.exit(options(opt))
stop()
}
it works but it doesn't return the OF back to optim and therefore I don't get the fitval info with estimated parameters.
Issue 2
Another issue is that the solver sometimes crashes and aborts the entire optimisation. I would like to harvest many solution sets for different initial guesses - so I need to handle failed simulations. probably related to issue 1.
Any advice would be very appreciated.
I am using the function plkhci from library Bhat to construct Profile-likelihood based confidence intervals and I got this warning:
Warning message: In dqstep(list(label = x$label, est = btrf(xt, x$low,
x$upp), low = x$low, : oops: unable to find stepsize, use default
when i run
r <- dfp(x,f=nlogf)
Can I ignore this warning as I still can get the output?
Following is the complete coding:
library(Bhat)
beta0<--8
beta1<-0.03
gamma<-0.0105
alpha<-0.05
n<-100
u<-runif(n)
u
x<-rnorm(n)
x
c<-rexp(100,1/1515)
c
t1<-(1/gamma)*log(1-((gamma/(exp(beta0+beta1*x)))*(log(1-u))))
t1
t<-pmin(t1,c)
t
delta<-1*(t1>c)
delta
length(delta)
cp<-length(delta[delta==1])/n
cp
delta[delta==1]<-ifelse(rbinom(length(delta[delta==1]),1,0.5),1,2)
delta
deltae<-ifelse(delta==0, 1,0)
deltar<-ifelse(delta==1, 1,0)
deltai<-ifelse(delta==2, 1,0)
dat=data.frame(t,delta, deltae,deltar,deltai,x)
dat$interval[delta==2] <- as.character(cut(dat$t[delta==2], breaks=seq(0, 600, 100)))
labs <- cut(dat$t[delta==2], breaks=seq(0, 600, 100))
dat$lower[delta==2]<-as.numeric( sub("\\((.+),.*", "\\1", labs) )
dat$upper[delta==2]<-as.numeric( sub("[^,]*,([^]]*)\\]", "\\1", labs) )
data0<-dat[which(dat$delta==0),]#uncensored data
data1<-dat[which(dat$delta==1),]#right censored data
data2<-dat[which(dat$delta==2),]#interval censored data
nlogf<-function(para)
{
b0<-para[1]
b1<-para[2]
g<-para[3]
e<-sum((b0+b1*data0$x)+g*data0$t+(1/g)*exp(b0+b1*data0$x)*(1-exp(g*data0$t)))
r<-sum((1/g)*exp(b0+b1*data1$x)*(1-exp(g*data1$t)))
i<-sum(log(exp((1/g)*exp(b0+b1*data2$x)*(1-exp(g*data2$lower)))-exp((1/g)*exp(b0+b1*data2$x)*(1-exp(g*data2$upper)))))
l<-e+r+i
return(-l)
}
x <- list(label=c("beta0","beta1","gamma"),est=c(-8,0.03,0.0105),low=c(-10,0,0),upp=c(10,1,1))
r <- dfp(x,f=nlogf)
x$est <- r$est
plkhci(x,nlogf,"beta0")
plkhci(x,nlogf,"beta1")
plkhci(x,nlogf,"gamma")
I am giving you a super long answer, but it will help you see that you can chase down your own error messages (most of the time, sometimes this means of looking at functions will not work). It is good to see what is happening inside a method when it throws an warning because sometimes it is fine and sometimes you need to fix your data.
This function is REALLY involved! You can look at it by typing dfp into the R command line (NO TRAILING PARENTHESES) and it will print out the whole function.
17 lines from the end, you will see an assignment:
del <- dqstep(x, f, sens = 0.01)
You can see that this calls the function dqstep, which is reflected in your warning.
You can see this function by typing dqstep into the command line of R again. In reading through this function, also long but not so tedious, there is this section of boolean logic:
if (r < 0 | is.na(r) | b == 0) {
warning("oops: unable to find stepsize, use default")
cat("problem with ", x$label[i], "\n")
break
}
This is the culprit, it returns the message you are getting. The line right above it spells out how r is calculated. You are feeding this function your default x from the prior function plus a sensitivity equations (which I assume dfp generates, it is huge and ugly, so I did not untangle all of it). When the previous nested function returns either an r value lower than Zero, and r value of NA or a b value of ZERO, that message is displayed.
The second error tells you that it was likely b==0 because b is in the denominator and it returned and infinity value, so NO STEP SIZE IS RETURNED FROM THIS NESTED FUNCTION to the variable del in dfp.
The step is fed into THIS equation:
h <- logit.hessian(x, f, del, dapprox = FALSE, nfcn)
which you can look into by typing logit.hessian into the R commandline.
When you do, you see that del is a step size in a logit scale, with a default value of del=rep(0.002, length(x$est))...which the function set for you because running the function dqstep returned no value.
So, you now get to decide if using that step size in the calculation of your confidence interval seems right or if there is a problem with your data which needs resolving to make this work better for you.
When I ran it, line by line, I got this message:
Error in if (denom <= 0) { : missing value where TRUE/FALSE needed
at this line of code:
r <- dfp(x,f=nlogf(x))
Which makes me think I was correct.
That is how I chase down issues I have with messages from packages when I get a message like yours.
I am using simulated annealing, as implemented in R's package GenSa (function GenSA), to search for values of input variables that result in "good values" (compared to some baseline) of a highly dimensional function. I noticed that setting maximum number of calls of the objective function has no effect on the running time. Am I doing something wrong or is this a bug?
Here is a modification of the example given in GenSA help file.
library(GenSA)
Rastrigin <- local({
index <- 0
function(x){
index <<- index + 1
if(index%%1000 == 0){
cat(index, " ")
}
sum(x^2 - 10*cos(2*pi*x)) + 10*length(x)
}
})
set.seed(1234)
dimension <- 1000
lower <- rep(-5.12, dimension)
upper <- rep(5.12, dimension)
out <- GenSA(lower = lower, upper = upper, fn = Rastrigin, control = list(max.call = 10^4))
Even though the max.call is specified to be 10,000, GenSA calls the objective function more than 46,000 times (note that the objective is called within a local environment in order to track the number of calls). The same problem rises when trying to specify the maximum running time via max.time.
This is an answer by the package maintainer :
max.call and max.time are soft limits that do not include local
searches that are performed before reaching these limits. The
algorithm does not stop the local search strategy loop before its end
and this may exceed the limitation that you have set but will stop
after that last search. We have designed the algorithm that way to
make sure that the algorithm isn't stopped in the middle of searching
valley. Such an option to stop anywhere will be implemented in the
next release of the package.
I'm relatively new in R and I would appreciated if you could take a look at the following code. I'm trying to estimate the shape parameter of the Frechet distribution (or inverse weibull) using mmedist (I tried also the fitdist that calls for mmedist) but it seems that I get the following error :
Error in mmedist(data, distname, start = start, fix.arg = fix.arg, ...) :
the empirical moment function must be defined.
The code that I use is the below:
require(actuar)
library(fitdistrplus)
library(MASS)
#values
n=100
scale = 1
shape=3
# simulate a sample
data_fre = rinvweibull(n, shape, scale)
memp=minvweibull(c(1,2), shape=3, rate=1, scale=1)
# estimating the parameters
para_lm = mmedist(data_fre,"invweibull",start=c(shape=3,scale=1),order=c(1,2),memp = "memp")
Please note that I tried many times en-changing the code in order to see if my mistake was in syntax but I always get the same error.
I'm aware of the paradigm in the documentation. I've tried that as well but with no luck. Please note that in order for the method to work the order of the moment must be smaller than the shape parameter (i.e. shape).
The example is the following:
require(actuar)
#simulate a sample
x4 <- rpareto(1000, 6, 2)
#empirical raw moment
memp <- function(x, order)
ifelse(order == 1, mean(x), sum(x^order)/length(x))
#fit
mmedist(x4, "pareto", order=c(1, 2), memp="memp",
start=c(shape=10, scale=10), lower=1, upper=Inf)
Thank you in advance for any help.
You will need to make non-trivial changes to the source of mmedist -- I recommend that you copy out the code, and make your own function foo_mmedist.
The first change you need to make is on line 94 of mmedist:
if (!exists("memp", mode = "function"))
That line checks whether "memp" is a function that exists, as opposed to whether the argument that you have actually passed exists as a function.
if (!exists(as.character(expression(memp)), mode = "function"))
The second, as I have already noted, relates to the fact that the optim routine actually calls funobj which calls DIFF2, which calls (see line 112) the user-supplied memp function, minvweibull in your case with two arguments -- obs, which resolves to data and order, but since minvweibull does not take data as the first argument, this fails.
This is expected, as the help page tells you:
memp A function implementing empirical moments, raw or centered but
has to be consistent with distr argument. This function must have
two arguments : as a first one the numeric vector of the data and as a
second the order of the moment returned by the function.
How can you fix this? Pass the function moment from the moments package. Here is complete code (assuming that you have made the change above, and created a new function called foo_mmedist):
# values
n = 100
scale = 1
shape = 3
# simulate a sample
data_fre = rinvweibull(n, shape, scale)
# estimating the parameters
para_lm = foo_mmedist(data_fre, "invweibull",
start= c(shape=5,scale=2), order=c(1, 2), memp = moment)
You can check that optimization has occurred as expected:
> para_lm$estimate
shape scale
2.490816 1.004128
Note however, that this actually reduces to a crude way of doing overdetermined method of moments, and am not sure that this is theoretically appropriate.