I am having some trouble using correctly the ForwardDiff package in Julia. I have managed to isolate my problem in the following chunk of code.
In short, I define the function:
using ForwardDiff
function likelihood(mu,X)
N = size(X,2)
# Calculate likelihood
aux = zeros(N)
for nn=1:N
aux[nn] = exp(-0.5 * (X[:,nn]-mu)' * (X[:,nn]-mu))[1]
end
# return log-likelihood
return sum(log(aux))
end
I then check if the function works:
# Check if function works at all
X = randn(2,3) # some random data
mu = [1.0;2.0] # arbitrary mean
#show likelihood(mu,X) # works fine for me
I then try to obtain the gradient using:
ForwardDiff.gradient( ARG -> likelihood(ARG, X), mu)
Unfortunately this fails and I see in my screen:
ERROR: MethodError: convert has no method matching
convert(::Type{Float64}, ::ForwardDiff.Dual{2,Float64}) This may have
arisen from a call to the constructor Float64(...), since type
constructors fall back to convert methods. Closest candidates are:
call{T<:AbstractFloat}(::Type{T<:AbstractFloat}, ::Real,
::RoundingMode{T}) call{T}(::Type{T}, ::Any)
convert(::Type{Float64}, ::Int8) ... in likelihood at none:10 in
anonymous at none:1
What am I doing wrong? Thanks, in advance.
I was just informed that this was a careless mistake on my side, though a bit hard to spot to the untrained eye.
The error is at the call to zeros:
aux = zeros(N)
Changing this to
aux = zeros(eltype(mu),N)
solves the problem. Hope this is useful to others.
Related
I have derived a survival function for a system of components (ignore the details of how this system is setup) and I am trying to maximize its expected, or more specifically, maximizing the expected value of the function:
surv_func = function(x,mu) = {(exp(-(x/(mu))^(1/3))*((1-exp(-(4/3)*x^(3/2)))+exp(-(-(4/3)*x^(3/2)))))*exp(-(x/(3-mu))^(1/3))}
and I am supposed (since the pdf including my tasks gives a hint about it) to use the function
optimize()
and the expected value for a function can be computed with
# Computes expected value of a the function "function"
E <- integrate(function, 0, Inf)
but my function depends on x and mu. The expected value could (obviously) be computed if the integral had no mu but instead only depended on x. For those interested, the mu comes from the fact that one of the components has a Weibull-distribution with parameters (1/3,mu) and the 3-mu comes from that has a Weibull-distribution with parameters (1/3,lambda). In the task there is a constraint mu + lambda = 3, so I tought substituting the lambda-parameter in the second Weibull-distribution with lambda = 3 - mu and trying to maximize this problem would yield not only mu, but also lambda.
If I try to, just for the sake of learing about R, compute the expected value using the code below (in the console window), it just gives me the following:
> E <- integrate(surv_func,0,Inf)
Error in (function (x, mu) : argument "mu" is missing, with no default
I am new to R and seem to be a little bit "slow" at learning. How can I approach this problem?
Here is my code:
using Plots
using SpecialFunctions
using QuadGK
kappa = 1
B = 1
xi = (kappa/B)^4
function correlation(x)
quadgk(q -> q * SpecialFunctions.besselj0(x*q)/(q^4 + xi), 0, 1e6)[1]/kappa
end
r = range(-20, 20, length = 1001)
plot(r, correlation(r))
I get an error on the Bessel function. I get that the argument is the problem and that it should be of the format ::BigFloat, ::Float16, or ::Float32, but I don't know how to do it. I tried to write x .* q instead, but the problem remains the same, I get the error:
ERROR: MethodError: no method matching besselj0(::StepRangeLen{Float64,Base.TwicePrecision{Float64},Base.TwicePrecision{Float64}})
Also, I'm searching for a way to write +infinity instead of 1e6.
Just replace correlation(r) with correlation.(r) in your code to use broadcasting, as is explained here.
The core of your problem that in Julia functions are not broadcastable by default - you usually have to opt-in to have it (especially when you work with numerical code). Here is a basic example:
julia> sin(1)
0.8414709848078965
julia> sin([1])
ERROR: MethodError: no method matching sin(::Array{Int64,1})
Closest candidates are:
sin(::BigFloat) at mpfr.jl:727
sin(::Missing) at math.jl:1197
sin(::Complex{Float16}) at math.jl:1145
...
Stacktrace:
[1] top-level scope at REPL[2]:1
julia> sin.([1])
1-element Array{Float64,1}:
0.8414709848078965
However, in your case the correlation function is quite expensive. In such a case I usually use ProgressMeter.jl to monitor the progress of the computations (it shows how long you can expect for the computations to finish). So you can write:
using ProgressMeter
result = #showprogress map(correlation, r)
and use the map function to apply correlation function to all elements of r (in this case the result will be the same as for broadcasting).
Finally, your computations will be much faster if you do not use global variables in quadgk. It is better to pass kappa and xi as arguments to the function like this:
function correlation(x, kappa, xi)
quadgk(q -> q * SpecialFunctions.besselj0(x*q)/(q^4 + xi), 0, 1e6)[1]/kappa
end
result = #showprogress map(x -> correlation(x, kappa, xi), r)
I am taking a numeric calculus class and we are not required to know any scilab programming except the very basic, which is taught through a booklet, since the class is mostly theoretical. I was reading the booklet and found this scilab code meant to find a root of a function through bissection method.
The problem is, I can't find a way to make it work. I tried to call it with bissecao(x,-1,1,0.1,40) however it didn't work.
The error I got was:
at line 3 of function bissecao ( E:\Downloads\bisseccao3.sce line 3 )
Invalid index.
As I highly doubt that the code itself isn't working, and I tried to search for anything I could spot that seemed wrong, to no avail, I guess I am probably calling it wrong, somehow.
The code is the following:
function p = bissecao(f, a, b, TOL, N)
i = 1
fa = f(a)
while (i <= N)
//iteraction of the bissection
p = a + (b-a)/2
fp = f(p)
//stop condition
if ((fp == 0) | ((b-a)/2 < TOL)) then
return p
end
//bissects the interval
i = i+1
if (fa * fp > 0) then
a = p
fa = fp
else
b = p
end
end
error ('Max number iter. exceded!')
endfunction
Where f is a function(I guess), a and b are the limits of the interval in which we will be iterating, TOL is the tolerance at which the program terminates close to a zero, and N is the maximum number of iteractions.
Any help on how to make this run is greatly appreciated.
Error in bissecao
The only error your bissecao function have is the call to return :
In a function return stops the execution of the function,
[x1,..,xn]=return(a1,..,an) stops the execution of the function and
put the local variables ai in calling environment under names xi.
So you should either call it without any argument (input our output) and the function will exit and return p.
Or you could call y1 = return(p) and the function will exit and p will be stored in y1.
It is better to use the non-arguments form return in functions to avoid changing values of variables in the parent/calling script/functions (possible side-effect).
The argument form is more useful when interactively debugging with pause:
In pause mode, it allows to return to lower level.
[x1,..,xn]=return(a1,..,an) returns to lower level and put the local
variables ai in calling environment under names xi.
Error in calling bissecao
The problem may come by your call: bissecao(x,-1,1,0.1,40) because you didn't defined x. Just fixing this by creating a function solves the problem:
function y=x(t)
y=t+0.3
enfunction
x0=bissecao(x,-1,1,0.1,40) // changed 'return p' to 'return'
disp(x0) // gives -0.3 as expected
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.
Suppose I want to integrate some function that involves sums and products of a few other user defined functions. Lets take an extremely simple example, it gives the same error.
integrate(f = sin + cos, lower=0, upper=1)
This yields "Error in sin + cos : non-numeric argument to binary operator" which I think is saying it doesn't make sense to just add functions together without passing them some sort of argument. So I am a bit stuck here. This thread poses what I think is a solution to a more complicated question, that can be applied here, but it seems long for such a simple task in this case. I'm actually kind of surprised that I am unable to find passing function arguments to functions in the help manual so I think I am not using the right terminology.
Just write your own function:
> integrate(f = function(x) sin(x) + cos(x), lower=0, upper=1)
1.301169 with absolute error < 1.4e-14
In this example I've used an anonymous function, but that's not necessary. The key is to write a function that represents whatever function you want to integrate over. In this case, the function should take a vector input and add the sin and cos of each element.
Equivalently, we could have done:
foo <- function(x){
sin(x) + cos(x)
}
integrate(f = foo, lower=0, upper=1)
This is an old question, but I recently struggled with it, so here is a simple example in case it helps others in the future. #joran's answer is still the best.
Define your first function: f1 <- function(x){return(x*2)}
Test it: f1(8) (expect 8*2=16); returns [1] 16
Define your second function: f2 <-function(f, y){return(f+y)}
Test it: f2(f=f1(8), y=1) (expect 8*2 = 16 +1 = 17); returns [1] 17