Is there a way to solve a poly() by declaring the variable? - scilab

I'm new with the scilab synthax, and I want to know if there is a way to
declare a poly, like p = 3x + 2, and use something like p(5) to get 17 as an answer. The reason to do so is that the poly synthax is a lot easier than defining the same expression over and over again.

In scilab polynom can be define with special function poly. In your case, this command will look like this:
x=poly(0,"x");
p = 3*x+2
for obtain value in point need use function horner:
horner(p,5)
and we get
ans =
17.

Related

Function taking Vectors and Scalars

I have a function that takes a vector
function foo(x::Vector{Int64})
x
end
How can I make it work also for scalars (i.e. turn them into one-element vectors)?
I know that I can do this:
foo(x::Int64) = foo([x])
but this stops being cool when there are more arguments, because you're writing multiple methods to achieve only one thing.
I think I something like foo(x::Union{Int64, Vector{Int64}}), but I don't know where or how it works or if it is the right thing to do.
Can anyone help?
You can make a helper function which either converts or does nothing. Then the main function can accept any combination:
_vec(x::Number) = [x]
_vec(x::AbstractVector) = x
function f(x, y, z) # could specify ::Union{Number, AbstractVector}
xv = _vec(x)
yv = _vec(y)
...
end
The ... could do the actual work, or could call f(xv, yv, zv) where another method f(x::AbstractVector, y::AbstractVector, z::AbstractVector) does the work --- whichever seems cleaner.
The main time this comes up is if the version of your function for vectors does the same thing for all of it's elements. In this case, what you want to do is define f(x::Int), and use broadcasting f.([1,2,3]) for the vector case.

How to use Simulated Annealing in R (GenSA) for a function with discrete variables with a few options without pattern?

I want to use Simulated Annealing. My objective function exist of multiple variables, for some of them there are only a few options possible. I saw the same question on Stack here:
How to use simulated annealing for a function with discrete paremeters?, but there was no answer but a reference to: How to put mathematical constraints with GenSA function in R.
I don't understand how to apply the advice from the second link to my situation (but I think the answer can be found there).
For example:
v <- c(50, 50, 25, 25)
lower <- c(0,0,0,20)
upper <- c(100,100,50,40)
out <- GenSA(v, lower = lower, upper = upper, fn = efficientFunction)
Assume that the fourth parameter, v[4], only can be in {20,25,30,35,40}. They suggested the use of Lagrange multipliers, hence, I was thinking of something like: lambda * ceil(v[4] / 5). Is this a good idea ?
But what can I do it the sample space of a variable does not have a nice pattern, for example third parameter, v[3], only can be in {0,21,33,89,100}. I don't understand why a Lagrange multiplier can help in this situation. Do I need to make the form of my parameters different that they follow a pattern or is there another option?
In case Lagrange multipliers are the only option, I'll end up with with 8 of these formulations in my objective. It seems to me that there is another option, but I don't know how!
With kind regards and thanks in advance,
Roos
With SA, you could start with a very simple neighbourhood sheme,
pick 1 of the parameters, and change it by selecting a new valid setting, 1 above, or 1 below the current one (we assume that they have a order, like I feel is your case).
There are no Lagrange multipliers involved in SA as I know. But there are many variations and maybe some with Constrainsts or other make use of them.

Function doesn't change value (R)

I have written a function that takes two arguments, a number between 0:16 and a vector which contains four parameter values.
The output of the function does change if I change the parameters in the vector, but it does not change if I change the number between 0:16.
I can add, that the function I'm having troubles with, includes another function (called 'pi') which takes the same arguments.
I have checked that the 'pi' function does actually change values if I change the value from 0:16 (and it does also change if I change the values of the parameters).
Firstly, here is my code;
pterm_ny <- function(x, theta){
(1-sum(theta[1:2]))*(theta[4]^(x))*exp((-1)*theta[4])/pi(x, theta)
}
pi <- function(x, theta){
theta[1]*1*(x==0)+theta[2]*(theta[3]^(x))*exp((-1)*(theta[3]))+(1-
sum(theta[1:2]))*(theta[4]^(x))*exp((-1)*(theta[4]))
}
Which returns 0.75 for pterm_ny(i,c(0.2,0.2,2,2)), were i = 1,...,16 and 0.2634 for i = 0, which tells me that the indicator function part in 'pi' does work.
With respect to raising a number to a certain power, I have been told that one should wrap the wished number in a 'I', as an example it would be like;
x^I(2)
I have tried to do that in my code, but that didn't help either.
I can't remember the argument for doing it, but I expect that it's to ensure that the number in parentheses is interpreted as an integer.
My end goal is to get 17 different values of the 'pterm' and to accomplish that, I was thinking of using the sapply function like this;
sapply(c(0:16),pterm_ny,theta = c(0.2,0.2,2,2))
I really hope that someone can point out what I'm missing here.
In advance, thank you!
You have a theta[4]^x term both in your main expression and in your pi() function; these are cancelling out, leaving the result invariant to changes in x ...
Also:
you might want to avoid using pi as your function name, as it's also a built-in variable (3.14159...) - this can sometimes cause confusion
the advice about using the "as is" function I() to protect powers is only relevant within formulas, e.g. as used in lm() (linear regression). (It would be used as I(x^2), not x^I(2)

How to make nonsymbolic plot_vector_field in sage?

I have a function f(x,y) whose outcome is random (I take mean from 20 random numbers depending on x and y). I see no way to modify this function to make it symbolic.
And when I run
x,y = var('x,y')
d = plot_vector_field((f(x),x), (x,0,1), (y,0,1))
it says it can't cast symbolic expression to real or rationa number. In fact it stops when I write:
a=matrix(RR,1,N)
a[0]=x
What is the way to change this variable to real numbers in the beginning, compute f(x) and draw a vector field? Or just draw a lot of arrows with slope (f(x),x)?
I can create something sort of like yours, though with no errors. At least it doesn't do what you want.
def f(m,n):
return m*randint(100,200)-n*randint(100,200)
var('x,y')
plot_vector_field((f(x,y),f(y,x)),(x,0,1),(y,0,1))
The reason is because Python functions immediately evaluate - in this case, f(x,y) was 161*x - 114*y, though that will change with each invocation.
My suspicion is that your problem is similar, the immediate evaluation of the Python function once and for all. Instead, try lambda functions. They are annoying but very useful in this case.
var('x,y')
plot_vector_field((lambda x,y: f(x,y), lambda x,y: f(y,x)),(x,0,1),(y,0,1))
Wow, I now I have to find an excuse to show off this picture, cool stuff. I hope your error ends up being very similar.

Using outer() with a multivariable function

Suppose you have a function f<- function(x,y,z) { ... }. How would you go about passing a constant to one argument, but letting the other ones vary? In other words, I would like to do something like this:
output <- outer(x,y,f(x,y,z=2))
This code doesn't evaluate, but is there a way to do this?
outer(x, y, f, z=2)
The arguments after the function are additional arguments to it, see ... in ?outer. This syntax is very common in R, the whole apply family works the same for instance.
Update:
I can't tell exactly what you want to accomplish in your follow up question, but think a solution on this form is probably what you should use.
outer(sigma_int, theta_int, function(s,t)
dmvnorm(y, rep(0, n), y_mat(n, lambda, t, s)))
This calculates a variance matrix for each combination of the values in sigma_int and theta_int, uses that matrix to define a dennsity and evaluates it in the point(s) defined in y. I haven't been able to test it though since I don't know the types and dimensions of the variables involved.
outer (along with the apply family of functions and others) will pass along extra arguments to the functions which they call. However, if you are dealing with a case where this is not supported (optim being one example), then you can use the more general approach of currying. To curry a function is to create a new function which has (some of) the variables fixed and therefore has fewer parameters.
library("functional")
output <- outer(x,y,Curry(f,z=2))

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