I am trying to create a function that computes the residuals of a system of equations using metaprogramming.
This is what I have tried so far (toy example):
function syst!(x::Vector, ou::Vector)
for i in 1:length(x)
eval(parse("ou[$i] = x[$i]^2 + x[$i]"))
end
return ou
end
However, when I try to compute the function, Julia says that the variable x is not defined. But if I include a println(parse("ou[$i] = x[$i]^2 + x[$i]")) I get the code that would be "typed" in the body of the function (sorry if I'm not using the correct technical CS terms, I come from the "scientific culture").
Anyways, it seems that the parseed x lives in another scope. How can I bring that parsed x to the scope of the function so that it represents the x from the arguments of syst!?
Bonus: I have a system of 700 equations and they are amenable to be "typed" using metaprogramming, what's the best way/technique to create a function that computes the residuals of the system? Was I on the right track?
Stefan's comment is right; in this specific example there is no need for metaprogramming. However, if you wanted to generate many lines similar to ou[i] = x[i]^2 + x[i] but different in complicated ways, you could generate them with a macro. See http://docs.julialang.org/en/release-0.4/manual/metaprogramming/. Macros expand to generated code "in place" as if you had typed it yourself, so variables can refer to the surrounding scope.
Related
I was remembering the haskell programming I learnt the last year and suddenly I had a little problem.
ghci> let test = [1,2,3,4]
ghci> test = drop 1 test
ghci> test
^CInterrupted.
I do not remember if it is possible.
Thanks!
test on the first line and test on the second line are not, in fact, the same variable. They're two different, separate, unrelated variables that just happen to have the same name.
Further, the concept of "saving in a variable" does not apply to Haskell. In Haskell, variables cannot be interpreted as "memory cells", which can hold values. Haskell's variables are more like mathematical variables - just names that you give to complex expressions for easier reasoning (well, this is a bit of an oversimplification, but good enough for now)
Consequently, variables in Haskell are immutable. You cannot change the value of a variable by assignment, like you can in many other languages. This property follows from interpreting the concept of "variable" in the mathematical sense, as described above.
Furthermore, definitions (aka "bindings") in Haskell are recursive. This means that the right side (the body) of a binding may refer to its left side. This is very handy for constructing infinite data structures, for example:
x = 42 : x
An infinite list of 42s
In your example, when you write test = drop 1 test, you're defining a list named test, which is completely unrelated to the list defined on the previous line, and which is equal to itself without the first element. It's only natural that trying to print such a list results in an infinite loop.
The bottom line is: you cannot do what you're trying to do. You cannot create a new binding, which shadows an existing binding, while at the same time references it. Just give it a different name.
Is there a function in Julia that is similar to the solver function in Excel where I can provide and equation, and it will solve for the unknown variable? If not, does anybody know the math behind Excel's solver function?
I am not expecting anybody to solve the equation, but if it helps:
Price = (Earnings_1/(1+r)^1)+(Earnings_2/(1+r)^2)++(Earnings_3/(1+r)^3)+(Earnings_4/(1+r)^4)+(Earnings_5/(1+r)^5)+(((Earnings_5)(RiskFreeRate))/((1+r)^5)(1-RiskFreeRate))
The known variables are: Price, All Earnings, and RiskFreeRate. I am just trying to figure out how to solve for r.
Write this instead as an expression f(r) = 0 by subtracting Price over to the other side. Now it's a rootfinding problem. If you only have one variable you're solving for (looks to be the case), then Roots.jl is a good choice.
fzero(f, a::Real, b::Real)
will search for a solution between a and b for example, and the docs have more choices for algorithms when you don't know a range to start with and only give an initial condition for example.
In addition, KINSOL in Sundials.jl is good when you know you're starting close to a multidimensional root. For multidimensional and needing some robustness to the initial condition, I'd recommend using NLsolve.jl.
There's nothing out of the box no. Root finding is a science in itself.
Luckily for you, your function has an analytic first derivative with respect to r. That means that you can use Newton Raphson, which will be extremely stable for your function.
I'm sure you're aware your function behaves badly around r = -1.
I started reading some of Haskell's documentation, and there's a fundamental concept I just don't understand. I read about it in other places as well, but I want to understand it once and for all.
In many places discussing functional programing, I keep reading that if the functions you're using are pure (have no side effects, and give same response for the same input at every call) then you can switch the order in which they are called when composing them, with it being guaranteed that the output of this composed call will remain the same regardless of the order.
For example, here is an entry from the Haskell Wiki:
Haskell is a pure language, which means that the result of any
function call is fully determined by its arguments. Pseudo-functions
like rand() or getchar() in C, which return different results on each
call, are simply impossible to write in Haskell. Moreover, Haskell
functions can't have side effects, which means that they can't effect
any changes to the "real world", like changing files, writing to the
screen, printing, sending data over the network, and so on. These two
restrictions together mean that any function call can be replaced by
the result of a previous call with the same parameters, and the
language guarantees that all these rearrangements will not change the
program result!
But when I fiddle with this idea I can quickly think of examples that contradict the statement above. For instance, let's say I have two functions (I will use pseudo code rather than Haskell):
x(a)->a+3
y(a)->a*3
z(a)->x(y(a))
w(a)->y(x(a))
Now, if we execute z and w, we get:
z(5) //gives 3*5+3=18
w(5) //gives (5+3)*3=24
That being so, I think I misunderstood the promised guarantee they speak about. Can anybody explain it to me?
When you compare x(y(a)) to y(x(a)), those two expressions are not equivalent because x and y aren't called with the same arguments in each. In the first expression x is called with the argument y(a) and y is called with the argument a. Whereas in the second y is called with x(a), not a, as its argument and x is called with a, not y(a). So: different arguments, (possibly) different results.
When people say that the order does not matter, they mean that in the following code:
a = f(x)
b = g(y)
you can switch the definition of a and b without affecting their values. That is it makes no difference whether f is called before g or vice versa. This is clearly not true for the following code:
a = getchar()
b = getchar()
If you switch a and b here, their values are switched as well, because getchar returns a (possibly) different character each time that it's called. So a purely functional language can't have a function exactly like getchar.
This question already has an answer here:
Julia: invoke a function by a given string
(1 answer)
Closed 6 years ago.
I know that you can call functions using their name as follows
f = x -> println(x)
y = :f
eval(:($y("hi")))
but this is slow since it is using eval is it possible to do this in a different way? I know it's easy to go the other direction by just doing symbol(f).
What are you trying to accomplish? Needing to eval a symbol sounds like a solution in search of a problem. In particular, you can just pass around the original function, thereby avoiding issues with needing to track the scope of f (or, since f is just an ordinary variable in your example, the possibility that it would get reassigned), and with fewer characters to type:
f = x -> println(x)
g = f
g("hi")
I know it's easy to go the other direction by just doing symbol(f).
This is misleading, since it's not actually going to give you back f (that transform would be non-unique). But it instead gives you the string representation for the function (which might happen to be f, sometimes). It is simply equivalent to calling Symbol(string(f)), since the combination is common enough to be useful for other purposes.
Actually I have found use for the above scenario. I am working on a simple form compiler allowing for the convenient definition of variational problems as encountered in e.g. finite element analysis.
I am relying on the Julia parser to do an initial analysis of the syntax. The equations entered are valid Julia syntax, but will trigger errors on execution because some of the symbols or methods are not available at the point of the problem definition.
So what I do is roughly this:
I have a type that can hold my problem description:
type Cmd f; a; b; end
I have defined a macro so that I have access to the problem description AST. I travers this expression and create a Cmd object from its elements (this is not completely unlike the strategy behind the #mat macro in MATLAB.jl):
macro m(xp)
c = Cmd(xp.args[1], xp.args[3], xp.args[2])
:($c)
end
At a later step, I run the Cmd. Evaluation of the symbols happens only at this stage (yes, I need to be careful of the evaluation context):
function run(c::Cmd)
xp = Expr(:call, c.f, c.a, c.b)
eval(xp)
end
Usage example:
c = #m a^b
...
a, b = 2, 3
run(c)
which returns 9. So in short, the question is relevant in at least some meta-programming scenarios. In my case I have to admit I couldn't care less about performance as all of this is mere preprocessing and syntactic sugar.
My problem (in Mathematica) is referring to variables given in a particular array and manipulating them in the following manner (as an example):
Inputs: vars={x,y,z}, system=some ODE like x^2+3*x*y+...etc
(note that I haven't actually created variables x y and z)
Aim:
To assign values to the variables in the list "var" with the intention of inputting these values into the system of ODEs. Then, once I am done, clear the values of the variables in the array vars so that it is in its original form {x,y,z} (and not something like {x,1,3} where y=1 and z=3). I want to do this by referring to the positional elements of vars (I aim not to know that x, y and z are the actual variables).
The reason why: I am trying to write a program that can have any number of variables and ODEs as defined by the user. Since the number of variables and the actual letters used for them are unknown, it is necessary to perform manipulations with the array itself.
Attempt:
A fixed number of variables is easy. For the arbitrary case, I have tried modules and blocks, but with no success. Consider the following code:
Clear[x,y,z,vars,svars]
vars={x,y,z}
svars=Map[ToString,vars]
Module[{vars=vars,svars=svars},
Symbol[svars[[1]]]//Evaluate=1
]
then vars={1,y,z} and not {x,y,z} after running this. I have done functional programming with lists, atoms etc. Thus is makes sense to me that vars is changed afterwards, because I have changed x and not vars. However, I cannot get "x" in the list of variables to remain local. Of course I could put in "x" itself, but that is particular to this specific case. I would prefer to put something like:
Clear[x,y,z,vars,svars]
vars={x,y,z}
svars=Map[ToString,vars]
Module[{vars=vars,svars=svars, vars[[1]]},
Symbol[svars[[1]]]//Evaluate=1
]
which of course doesn't work because vars[[1]] is not a symbol or an assignment to a symbol.
Other possibilities:
I found a function
assignToName[name_String, value_] :=
ToExpression[name, InputForm, Function[var, var = value, HoldAll]]
which looked promising. Basically name_String is the name of the variable and value is its new value. I attempted to do:
vars={x,y,z}
svars=Map[ToString,vars]
vars[[1]]=//Evaluate=1
assignToName[svars[[1]],svars[[1]]]
but then something likeD[x^2, vars[[1]]] doesn't work (x is not a valid variable).
If I am missing something, or if perhaps I am going down the wrong path, I'm open to trying other things.
Thanks.
I can't say that I followed your train(s) of thought very well, so these are fragments which might help you to answer your own questions than a coherent and fully-formed answer. But to answer your final 'question', I think you may be going down some wrong path(s).
In passing, note that evaluating the expression
vars = {x,y,z}
does in fact define those three variables though it doesn't define any rewrite rules (such as values) for them.
Given a polynomial poly you can extract the variables in it with the function Variables[poly] so something like
Variables[x^2+3*x*y]
should return
{x,y}
Note that I write 'should' rather than does because I don't have Mathematica on this machine so my syntax may be a bit wonky. Note also that your example ODE is nothing of the sort but it strikes me that you can probably write a wrapper to manipulate an ODE into a form from which Variables can extract the variables. Mathematica offers a lot of other functions for picking expressions apart and re-assembling them, follow the trails from Variables. It often allows the use of functions defined on Lists on expressions with other heads too so it's always worth experimenting a bit.
There are a couple of widely applicable ways to avoid setting values of variables in Mathematica. For instance, you could write
x^2+3*x*y/.{x->2,y->3}
which will evaluate to
22
but not set values for x and y. This is a very simple example of using (sets of) replacement rules for temporary assignment of values to variables
The other way to avoid setting values for variables is to define functions using Modules or Blocks both of which define their own contexts. The documentation will tell you all about these two and the differences between them.
I can't help thinking that all your clever tricks using Symbol, ToExpression and ToString are a bit beside the point. Spend some time familiarising yourself with Mathematica's in-built functionality before going further down that route, you may well find you don't need to.
Finally, writing, in any language, expressions such as
vars=vars,svars=svars
will lead to madness. It may be syntactically correct, you may even be able to decrypt the semantics when you first write code like that, but in a week's time you will curse your younger self for writing it.