I can't seem to find a simple answer to this seemingly simple SML question. I have the code:
fun inde(x, y, L) = if null L then nil else
if x=hd(L) then y+1::inde(x,y+1,tl L) else
inde(x,y+1,tl L);
I want y to be a variable outside the function, so it'll be inde(x,L) but have the y still count properly. When I declare it outside the function (to 0), when the function is recursively called, it resets to 0.
If you were to run this current function, it'd produce a list of where ever x is in the list (L).
so inde(1,0,[1,2,2,1,1]) would produce [1,4,5]
Idiomatic structure when using a functional programming style is to define an inner function that takes arguments that are of interest to the programmer, but not the user and then to call the inner function from the main function:
fun inde(x : int, L) =
let
fun inner(list1, list2, y : int) =
if null list1
then List.rev list2
else
if x = hd list1
then
inner(tl list1, y::list2, y + 1)
else
inner(tl list1, list2, y +1)
in
inner(L,[],1)
end
In the example function:
inner uses four values: the local variables list1,list2, and y. It also uses x from the enclosing scope.
inner builds (conses up) the list that will be returned using list2. It reverses the list with a call to List.rev from the SML Basis Library. This adds O(n) to the execution time.
The last part of the let...in...end construct: inner(L,[],1) is called "the trampoline" because the code gets all the way to the bottom of the source file and then bounces off it to start execution. It's a standard pattern.
Note that I started iterating with y equal to 1, rather than 0. Starting at zero wasn't getting anything done in the original file.
Related
Lets assume the following function
val foo : int -> int -> bool -> int
let foo x y b =
let x,y = if x > y then y,x else x,y in
let rec loop x y b =
if x >= then x
else if b then loop (x+1) y (not b)
else loop x (y-1) (not b)
in
loop x y b
I still don't quite understand the concept of the "in".
Does the line mean "let x,y = ... in" that it is executed immediately or only when you "Call" it? And when i dont need to call it why do i need the last line loop x y b?
Thanks in advance :)
in is just part of the OCaml syntax - let PATTERN = EXPR1 in EXPR2 is an expression which binds the result of EXPR1 to PATTERN and then evaluates EXPR2 with the new bindings present. In some languages like F# and Haskell, you don't (always) need in - it's inferred from the indentation. OCaml syntax is indentation insensitive which requires it to have an explicit in.
Does the line mean "let x,y = ... in" that it is executed immediately or only when you "Call" it?
It's evaluated immediately.
And when i dont need to call it why do i need the last line loop x y b?
In this code, the previous line defines a function named loop with 3 arguments, and then later you call the function with the arguments x y b.
This function is supposed to return a function handle to the nested function inside, but if the variable x is set to a negative value in the outer function, it doesn't work.
The inner nested function is just a constant function returning the value of the variable x that is set in the outer function.
function t=test(x)
x=-1;
function y=f()
y=x;
endfunction
t=#f;
endfunction
If I try to evaluate the returned function, e.g. test()(3), I get an error about x being undefined. The same happens if x is defined as a vector with at least one negative entry or if x is argument of the function and a negative default value is used for evaluation. But if I instead define it as some nonnegative value
function t=test(x)
x=1;
function y=f()
y=x;
endfunction
t=#f;
endfunction,
then the returned function works just fine. Also if I remove the internal definition of x and give the value for x as an argument to the outer function (negative or not), like
function t=test(x)
function y=f()
y=x;
endfunction
t=#f;
endfunction
and then evaluate e.g. test(-1)(3), the error doesn't occur either. Is this a bug or am misunderstanding how function handles or nested functions work?
The Octave documentation recommends using subfunctions instead of nested functions, but they cannot access the local variables of their parent function and I need the returned function to depend on the input of the function returning it. Any ideas how to go about this?
This is a bug that was tracked here:
https://savannah.gnu.org/bugs/?func=detailitem&item_id=60137
Looks like it was fixed and will be gone in the next release.
Also, to explain the different behavior of negative and positive numbers: I experimented a bit, and no variable that is assigned a computed value is being captured:
function t=tst()
x = [5,3;0,0]; # captured
y = [5,3;0,0+1]; # not captured
z = x + 1; # not captured
function y=f()
endfunction
t=#f;
endfunction
>> functions(tst)
ans =
scalar structure containing the fields:
function = f
type = nested
file =
workspace =
{
[1,1] =
scalar structure containing the fields:
t = #f
x =
5 3
0 0
}
The different behavior of negative and positive numbers are probably caused by the minus sign - before the numbers being treated as a unary operator (uminus).
As of octave version 5.2.0 the nested function handles were not supported at all. I'm going to guess that is the novelty of the version 6.
In octave functions are not variables, the engine compiles\translates them at the moment of reading the file. My guess would be that behavior you are observing is influenced by your current workspace at the time of function loading.
The common way for doing what you are trying to do was to generate the anonymous (lambda) functions:
function t = test1(x=-1)
t = #()x;
end
function t = test2(x=-1)
function s = calc(y,z)
s = y + 2*z;
end
t = #(a=1)calc(a,x);
end
Note that default parameters for the generated function should be stated in lambda definition. Otherwise if you'd call it like test2()() it would not know what to put into a when calling calc(a,x).
If you are trying to create a closure (a function with associated state), octave has limited options for that. In such a case you could have a look at octave's object oriented functionality. Classdef might be useful for quick solutions.
Unlike Matlab, Octave Symbolic has no piecewise function. Is there a work around? I would like to do something like this:
syms x
y = piecewise(x0, 1)
Relatedly, how does one get pieces of a piecewise function? I ran the following:
>> int (exp(-a*x), x, 0, t)
And got the following correct answer displayed and stored in a variable:
t for a = 0
-a*t
1 e
- - ----- otherwise
a a
But now I would like to access the "otherwise" part of the answer so I can factor it. How do I do that?
(Yes, I can factor it in my head, but I am practicing for when more complicated expressions come along. I am also only really looking for an approach using symbolic expressions -- even though in any single case numerics may work fine, I want to understand the symbolic approach.)
Thanks!
Matlab's piecewise function seems to be fairly new (introduced in 2016b), but it basically just looks like a glorified ternary operator. Unfortunately I don't have 2016 to check if it performs any checks on the inputs or not, but in general you can recreate a 'ternary' operator in octave by indexing into a cell using logical indexing. E.g.
{#() return_A(), #() return_B(), #() return_default()}([test1, test2, true]){1}()
Explanation:
Step 1: You put all the values of interest in a cell array. Wrap them in function handles if you want to prevent them being evaluated at the time of parsing (e.g. if you wanted the output of the ternary operator to be to produce an error)
Step 2: Index this cell array using logical indexing, where at each index you perform a logical test
Step 3: If you need a 'default' case, use a 'true' test for the last element.
Step 4: From the cell (sub)array that results from above, select the first element and 'run' the resulting function handle. Selecting the first element has the effect that if more than one tests succeed, you only pick the first result; given the 'default' test will always succeed, this also makes sure that this is not picked unless it's the first and only test that succeeds (which it does so by default).
Here are the above steps implemented into a function (appropriate sanity checks omitted here for brevity), following the same syntax as matlab's piecewise:
function Out = piecewise (varargin)
Conditions = varargin(1:2:end); % Select all 'odd' inputs
Values = varargin(2:2:end); % Select all 'even' inputs
N = length (Conditions);
if length (Values) ~= N % 'default' case has been provided
Values{end+1} = Conditions{end}; % move default return-value to 'Values'
Conditions{end} = true; % replace final (ie. default) test with true
end
% Wrap return-values into function-handles
ValFuncs = cell (1, N);
for n = 1 : N; ValFuncs{n} = #() Values{n}; end
% Grab funhandle for first successful test and call it to return its value
Out = ValFuncs([Conditions{:}]){1}();
end
Example use:
>> syms x t;
>> F = #(a) piecewise(a == 0, t, (1/a)*exp(-a*t)/a);
>> F(0)
ans = (sym) t
>> F(3)
ans = (sym)
-3⋅t
ℯ
─────
9
I have created a function that returns the magnitude of a vector.the output is 360x3 dimension matrix. the input is 360x2.
Everything works fine outside the function. how do i get it to work ?
clc
P_dot_ij_om_13= rand(360,2); // 360x2 values of omega in vectors i and j
//P_dot_ij_om_13(:,3)=0;
function [A]=mag_x(A)
//b="P_dot_ijOmag_"+ string(k);
//execstr(b+'=[]'); // declare indexed matrix P_dot_ijOmag_k
//disp(b)
for i=1:1:360
//funcprot(0);
A(i,3)=(A(i,2)^2+A(i,1)^2)^0.5; //calculates magnitude of i and j and adds 3rd column
disp(A(i,3),"vector magnitude")
end
funcprot(1);
return [A] // should return P_dot_ijOmag_k in the variable browser [360x3 dim]
endfunction
mag_x(P_dot_ij_om_13);
//i=1;
//P_dot_ij_om_13(i,3)= (P_dot_ij_om_13(i,2)^2+P_dot_ij_om_13(i,1)^2)^0.5;// example
You never assigned mag_x(P_dot_ij_om_13) to any variable, so the output of this function disappears into nowhere. The variable A is local to this function, it does not exist outside of it.
To have the result of calculation available, assign it to some variable:
res = mag_x(P_dot_ij_om_13)
or A = mag_x(P_dot_ij_om_13) if you want to use the same name outside of the function as was used inside of it.
By the way, the Scilab documentation discourages the use of return, as it leads to confusion. The Scilab / Matlab function syntax is different from the languages in which return specifies the output of a function:
function y = sq(x)
y = x^2
endfunction
disp(sq(3)) // displays 9
No need for return here.
How does returning functions work in Lua? I'm trying to wrap my head around this derivative function but I'm having trouble understanding it.
function derivative(f, delta)
delta = delta or 1e-4
return function(x)
return (f(x + delta) - f(x))/delta
end
end
Any help is appreciated, thanks.
First see here.
Shortly, functions are first-class citizens and you an store them in variable and return from functions.
Second. In your example there is a closure to be created. It will have upvalues f and delta, wich can be used in inner function.
So when you call your derivative function, new closure will be created with copy of f and delta. And you can call it later like any other function
local d = derivative(f, 1e-6)
d(10)
EDIT: "Answer on but I'm having trouble understanding how the x argument is treated in the anonymous function in my example"
Each function have a signature, number of formal attributes, it will get.
In Lua you can call function with any number of arguments. Let's consider an example.
local a = function(x1, x2) print(x1, x2) end
a(1) // 1, nil
a(1, 2) // 1, 2
a(1, 2, 3) // 1, 2
When you call function in variable a, each given argument value, one by one will be matched with function argumentList. In 3-rd example 1 will be assigned to x1, 2 to x2, 3 will be thrown away. In term's of vararg function smth like this will be performed.
function a(...)
local x1 = (...)[1]
local x2 = (...)[2]
// Body
end
In your example x is treated as inner function argument, will be visible inside it, and initialized when you call your inner function instance.
f and delta will be unique for each function instance, as I mentioned above.
Hope my clumsy explanations will hit their goal and help you a little.