Why does this does not work for assigning a function:
F(t) := Matrix(matrixDE(A, t)[1])
While using a label to reference the output works:
Matrix(matrixDE(A, t)[1]) (1)
F1(t) := (1)
matrixDE(A, t)[1] (2)
F2(t) := Matrix((2))
It seems the indexing operation [1] is what poses the problem, but I don't understand the mechanic behind this.
I found this question but it did not make me any wiser...
Edit:
I ended up getting my desired effect using eval and :
F(x) := eval(Matrix(matrixDE(A, t)[1]), t = x)
The price to pay being that the argument of Fis named x instead of t.
The problem has nothing to with the indexing. The problem is exactly what you said in your title: You cannot (properly) assign a Maple function by directly using an (unevalauted) expression; rather, you must reference the output (or evaluation).
More specifically, it depends on the desired order of these two operations: evaluation of a parameterized expression and substitution of values for the parameter. If you want to first evaluate the expression with a symbolic parameter (your t in this case), then the command to use is unapply:
F:= unapply(Matrix(matrixDE(A, t)[1]), t);
Related
i see that fix documentation says:
http://www.minizinc.org/doc-lib/doc-builtins-reflect.html#Ifunction-dd-T-cl-fix-po-var-opt-dd-T-cl-x-pc
function array [$U] of $T: fix(array [$U] of var opt $T: x)
Check if the value of every element of the array x is fixedat this point in evaluation. If all are fixed, return an array of their values, otherwise abort.
I am thinking it can be used to coerce a var to a par.
Here is the code.
array [1..num] of var int: value ;
%% generate random numbers from 0..num-1, this should fix the value of the var "value" or so i think
constraint forall(i in index_set(value))(let {int:temp_value=discrete_distribution([1|i in index_set(value)]); } in value[i]=trace(show(temp_value)++"\n", temp_value));
%%% this i was expecting to work, as "value" elements are fixed above
array [1..num] of int:value__ =[ trace(show(fix(value[i])), fix(value[i])) | i in index_set(value)] ;
But i get:
MiniZinc: evaluation error:
with i = 1
in call 'trace'
in call 'fix'
expression is not fixed
My questions are:
1) I think i should expect this error as minizinc is not sequential execution language?
2) Examples of fix in user guide is only where output statement is used. Is it the only place to use fix?
3) How would i coerce a var to a par?
By the way I am trying this var to par conversion because i am having problem with array generator expression. Here is the code
int:num__=200;
int:seed=134;
int: two_m=2097184;
%% prepare weights for generating numbers form 1..(two_m div 64), basically same weight
array [1..(two_m div 64)] of int: value_6_wt= [seed+1 | i in 1..(two_m div 64)] ;
%% generate numbers. this dose not work gives out
%% in variable declaration for 'value6'
%% parameter value out of range
array [1..num__] of int : value6 = [ discrete_distribution(value_6_wt) | j in 1..num__];
In the MiniZinc language the difference between a parameter and a variable is only the fact that a parameter must have a value at compile time. Within the compiler we turn as many variables into parameters as we can. This saves the solver from having to do some work. When we know that a variable has been turned into a parameter, then we can use the fix function to convince the type system that we really can use this variable as a parameter and see its value.
The problem here however is that fix is defined to abort when the variable is not fixed to one value. If no testing is done, this requires some (magic/)knowledge about the compilation process. In your case it seems that the second array is evaluated before the optimisation stage, in which all aliasing is resolved. This is the reason why it does not work. (This is indeed one of the things that is a consequence of a declarative language)
Although fix might only be used in the output statements in the examples (where it's guaranteed to work), it is used in many locations in the MiniZinc libraries. If we for example look at the library that is used for MIP solvers, there are many constraints that can be encoded more efficiently if one of the arguments is a parameter. Therefore, you will often see that the a constraint in this library first tests its arguments with is_fixed, and then use a better encoding if this returns true.
The output statement and when is_fixed returns true will both give the guarantee that a variable is fixed and ensure that the compilation doesn't abort. There is no other way to coerce a variable to a parameter, but unless you are dealing with dependant predicate definitions, you can just trust the MiniZinc compiler to ensure that the resulting FlatZinc will contain a parameter instead of a variable.
For example I have a 1*30 structure a.field, when I type a(:).field in command window it just iteratively display a(1).field, a(2).field,... However, when I was trying to assign a(:).field to another variable b, what b get is just a(1).field.
BTW, if I attampt to pass a(:).field to a function, Matlab just throws an error "too many input arguments".
What is the mechanism behind? My guess is that matlab threat colon equivlant to the first element during assignment, is that true?
You need to add brackets, otherwise matlab don't understand that your trying to store an array:
b = [a(:).field]
Another option that provide similar result:
b = horzcat(a(:).field)
Just having a little difficulty with the syntax of matlab functions;
function f = fact(x)
if x == 1
return
else
f = 1 - x*(fact(x-1))
end
end
When calling this function in the command window with the argument 10 I receive the error
Undefined function 'fact' for input arguments of type 'double'.
Error in recursion (line 6)
f = 1 - x*(fact(x-1))
I've had a look around and solutions for the first revolve around the pathing of the m-file which doesn't seem to be a problem as other files in the same directory run fine,
The second I'm not sure why the error in line 6 occurs, my guess is it has something to do with the variable and function names.
As a side question, are both these end statements necessary?
Thanks!
The most obvious error is your function filename. You have a function called fact defined in your code but you named your file recursion. Make sure that both your function name and the filename are both called fact.
If you were to name your file as recursion, then make the function name defined in your code as fact, this is what would happen if you tried calling your code:
>> f = recursion(10);
Undefined function 'fact' for input arguments of type 'double'.
Error in recursion (line 6)
f = 1 - x*(fact(x-1));
... look familiar?
As such make sure your filename and your function name are named the same. In fact, in the MATLAB editor, it should automatically give you an error saying that both of these are not the same.
There is also another error in your code. The base case is not defined properly. Always remember when you are writing recursive algorithms is that eventually the function is going to return... and that's when you hit the base case. We can see here that it is when x = 1. When x = 1, you're supposed to assign something to f which is the output. You are simply exiting the function, and so when x becomes 1, your code will spit out an error saying that f was not assigned when the function finishes. As such, you need to figure out what your base case is. I'm going to assume that your base case (when x = 1) is going to equal 0. You will obviously need to change this as I don't know what your code is actually computing. Basically, you need to do this:
function f = fact(x)
if x == 1
f = 0; %// Base case needs to change according to your specs
else
f = 1 - x*(fact(x-1))
end
end
When I do this, I get the following output when x = 10
>> f = fact(10);
f =
1334961
I don't get an error when I run this code now. Also, check to see if you have any variables named fact in your workspace. When this happens, you are in fact shadowing over your function with a variable, so it is actually trying to access the variable called fact instead. As such, try clearing your workspace by doing clear all;, then try this code again.
One warning
If you were to specify x to be 0 or negative, this function will never stop. As such, you need to provide some check and perform the proper action when this happens. Also, you need to make sure that you specify what type of inputs are accepted for x. Judging from the context, x are positive integers only. As what #Glen_b has noted, should you provide any number that isn't a positive integer, this function will never stop as x will never equal 1 down the recursion pipeline.
To answer your optional question
The first end statement is required to end the if statement. The second end statement isn't required, but it's good practice anyway. However, if you have multiple functions defined inside your function file, then yes it is most definitely required to properly signify that the end of that function is defined there. However, you don't need it if you're only writing one function per file, but I would recommend keeping it there as it's good practice.
Suppose one needs to select the real solutions after solving some equation.
Is this the correct and optimal way to do it, or is there a better one?
restart;
mu := 3.986*10^5; T:= 8*60*60:
eq := T = 2*Pi*sqrt(a^3/mu):
sol := solve(eq,a);
select(x->type(x,'realcons'),[sol]);
I could not find real as type. So I used realcons. At first I did this:
select(x->not(type(x,'complex')),[sol]);
which did not work, since in Maple 5 is considered complex! So ended up with no solutions.
type(5,'complex');
(* true *)
Also I could not find an isreal() type of function. (unless I missed one)
Is there a better way to do this that one should use?
update:
To answer the comment below about 5 not supposed to be complex in maple.
restart;
type(5,complex);
true
type(5,'complex');
true
interface(version);
Standard Worksheet Interface, Maple 18.00, Windows 7, February
From help
The type(x, complex) function returns true if x is an expression of the form
a + I b, where a (if present) and b (if present) are finite and of type realcons.
Your solutions sol are all of type complex(numeric). You can select only the real ones with type,numeric, ie.
restart;
mu := 3.986*10^5: T:= 8*60*60:
eq := T = 2*Pi*sqrt(a^3/mu):
sol := solve(eq,a);
20307.39319, -10153.69659 + 17586.71839 I, -10153.69659 - 17586.71839 I
select( type, [sol], numeric );
[20307.39319]
By using the multiple argument calling form of the select command we here can avoid using a custom operator as the first argument. You won't notice it for your small example, but it should be more efficient to do so. Other commands such as map perform similarly, to avoid having to make an additional function call for each individual test.
The types numeric and complex(numeric) cover real and complex integers, rationals, and floats.
The types realcons and complex(realcons) includes the previous, but also allow for an application of evalf done during the test. So Int(sin(x),x=1..3) and Pi and sqrt(2) are all of type realcons since following an application of evalf they become floats of type numeric.
The above is about types. There are also properties to consider. Types are properties, but not necessarily vice versa. There is a real property, but no real type. The is command can test for a property, and while it is often used for mixed numeric-symbolic tests under assumptions (on the symbols) it can also be used in tests like yours.
select( is, [sol], real );
[20307.39319]
It is less efficient to use is for your example. If you know that you have a collection of (possibly non-real) floats then type,numeric should be an efficient test.
And, just to muddy the waters... there is a type nonreal.
remove( type, [sol], nonreal );
[20307.39319]
The one possibility is to restrict the domain before the calculation takes place.
Here is an explanation on the Maplesoft website regarding restricting the domain:
4 Basic Computation
UPD: Basically, according to this and that, 5 is NOT considered complex in Maple, so there might be some bug/error/mistake (try checking what may be wrong there).
For instance, try putting complex without quotes.
Your way seems very logical according to this.
UPD2: According to the Maplesoft Website, all the type checks are done with type() function, so there is rather no isreal() function.
Is there an idiomatic way in Matlab to bind the value of an expression to the nth return value of another expression?
For example, say I want an array of indices corresponding to the maximum value of a number of vectors stored in a cell array. I can do that by
function I = max_index(varargin)
[~,I]=max(varargin{:});
cellfun(#max_index, my_data);
But this requires one to define a function (max_index) specific for each case one wants to select a particular return value in an expression. I can of course define a generic function that does what I want:
function y = nth_return(n,fun,varargin)
[vals{1:n}] = fun(varargin{:});
y = vals{n};
And call it like:
cellfun(#(x) nth_return(2,#max,x), my_data)
Adding such functions, however, makes code snippets less portable and harder to understand. Is there an idiomatic to achieve the same result without having to rely on the custom nth_return function?
This is as far as I know not possible in another way as with the solutions you mention. So just use the syntax:
[~,I]=max(var);
Or indeed create an extra function. But I would also suggest against this. Just write the extra line of code, in case you want to use the output in another function. I found two earlier questions on stackoverflow, which adress the same topic, and seem to confirm that this is not possible.
Skipping outputs with anonymous function in MATLAB
How to elegantly ignore some return values of a MATLAB function?
The reason why the ~ operator was added to MATLAB some versions ago was to prevent you from saving variables you do not need. If there would be a syntax like the one you are searching for, this would not have been necessary.