I would like to implement a simple concatenative language (aka Joy or Factor) as a DSL in Julia and I am troubled how to optimally represent the stack.
The stack, which represents both data and program code, should be able to hold a sequence of items of different types. In the simplest case Ints, Symbols and, recursively again, stacks (to represent quoted code). The program will then heavily use push! and pop! to shuffle values between different such stacks.
One obvious implementation in Julia, which works but runs rather slow, is to use cell arrays. For example, the following Joy stack [ 1 [ 1 2 +] i + ] (which evaluates to [4]) can be implemented in Julia as
stack = Any[:+,:i,Any[:+,2,1],1]. My typical code then looks like this:
x = pop!(callstack)
if isa(x,Int)
push!(x,datastack)
elseif isa(x,Symbol)
do_stuff(x,datastack)
end
This, however, runs really slow and uses huge memory allocations, probably because such code is not typestable (which is a big performance bottleneck in Julia).
Using C, I would represent the stack compactly as an array (or alternatively as a linked list) of a union:
typedef union Stackelem{
int val;
char *sym;
union Stackelem *quote;
} Stackelem;
Stackelem stack[n];
But how can I achieve such a compact representation of the heterogeneous stack in Julia, and how I avoid the type instability?
This is one way, another way would be to represent args with type Vector{Any}:
julia> immutable Exp
head::Symbol
args::Tuple
end
julia> q = Exp(:+, (1, Exp(:-, (3, 4))))
Exp(:+,(1,Exp(:-,(3,4))))
edit: Another way to represent it might be:
immutable QuoteExp{T} ; vec::Vector{T} ; end
typealias ExpTyp Union{QuoteExp, Int, Symbol}
typealias Exp QuoteExp{ExpTyp}
and then you can do the following:
julia> x = Exp(ExpTyp[:+, 1, 2])
QuoteExp{Union{Int64,QuoteExp{T},Symbol}}(Union{Int64,QuoteExp{T},Symbol}[:+,1,2])
julia> x.vec[1]
:+
julia> x.vec[2]
1
julia> x.vec[3]
2
julia> push!(x.vec,:Scott)
4-element Array{Union{Int64,QuoteExp{T},Symbol},1}:
:+
1
2
:Scott
julia> x.vec[4]
:Scott
Related
I would like to have an algorithm for a n=a (where a is any number that I choose), such that
Julia understands that I declare the monomial variables x_{1}...x_{a} with #polyvar
Which package do you recommend me to do this?TypedPolynomials?
Thank you for your help
Normally you almost always use Vectors for things like this:
julia> a=3
3
julia> x = [n for n in 1:a]
3-element Vector{Int64}:
1
2
3
julia> x[2]
2
It is very hard to think about a scenario when it is not the best idea.
However, if you really need you can do a macro. The rule of thumb with Julia macros is "if you do not know how to write it perhaps you do not need it". Anyway here it is:
julia> macro var(n, initval)
exs = [ :( $(Symbol("x_",i)) = $initval ) for i in 1:n ]
esc(Expr(:block, exs...))
end
#var (macro with 1 method)
julia> #var(5, 77)
77
julia> x_4
77
When would I use vector-of to create a vector, instead of the vector function. Is the guideline to use vector most of the time and only for performance reason switch to vector-of?
I could not find good info on when to use vector-of.
vector-of is used for creating a vector of a single primitive type, :int, :long, :float, :double, :byte, :short, :char, or :boolean. It doesn't allow other types as it stores the values unboxed internally. So, if your vector need to include other types than those primitive types, you cannot use vector-of. But if you are sure that the vector will have data of a single primitive type, you can use vector-of for better performance.
user=> (vector-of :int 1 2 3 4 5)
[1 2 3 4 5]
user=> (vector-of :double 1.0 2.0)
[1.0 2.0]
user=> (vector-of :string "hello" "world")
Execution error (IllegalArgumentException) at user/eval5 (REPL:1).
Unrecognized type :string
As you can see, you should specify primitive type as an argument.
vector can be used to create a vector of any type.
user=> (vector 1 2.0 "hello")
[1 2.0 "hello"]
You can put any type when you use vector.
Also, there's another function vec, which is used for creating a new vector containing the contents of coll.
user=> (vec '(1 2 3 4 5))
[1 2 3 4 5]
Usually, you can get the basic information of a function/macro from the repl, like the following.
user=> (doc vector-of)
-------------------------
clojure.core/vector-of
([t] [t & elements])
Creates a new vector of a single primitive type t, where t is one
of :int :long :float :double :byte :short :char or :boolean. The
resulting vector complies with the interface of vectors in general,
but stores the values unboxed internally.
Optionally takes one or more elements to populate the vector.
Reference:
https://clojuredocs.org/clojure.core/vector-of
https://clojuredocs.org/clojure.core/vector
https://clojuredocs.org/clojure.core/vec
Nobody really ever uses vector-of. If you don't super care about performance, vector is fine, and if you do super care about performance you usually want a primitive array or some other java type. Honestly I would expect occasional weird snags when passing a vector-of to anything that expects an ordinary vector or sequence - maybe it works fine, but it's just such a rare thing to see that it wouldn't surprise me if it caused issues.
while working in julia programming, for creating an array instead of using a=[1:1:20...] i used a=[1:1:20] and it created an array saying "1-element Array{StepRange{Int64,Int64},1}".
What does this "1-element Array{StepRange{Int64,Int64},1}" mean? what StepRange means?
From the documentation of StepRange (type ?StepRange in the Julia REPL to see this):
StepRange{T, S} <: OrdinalRange{T, S}
Ranges with elements of type T with spacing of type S. The step
between each element is constant, and the range is defined in terms
of a start and stop of type T and a step of type S. Neither T nor S
should be floating point types. The syntax a:b:c with b > 1 and a,
b, and c all integers creates a StepRange.
So, for example
julia> typeof(1:1:20)
StepRange{Int64,Int64}
and
julia> [1:1:20]
1-element Array{StepRange{Int64,Int64},1}:
1:1:20
thus constructs a Vector (1D Array) containing one StepRange. If you want to materialize the lazy StepRange I would recommend collect(1:1:20) instead of using splatting ([1:1:20...]).
You can access start / step / stop fields of a StepRange using:
julia> r = 1:1:20
julia> r.start
1
julia> r.stop
20
julia> r.step
1
I've run across a little problem when trying to solve a Project Euler problem in Julia. I've basically written a recursive function which produces fractions with increasingly large numerators and denominators. I don't want to post the code for obvious reasons, but the last few fractions are as follows:
1180872205318713601//835002744095575440
2850877693509864481//2015874949414289041
6882627592338442563//4866752642924153522
At that point I get an OverflowError(), presumably because the numerator and/or denominator now exceeds 19 digits. Is there a way of handling 'Big' fractions in Julia (i.e. those with BigInt-type numerators and denominators)?
Addendum:
OK, I've simplified the code and disguised it a bit. If anyone wants to wade through 650 Project Euler problems to try to work out which question it is, good luck to them – there will probably be around 200 better solutions!
function series(limit::Int64, i::Int64=1, n::Rational{Int64}=1//1)
while i <= limit
n = 1 + 1//(1 + 2n)
println(n)
return series(limit, i + 1, n)
end
end
series(50)
If I run the above function with, say, 20 as the argument it runs fine. With 50 I get the OverflowError().
Julia defaults to using machine integers. For more information on this see the FAQ: Why does Julia use native machine integer arithmetic?.
In short: the most efficient integer operations on any modern CPU involves computing on a fixed number of bits. On your machine, that's 64 bits.
julia> 9223372036854775805 + 1
9223372036854775806
julia> 9223372036854775805 + 2
9223372036854775807
julia> 9223372036854775805 + 3
-9223372036854775808
Whoa! What just happened!? That's definitely wrong! It's more obvious if you look at how these numbers are represented in binary:
julia> bitstring(9223372036854775805 + 1)
"0111111111111111111111111111111111111111111111111111111111111110"
julia> bitstring(9223372036854775805 + 2)
"0111111111111111111111111111111111111111111111111111111111111111"
julia> bitstring(9223372036854775805 + 3)
"1000000000000000000000000000000000000000000000000000000000000000"
So you can see that those 63 bits "ran out of space" and rolled over — the 64th bit there is called the "sign bit" and signals a negative number.
There are two potential solutions when you see overflow like this: you can use "checked arithmetic" — like the rational code does — that ensures you don't silently have this problem:
julia> Base.Checked.checked_add(9223372036854775805, 3)
ERROR: OverflowError: 9223372036854775805 + 3 overflowed for type Int64
Or you can use a bigger integer type — like the unbounded BigInt:
julia> big(9223372036854775805) + 3
9223372036854775808
So an easy fix here is to remove your type annotations and dynamically choose your integer types based upon limit:
function series(limit, i=one(limit), n=one(limit)//one(limit))
while i <= limit
n = 1 + 1//(1 + 2n)
println(n)
return series(limit, i + 1, n)
end
end
julia> series(big(50))
#…
1186364911176312505629042874//926285732032534439103474303
4225301286417693889465034354//3299015554385159450361560051
I was delighted to learn that Julia allows a beautifully succinct way to form inner products:
julia> x = [1;0]; y = [0;1];
julia> x'y
1-element Array{Int64,1}:
0
This alternative to dot(x,y) is nice, but it can lead to surprises:
julia> #printf "Inner product = %f\n" x'y
Inner product = ERROR: type: non-boolean (Array{Bool,1}) used in boolean context
julia> #printf "Inner product = %f\n" dot(x,y)
Inner product = 0.000000
So while i'd like to write x'y, it seems best to avoid it, since otherwise I need to be conscious of pitfalls related to scalars versus 1-by-1 matrices.
But I'm new to Julia, and probably I'm not thinking in the right way. Do others use this succinct alternative to dot, and if so, when is it safe to do so?
There is a conceptual problem here. When you do
julia> x = [1;0]; y = [0;1];
julia> x'y
0
That is actually turned into a matrix * vector product with dimensions of 2x1 and 1 respectively, resulting in a 1x1 matrix. Other languages, such as MATLAB, don't distinguish between a 1x1 matrix and a scalar quantity, but Julia does for a variety of reasons. It is thus never safe to use it as alternative to the "true" inner product function dot, which is defined to return a scalar output.
Now, if you aren't a fan of the dots, you can consider sum(x.*y) of sum(x'y). Also keep in mind that column and row vectors are different: in fact, there is no such thing as a row vector in Julia, more that there is a 1xN matrix. So you get things like
julia> x = [ 1 2 3 ]
1x3 Array{Int64,2}:
1 2 3
julia> y = [ 3 2 1]
1x3 Array{Int64,2}:
3 2 1
julia> dot(x,y)
ERROR: `dot` has no method matching dot(::Array{Int64,2}, ::Array{Int64,2})
You might have used a 2d row vector where a 1d column vector was required.
Note the difference between 1d column vector [1,2,3] and 2d row vector [1 2 3].
You can convert to a column vector with the vec() function.
The error message suggestion is dot(vec(x),vec(y), but sum(x.*y) also works in this case and is shorter.
julia> sum(x.*y)
10
julia> dot(vec(x),vec(y))
10
Now, you can write x⋅y instead of dot(x,y).
To write the ⋅ symbol, type \cdot followed by the TAB key.
If the first argument is complex, it is conjugated.
Now, dot() and ⋅ also work for matrices.
Since version 1.0, you need
using LinearAlgebra
before you use the dot product function or operator.