I just stumbled upon a strange behaviour while working with macros in Julia. I want to escape all elements in a vector and approched this with the map function. When I store the result back in a vector, the escape function is seems not to be evaluated after the macro call, but when the output of the map function gets stored into variables, the expected behaviour can be seen.
Stored in the vector the elements look similar to a single escaped symbol.
input = [:a, :b]
escaped_input = map(x -> esc(x),input)
println(escaped_input)
>>> Expr[:($(Expr(:escape, :a))), :($(Expr(:escape, :b)))]
println(esc(:a))
>>> :($(Expr(:escape, :a)))
But when the elements are accessed inside a macro in the quote, the escape function seems to not get executed:
a = 1
b = 2
macro escapeVector(testVec...)
escapedVec = map(x -> esc(x), collect(testVec))
quote
map(x -> println(x), $escapedVec)
println($escapedVec)
end
end
macro escapeVars(testVec...)
esc_a, esc_b = map(x -> esc(x), collect(testVec))
quote
println($esc_a)
println($esc_b)
end
end
#escapeVector a b
#escapeVars a b
This code produces the following output:
>>> $(Expr(:escape, :a))
>>> $(Expr(:escape, :b))
>>> Expr[:($(Expr(:escape, :a))), :($(Expr(:escape, :b)))]
>>> 1
>>> 2
Is there a way to trigger the escape inside the vector, so that the variables can be accessed or do I miss something obvious here?
Related
In my equations we have many expressions with a^2, and so on. I would like to map "²" to ^2, to obtain something like that:
julia> a² == a^2
true
The above is not however a legal code in Julia. Any idea on how could I implement it ?
Here is a sample macro #hoo that does what you requested in a simplified scenario (since the code is long I will start with usage).
julia> x=5
5
julia> #hoo 3x² + 4x³
575
julia> #macroexpand #hoo 2x³+3x²
:(2 * Main.x ^ 3 + 3 * Main.x ^ 2)
Now, let us see the macro code:
const charsdict=Dict(Symbol.(split("¹²³⁴⁵⁶⁷⁸⁹","")) .=> 1:9)
const charsre = Regex("[$(join(String.(keys(charsdict))))]")
function proc_expr(e::Expr)
for i=1:length(e.args)
el = e.args[i]
typeof(el) == Expr && proc_expr(el)
if typeof(el) == Symbol
mm = match(charsre, String(el))
if mm != nothing
a1 = Symbol(String(el)[1:(mm.offset-1)])
a2 = charsdict[Symbol(mm.match)]
e.args[i] = :($a1^$a2)
end
end
end
end
macro hoo(expr)
typeof(expr) != Expr && return expr
proc_expr(expr)
expr
end
Of course it would be quite easy to expand this concept into "pure-math" library for Julia.
I don't think that there is any reasonable way of doing this.
When parsing your input, Julia makes no real difference between the unicode character ² and any other characters you might use in a variable name. Attempting to make this into an operator would be similar to trying to make the suffix square into an operator
julia> asquare == a^2
The a and the ² are not parsed as two separate things, just like the a and the square in asquare would not be.
a^2, on the other hand, is parsed as three separate things. This is because ^ is not a valid character for a variable name and it is therefore parsed as an operator instead.
I'm writing a genetic program in order to test the fitness of randomly generated expressions. Shown here is the function to generate the expression as well a the main function. DIV and GT are defined elsewhere in the code:
function create_single_full_tree(depth, fs, ts)
"""
Creates a single AST with full depth
Inputs
depth Current depth of tree. Initially called from main() with max depth
fs Function Set - Array of allowed functions
ts Terminal Set - Array of allowed terminal values
Output
Full AST of typeof()==Expr
"""
# If we are at the bottom
if depth == 1
# End of tree, return function with two terminal nodes
return Expr(:call, fs[rand(1:length(fs))], ts[rand(1:length(ts))], ts[rand(1:length(ts))])
else
# Not end of expression, recurively go back through and create functions for each new node
return Expr(:call, fs[rand(1:length(fs))], create_single_full_tree(depth-1, fs, ts), create_single_full_tree(depth-1, fs, ts))
end
end
function main()
"""
Main function
"""
# Define functional and terminal sets
fs = [:+, :-, :DIV, :GT]
ts = [:x, :v, -1]
# Create the tree
ast = create_single_full_tree(4, fs, ts)
#println(typeof(ast))
#println(ast)
#println(dump(ast))
x = 1
v = 1
eval(ast) # Error out unless x and v are globals
end
main()
I am generating a random expression based on certain allowed functions and variables. As seen in the code, the expression can only have symbols x and v, as well as the value -1. I will need to test the expression with a variety of x and v values; here I am just using x=1 and v=1 to test the code.
The expression is being returned correctly, however, eval() can only be used with global variables, so it will error out when run unless I declare x and v to be global (ERROR: LoadError: UndefVarError: x not defined). I would like to avoid globals if possible. Is there a better way to generate and evaluate these generated expressions with locally defined variables?
Here is an example for generating an (anonymous) function. The result of eval can be called as a function and your variable can be passed as parameters:
myfun = eval(Expr(:->,:x, Expr(:block, Expr(:call,:*,3,:x) )))
myfun(14)
# returns 42
The dump function is very useful to inspect the expression that the parsers has created. For two input arguments you would use a tuple for example as args[1]:
julia> dump(parse("(x,y) -> 3x + y"))
Expr
head: Symbol ->
args: Array{Any}((2,))
1: Expr
head: Symbol tuple
args: Array{Any}((2,))
1: Symbol x
2: Symbol y
typ: Any
2: Expr
[...]
Does this help?
In the Metaprogramming part of the Julia documentation, there is a sentence under the eval() and effects section which says
Every module has its own eval() function that evaluates expressions in its global scope.
Similarly, the REPL help ?eval will give you, on Julia 0.6.2, the following help:
Evaluate an expression in the given module and return the result. Every Module (except those defined with baremodule) has its own 1-argument definition of eval, which evaluates expressions in that module.
I assume, you are working in the Main module in your example. That's why you need to have the globals defined there. For your problem, you can use macros and interpolate the values of x and y directly inside the macro.
A minimal working example would be:
macro eval_line(a, b, x)
isa(a, Real) || (warn("$a is not a real number."); return :(throw(DomainError())))
isa(b, Real) || (warn("$b is not a real number."); return :(throw(DomainError())))
return :($a * $x + $b) # interpolate the variables
end
Here, #eval_line macro does the following:
Main> #macroexpand #eval_line(5, 6, 2)
:(5 * 2 + 6)
As you can see, the values of macro's arguments are interpolated inside the macro and the expression is given to the user accordingly. When the user does not behave,
Main> #macroexpand #eval_line([1,2,3], 7, 8)
WARNING: [1, 2, 3] is not a real number.
:((Main.throw)((Main.DomainError)()))
a user-friendly warning message is provided to the user at parse-time, and a DomainError is thrown at run-time.
Of course, you can do these things within your functions, again by interpolating the variables --- you do not need to use macros. However, what you would like to achieve in the end is to combine eval with the output of a function that returns Expr. This is what the macro functionality is for. Finally, you would simply call your macros with an # sign preceding the macro name:
Main> #eval_line(5, 6, 2)
16
Main> #eval_line([1,2,3], 7, 8)
WARNING: [1, 2, 3] is not a real number.
ERROR: DomainError:
Stacktrace:
[1] eval(::Module, ::Any) at ./boot.jl:235
EDIT 1. You can take this one step further, and create functions accordingly:
macro define_lines(linedefs)
for (name, a, b) in eval(linedefs)
ex = quote
function $(Symbol(name))(x) # interpolate name
return $a * x + $b # interpolate a and b here
end
end
eval(ex) # evaluate the function definition expression in the module
end
end
Then, you can call this macro to create different line definitions in the form of functions to be called later on:
#define_lines([
("identity_line", 1, 0);
("null_line", 0, 0);
("unit_shift", 0, 1)
])
identity_line(5) # returns 5
null_line(5) # returns 0
unit_shift(5) # returns 1
EDIT 2. You can, I guess, achieve what you would like to achieve by using a macro similar to that below:
macro random_oper(depth, fs, ts)
operations = eval(fs)
oper = operations[rand(1:length(operations))]
terminals = eval(ts)
ts = terminals[rand(1:length(terminals), 2)]
ex = :($oper($ts...))
for d in 2:depth
oper = operations[rand(1:length(operations))]
t = terminals[rand(1:length(terminals))]
ex = :($oper($ex, $t))
end
return ex
end
which will give the following, for instance:
Main> #macroexpand #random_oper(1, [+, -, /], [1,2,3])
:((-)([3, 3]...))
Main> #macroexpand #random_oper(2, [+, -, /], [1,2,3])
:((+)((-)([2, 3]...), 3))
Thanks Arda for the thorough response! This helped, but part of me thinks there may be a better way to do this as it seems too roundabout. Since I am writing a genetic program, I will need to create 500 of these ASTs, all with random functions and terminals from a set of allowed functions and terminals (fs and ts in the code). I will also need to test each function with 20 different values of x and v.
In order to accomplish this with the information you have given, I have come up with the following macro:
macro create_function(defs)
for name in eval(defs)
ex = quote
function $(Symbol(name))(x,v)
fs = [:+, :-, :DIV, :GT]
ts = [x,v,-1]
return create_single_full_tree(4, fs, ts)
end
end
eval(ex)
end
end
I can then supply a list of 500 random function names in my main() function, such as ["func1, func2, func3,.....". Which I can eval with any x and v values in my main function. This has solved my issue, however, this seems to be a very roundabout way of doing this, and may make it difficult to evolve each AST with each iteration.
Hello i trying create converter method from Disct to Vector in Julia language.
But i receive error, with i can't understand
ERROR: TypeError: Tuple: in parameter, expected Type{T}, got Dict{AbstractString,Int64}
My code
type Family
name::UTF8String
value::Int
end
function convertToVector(a1::Dict{AbstractString, Int64}())
A::Vector{Node}
for k in sort(collect(keys(a1)))
push!(A, Family(a1[k] , k))
end
return A
end
Any idea hot to change convertToVector method ?
There were several typos in the above code, but I think this should work:
# No () after the type of a1
# Also, see comment, better to parameterize function, use concrete type for Dict
function convertToVector{T<:AbstractString}(a1::Dict{T, Int64})
# This is how you create an empty vector to hold Family objects
A = Vector{Family}()
for k in sort(collect(keys(a1)))
# The values passed to the Family constructor were backwards
push!(A, Family(k, a1[k]))
end
A
end
Another way (probably not very quick):
julia> dict = Dict("fred" => 3, "jim" => 4)
Dict{ASCIIString,Int64} with 2 entries:
"fred" => 3
"jim" => 4
julia> Vector{Family}(map(f -> Family(f...), map(x -> collect(x), dict)))
2-element Array{Family,1}:
Family("fred",3)
Family("jim",4)
Perhaps I've been using too much Lisp recently...
Is there a possibility to construct dictionary with tuple values in Julia?
I tried
dict = Dict{Int64, (Int64, Int64)}()
dict = Dict{Int64, Tuple(Int64, Int64)}()
I also tried inserting tuple values but I was able to change them after so they were not tuples.
Any idea?
Edit:
parallel_check = Dict{Any, (Any, Any)}()
for i in 1:10
dict[i] = (i+41, i+41)
end
dict[1][2] = 1 # not able to change this way, setindex error!
dict[1] = (3, 5) # this is acceptable. why?
The syntax for tuple types (i.e. the types of tuples) changed from (Int64,Int64) in version 0.3 and earlier to Tuple{Int64,Int64} in 0.4. Note the curly braces, not parens around Int64,Int64. You can also discover this at the REPL by applying the typeof function to an example tuple:
julia> typeof((1,2))
Tuple{Int64,Int64}
So you can construct the dictionary you want like this:
julia> dict = Dict{Int64,Tuple{Int64,Int64}}()
Dict{Int64,Tuple{Int64,Int64}} with 0 entries
julia> dict[1] = (2,3)
(2,3)
julia> dict[2.0] = (3.0,4)
(3.0,4)
julia> dict
Dict{Int64,Tuple{Int64,Int64}} with 2 entries:
2 => (3,4)
1 => (2,3)
The other part of your question is unrelated, but I'll answer it here anyway: tuples are immutable – you cannot change one of the elements in a tuple. Dictionaries, on the other hand are mutable, so you can assign an entirely new tuple value to a slot in a dictionary. In other words, when you write dict[1] = (3,5) you are assigning into dict, which is ok, but when you write dict[1][2] = 1 you are assigning into the tuple at position 1 in dict which is not ok.
Suppose I have a Dict defined as follows:
x = Dict{AbstractString,Array{Integer,1}}("A" => [1,2,3], "B" => [4,5,6])
I want to convert this to a DataFrame object (from the DataFrames module). Constructing a DataFrame has a similar syntax to constructing a dictionary. For example, the above dictionary could be manually constructed as a data frame as follows:
DataFrame(A = [1,2,3], B = [4,5,6])
I haven't found a direct way to get from a dictionary to a data frame but I figured one could exploit the syntactic similarity and write a macro to do this. The following doesn't work at all but it illustrates the approach I had in mind:
macro dict_to_df(x)
typeof(eval(x)) <: Dict || throw(ArgumentError("Expected Dict"))
return quote
DataFrame(
for k in keys(eval(x))
#eval ($k) = $(eval(x)[$k])
end
)
end
end
I also tried writing this as a function, which does work when all dictionary values have the same length:
function dict_to_df(x::Dict)
s = "DataFrame("
for k in keys(x)
v = x[k]
if typeof(v) <: AbstractString
v = string('"', v, '"')
end
s *= "$(k) = $(v),"
end
s = chop(s) * ")"
return eval(parse(s))
end
Is there a better, faster, or more idiomatic approach to this?
Another method could be
DataFrame(Any[values(x)...],Symbol[map(symbol,keys(x))...])
It was a bit tricky to get the types in order to access the right constructor. To get a list of the constructors for DataFrames I used methods(DataFrame).
The DataFrame(a=[1,2,3]) way of creating a DataFrame uses keyword arguments. To use splatting (...) for keyword arguments the keys need to be symbols. In the example x has strings, but these can be converted to symbols. In code, this is:
DataFrame(;[Symbol(k)=>v for (k,v) in x]...)
Finally, things would be cleaner if x had originally been with symbols. Then the code would go:
x = Dict{Symbol,Array{Integer,1}}(:A => [1,2,3], :B => [4,5,6])
df = DataFrame(;x...)