Accordingly to the standard function names, Expr type is said to follow this simple structure:
julia> names(Expr)
3-element Array{Symbol,1}:
:head
:args
:typ
Therefore, I expect to be able to build an Expr from a direct call of the Expr constructor with this following trivial method, from another Expr:
julia> exp1 = :(x+y);
julia> exp2 = Expr(exp1.head,exp1.args,exp1.typ)
:(({:+,:x,:y})(Any))
But the resulting exp2 does not contain the same args as exp1; Expr build the new args from exp1.args and exp1.typ :
julia> exp2.args
2-element Array{Any,1}:
{:+,:x,:y}
Any
vs
julia> exp1.args
3-element Array{Any,1}:
:+
:x
:y
What did I do wrong?
Edit:
Following the Varargs section of the documentation, it is also possible to splice an iterable object into a function call. So in my case, it simply goes:
julia> Expr(:call,exp1.args...)
:(x+y)
The Expr() constructor doesn't map the inputs directly to the fields of the Expr type. Rather the first input becomes the head and the rest of the inputs become the args. So to construct the expression x + y using the Expr constructor you would do:
Expr(:call, :(+), :x, :y)
Related
macro test1(name,arg)
println(arg.args[2])
typeof(arg.args[2])
end
#test1 test1 (
(arg1, (:max=>10))
)
I have the macro above and I'm trying to pass a Dict parameter as the second parameter. But the problem is that I cannot access it. The macro will always interpret the second parameter as an expression, and when I try to access arg.args[2].args it shows:
Vector{Any} (alias for Array{Any, 1})
so I don't know how to pass the Dict as it is.
I just want to get the second argument like:
Dict{Symbol, Int64} with 1 entry:
:max => 10
This is because macros work on code before the code is compiled. Source code is first parsed to Symbols, literals (integers, floats, strings, etc), or Expr (expressions). At this point, all expressions contain only these three things.** After the macro is done and returns an expression, that expression is compiled into runtime code where more complicated objects like Dicts can exist.
The code below illustrates the difference before and after compiling. Note how 1+5 and Dict() were expressions in the macro body, but is afterward evaluated to an Int64 and a Dict.
# splat arbitrary number of Expr
macro peekExpr(expr1, expr2, expr3tuple...)
println(typeof(expr1), " ", expr1)
println(typeof(expr2), " ", expr2)
println(typeof(expr3tuple), " ", expr3tuple)
:($expr1, $expr2, $expr3tuple)
end
evaluated = #peekExpr 1+5 Dict() Int64 10 max::Int64
#= printout
Expr 1 + 5
Expr Dict()
Tuple{Symbol,Int64,Expr} (:Int64, 10, :(max::Int64))
=#
for item in evaluated println(typeof(item), " ", item) end
#= printout
Int64 6
Dict{Any, Any} Dict{Any, Any}()
Tuple{Symbol,Int64,Expr} (:Int64, 10, :(max::Int64))
=#
**PS: Bear in mind that Expr can contain other objects if you interpolate runtime objects into them (x = Dict(); :(a in $x).args[end] vs :(a in Dict()).args[end]). It's just that macros do not work at a phase where it can access runtime objects. #peekExpr $x will only see a Expr(:$, :x).
I have an Expr
:([4.01673077 7.62137432 8.74459436 10.14775067 12.57379884 14.39781963; 17.00925227 19.2273262])
and I wish to transfer it into a Vector{Float64}, but calling eval on this directly does not seem to work.
How can I do this?
You can either extract the data from Expr (faster and safe) or just eval it (slower, unsafe, easier code).
julia> a = :([12.57379884 14.39781963; 17.00925227 19.2273262])
:([12.57379884 14.39781963; 17.00925227 19.2273262])
julia> [Float64.(vals) for vals in getfield.(a.args, :args)]
2-element Vector{Vector{Float64}}:
[12.57379884, 14.39781963]
[17.00925227, 19.2273262]
julia> eval(a)
2×2 Matrix{Float64}:
12.5738 14.3978
17.0093 19.2273
Your matrix is wrongly formed and hence only the first solution will work with it.
Julia has the setter functions setproperty! and setfield! and the getter functions getproperty and getfield that operate on structs. What is the difference between properties and fields in Julia?
For example, the following seems to indicate that they do the same thing:
julia> mutable struct S
a
end
julia> s = S(2)
S(2)
julia> getfield(s, :a)
2
julia> getproperty(s, :a)
2
julia> setfield!(s, :a, 3)
3
julia> s
S(3)
julia> setproperty!(s, :a, 4)
4
julia> s
S(4)
fields are simply the "components" of a struct. The struct
struct A
b
c::Int
end
has the fields b and c. A call to getfield returns the object that is bound to the field:
julia> a = A("foo", 3)
A("foo", 3)
julia> getfield(a, :b)
"foo"
In early versions of Julia, the syntax a.b used to "lower", i.e. be the same as, writing getfield(a, :b). What has changed now is that a.b lowers to getproperty(a, :b) with the default fallback
getproperty(a::Type, v::Symbol) = getfield(a, v)
So by default, nothing has changed. However, authors of structs can overload getproperty (it is not possible to overload getfield) to provide extra functionality to the dot-syntax:
julia> function Base.getproperty(a::A, v::Symbol)
if v == :c
return getfield(a, :c) * 2
elseif v == :q
return "q"
else
return getfield(a, v)
end
end
julia> a.q
"q"
julia> getfield(a, :q)
ERROR: type A has no field q
julia> a.c
6
julia> getfield(a, :c)
3
julia> a.b
"foo"
So we can add extra functionality to the dot syntax (dynamically if we want). As a concrete example where this is useful is for the package PyCall.jl where you used to have to write pyobject[:field] while it is possible now to implement it such that you can write pyobject.field.
The difference between setfield! and setproperty! is analogous to the difference between getfield and getproperty, explained above.
In addition, it is possible to hook into the function Base.propertynames to provide tab completion of properties in the REPL. By default, only the field names will be shown:
julia> a.<TAB><TAB>
b c
But by overloading propertynames we can make it also show the extra property q:
julia> Base.propertynames(::A) = (:b, :c, :q)
julia> a.<TAB><TAB>
b c q
I followed the documentation of julia:
julia> :(a in (1,2,3))
:($(Expr(:in, :a, :((1,2,3)))))
Now that :(a in (1,2,3))==:($(Expr(:in, :a, :((1,2,3))))), why does julia express this expression in this way？ And what does $ exactly means? It seems to me that $ just evaluates the next expression in a global scope. I found the documentation unclear about this.
The reason :(a in (1,2,3)) is displayed awkwardly as :($(Expr(...))) is because the show function for Expr typed objects (show_unquoted in show.jl) does not understand the in infix operator and fallbacks into a generic printing format.
Essentially it is the same as :(1 + 1) except that show_unquoted recognizes + as an infix operator and formats it nicely.
In any case, :(...) and $(...) are inverse operators in some sense, so :($(..thing..)) is exactly like ..thing.., which in this case is Expr(:in,:a,:((1,2,3))).
One can see this weirdness in :(1+1) for example. The output is of Expr type, as typeof(:(1+1))==Expr confirms. It is actually Expr(:+,1,1), but typing Expr(:+,1,1) on the REPL will show :($(Expr(:+,1,1))) - the generic formatting style of Expr typed objects.
Fixing show.jl to handle in could be a nice change. But the issue is harmless and concerns display formatting.
$ is the interpolation command, Julia use this notation to interpolate Strings as well as Expression:
julia> a=1;
julia> "test $a" # => "test 1"
julia> :(b+$a) # => :(b + 1)
When you type a command in Julia REPL, it tries to evaluates the command and if the code do not have ; char at the end it prints the result, so it's more related to printing functions, that what will be seen on REPL, when a command executes.
so if you want to see the real contents of a variable one possibility is to use dump function:
julia> dump(:(a+b))
Expr
head: Symbol call
args: Array(Any,(3,))
1: Symbol +
2: Symbol a
3: Symbol b
typ: Any
julia> dump(:(a in b))
Expr
head: Symbol in
args: Array(Any,(2,))
1: Symbol a
2: Symbol b
typ: Any
It's clear from above tests, that both expressions use a common data structure of Expr with head, args and typ without any $ inside.
Now try to evaluate and print result:
julia> :(a in b)
:($(Expr(:in, :a, :b)))
julia> :(a+b)
:(a + b)
We already know that both command create a same structure but REPL can't show the result of :(a in b) better that an Expr of result of another Expr and it's why there in a $ inside. But when dealing with :(a+b), REPL do more intelligently and understands that this:
Expr
head: Symbol call
args: Array(Any,(3,))
1: Symbol +
2: Symbol a
3: Symbol b
typ: Any
is equal to :(a+b).
Given a dictionary of values,
values = {:A => 3, :B => 1}
turn an (arbitrary) expression like
expr = :(2*A)
into a function foo(values) that evaluates the expression, so in this case foo(values) = 6. The resulting function will be called millions of times, so speed is an important consideration. I am happy to adopt a slightly different approach if necessary, as long as it can be automatised.
Things I tried:
The conversion using convert(Function, expr), as suggested here. Fails for me (Julia 0.3.8-pre):
convert has no method matching convert(::Type{Function}, ::Expr)
Using #eval one can do
#eval foo(A) = $(expr)
and then call foo(values[:A]), but that would require knowing that expr depends on A (and only on A).
I wrote a function find_vars(exp) to return the symbols in expr (in this case [:A]), but couldn't find how to use them in the #eval approach.
Base.Cartesian has an unexported function lreplace which may be what you're after. Then you can do something like:
julia> values = Dict(:A=>3, :B=>1)
Dict{Symbol,Int64} with 2 entries:
:B => 1
:A => 3
julia> import Base.Cartesian.lreplace
julia> expr = :(2*A)
:(2A)
julia> function lreplace_all(expr, d)
for (k, v) in d
expr = lreplace(expr, k, v)
end
expr
end
lreplace_all (generic function with 1 method)
julia> lreplace_all(expr, values)
:(2 * 3)
julia> #eval foo(A) = $(lreplace_all(:(2A), values))
foo (generic function with 1 method)
julia> foo(1)
6
Although, since A is defined by the values dict, it makes more sense to define foo as a zero-argument function (unless I've missed something).
EDIT:
After rereading your question it seems like you want to pass in the actual dictionary to the function rather than have the values available at compile time as I've done above. In that case, we have get a little creative:
First we need an lreplace like function that will work with expressions which is easy enough
julia> dictreplace!(ex, s, v) = ex
dictreplace! (generic function with 1 method)
julia> dictreplace!(ex::Symbol, s, v) = s == ex ? v : ex
dictreplace! (generic function with 2 methods)
julia> function dictreplace!(ex::Expr, s, v)
for i=1:length(ex.args)
ex.args[i] = dictreplace!(ex.args[i], s, v)
end
ex
end
dictreplace! (generic function with 3 methods)
julia> dictreplace(ex, s, v) = dictreplace!(copy(ex), s, v)
dictreplace (generic function with 1 method)
Now we want to replace every occurence of a symbol in our dict keys with a dictionary lookup
julia> function dictreplace_all(expr, kys, dsym)
for k in kys
expr = dictreplace(expr, k, :($(dsym)[$(QuoteNode(k))]))
end
expr
end
dictreplace_all (generic function with 1 method)
julia> dictreplace_all(:(2A), keys(values), :d)
:(2 * d[:A])
julia> #eval foo(args) = $(dictreplace_all(:(2A), keys(values), :args))
foo (generic function with 1 method)
julia> values[:A] = -99
-99
julia> foo(values)
-198
Thanks to the solution by #ptb and another metaprogramming question I found a simpler yet slower solution:
function foo(values, expr)
expr = quote
A = values[:A]
B = values[:B]
return $(expr)
end
eval(expr)
end
Reading in the values from the dictionary can also be done programmatically by replacing the inner evaluation by
$([:($k = $v) for (k, v) in values]...)
return $(expr)