Generate functions inside a function from string

[Chong_Wang]

Thank you.I’ll try the package next time I do symbolic calculation.

[wuhuagumu]

This is really good, 1 allocation more and almost comparable time.

using SymPy
using SyntaxTree

@vars x

expr=integrate(sin(x))
f=eval(parse("x->$(repr(expr))"))
h=SyntaxTree.genfun(parse("$(repr(expr))"),[:x])

@time f(1.0)
@time h(1.0)
#   0.000003 seconds (5 allocations: 176 bytes)
#   0.000005 seconds (6 allocations: 192 bytes)

[chakravala]

@wuhuagumu

h=SyntaxTree.genfun(parse(“x->$(repr(expr))”),[:x])

Note that the input to SyntaxTree.genfun doesn’t need to be a function, it should only be the expression of the function that is added, so you might want to use parse("$(repr(expr))") instead, since it automatically turns it into the function using the arguments from the list.

[wuhuagumu]

Any consequences for invokelatest?

[wuhuagumu]

My fault, sorry. I noticed this and forgot to modify it. Changed.

[StefanKarpinski]

It’s slow and can’t be inlined because it has to do a full dynamic dispatch.

[rdeits]

Just as a note, those functions are fast enough that @time is not going to be giving you a reliable or informative benchmark (you’re only running the function once, possibly also measuring compilation time, and you’re operating at global scope, which is slow). BenchmarkTools.jl solves all of those problems for you by running the function multiple times and gathering statistics. In general, @time is probably good enough for something which takes a few seconds, but for a function that takes less than a second you really want to use BenchmarkTools.

Using @btime from BenchmarkTools:

julia> using BenchmarkTools

julia> @btime subs($expr, $x => 1.0)
  139.558 μs (79 allocations: 2.34 KiB)
-0.540302305868140

julia> @btime $f(1.0)
  11.564 ns (0 allocations: 0 bytes)
-0.5403023058681398

julia> @btime $g(1.0)
  115.019 ns (2 allocations: 32 bytes)
-0.5403023058681398

Besides, I thought building something from string should not be a very uncommon thing…

Actually it’s not common in Julia precisely because it’s much easier to manipulate expressions directly in Julia compared to other programming languages. That means that we can treat Julia code as objects in Julia, manipulating or modifying them in code, without ever having to treat them as strings. For example, I can construct the expression x + y by quoting it:

julia> :(x + y)
:(x + y)

Or I can construct an Expression type by hand:

julia> Expr(:call, :+, :x, :y)
:(x + y)

and get the same result.

I can also modify an expression to get a new expression:

julia> ex = Expr(:call, :+, :x, :y)
:(x + y)

julia> ex.args[1] = :-  # change that + to a -
:-

julia> ex
:(x - y)

In the above case I modified a Julia expression without having to do any string manipulation or invoking of parse().

[wuhuagumu]

But you have to eval these expressions to generate a function. Here my main concern is how to construct a function from expressions without calling eval. String is not the most difficult part as parsing can do it perfectly.

[jandehaan]

I can’t tell from the discussion what exactly you are trying to accomplish (besides not wanting to use eval). But here is a stab in a somewhat different direction.

macro foo(partial_expr)
    expr = :(x->x)
    expr.args[2] = partial_expr
    expr
end

function bar()
        (@foo(x+2)(3), @foo(x*2)(3), @foo(x^2)(3))
end

julia> bar()
(5, 6, 9)

[Chong_Wang]

I created a separate thread here summarizing generally what I want:

How do I create a function from an expression

Suppose I have expr = :(sin(x)+cos(x)) How can I get a function looking like x->sin(x)+cos(x) I have tried to write such a macro macro expr2fn(expr, arg) return :($arg -> $(expr)) end but it does not work julia> f = @expr2fn expr x (::#63) (generic function with 1 method) julia> f(1.0) :(sin(x) + cos(x)) Not sure what is happening here.

[Chong_Wang]

I believe you posted in the wrong thread. To avoid confusion, I suggest you delete this post and post again in the correct thread: How do I create a function from an expression

[chakravala]

@Chong_Wang

the function produced f is type unstable

The type instability is because of the Base.invokelatest, if you want to avoid it, you need to specify the output type like this Base.invokelatest(...)::Float64, but you already need to know your type.

Would be great if Base.invokelatest could be type stable.

[kristoffer.carlsson]

Would be great if Base.invokelatest could be type stable.

The reason why invokelatest works at all is also what prevents it to be type stable.

[chakravala]

@Chong_Wang

the function produced f is type unstable?

Alright, so I have fixed this issue with a new implementation of genfun in SyntaxTree package

using SyntaxTree

f = SyntaxTree.genfun(:(-cos(x)), [:(x::Float64)], Float64)

julia> @btime $f(1.0)
  148.596 ns (2 allocations: 32 bytes)
-0.5403023058681398

while with the old version I get

julia> @btime $f(1.0)
  199.548 ns (2 allocations: 32 bytes)
-0.5403023058681398

So there is an improvement by adding the type stability option into SyntaxTree.genfun.

"""
    genfun(expr, args::Array, typ=Any)

Returns an anonymous function based on the given `expr` and `args`.
"""
function genfun(expr,args::Array,typ=Any)
    gs = gensym()
    eval(Expr(:function,Expr(:call,gs,args...),expr))
    list = Symbol[]
    for arg ∈ args
        push!(list,typeof(arg) == Expr ? arg.args[1] : arg)
    end
    eval(:($(Expr(:tuple,list...))->Base.invokelatest($gs,$(list...))::$typ))
end

This now appends ::DataType at the end of the Base.invokelatest call.

[chakravala]

The reason why invokelatest works at all is also what prevents it to be type stable.

In my new version of SyntaxTree.genfun described in previous post,

julia> @code_warntype f(1.0)
Variables:
  #self# <optimized out>
  x::Float64

Body:
  begin 
      return (Core.typeassert)((Core._apply_latest)(SyntaxTree.##713, (Core.tuple)(x::Float64)::Tuple{Float64})::Any, Float64)::Float64
  end::Float64

So what is the difference between Base.invokelatest and Core._apply_latest?

[kristoffer.carlsson]

Let’s rewrite that as

x = (Core._apply_latest)(SyntaxTree.##713, (Core.tuple)(x::Float64)::Tuple{Float64})::Any
y = (Core.typeassert)(x, Float64)::Float64

So __apply_latest infers to Any and then you have a type assert that errors if the result is not a Float64 and from there on we can assume that y is a Float64.

[chakravala]

Hmmm, after restarting my REPL (was using Revise) the performance for both versions is the same

julia> g = SyntaxTree.genfun(:x,[:x])

julia> @btime $g(1.0)
  106.034 ns (2 allocations: 32 bytes)
1.0

julia> f = SyntaxTree.genfun(:x,[:(x::Float64)], Float64)
(::#15) (generic function with 1 method)

julia> @btime $f(1.0)
  109.066 ns (2 allocations: 32 bytes)
1.0

So it seems that the type assertion for Any vs. Float64 actually does not make a difference, but the new version of genfun seems to be faster than the original version anyway…

[kristoffer.carlsson]

The type assert will only be beneficial if you use the result of the function in the same function as it was called.
Just tacking on a type assert will not make the call itself faster (in fact a bit slower since the type has to be checked).

[chakravala]

Ah, that makes sense, thanks for clarifying. With SyntaxTree.genfun you will have both options.

[chakravala]

As discussed in the other thread, in the SyntaxTree package there is now an updated genfun and @genfun for creating functions, and it now does not require Base.invokelatest, so is faster

How do I create a function from an expression

Actually, I have just figured out to do this macro without Base.invokelatest and updated SyntaxTree : julia> using SyntaxTree, BenchmarkTools julia> g = genfun(:(-(cos(x))),[:x]) (::#3) (generic function with 1 method) julia> @btime $g(1.0) 40.790 ns (0 allocations: 0 bytes) -0.5403023058681398 julia> f = @genfun -cos(x) [x] (::#3) (generic function with 1 method) julia> @btime $f(1.0) 40.801 ns (0 allocations: 0 bytes) -0.5403023058681398 Now the evaluation is much faster because it i…