I am a newbie to Julia and Flux with some experience in Tensorflow Keras and python. I tried to use the Flux.withgradient command to write a user-defined training function with more flexibility. Here is the training part of my code:
loss, grad = Flux.withgradient(modelDQN.evalParameters) do
qEval = modelDQN.evalModel(evalInput)
Flux.mse(qEval, qTarget)
end
Flux.update!(modelDQN.optimizer, modelDQN.evalParameters, grad)
This code works just fine. But if I put the command qEval = modelDQN.evalModel(evalInput) outside the do end loop, as follows:
qEval = modelDQN.evalModel(evalInput)
loss, grad = Flux.withgradient(modelDQN.evalParameters) do
Flux.mse(qEval, qTarget)
end
Flux.update!(modelDQN.optimizer, modelDQN.evalParameters, grad)
The model parameters will not be updated. As far as I know, the do end loop works as an anonymous function that takes 0 arguments. Then why do we need the command qEval = modelDQN.evalModel(evalInput) inside the loop to get the model updated?
The short answer is that anything to be differentiated has to happen inside the (anonymous) function which you pass to gradient (or withgradient), because this is very much not a standard function call -- Zygote (Flux's auto-differentiation library) traces its execution to compute the derivative, and can't transform what it can't see.
Longer, this is Zygote's "implicit" mode, which relies on global references to arrays. The simplest use is something like this:
julia> using Zygote
julia> x = [2.0, 3.0];
julia> g = gradient(() -> sum(x .^ 2), Params([x]))
Grads(...)
julia> g[x] # lookup by objectid(x)
2-element Vector{Float64}:
4.0
6.0
If you move some of that calculation outside, then you make a new array y with a new objectid. Julia has no memory of where this came from, it is completely unrelated to x. They are ordinary arrays, not a special tracked type.
So if you refer to y in the gradient, Zygote cannot infer how this depends on x:
julia> y = x .^ 2 # calculate this outside of gradient
2-element Vector{Float64}:
4.0
9.0
julia> g2 = gradient(() -> sum(y), Params([x]))
Grads(...)
julia> g2[x] === nothing # represents zero
true
Zygote doesn't have to be used in this way. It also has an "explicit" mode which does not rely on global references. This is perhaps less confusing:
julia> gradient(x1 -> sum(x1 .^ 2), x) # x1 is a local variable
([4.0, 6.0],)
julia> gradient(x1 -> sum(y), x) # sum(y) is obviously indep. x1
(nothing,)
julia> gradient((x1, y1) -> sum(y1), x, y)
(nothing, Fill(1.0, 2))
Flux is in the process of changing to use this second form. On v0.13.9 or later, something like this ought to work:
opt_state = Flux.setup(modelDQN.optimizer, modelDQN) # do this once
loss, grads = Flux.withgradient(modelDQN.model) do m
qEval = m(evalInput) # local variable m
Flux.mse(qEval, qTarget)
end
Flux.update!(opt_state, modelDQN.model, grads[1])
First post here, thanks for reading!
Problem: I have a Vector{String} - call it A - where each element is a part of an equation, e.g. the first element of A is "x[1] - (0.8*x[1])". I would like to write a macro that takes as arguments i) a String - call it fn_name - with the name of a function, ii) the vector A, and returns a function named fn_name which looks like
function fn_name(f, x)
f[1] = x[1] - (0.8*x[1])
f[2] = (exp(x[4]) - 0.8*exp(x[3]))^(-1.1) - (0.99*(exp(x[4]) - 0.8*exp(x[4]))^(-1.1)*(1.0 - 0.025 + 0.30*exp(x[1])*exp(x[2])^(0.30 - 1.0)))
f[3] = exp(x[2]) - ((1.0 - 0.025)*exp(x[2]) + exp(x[1])*exp(x[2])^0.30 - exp(x[4]))
f[4] = x[3] - (x[4])
end
where each rhs is one element of
A = ["x[1] - (0.8*x[1])", "(exp(x[4]) - 0.8*exp(x[3]))^(-1.1) - (0.99*(exp(x[4]) - 0.8*exp(x[4]))^(-1.1)*(1.0 - 0.025 + 0.30*exp(x[1])*exp(x[2])^(0.30 - 1.0)))", "exp(x[2]) - ((1.0 - 0.025)*exp(x[2]) + exp(x[1])*exp(x[2])^0.30 - exp(x[4]))", "x[3] - (x[4])"]
What I tried: my best attempt at solving the problem is the following
macro make_fn(fn_name, A)
esc(quote
function $(Symbol(fn_name))(f, x)
for i = 1:length($(A))
f[$i] = Meta.parse($(A)[$i])
end
end
end)
end
which however doesn't work: when I run #make_fn("my_name", A) I get the error LoadError: UndefVarError: i not defined.
I find it quite hard to wrap my head around Julia metaprogramming, and while I'd be very happy to avoid using it, I think for this problem it's unavoidable.
Can you please help me understand where my mistake is?
Thanks
Macros in this case are not only avoidable, but even inapplicable, unless A is literal known at compile time.
I can provide a solution using eval and some closures:
julia> function make_fn2(A)
Af = [#eval(x -> $(Meta.parse(expr))) for expr in A]
function (f, x)
for i in eachindex(A, f)
f[i] = Af[i](x)
end
return f
end
end
make_fn2 (generic function with 1 method)
julia> fn_name = make_fn2(A)
#46 (generic function with 1 method)
julia> fn_name(zeros(4), [1,2,3,4])
4-element Array{Float64,1}:
0.19999999999999996
-0.06594092302655707
49.82984401122239
-1.0
with the restrictions that
eval will evaluate the expressions in a global scope of the module where this is defined (so it is potentially a different scope that a scope of the calling function), and
the newly created function will work only if you first return to global scope (i.e. it will not work if you try to run it within a function in which you have created it).
But I'd really recommend thinking about a better input format than strings.
I'm trying to build a function that will output an expression to be assigned to a new in-memory function. I might be misinterpreting the capability of metaprogramming but, I'm trying to build a function that generates a math series and assigns it to a function such as:
main.jl
function series(iter)
S = ""
for i in 1:iter
a = "x^$i + "
S = S*a
end
return chop(S, tail=3)
end
So, this will build the pattern and I'm temporarily working with it in the repl:
julia> a = Meta.parse(series(4))
:(x ^ 1 + x ^ 2 + x ^ 3 + x ^ 4)
julia> f =eval(Meta.parse(series(4)))
120
julia> f(x) =eval(Meta.parse(series(4)))
ERROR: cannot define function f; it already has a value
Obviously eval isn't what I'm looking for in this case but, is there another function I can use? Or, is this just not a viable way to accomplish the task in Julia?
The actual error you get has to do nothing with metaprogramming, but with the fact that you are reassigning f, which was assigned a value before:
julia> f = 10
10
julia> f(x) = x + 1
ERROR: cannot define function f; it already has a value
Stacktrace:
[1] top-level scope at none:0
[2] top-level scope at REPL[2]:1
It just doesn't like that. Call either of those variables differently.
Now to the conceptual problem. First, what you do here is not "proper" metaprogramming in Julia: why deal with strings and parsing at all? You can work directly on expressions:
julia> function series(N)
S = Expr(:call, :+)
for i in 1:N
push!(S.args, :(x ^ $i))
end
return S
end
series (generic function with 1 method)
julia> series(3)
:(x ^ 1 + x ^ 2 + x ^ 3)
This makes use of the fact that + belongs to the class of expressions that are automatically collected in repeated applications.
Second, you don't call eval at the appropriate place. I assume you meant to say "give me the function of x, with the body being what series(4) returns". Now, while the following works:
julia> f3(x) = eval(series(4))
f3 (generic function with 1 method)
julia> f3(2)
30
it is not ideal, as you newly compile the body every time the function is called. If you do something like that, it is preferred to expand the code once into the body at function definition:
julia> #eval f2(x) = $(series(4))
f2 (generic function with 1 method)
julia> f2(2)
30
You just need to be careful with hygiene here. All depends on the fact that you know that the generated body is formulated in terms of x, and the function argument matches that. In my opinion, the most Julian way of implementing your idea is through a macro:
julia> macro series(N::Int, x)
S = Expr(:call, :+)
for i in 1:N
push!(S.args, :($x ^ $i))
end
return S
end
#series (macro with 1 method)
julia> #macroexpand #series(4, 2)
:(2 ^ 1 + 2 ^ 2 + 2 ^ 3 + 2 ^ 4)
julia> #series(4, 2)
30
No free variables remaining in the output.
Finally, as has been noted in the comments, there's a function (and corresponding macro) evalpoly in Base which generalizes your use case. Note that this function does not use code generation -- it uses a well-designed generated function, which in combination with the optimizations results in code that is usually equal to the macro-generated code.
Another elegant option would be to use the multiple-dispatch mechanism of Julia and dispatch the generated code on type rather than value.
#generated function series2(p::Val{N}, x) where N
S = Expr(:call, :+)
for i in 1:N
push!(S.args, :(x ^ $i))
end
return S
end
Usage
julia> series2(Val(20), 150.5)
3.5778761722367333e43
julia> series2(Val{20}(), 150.5)
3.5778761722367333e43
This task can be accomplished with comprehensions. I need to RTFM...
https://docs.julialang.org/en/v1/manual/arrays/#Generator-Expressions
In my equations we have many expressions with a^2, and so on. I would like to map "²" to ^2, to obtain something like that:
julia> a² == a^2
true
The above is not however a legal code in Julia. Any idea on how could I implement it ?
Here is a sample macro #hoo that does what you requested in a simplified scenario (since the code is long I will start with usage).
julia> x=5
5
julia> #hoo 3x² + 4x³
575
julia> #macroexpand #hoo 2x³+3x²
:(2 * Main.x ^ 3 + 3 * Main.x ^ 2)
Now, let us see the macro code:
const charsdict=Dict(Symbol.(split("¹²³⁴⁵⁶⁷⁸⁹","")) .=> 1:9)
const charsre = Regex("[$(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.
Why is
julia> collect(partitions(1,2))
0-element Array{Any,1}
returned instead of
2-element Array{Any,1}:
[0,1]
[1,0]
and do I really have to
x = collect(partitions(n,m));
y = Array(Int64,length(x),length(x[1]));
for i in 1:length(x)
for j in 1:length(x[1])
y[i,j] = x[i][j];
end
end
to convert the result to a two-dimensional array?
From the wikipedia:
In number theory and combinatorics, a partition of a positive integer n, also called an integer partition, is a way of writing n as a sum of positive integers.
For array conversion, try:
julia> x = collect(partitions(5,3))
2-element Array{Any,1}:
[3,1,1]
[2,2,1]
or
julia> x = partitions(5,3)
Base.FixedPartitions(5,3)
then
julia> hcat(x...)
3x2 Array{Int64,2}:
3 2
1 2
1 1
Here's another approach to your problem that I think is a little simpler, using the Combinatorics.jl library:
multisets(n, k) = map(A -> [sum(A .== i) for i in 1:n],
with_replacement_combinations(1:n, k))
This allocates a bunch of memory, but I think your current approach does too. Maybe it would be useful to make a first-class version and add it to Combinatorics.jl.
Examples:
julia> multisets(2, 1)
2-element Array{Array{Int64,1},1}:
[1,0]
[0,1]
julia> multisets(3, 5)
21-element Array{Array{Int64,1},1}:
[5,0,0]
[4,1,0]
[4,0,1]
[3,2,0]
[3,1,1]
[3,0,2]
[2,3,0]
[2,2,1]
[2,1,2]
[2,0,3]
⋮
[1,2,2]
[1,1,3]
[1,0,4]
[0,5,0]
[0,4,1]
[0,3,2]
[0,2,3]
[0,1,4]
[0,0,5]
The argument order is backwards from yours to match mathematical convention. If you prefer the other way, that can easily be changed.
one robust solution can be achieved using lexicographic premutations generation algorithm, originally By Donald Knuth plus classic partitions(n).
that is lexicographic premutations generator:
function lpremutations{T}(a::T)
b=Vector{T}()
sort!(a)
n=length(a)
while(true)
push!(b,copy(a))
j=n-1
while(a[j]>=a[j+1])
j-=1
j==0 && return(b)
end
l=n
while(a[j]>=a[l])
l-=1
end
tmp=a[l]
a[l]=a[j]
a[j]=tmp
k=j+1
l=n
while(k<l)
tmp=a[k]
a[k]=a[l]
a[l]=tmp
k+=1
l-=1
end
end
end
The above algorithm will generates all possible unique
combinations of an array elements with repetition:
julia> lpremutations([2,2,0])
3-element Array{Array{Int64,1},1}:
[0,2,2]
[2,0,2]
[2,2,0]
Then we will generate all integer arrays that sum to n using partitions(n) (forget the length of desired arrays m), and resize them to the lenght m using resize_!
function resize_!(x,m)
[x;zeros(Int,m-length(x))]
end
And main function looks like:
function lpartitions(n,m)
result=[]
for i in partitions(n)
append!(result,lpremutations(resize_!(i, m)))
end
result
end
Check it
julia> lpartitions(3,4)
20-element Array{Any,1}:
[0,0,0,3]
[0,0,3,0]
[0,3,0,0]
[3,0,0,0]
[0,0,1,2]
[0,0,2,1]
[0,1,0,2]
[0,1,2,0]
[0,2,0,1]
[0,2,1,0]
[1,0,0,2]
[1,0,2,0]
[1,2,0,0]
[2,0,0,1]
[2,0,1,0]
[2,1,0,0]
[0,1,1,1]
[1,0,1,1]
[1,1,0,1]
[1,1,1,0]
The MATLAB script from http://www.mathworks.com/matlabcentral/fileexchange/28340-nsumk actually behaves the way I need, and is what I though that partitions() would do from the description given. The Julia version is
# k - sum, n - number of non-negative integers
function nsumk(k,n)
m = binomial(k+n-1,n-1);
d1 = zeros(Int16,m,1);
d2 = collect(combinations(collect((1:(k+n-1))),n-1));
d2 = convert(Array{Int16,2},hcat(d2...)');
d3 = ones(Int16,m,1)*(k+n);
dividers = [d1 d2 d3];
return diff(dividers,2)-1;
end
julia> nsumk(3,2)
4x2 Array{Int16,2}:
0 3
1 2
2 1
3 0
using daycaster's lovely hcat(x...) tidbit :)
I still wish there would be a more compact way of doing this.
The the first mention of this approach seem to be https://au.mathworks.com/matlabcentral/newsreader/view_thread/52610, and as far as I can understand it is based on the "stars and bars" method https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)