How could I create variables from keys and values of a dictionary? Is the following extract (like PHP) possible?
julia> d = Dict(
"key1" =>111,
"key2" =>222,
"key3" =>333
);
julia> extract(d)
julia> key1, key2, key3
(111,222,333)
k = collect(keys(d))
v = collect(values(d))
Both keys and values return iterators.
collect then produces an array.
But note that you often do not need to do this and can just iterate through the dictionary using
for (k, v) in d
It is possible to introduce new variables into the global scope with eval:
julia> x = 1
1
julia> function testeval()
eval(:(x = 5))
return x
end
testeval (generic function with 1 method)
julia> testeval()
5
julia> x # the global x has changed!
5
An extract function could look like this:
julia> function extract(d)
expr = quote end
for (k, v) in d
push!(expr.args, :($(Symbol(k)) = $v))
end
eval(expr)
return
end
julia> extract(d)
julia> key1, key2, key3
(111,222,333)
Note that every module has its own global scope. Therefore, this will introduce the variables into the scope of the module where the extract function is defined, i.e., into the Main module if defined at the REPL as in the example.
You should be very careful when using eval and first consider other approaches, e.g, the ones mentioned by David P. Sanders and Dan Getz.
Related
I want to create an empty lsit and gardually fill that out with tuples. I've tried the following and each returns an error. My question is: how to append or add and element to an empty array?
My try:
A = []
A.append((2,5)) # return Error type Array has no field append
append(A, (2,5)) # ERROR: UndefVarError: append not defined
B = Vector{Tuple{String, String}}
# same error occues
You do not actually want to append, you want to push elements into your vector. To do that use the function push! (the trailing ! indicates that the function modifies one of its input arguments. It's a naming convention only, the ! doesn't do anything).
I would also recommend creating a typed vector instead of A = [], which is a Vector{Any} with poor performance.
julia> A = Tuple{Int, Int}[]
Tuple{Int64, Int64}[]
julia> push!(A, (2,3))
1-element Vector{Tuple{Int64, Int64}}:
(2, 3)
julia> push!(A, (11,3))
2-element Vector{Tuple{Int64, Int64}}:
(2, 3)
(11, 3)
For the vector of string tuples, do this:
julia> B = Tuple{String, String}[]
Tuple{String, String}[]
julia> push!(B, ("hi", "bye"))
1-element Vector{Tuple{String, String}}:
("hi", "bye")
This line in your code is wrong, btw:
B = Vector{Tuple{String, String}}
It does not create a vector, but a type variable. To create an instance you can write e.g. one of these:
B = Tuple{String, String}[]
B = Vector{Tuple{String,String}}() # <- parens necessary to construct an instance
It can also be convenient to use the NTuple notation:
julia> NTuple{2, String} === Tuple{String, String}
true
julia> NTuple{3, String} === Tuple{String, String, String}
true
The code at the end of this post constructs a function which is bound to the variables of a given dictionary. Furthermore, the function is not bound to the actual name of the dictionary (as I use the Ref() statement).
An example:
julia> D = Dict(:x => 4, :y => 5)
julia> f= #mymacro4(x+2y, D)
julia> f()
14
julia> DD = D
julia> D = nothing
julia> f()
14
julia> DD[:x] = 12
julia> f()
22
Now I want to be able to construct exactly the same function when I only have access to the expression expr = :(x+2y).
How do I do this? I tried several things, but was not able to find a solution.
julia> f = #mymacro4(:(x+2y), D)
julia> f() ### the function evaluation should also yield 14. But it yields:
:(DR.x[:x] + 2 * DR.x[:y])
(I actually want to use it within another macro in which the dictionary is automatically created. I want to store this dictionary and the function within a struct, such that I'm able to call this function at a later point in time and manipulate the objects in the dictionary. If necessary, I may post the complete example and explain the complete problem.)
_freevars2(literal) = literal
function _freevars2(s::Symbol)
try
if typeof(eval(s)) <: Function
return s
else
return Meta.parse("DR.x[:$s]")
end
catch
return Meta.parse("DR.x[:$s]")
end
end
function _freevars2(expr::Expr)
for (it, s) in enumerate(expr.args)
expr.args[it] = _freevars2(s)
end
return expr
end
macro mymacro4(expr, D)
expr2 = _freevars2(expr)
quote
let DR = Ref($(esc(D)))
function mysym()
$expr2
end
end
end
end
I would expect the global variables m and n get the first and second dimension size. However it seeems incorrect:
julia> x=rand(3,3)
3×3 Array{Float64,2}:
0.680079 0.929336 0.267358
0.874437 0.625239 0.804478
0.92407 0.737254 0.443433
julia> m,n = size(x);
julia> m,n
(3,3)
julia> global m,n = size(x);
julia> m,n
(3,(3,3))
Why is this behaviour of assigning 2 variables to a tuple different when we add the global key word?
It appears to split the statement at the comma. If you just do
x = rand(3,3)
global m,n = size(x)
in a fresh session, you will see that m is not defined (in your example it is defined from the statement just before). global takes the m out of the assignment and assigns the result to n.
I think you have to
global m = size(x,1)
global n = size(x,2)
I can unpack a tuple. I'm trying to write a function (or macro) that would unpack a subset of these from an instance of the type-constructor Parameters(). That is, I know how to do:
a,b,c = unpack(p::Parameters)
But I would like to do something like this:
b,c = unpack(p::Parameters, b,c)
or maybe even lazier:
unpack(p::Parameters, b, c)
This is to avoid writing things like:
function unpack_all_oldstyle(p::Parameters)
a=p.a; b=p.b; c=p.c; ... z=p.z;
return a,b,c,...,z
end
There's something wrong with my approach, but hopefully there is a fix.
In case it wasn't clear from the wording of my question, I'm a total ignoramus. I read about unpacking the ellipsis here: how-to-pass-tuple-as-function-arguments
"module UP tests Unpacking Parameters"
module UP
struct Parameters
a::Int64
b::Int64
c::Int64
end
"this method sets default parameters and returns a tuple of default values"
function Parameters(;
a::Int64 = 3,
b::Int64 = 11,
c::Int64 = 101
)
Parameters(a, b, c)
end
"this function unpacks all parameters"
function unpack_all(p::Parameters)
return p.a, p.b, p.c
end
"this function tests the unpacking function: in the body of the function one can now refer to a rather than p.a : worth the effort if you have dozens of parameters and complicated expressions to compute, e.g. type (-b+sqrt(b^2-4*a*c))/2/a instead of (-p.b+sqrt(p.b^2-4*p.a *p.c))/2/p.a"
function unpack_all_test(p::Parameters)
a, b, c = unpack_all(p)
return a, b, c
end
"""
This function is intended to unpack selected parameters. The first, unnamed argument is the constructor for all parameters. The second argument is a tuple of selected parameters.
"""
function unpack_selected(p::Parameters; x...)
return p.x
end
function unpack_selected_test(p::Parameters; x...)
x = unpack_selected(p, x)
return x
end
export Parameters, unpack_all, unpack_all_test, unpack_selected, unpack_selected_test
end
p = UP.Parameters() # make an instance
UP.unpack_all_test(p)
## (3,11,101) ## Test successful
UP.unpack_selected_test(p, 12)
## 12 ## intended outcome
UP.unpack_selected_test(p, b)
## 11 ## intended outcome
UP.unpack_selected_test(p, c, b, a)
## (101,11,3) ## intended outcome
There already exists one: Parameters.jl.
julia> using Parameters
julia> struct Params
a::Int64
b::Int64
c::Int64
end
julia> #unpack a, c = Params(1,2,3)
Params(1,2,3)
julia> a,c
(1,3)
julia> #with_kw struct Params
a::Int64 = 3
b::Int64 = 11
c::Int64 = 101
end
julia> #unpack c,b,a = Params()
Params
a: Int64 3
b: Int64 11
c: Int64 101
julia> c,b,a
(101,11,3)
BTW, you can fix your unpack_selected by:
unpack_selected(p::Parameters, fields...) = map(x->getfield(p, x), fields).
# note that, the selected field names should be Symbol here
julia> unpack_selected(p, :b)
(11,)
julia> unpack_selected(p, :c, :b, :a)
(101,11,3)
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)