I have a module with an enum defined in it.
module myModule
#enum type A B B C D
end
type1 = myModule.A
Now I want to declare an instance of this enum type but I only have a string specifying which type it is. I tried the following:
str = "B"
type2 = eval(:(myModule.Symbol($str)))
But I get a warning message which I do not quite understand:
WARNING: replacing module myModule.
and the type of type2 is also just a Symbol.
Probably the simplest way is to use getproperty:
julia> module myModule
#enum type A B C D
end
Main.myModule
julia> str = "B";
julia> getproperty(myModule, Symbol(str))
B::type = 1
Alternatively, you could create your expression as a string and then parse and evaluate it:
julia> eval(Meta.parse(string("myModule.", str)))
B::type = 1
Or, the same thing, but with string interpolation instead of using the string function:
julia> eval(Meta.parse("myModule.$str"))
B::type = 1
Note that the syntax myModule.Symbol(str) is not equivalent to myModule.B. It looks like that syntax really just calls Symbol(str) in the global scope. For example, try the following:
julia> myModule.length([1, 2, 3])
3
julia> #code_lowered myModule.length([1, 2, 3])
CodeInfo(
1 ─ %1 = (Base.arraylen)(a)
└── return %1
)
haskey() and in() functions are very useful to test the content of dictionaries in Julia :
julia> dict = Dict("a" => 1, "b" => 2, "c" => 3, "d" => 4, "e" => 5)
Dict{String,Int64} with 5 entries:
"c" => 3
"e" => 5
"b" => 2
"a" => 1
"d" => 4
julia> haskey(dict, "a")
true
julia> in(("a" => 1), dict)
true
but I was surprised by their behavior with complex keys :
julia> immutable MyT
A::String
B::Int64
end
julia> a = Dict(MyT("Tom",191)=>1,MyT("Bob",20)=>1,MyT("Jo",315)=>1,MyT("Luc",493)=>1)
Dict{MyT,Int64} with 4 entries:
MyT("Tom",191) => 1
MyT("Jo",315) => 1
MyT("Bob",20) => 1
MyT("Luc",493) => 1
julia> keys(a)
Base.KeyIterator for a Dict{MyT,Int64} with 4 entries. Keys:
MyT("Tom",191)
MyT("Jo",315)
MyT("Bob",20)
MyT("Luc",493)
julia> haskey(a, MyT("Tom",191))
false
julia> in((MyT("Tom",191) => 1), a)
false
What I did wrong ?
Thank you very much for your comments !
Thanks to #Michael K. Borregaard, I can propose this solution :
a = Dict{MyT, Int64}()
keyArray = Array{MyT,1}()
keyArray = [MyT("Tom",191),MyT("Bob",20),MyT("Jo",315),MyT("Luc",493)]
for i in keyArray
a[i] = 1
end
println(a)
# Dict(MyT("Tom",191)=>1,MyT("Tom",191)=>1,MyT("Luc",493)=>1,MyT("Jo",315)=>1,MyT("Luc",493)=>1,MyT("Bob",20)=>1,MyT("Jo",315)=>1,MyT("Bob",20)=>1)
keyArray[1] # MyT("Tom",191)
haskey(a, keyArray[1]) # true
But I have to store keys in a separate array. This means that can't warranty the unicity of the keys which is the strength of the dictionaries and why I choose to use it :(
So I have to use another step :
unique(keyArray)
Another better solution :
function CompareKeys(k1::MyT, k2::MyT)
if k1.A == k2.A && k1.B == k2.B
return true
else
return false
end
end
function ExistKey(k::MyT, d::Dict{MyT, Int64})
for i in keys(d)
if CompareKeys(k, i)
return true
end
end
return false
end
a = Dict(MyT("Tom",191)=>1,MyT("Bob",20)=>1,MyT("Jo",315)=>1,MyT("Luc",493)=>1)
ExistKey(MyT("Tom",192),a) # false
ExistKey(MyT("Tom",191),a) # true
Compared to Julia, Go is more straightforward for this problem :
package main
import (
"fmt"
)
type MyT struct {
A string
B int
}
func main() {
dic := map[MyT]int{MyT{"Bob", 10}: 1, MyT{"Jo", 21}: 1}
if _, ok := dic[MyT{"Bob", 10}]; ok {
fmt.Println("key exists")
}
}
// answer is "key exists"
You just need to teach your MyT type that you want it to consider equality in terms of its composite fields:
julia> immutable MyT
A::String
B::Int64
end
import Base: ==, hash
==(x::MyT, y::MyT) = x.A == y.A && x.B == y.B
hash(x::MyT, h::UInt) = hash(x.A, hash(x.B, hash(0x7d6979235cb005d0, h)))
julia> a = Dict(MyT("Tom",191)=>1,MyT("Bob",20)=>1,MyT("Jo",315)=>1,MyT("Luc",493)=>1)
Dict{MyT,Int64} with 4 entries:
MyT("Jo", 315) => 1
MyT("Luc", 493) => 1
MyT("Tom", 191) => 1
MyT("Bob", 20) => 1
julia> haskey(a, MyT("Tom",191))
true
julia> in((MyT("Tom",191) => 1), a)
true
There are lots of good answers here, I'd just like to add a subtlety: this is partly because == calls === rather than recursively calling == when checking for structural equality, and partly because equal (==) strings are not generally identical (===) currently. Specifically, the fact that MyT("foo", 1) != MyT("foo", 1) is because "foo" !== "foo".
Strings are only "immutable by convention" – they are technically mutable, but Julia doesn't expose APIs for mutating them and encourages you not to mutate them. You can, however, access their underlying bytes and mutate that, which allows you to write a program that distinguishes two strings by getting by mutating one and not the other. That means that they cannot be === in the sense of Henry Baker's "EGAL" predicate (also here). If you have an immutable type with only "primitive" type fields, then this does not happen:
julia> immutable MyT2 # `struct MyT2` in 0.6
A::Float64
B::Int
end
julia> x = MyT2(1, 1)
MyT2(1.0, 1)
julia> y = MyT2(1, 1)
MyT2(1.0, 1)
julia> x == y
true
julia> x === y
true
I have already proposed that we change this and have == recursively call ==. This should be fixed, someone just needs to do the work. Moreover, in Julia 1.0 we could make Strings truly immutable rather than merely immutable by convention, and therefore have "foo" === "foo" be true. I've created an issue to discuss and track this change.
You're creating a new object in the haskey call. But two objects created by MyT("Tom", 191) are just two different MyT objects with the same field values.
Instead, do
key1 = MyT("Tom", 191)
a = Dict(key1 => 1)
haskey(a, key1)
see also
key2 = MyT("Tom", 191)
key1 == key2 # false
A julia-ideomatic way to deal with this would be to define an == method for MyT objects, so two objects are equal if they have the same field values. That would allow you to use them like you do.
It depends whether you need the type to be complex. Another easy and performant way to do what you want is to use a Tuple as the key:
a = Dict(("Tom", 191) => 1)
haskey(a, ("Tom", 191)) # true
a[("Tom", 191)] # 1
My approach would be similar to Matt's, but a bit simpler(?). Tuples are perfectly valid dictionary keys, so I would simply overload the relevant functions to convert your type back and forth to a tuple:
julia> immutable M; A::String; B::Int64; end
julia> import Base: =>, haskey, in
julia> =>(a::M, b) = (a.A, a.B)=>b
julia> haskey(a::Dict, b::M) = haskey(a, (b.A, b.B))
julia> in(a::Pair{M, Int64}, b::Int64) = in((a.first.A,a.first.B)=>a.second,b)
julia> a = Dict(M("Dick", 10)=>1, M("Harry", 20)=>2)
Dict{Tuple{String,Int64},Int64} with 2 entries:
("Dick", 10) => 1
("Harry", 20) => 2
julia> haskey(a, M("Dick", 10))
true
julia> in(M("Dick", 10)=>1, a)
true
"Compared to Julia, Go is more straightforward for this problem"
True. It also happens to be more error-prone (depending on your perspective). If you wanted to differentiate between two objects (used as keys) that do not correspond to the same object in memory, then Go's approach of simply testing 'value equality' would have landed you in trouble here (though one could argue 'value equality' generally makes more sense when comparing 'keys').
When I tried to do:
d = {1:2, 3:10, 6:300, 2:1, 4:5}
I get the error:
syntax: { } vector syntax is discontinued
How to initialize a dictionary in Julia?
The {} syntax has been deprecated in julia for a while now. The way to construct a dict now is:
Given a single iterable argument, constructs a Dict whose key-value pairs are taken from 2-tuples (key,value) generated by the argument.
julia> Dict([("A", 1), ("B", 2)])
Dict{String,Int64} with 2 entries:
"B" => 2
"A" => 1
Alternatively, a sequence of pair arguments may be passed.
julia> Dict("A"=>1, "B"=>2)
Dict{String,Int64} with 2 entries:
"B" => 2
"A" => 1
(as quoted from the documentation, which can be obtained by pressing ? in the terminal to access the "help" mode, and then type Dict)
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...)