Nested List Comprehensions in Julia
In Python I Can Do Nested List Comprehensions, for Instance I Can Flatten the Following Array Thus: a = [[1,2,3], [4,5,6]] [I for Arr in a for I in Arr] to Get...
In python I can do nested list comprehensions, for instance I can flatten the following array thus:
a = [[1,2,3],[4,5,6]]
[i for arr in a for i in arr]
to get [1,2,3,4,5,6]
If I try this syntax in Julia I get:
julia> a
([1,2,3],[4,5,6],[7,8,9])
julia> [i for arr in a for i in arr]
ERROR: syntax: expected ]
Are nested list comprehensions in Julia possible?
5 Answers
This feature has been added in julia v0.5:
julia> a = ([1,2,3],[4,5,6],[7,8,9])
([1,2,3],[4,5,6],[7,8,9])
julia> [i for arr in a for i in arr]
9-element Array{Int64,1}:
1
2
3
4
5
6
7
8
9
List comprehensions work a bit differently in Julia:
> [(x,y) for x=1:2, y=3:4]
2x2 Array{(Int64,Int64),2}:
(1,3) (1,4)
(2,3) (2,4)
If a=[[1 2],[3 4],[5 6]] was a multidimensional array, vec would flatten it:
> vec(a)
6-element Array{Int64,1}:
1
2
3
4
5
6
Since a contains tuples, this is a bit more complicated in Julia. This works, but likely isn't the best way to handle it:
function flatten(x, y)
state = start(x)
if state==false
push!(y, x)
else
while !done(x, state)
(item, state) = next(x, state)
flatten(item, y)
end
end
y
end
flatten(x)=flatten(x,Array(Any, 0))
Then, we can run:
> flatten([(1,2),(3,4)])
4-element Array{Any,1}:
1
2
3
4
You can get some mileage out of using the splat operator with the array constructor here (transposing to save space)
julia> a = ([1,2,3],[4,5,6],[7,8,9])
([1,2,3],[4,5,6],[7,8,9])
julia> [a...]'
1x9 Array{Int64,2}:
1 2 3 4 5 6 7 8 9
Any reason why you're using a tuple of vectors? It's much simpler with arrays, as Ben has already shown with vec. But you can also use comprehensions pretty simply in either case:
julia> a = ([1,2,3],[4,5,6],[7,8,9]);
julia> [i for i in hcat(a...)]
9-element Array{Any,1}:
1
2
⋮
The expression hcat(a...) "splats" your tuple and concatenates it into an array. But remember that, unlike Python, Julia uses column-major array semantics. You have three column vectors in your tuple; is that what you intend? (If they were row vectors — delimited by spaces — you could just use [a...] to do the concatenation). Arrays are iterated through all elements, regardless of their dimensionality.
Don't have enough reputation for comment so posting a modification @ben-hammer. Thanks for the example of flatten(), it was helpful to me.
But it did break if the tuples/arrays contained strings. Since strings are iterables the function would further break them down to characters. I had to insert condition to check for ASCIIString to fix that. The code is below
function flatten(x, y)
state = start(x)
if state==false
push!(y, x)
else
if typeof(x) <: String
push!(y, x)
else
while (!done(x, state))
(item, state) = next(x, state)
flatten(item, y)
end
end
end
y
end
flatten(x)=flatten(x,Array(Any, 0))
Sarah Jenkins is a veteran tech journalist with over 12 years of experience covering artificial intelligence, mobile innovations, and digital ethics. Her insights have appeared in leading technology publications worldwide.
