Sorting a vector of vectors by multiple elements in Julia

I've had good read of the 'Sorting Functions' section of the Julia manual, and had a look at some of the similar questions that have already been asked on this board, but I don't think I've quite found the answer to my question. Apologies if I've missed something.

Essentially I have a vector of vectors, with the enclosed vectors containing integers. For the purposes of the example, each enclosed vector contains 3 integers, but it could be any number. I want to sort the enclosed vectors by the first element, then by the second element, then by the third element etc.

Let's start with the vector:

v = [[3, 6, 1], [2, 2, 6], [1, 5, 9], [2, 1, 8], [3, 7, 9],
[1, 1, 2], [2, 2, 2], [3, 6, 2], [1, 2, 5], [1, 5, 6],
[3, 7, 4], [2, 1, 4], [2, 2, 1], [3, 1, 2], [1, 2, 8]]


And continue with what I'm actually looking for:

v = [[1, 1, 2], [1, 2, 5], [1, 2, 8], [1, 5, 6], [1, 5, 9],
[2, 1, 4], [2, 1, 8], [2, 2, 1], [2, 2, 2], [2, 2, 6],
[3, 1, 2], [3, 6, 1], [3, 6, 2], [3, 7, 4], [3, 7, 9]]


So there should be no requirement for rocket science.

I can easily sort the vector by the first element of the enclosed vectors by one of two ways:

v = sort(v, lt = (x, y) -> isless(x[1], y[2]))


or:

v = sort(v, by = x -> x[1])


Both these methods produce the same answer:

v = [[1, 5, 9], [1, 1, 2], [1, 2, 5], [1, 5, 6], [1, 2, 8],
[2, 2, 6], [2, 1, 8], [2, 2, 2], [2, 1, 4], [2, 2, 1],
[3, 6, 1], [3, 7, 9], [3, 6, 2], [3, 7, 4], [3, 1, 2]]


So, as you can see, I have sorted by the first element of the enclosed vectors, but not by the subsequent elements.

So, to come back to the question in the title, is there a method of sorting by multiple elements using the sort() function?

sort()

I can actually get what I want using loops:

for i = 3:-1:1
v = sort(v, lt = (x, y) -> isless(x[i], y[i]))
end


or:

for i = 3:-1:1
v = sort(v, by = x -> x[i])
end


However, I don't want to re-invent the wheel, so if there's a way of doing it within the sort() function I'd love to learn about it.

sort()

1 Answer
1

You can use lexless function as lt keyword argument that does exactly what you want if I understood your question correctly:

lexless

lt

julia> sort(v, lt=lexless)
15-element Array{Array{Int64,1},1}:
[1, 1, 2]
[1, 2, 5]
[1, 2, 8]
[1, 5, 6]
[1, 5, 9]
[2, 1, 4]
[2, 1, 8]
[2, 2, 1]
[2, 2, 2]
[2, 2, 6]
[3, 1, 2]
[3, 6, 1]
[3, 6, 2]
[3, 7, 4]
[3, 7, 9]


EDIT: I have just checked that this is a solution for Julia 0.6. In Julia 0.7 you can simply write:

julia> sort(v)
15-element Array{Array{Int64,1},1}:
[1, 1, 2]
[1, 2, 5]
[1, 2, 8]
[1, 5, 6]
[1, 5, 9]
[2, 1, 4]
[2, 1, 8]
[2, 2, 1]
[2, 2, 2]
[2, 2, 6]
[3, 1, 2]
[3, 6, 1]
[3, 6, 2]
[3, 7, 4]
[3, 7, 9]


I'm still running 0.6.2, and your first method works fine. I knew there had to be a simple solution, though I've just gone back to the manual and can't find a reference to lt=lexless within the 'Sorting Functions' section. Many thanks! It's also nice to see the improvement in v 0.7.
– Mark Birtwistle
Jun 29 at 9:41

lt=lexless

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.