Sorting by two values (basic sorting)

New to Julia
,
1

This is probably fairly simple. I want to sort a set of tuples first by the second value, and then by the first value, like this:

julia> a = [ rand(1:3,2) for i in 1: 5 ]
5-element Array{Array{Int64,1},1}:
 [2, 1]
 [3, 2]
 [3, 2]
 [1, 1]
 [2, 1]

julia> sort!(a, by = x -> x[2])
5-element Array{Array{Int64,1},1}:
 [2, 1]
 [1, 1]
 [2, 1]
 [3, 2]
 [3, 2]

julia> sort!(a, by = x -> x[1])
5-element Array{Array{Int64,1},1}:
 [1, 1]
 [2, 1]
 [2, 1]
 [3, 2]
 [3, 2]

The result is perfectly fine. However, I am somewhat unsure if the result of the second sorting might be dependent on the sorting algorithm used. Am I safe with that?

In this case I really don’t care about performance, but I wonder if there is a special function or syntax for this kind of thing.

2

It is dependent on the algorithm, but the default Julia sort algorithm is stable, which means that equal keys will be in the same order after sort as before. You can supply a function to sort that gives the relative order of two elements. See the lt keyword argument.

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3

Yes, of course, that is much smarter.

4

Those aren’t tuples, they’re vectors, and they already sort lexicographically. Thus, we can generically have by return the list of keys in order to compare:

julia> sort!(a, by=x -> (x[1], x[2]))

Which in this case is also just:

julia> sort!(a, by=identity)

or just

julia> sort!(a)
8 Likes
5

Yes, sorry, in the MWE I have used vectors, but my real case is with tuples. Good to know that, anyway.

6

Yes, but in this case, OP wants:

sort!(a, by = x -> (x[2], x[1]))
7

Actually there is a isless function defined for tuples that does what I want, it seems:

julia> a = [ (rand(1:3),rand(1:3)) for i in 1: 5 ]
5-element Array{Tuple{Int64,Int64},1}:
 (1, 3)
 (1, 3)
 (1, 2)
 (3, 3)
 (2, 1)

julia> sort!(a)
5-element Array{Tuple{Int64,Int64},1}:
 (1, 2)
 (1, 3)
 (1, 3)
 (2, 1)
 (3, 3)

(I had to define a custom one because in my case the tuple was something computed from a struct, but that is another story).

8

Yes, that’s what sort!(a, by = x -> (x[2], x[1])) uses

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