Filter and reorder pairs of elements?

General Usage
1

Hi everybody, this is an algorithmic problem:
i have a set of pairs of elements, from the same group,the members of the group are integers, the first member of a pair is an element of the group, and the second one is a group of elements of the same mother group. distinct from the first member of the pair

group = 1:9 #this element contains all possible unique elements
pairs = ((1,[2,3,4]),
             (2,[5,7,8]),
             (5,[8]),
             (6,[8])

I’m looking for a function to reorder and filter the first element of the group and the second members of the pairs, an example:

subgroup= [8,5,1,2]
pairs = ((1,[2,3,4]),
             (2,[5,7,8]),
             (5,[8]),
             (6,[8])
julia>reorder!(pairs,subgroup)
pairs =  ((5,[8])
               (1,[2])
               (2,[8,5]))

there is a function that can do that in julia? or the solution is a loop of loops? there is no need for optimizations, as the number of elements is low. there is a function that does something similar, but just for a vector:

group = 1:21
pairs= 1:3 
subgroup = [3,1,5]
julia> intersect(subgroup,pairs)
2-element Array{Int64,1}:
 3
 1
2

Sorry, I find the description of the problem hard to understand.

It is unlikely that there is a function specifically for this, so you would have to build from existing parts. sort, with lt and by arguments, should be useful, and also filter. Alternatively, if you can write your algorithm using loops, that should be fine too.

1 Like
3

Yes, I think I have a solution now. First I intersect see branches, then I reorder the first level and finally I reconstruct the structure. I’ll post it here when I write it, basically is an alternate way to iterate over few elements of a matrix, and the numbers are the indices, but the order or amount of those indices can vary over time, I’m trying to do the equivalent of:
A=rand(21,21)
n=[3, 2,5]
B=A[n,n]
But with a tree, I learned