Groupe tuple with 2 values

[wdesaintjean]

Hello,
I need help about en operation on channel:

I have this channel:

[5Y_siGL2_B1, group1]
[5Y_siGL2_B2, group1]
[5Y_siGL2_B3, group2]
[5Y_siDDX5_17_B1, group2]
[5Y_siDDX5_17_B2, group3]
[5Y_siDDX5_17_B3, group3]

and I want them to be grouped by the second item (here group1, group2, group3), but I want a group that contain 2 values of the second item, given by this other channel.

[group1, group2]
[group2, group3]
[group1, group3]

I know I can group them by the first item with .groupTuple(by: 1), but I obtain this channel:

[5Y_siGL2_B1, 5Y_siGL2_B2], group1]
[5Y_siDDX5_17_B1, 5Y_siGL2_B3], group2]
[5Y_siDDX5_17_B2, 5Y_siDDX5_17_B3], group3]

The results I want is more like this:

[5Y_siGL2_B1, 5Y_siGL2_B2, 5Y_siGL2_B3, 5Y_siDDX5_17_B1], [group1, group2]]
[5Y_siGL2_B3, 5Y_siDDX5_17_B1, 5Y_siGL2_B3, 5Y_siDDX5_17_B2, 5Y_siDDX5_17_B3], [group2, group3]]
[5Y_siGL2_B1, 5Y_siGL2_B2, 5Y_siDDX5_17_B2, 5Y_siDDX5_17_B3], [group1, group3]]

Is it possible with simple operators?
Thank you for your help.

[bentsherman]

It's possible, but it ain't pretty

inputs_ch = Channel.of(
    tuple('5Y_siGL2_B1', 'group1'),
    tuple('5Y_siGL2_B2', 'group1'),
    tuple('5Y_siGL2_B3', 'group2'),
    tuple('5Y_siDDX5_17_B1', 'group2'),
    tuple('5Y_siDDX5_17_B2', 'group3'),
    tuple('5Y_siDDX5_17_B3', 'group3')
)

inputs_ch

    // generate pairwise combinations
    .combine(inputs_ch)

    // filter out duplicates
    .filter { _vleft, kleft, _vright, kright ->
        kleft < kright
    }

    // rearrange some things
    .map { vleft, kleft, vright, kright ->
        tuple([vleft, vright].toSet(), [kleft, kright].toSet())
    }

    // group by pair
    .groupTuple(by: 1)

    // flatten all the pair sets from map() into one big set
    .map { vpairs, key ->
        tuple(vpairs.flatten().toSet(), key)
    }

    // all done
    .view()

Probably a few different ways to do it, but fundamentally I think you have to (1) generate the pairwise combinations, (2) filter out the duplicates, and (3) group as normal, with some glue logic here and there to make it work

[wdesaintjean]

Thank you, I will try.

Have a good day!