All possible combinations within channel no overlap

[davidmasp]

Hi,

I am trying to combine a channel with itself to get all possible meaningful comparisons.

Should be something like this:

cheers = Channel.from 'Bonjour', 'Ciao', 'Hello', 'Hola'

cheers.into{cheers_a; cheers_b}
cheers_a.combine(cheers_b).set{cheers_comb}

// remove self-self
cheers_comb.filter{ it[0] != it[1] }.set{cheers_comb2}

cheers_comb2.view()

which returns something like this:

[Ciao, Bonjour]
[Hello, Bonjour]
[Hola, Bonjour]
[Bonjour, Ciao]
[Hello, Ciao]
[Hola, Ciao]
[Bonjour, Hello]
[Ciao, Hello]
[Hola, Hello]
[Bonjour, Hola]
[Ciao, Hola]
[Hello, Hola]

However, I would also like to remove the pairs which are already repeated. Meaning, from [Ciao, Bonjour] and [Bonjour, Ciao] only keep one.

Is there a way to do this somehow? Or could anyone guide me in how would be a nice way to approach it?

Much appreciated!

[davidmasp]

Okay, I think this seems to be working but I am not sure how to scale it for a more complex channel.

cheers_comb.filter{ it[0] != it[1] }
           .map{it = it.sort()}
           .unique()
           .set{cheers_comb2}

[billw1955]

You could add a closure to sort to pick only the feature of each 'cheer' to sort on, but branch might be clearer code.

workflow {
    cheers = Channel.of([[id:'Bonjour'], 'payload Z'],
                        [[id:'Ciao'], 'payload Y'],
                        [[id: 'Hello'], 'payload X'],
                        [[id:'Hola'], 'payload W'],
                        )
    def payloadLength = 1 // N.B. combine flattens

    cheers.combine(cheers)
        .set{cheers_comb}

    cheers_comb
        .branch { v ->
            equal: v[0]['id'] == v[payloadLength + 1]['id'] 
            lower: v[0]['id'] < v[payloadLength + 1]['id']
        }
        .set{cheers_comb2}

    cheers_comb2.lower.view()
}