Match branched lists ?! List management problem

Hey guys,

I have a list management issue.

I have two lists, X and Y:

X is made of {0;0} (N=1), {0,1}(N=1)...{0,15}(N=1), {1;0}(N=1), {1,1}(N=1)...{1,15}(N=1). While Y is made of {0}(N=6), {1}(N=15).

What I need to do, is to cycle through all the branched items of the X lists with every single item of the corresponding list of Y.

So:

{0,0}(N=1) with 1st item of Y{0},

{0,1}(N=1) with 1st item of Y{0}

...

{0,15}(N=1) with 1st item of Y{0} (and this is the first cycle).

Then (cycle 2)

{0,0}(N=1) with 2nd item of Y{0},

{0,1}(N=1) with 2nd item of Y{0}

...

{0,15}(N=1) with 2nd item of Y{0}

Then do the same thing with the X list {1,0} .. {1,15} with the corresponding Y{1}

As a result I need to have the following list structure:

{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,

{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.

If I use the component "shift path" (as suggested in another discussion) it will match the items per item-order instead of cycling through all the items of the first sublist...

Hope this makes sense.

Any suggestion?

Replies to This Discussion

Permalink Reply by ng5 Alex on August 7, 2014 at 8:25pm

Hello

i am trying to understand what you mean when you say cycle. you get the first branch {0;0} of the x tree, which is a value since it has N=1 and you want to do something with the first branch {0} of the y tree which has 6 values. and then the same with the others. your end result is a tree that the {0;?} branches have 6 values each. so what happens to first value? you don't merge them, because then you would have 7 values in each branch. so how do you filter out one value? smaller larger comparison?

maybe i am missing something obvious.

cheers

alex

edit or you want for example to create a line from {0;0} N=1 to the all the others in {0} N=6 ?

Permalink Reply by lv2311 on August 7, 2014 at 8:48pm

Hi Alex,

Thanks for the reply. With "cycle" I mean the application of each item of the Ylists to all the items of its corresponding branch in Xlist. So if in the Y list in branch {0} I have 6 items, then the each element of the corresponding X branch list will be repeated 6 times >> basically the output list would have the same structure of the input Xlist but instead of having only one item per branch I would have 6 items per branch >>

{0,0}N=6, {0,1}N=6, {0,2}N=6 ... {0,15}N=6,

{1,0}N=15, {1,1}N=15, {1,2}N=15 ... {0,15}N=15.

I made a quick example of my problem, here attached.

Attachments:

forum_listMatching.gh, 15 KB
listMatching.JPG, 74 KB

Permalink Reply by ng5 Alex on August 8, 2014 at 12:45pm

i cant think of a better way at the moment, but surely there must be one.

Permalink Reply by Arend on August 8, 2014 at 2:33pm

It this what you mean?

Attachments:

forum_listMatching-repeat-branch.gh, 13 KB

Permalink Reply by lv2311 on August 8, 2014 at 3:11pm

Yes Arend!

That's exactly what I need! I managed to make a workaround by duplicating the data of one list and stacking the other one, but this one is def quicker!

Thank you ng5 Alex as well for your time

cheers

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