Partition Data

Hello, Someone else considers it necessary? Partition Data by prepending a series according the number of partitions.
See the attachment where it can be applied in a more simple way, I know I can use ReplacePath or pathmapper but I think could be an additional component like PartitionList where you can control the SIZE of partitions.

Regards.

Attachments:

131104_PartitionDATA.gh, 281 KB

Replies to This Discussion

Permalink Reply by David Rutten on November 4, 2013 at 7:58pm

I'm confused. What?

David Rutten

david@mcneel.com

Seattle, WA

Permalink Reply by Michael Pryor on November 4, 2013 at 8:06pm

I think he is saying a component like replace path but by partition size input?

so you input data structure say

{0;0}

{0;1}

{0;2}

{1;0}

{1;1}

{2;0}

{3;0}

{3;1}

{3;2}

{3;3}

with a partition of say

n the output would be

{0;0}

{0;1}

{1;0}

{1;1}

{1;2}

{1;3}

{1;4}

{2;0}

{2;1}

{2;2}

that's at least what i understood

Permalink Reply by David Rutten on November 4, 2013 at 8:43pm

So the paths in the resulting tree have nothing whatsoever to do with the original paths, it's just {N;i} where N refers to the index of each partition size number and i is incremented every time from zero?

David Rutten

david@mcneel.com

Seattle, WA

Permalink Reply by Erick Vásquez on November 4, 2013 at 9:35pm

Yes David, you are right, The idea comes from my example, I wrote a "FlipMatrix" and I thought there could be a shorter way to partition the data.

Permalink Reply by Michael Pryor on November 4, 2013 at 9:48pm

The example should be,

{0}

{1}

{2}

{3}

{4}

{5}

{6}

{7}

{8}

{9}

with a partition of

the output would be

{0;0}

{0;1}

{1;0}

{1;1}

{1;2}

{1;3}

{1;4}

{2;0}

{2;1}

{2;2}

if this does get made I would say partition branches makes more sense than partition data as a name.

Permalink Reply by David Rutten on November 4, 2013 at 10:00pm

Ok I think I understand now the logic of this mapping. I've never encountered the need to do this myself, under what sort of circumstances is this a useful operation to have?

David Rutten

david@mcneel.com

Seattle, WA

Permalink Reply by Erick Vásquez on November 4, 2013 at 10:41pm

was just an idea and so my question was open if someone was found with this type of situation, but now that I see it in detail only serves inside the whole process of my "FlipMatrix".

So my apologies for the disturbances.

Permalink Reply by Michael Pryor on November 4, 2013 at 11:44pm

Ah, so maybe some native "flip complex matrix" could be a useful component. Usually I just shift the path then flip.

Permalink Reply by Michael Pryor on November 4, 2013 at 11:30pm

Im not sure, was just interpreting what he is saying. Usually the situation I run into is having to go from the second list type to the first of my description above which is an easy task.

Permalink Reply by Erick Vásquez on November 4, 2013 at 11:40pm

for sure, ShiftPath is most common ;) regards!

Permalink Reply by David Stasiuk on November 5, 2013 at 12:50am

I wonder if the "Flip Last Matrix" from Tree Sloth might be useful for you? It retains all of the data structure to the left of the last path, and then flips the item index and last path index only.

Attachments: