Grasshopper

algorithmic modeling for Rhino

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.

E.

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I'm confused. What?

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David Rutten

david@mcneel.com

Seattle, WA

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 

2

5

3

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

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

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.

The example should be,

{0}

{1}

{2}

{3}

{4}

{5}

{6}

{7}

{8}

{9}

with a partition of

2

5

3

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. 

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

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.

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

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. 

for sure, ShiftPath is most common ;) regards!

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.

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