lections import defaultdict
example_list = [[1,2,4], [1,2,3], [3,4,3], [1,2,3], [5,7,1], [3,4,3], [5,7,1], [1,2,4], [9,4,9], [9,3,9], [9,4,9]]
d = defaultdict(list)
for item in example_list: d[tuple(item)].append(item)
groupedlist = sorted(d[x] for x in d)
print groupedlist
# Returns [[[1, 2, 3], [1, 2, 3]], [[1, 2, 4], [1, 2, 4]], [[3, 4, 3], [3, 4, 3]], [[5, 7, 1], [5, 7, 1]], [[9, 3, 9]], [[9, 4, 9], [9, 4, 9]]]
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However working with python inside GH the code fails as (I'm assuming the id is included in the list ie.
[<Rhino.Geometry.Point3d object at 0x0000000000000BD0 [1611664.12669822,2952929.94466619,5788.70761766478]>]
If I print my object[0] i get just the vector, i.e [1611664.12669822,2952929.94466619,5788.70761766478]
However if i append my object, (for i in xrange (len(X)... list.append(my object[x])
The id included!?
[<Rhino.Geometry.Point3d object at 0x0000000000000BD0 [1611664.12669822,2952929.94466619,5788.70761766478]>]
Perhaps I'm over complicating this and should just use GH to group objects with the same position...
Also sometimes when getting vectors I receive long numbers such as.. 9.0122222900391e(10), and another times I don't.... Same file, script, geometry etc..
I'm quite a noobie in python for GH so any help or pointers are welcome...
Thank you in advance!
…
53 → 53 → 63 → 74 → 74 → 84 → 9
As you can see from the above list the connection sequence comes in waves of three, where each group of similar indices on the left is associated with a group of three incrementing indices on the right.
Some combination of Series components will probably generate this list, but it'll only work for the first ring, the second one will need a different connection pattern. It is perhaps better to just encode the integer pairs by hand. But then you cannot change your mind about the number of sides later.…
Added by David Rutten at 10:39am on October 21, 2015
ents will do or which components will be available.
My problem arises because I want to obtain a list such as the following:
{{6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6}, {5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5}, {4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4}, {3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3}, {2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2}, {1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1}}
Which displayed as a matrix is:
If it were possible to combine GH operations (series, shift list, replace string...) with matrices I think it would be quite powerful. A matrix to list component like those available on scientific calculators, would then translate the matrix to list.
For me, matrices come in handy when dealing with surface patterns.
…
Added by Jesus Galvez at 6:46am on November 26, 2012
ards to the number before the start number...
i.e. 9, 0, 1, 2, 3, 4, 5, 6, 7, 8
then it will need to repeat this pattern (continuing to count upwards) and the repeat number is based on a slider (for example 3 in the case illustrated below):
9, 0, 1, 2, 3, 4, 5, 6, 7, 8
19, 10, 11, 12, 13, 14, 15, 16, 17, 18,
29, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29,
if anyone has any ideas on how to create this series it would be great
M.…
the curves on surface issue it's solved seting flatten to the surface control point output. Still didnt know how to group points like:
1;1, 2;2, 3;3.....
1;2, 2;3, 3;4....
1;3, 2;4, 3;5...
....
Hi,
I want to divide curve with distance between points so it will be like this:
1--2---3----4-----5------6-------7-----, ...
with values in range 1 to 50, must be simple but im stuck..
tnx