Glad to hear that it's working out for you!
And don't worry for not understanding how the `sort_uvs_by_shells` function
works, it's my fault for writing it in a not so readable way.
I'll give a go at decomposing and describing each of its operations. It's a
good exercise for me as I'm usually pretty bad at explaining things :)
def sort_uvs_by_shells(uvs, shells):
zipped = zip(shells, uvs)
ordered = sorted(zipped)
groupped = itertools.groupby(ordered, key=lambda x: x[0])
return ((group[0], tuple(x[1] for x in group[1])) for group in
groupped)
First, the `zip` function. It actually has a good fitting name—imagine two
sequences of a same length, say `left = ['a', 'b', 'c']` and `right = [1,
2, 3]`. Calling `zip(left, right)` will return a new list, still of the
same length, where each corresponding element from the left and right lists
are joined together in a tuple. So the result here would be `[('a', 1),
('b', 2), ('c', 3)]`.
In our case, we use `zip` to group each UV to its shell, and the result we
get form the zip is something following the idea of `[('shell_0', 'uv_1'),
('shell_0', 'uv_2'), ('shell_1', 'uv_4'), ('shell_0', 'uv_5'), ...]`.
With this being done, we can now keep moving on grouping together all the
UVs for a same shell. There's another neat function to do just this: `
itertools.groupby`. If we were to do it on a list such as `['a', 'a', 'a',
'a', 'b', 'b', 'a', 'a', 'a']`, then it would return an iterator of tuples
where the first tuple would represent the first 4 letters 'a', the second
tuple the 2 following letters 'b', and finally the last tuple the last 3
letters 'a'. So it only groups elements which are contiguous in the list,
and it happens that our shell/uvs list is not in an ideal state for such a
grouping (in the example above, we've got 2 'shell_0', then 1 'shell_1',
then another 'shell_0').
That's why we need to order first the result of the zip operation so all
the shell/uvs pairs which belong to the same shell are listed next to each
other. The built-in `sorted` function does just that. The sorting works by
comparing corresponding values of each tuple, starting by the first one.
For example `(4, 3) > (3, 5)` so `(3, 5)` will be end up being before `(4,
3)`. In other words, putting the shells and uvs items in this order when
doing the zip operation made it easy for us here. Otherwise we could still
have used either the `cmp` or the `key` parameters from the `sorted`
function.
Now the `ordered` list is ready to be grouped. The second parameter of the `
itertools.groupby` function is being passed a lambda (an anonymous
function) to tell the function which element from our shell/uvs pair should
be used for the grouping—that's the shell item, which is index 0.
And finally, what's left is to output the result after extracting the data
from the `itertools.groupby` function.
On 22 August 2016 at 10:45, Aren Voorhees <[email protected]> wrote:
> Thanks to both of you for your rapid responses! Using Christopher's
> example I was able to get it working to where it could sort the uvs by
> their shell (I must confess I don't exactly understand how the
> sort_uvs_by_shells function works, but I'll study up on it and get it
> figured out in my head). Thanks again, this will allow me to finally
> finish up this tool I've been trying to work on!
>
> On Thu, Aug 18, 2016 at 12:02 AM, Justin Israel <[email protected]>
> wrote:
>
>>
>>
>> On Thu, Aug 18, 2016 at 4:51 PM Christopher Crouzet <
>> [email protected]> wrote:
>>
>>> Aren asked for “any number of resulting lists”, which I interpreted as
>>> him wanting one list of UVs per shell, whereas the sorted zip version
>>> returns a single list of ordered UVs with no way to figure out from which
>>> shell they belong to. But that wouldn't be the first time that I
>>> misunderstand something! :)
>>>
>>
>> No I see what you mean. There was that extra sentence that I didn't read
>> which explained that. I only saw it as sorting one list by another list :-)
>>
>>
>>>
>>>
>>> On 18 August 2016 at 11:41, Justin Israel <[email protected]>
>>> wrote:
>>>
>>>>
>>>>
>>>> On Thu, Aug 18, 2016 at 4:40 PM Justin Israel <[email protected]>
>>>> wrote:
>>>>
>>>>> On Thu, Aug 18, 2016 at 4:34 PM Christopher Crouzet <
>>>>> [email protected]> wrote:
>>>>>
>>>>>> If I'm understanding it right, try this:
>>>>>>
>>>>>> import itertools
>>>>>>
>>>>>> def sort_uvs_by_shells(uvs, shells):
>>>>>> groups = itertools.groupby(sorted(zip(shells, uvs)), key=lambda
>>>>>> x: x[0])
>>>>>> return ((group[0], [x[1] for x in group[1]]) for group in groups)
>>>>>> # dictionary version:
>>>>>> # return {group[0]: [x[1] for x in group[1]] for group in groups}
>>>>>>
>>>>>> uvs = list(range(16))
>>>>>> shells = [0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1]
>>>>>> result = sort_uvs_by_shells(uvs, shells)
>>>>>> for shell, uvs in result:
>>>>>> # dictionary version:
>>>>>> # for shell, uvs in result.iteritems():
>>>>>> print("uvs for shell %d: %s" % (shell, uvs))
>>>>>>
>>>>>
>>>>> Nice example. Could this even be shortened to something like this? (I
>>>>> have simplified the data for example purposes)
>>>>>
>>>>> from itertools import izip
>>>>>
>>>>> l1 = ['a', 'b', 'c']
>>>>> l2 = [3, 2 ,1]
>>>>> print [i[1] for i in sorted(izip(l2, l1))]# ['c', 'b', 'a']
>>>>>
>>>>>
>>>>>
>>>>>
>>>> Correction: Forget the use of itertool.izip. It is not necc here.
>>>>
>>>> [i[1] for i in sorted(zip(l2, l1))]
>>>>
>>>>
>>>>
>>>>> Justin
>>>>>
>>>>>
>>>>>>
>>>>>> On 18 August 2016 at 11:02, Aren Voorhees <[email protected]> wrote:
>>>>>>
>>>>>>> I realized that subject doesn't sound like it makes any sense - but
>>>>>>> here is what I'm trying to do: I am generating two lists...one looks
>>>>>>> something like this -
>>>>>>>
>>>>>>> [MeshUV(u'pCubeShape1.map[0]'), MeshUV(u'pCubeShape1.map[1]'),
>>>>>>> MeshUV(u'pCubeShape1.map[2]'), MeshUV(u'pCubeShape1.map[3]'),
>>>>>>> MeshUV(u'pCubeShape1.map[4]'), MeshUV(u'pCubeShape1.map[5]'),
>>>>>>>
>>>>>>> MeshUV(u'pCubeShape1.map[6]'), MeshUV(u'pCubeShape1.map[7]'),
>>>>>>> MeshUV(u'pCubeShape1.map[8]'), MeshUV(u'pCubeShape1.map[9]'),
>>>>>>> MeshUV(u'pCubeShape1.map[10]'), MeshUV(u'pCubeShape1.map[11]'),
>>>>>>>
>>>>>>> MeshUV(u'pCubeShape1.map[12]'), MeshUV(u'pCubeShape1.map[13]'),
>>>>>>> MeshUV(u'pCubeShape1.map[14]'), MeshUV(u'pCubeShape1.map[15]')]
>>>>>>>
>>>>>>>
>>>>>>> The other is something like this: [0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0,
>>>>>>> 0, 0, 0, 0, 1]
>>>>>>>
>>>>>>>
>>>>>>> The two lists correspond to each other, so I'd like to sort the
>>>>>>> elements in the first list by the matching integer in the 2nd list. In
>>>>>>> this example there would only be two resulting lists of uvs, but
>>>>>>> ultimately
>>>>>>> there could be any number of resulting lists.
>>>>>>>
>>>>>>>
>>>>>>> BTW, the point of this is to sort the uvs of a given mesh into their
>>>>>>> different shells so I can perform an operation on all the shells of a
>>>>>>> mesh.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Thanks!
>>>>>>>
>>>>>>> Aren
>>>>>>>
>>>>>>> --
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>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Christopher Crouzet
>>>>>> *http://christophercrouzet.com* <http://christophercrouzet.com>
>>>>>>
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>>>
>>> --
>>> Christopher Crouzet
>>> *http://christophercrouzet.com* <http://christophercrouzet.com>
>>>
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