David Crisp wrote:
>>>>> np.array([np.arange(9)//3, np.arange(9)%3, a.flatten()],
>> dtype=object).transpose()
>> array([[0, 0, x],
>> [0, 1, o],
>> [0, 2, o],
>> [1, 0, o],
>> [1, 1, x],
>> [1, 2, x],
>> [2, 0, o],
>> [2, 1, x],
>> [2, 2, o]], dtype=object)
>>
>> If that's not good enough you may also ask on the numpy mailing list.
>
> Thanks Peter,
>
> That appears to do what I want, in a way. How does this work if you
> have a matrix which is of variable size? For instance, some of my
> data will create a 10 by 10 matrix but some will create a 40 by 40
> matrix, Or for that matter any size. I notice your example
> specifically states there will be 9 outputs ( tupples? ) what if I
> want to say "just create as many tuples as you need to use to
> transpose the data"
You can find out the size of the matrix with the shape attribute:
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> a.shape
(3, 4)
Use that to calculate the values needed to replace the constants in my
previous post. Try to make do without the spoiler below!
>>> def process(a, dtype=object):
... x, y = a.shape
... n = x*y
... return np.array([np.arange(n)//y, np.arange(n)%y, a.flatten()],
dtype=dtype).transpose()
...
>>> a = np.arange(12).reshape(3, 4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> process(a, int)
array([[ 0, 0, 0],
[ 0, 1, 1],
[ 0, 2, 2],
[ 0, 3, 3],
[ 1, 0, 4],
[ 1, 1, 5],
[ 1, 2, 6],
[ 1, 3, 7],
[ 2, 0, 8],
[ 2, 1, 9],
[ 2, 2, 10],
[ 2, 3, 11]])
>>> b = np.array(list(
... "xoo"
... "oxx"
... "oxo")).reshape(3, 3)
>>> process(b, object)
array([[0, 0, x],
[0, 1, o],
[0, 2, o],
[1, 0, o],
[1, 1, x],
[1, 2, x],
[2, 0, o],
[2, 1, x],
[2, 2, o]], dtype=object)
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