On 6 April 2016 at 17:26, Heli <heml...@gmail.com> wrote:
>
> Thanks for your replies. I have a question in regard with my previous
> question. I have a file that contains x,y,z and a value for that coordinate
> on each line. Here I am giving an example of the file using a numpy array
> called f.
>
> f=np.array([[1,1,1,1],
> [1,1,2,2],
> [1,1,3,3],
> [1,2,1,4],
> [1,2,2,5],
...
> [3,2,3,24],
> [3,3,1,25],
> [3,3,2,26],
> [3,3,3,27],
> ])
>
> then after tranposing f, I get the x,y and z coordinates:
> f_tranpose=f.T
> x=np.sort(np.unique(f_tranpose[0]))
> y=np.sort(np.unique(f_tranpose[1]))
> z=np.sort(np.unique(f_tranpose[2]))
You don't actually need to transpose the matrix to get a column as a 1D array:
>>> a = np.array([[1,2,3],[4,5,6]])
>>> a
array([[1, 2, 3],
[4, 5, 6]])
>>> a[0]
array([1, 2, 3])
>>> a[0,:] # 1st row
array([1, 2, 3])
>>> a[:,0] # 1st column
array([1, 4])
(Not that there's a correctness/performance difference or anything I
just think that asking for the column is clearer than asking for the
row of the transpose.)
> Then I will create a 3D array to put the values inside. The only way I see to
> do this is the following:
> arr_size=x.size
> val2=np.empty([3, 3,3])
>
> for sub_arr in f:
> idx = (np.abs(x-sub_arr[0])).argmin()
> idy = (np.abs(y-sub_arr[1])).argmin()
> idz = (np.abs(z-sub_arr[2])).argmin()
> val2[idx,idy,idz]=sub_arr[3]
>
> I know that in the example above I could simple reshape f_tranpose[3] to a
> three by three by three array, but in my real example the coordinates are not
> in order and the only way I see to do this is by looping over the whole file
> which takes a lot of time.
So it's easy if the array is in order like your example? Why not sort
the array into order then? I think lexsort will do what you want:
http://docs.scipy.org/doc/numpy-1.10.1/reference/generated/numpy.lexsort.html
http://stackoverflow.com/questions/8153540/sort-a-numpy-array-like-a-table
--
Oscar
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