Re: [Numpy-discussion] advanced indexing question
On Mi, 2015-02-04 at 07:22 +, David Kershaw wrote: The numpy reference manual, array objects/indexing/advance indexing, says: Advanced indexing always returns a copy of the data (contrast with basic slicing that returns a view). If I run the following code: import numpy as np d=range[2] x=np.arange(36).reshape(3,2,3,2) y=x[:,d,:,d] y+=1 print x x[:,d,:,d]+=1 print x then the first print x shows that x is unchanged as it should be since y was a copy, not a view, but the second print x shows that all the elements of x with 1st index = 3rd index are now 1 bigger. Why did the left side of x[:,d,:,d]+=1 act like a view and not a copy? Python has a mechanism both for getting an item and for setting an item. The latter will end up doing this (python already does this for us): x[:,d,:,d] = x[:,d,:,d] + 1 so there is an item assignment going on (__setitem__ not __getitem__) - Sebastian Thanks, David ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion signature.asc Description: This is a digitally signed message part ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] advanced indexing question
Sebastian Berg sebastian at sipsolutions.net writes: Python has a mechanism both for getting an item and for setting an item. The latter will end up doing this (python already does this for us): x[:,d,:,d] = x[:,d,:,d] + 1 so there is an item assignment going on (__setitem__ not __getitem__) - Sebastian ___ NumPy-Discussion mailing list NumPy-Discussion at scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion Thanks for the prompt help Sebastian, So can I use any legitimate ndarray indexing selection object, obj, in x.__setitem__(obj,y) and as long as y's shape can be broadcast to x[obj]'s shape it will always set the appropriate elements of x to the corresponding elements of y? ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] advanced indexing question
The numpy reference manual, array objects/indexing/advance indexing, says: Advanced indexing always returns a copy of the data (contrast with basic slicing that returns a view). If I run the following code: import numpy as np d=range[2] x=np.arange(36).reshape(3,2,3,2) y=x[:,d,:,d] y+=1 print x x[:,d,:,d]+=1 print x then the first print x shows that x is unchanged as it should be since y was a copy, not a view, but the second print x shows that all the elements of x with 1st index = 3rd index are now 1 bigger. Why did the left side of x[:,d,:,d]+=1 act like a view and not a copy? Thanks, David ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] Advanced indexing question - subset of data cube as a 2D array.
In [85]: bi = (f.bolo_indices[np.newaxis,:]+
ones([7751,1])).astype('int')
In [86]: whc = (whscan[:,np.newaxis] + ones([1,107])).astype('int')
In [87]: array2d[whc,bi] = temp2d
I thought this had worked, but the values didn't seem to be going to the
right places when I re-examined them.
I think I answered my own question: if I'd used zeros() instead of
ones() it would have worked fine. I don't know why I tried to use
ones().
Adam
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Re: [Numpy-discussion] Advanced indexing question - subset of data cube as a 2D array.
On Wednesday 29 October 2008 01:44:06 Adam wrote: In [62]: temp2d = reshape(array3d,[23*337,107]) In [63]: temp2d2 = zeros([23*337,144]) In [64]: temp2d2[:,f.bolo_indices] = temp2d In [65]: array2d[whscan,:] = temp2d2 This works, but it feels wrong to me: I think there should be a way to do this by directly indexing array2d with two numpy arrays Does numpy.ix_ not fit your use case? In [2]: y = arange(15).reshape((3,5)) In [3]: y Out[3]: array([[ 0, 1, 2, 3, 4], [ 5, 6, 7, 8, 9], [10, 11, 12, 13, 14]]) In [4]: y[ ix_( array([1,2]), array([1,3,4]) ) ] Out[4]: array([[ 6, 8, 9], [11, 13, 14]]) Regards, Ravi ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] Advanced indexing question - subset of data cube as a 2D array.
Hi numpy group,
I have a problem I know there is an elegant solution to, but I can't
wrap my head around the right way to do the indexing.
The problem:
I have a 2D array that has been chopped up into 3 dimensions - it was [ time
X detectors ], it is now [ scans X time X detectors ]. During the chopping,
some of the time points and detector points have been removed, so the 3D
array contains only a subset of the data in the 2D array. I'd like to
restore the 3D array back to the shape of the original 2D array b/c it's
being stored in a netCDF file that is not flexible.
My solution:
In [58]: array2d.shape
Out[58]: (11008, 144)
In [59]: array3d.shape
Out[59]: (23, 337, 107)
In [60]: whscan.shape
Out[60]: (7751,)
In [61]: 23*337
Out[61]: 7751
In [62]: temp2d = reshape(array3d,[23*337,107])
In [63]: temp2d2 = zeros([23*337,144])
In [64]: temp2d2[:,f.bolo_indices] = temp2d
In [65]: array2d[whscan,:] = temp2d2
This works, but it feels wrong to me: I think there should be a way to do
this by directly indexing array2d with two numpy arrays
In the process of asking this question, I might have come up with the answer
(courtesy Stefan at http://mentat.za.net/):
In [85]: bi = (f.bolo_indices[np.newaxis,:]+ones([7751,1])).astype('int')
In [86]: whc = (whscan[:,np.newaxis] + ones([1,107])).astype('int')
In [87]: array2d[whc,bi] = temp2d
I thought this had worked, but the values didn't seem to be going to the
right places when I re-examined them.
Thanks,
Adam
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