Re: [Numpy-discussion] indexing with booleans without making a copy?
8/09/10 @ 15:35 (-0400), thus spake Anne Archibald: > 2010/9/8 Ernest Adrogué : > > I have a sorted, flat array: > > > > In [139]: a =np.array([0,1,2,2,2,3]) > > > > Basically, I want views of the areas where there > > are repeated numbers (since the array is sorted, they > > will be contiguous). > > > > But, of course, to find the numbers that are repeated > > I have to use comparison operations that return > > boolean arrays, so I suppose the problem is converting > > the boolean array into a slice. > > Well, you're going to have to do some allocation, but how's this? Use > unique1d to get an array of unique values. Then use searchsorted > twice, once to find the leftmost end of each hunk, and once to find > the rightmost end of each hunk. I like it. Thanks :) > > Anne > ___ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] indexing with booleans without making a copy?
2010/9/8 Ernest Adrogué : > Hello, > > I have a sorted, flat array: > > In [139]: a =np.array([0,1,2,2,2,3]) > > Basically, I want views of the areas where there > are repeated numbers (since the array is sorted, they > will be contiguous). > > But, of course, to find the numbers that are repeated > I have to use comparison operations that return > boolean arrays, so I suppose the problem is converting > the boolean array into a slice. > > This is what I have come up with: > > In [146]: np.flatnonzero(a==2) > Out[146]: array([2, 3, 4]) > > In [147]: b=np.flatnonzero(a==2) > > In [148]: b.min() > Out[148]: 2 > > In [149]: b.max() > Out[149]: 4 > > In [150]: a[b.min():b.max()+1] > Out[150]: array([2, 2, 2]) > > If you know a more straightforward way of doing this, > I'll be glad to know... np.nonzero(np.diff(a)) add boundaries index zero and end make two arrays one with starting points, one with endpoints. Josef > > Cheers :) > Ernest > ___ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] indexing with booleans without making a copy?
2010/9/8 Ernest Adrogué : > I have a sorted, flat array: > > In [139]: a =np.array([0,1,2,2,2,3]) > > Basically, I want views of the areas where there > are repeated numbers (since the array is sorted, they > will be contiguous). > > But, of course, to find the numbers that are repeated > I have to use comparison operations that return > boolean arrays, so I suppose the problem is converting > the boolean array into a slice. Well, you're going to have to do some allocation, but how's this? Use unique1d to get an array of unique values. Then use searchsorted twice, once to find the leftmost end of each hunk, and once to find the rightmost end of each hunk. Anne ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] indexing with booleans without making a copy?
Hello, I have a sorted, flat array: In [139]: a =np.array([0,1,2,2,2,3]) Basically, I want views of the areas where there are repeated numbers (since the array is sorted, they will be contiguous). But, of course, to find the numbers that are repeated I have to use comparison operations that return boolean arrays, so I suppose the problem is converting the boolean array into a slice. This is what I have come up with: In [146]: np.flatnonzero(a==2) Out[146]: array([2, 3, 4]) In [147]: b=np.flatnonzero(a==2) In [148]: b.min() Out[148]: 2 In [149]: b.max() Out[149]: 4 In [150]: a[b.min():b.max()+1] Out[150]: array([2, 2, 2]) If you know a more straightforward way of doing this, I'll be glad to know... Cheers :) Ernest ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
