Re: deleting certain entries in numpy array

[Robert Kern]

On 2009-07-01 09:51, Sebastian Schabe wrote:

Hello everybody,

I'm new to python and numpy and have a little/special problem:

You will want to ask numpy questions on the numpy mailing list.

  http://www.scipy.org/Mailing_Lists

I have an numpy array which is in fact a gray scale image mask, e.g.:

mask =
array([[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 255, 255, 255, 0, 0, 255, 0],
[ 0, 0, 255, 255, 255, 0, 0, 255, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0]], dtype=uint8)

and I have another array of (y, x) positions in that mask (first two
values of each row):

pos =
array([[ 3., 2., 0., 0.],
[ 3., 4., 0., 0.],
[ 5., 2., 0., 0.],
[ 5., 4., 0., 0.],
[ 6., 2., 0., 0.],
[ 6., 7., 0., 0.],
[ 0., 0., 0., 0.],
[ 8., 8., 0., 0.]])

and now I only want to keep all lines from 2nd array pos with those
indices that are nonzero in the mask, i.e. line 3-6 (pos[2]-pos[5]).

F.e. line 4 in pos has the values (5, 4) and mask[5][4] is nonzero, so I
want to keep it. While line 2 (pos[1]) has the values (4, 6) and
mask[4][6] is zero, so shall be discarded.

I want to avoid a for loop (if possible!!!) cause I think (but don't
know) numpy array are handled in another way. I think numpy.delete is
the right function for discarding the values, but I don't know how to
build the indices.

First, convert the pos array to integers, and just the columns with indices in 
them:

  ipos = pos[:,:2].astype(int)

Now check the values in the mask corresponding to these positions:

  mask_values = mask[ipos[:,0], ipos[:,1]]

Now extract the rows from the original pos array where mask_values is nonzero:

  result = pos[mask_values != 0]

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
 that is made terrible by our own mad attempt to interpret it as though it had
 an underlying truth."
  -- Umberto Eco

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