carving 2D numpy array by index

Question

How do i "carve" or mask a 2D numpy array according to an index formula? I don't care what the element value is, only its position in the array.

For example, given an mxm array, how do I extract all elements whose address conforms to

for i in range(0,m):
    for j in range(0,m):
        if j-i-k>=0:
            A[i,j] = 1
        elif j-p-k>=0:
            A[i,j] = 1
        elif i-k>=0:
            A[i,j] = 1
        else:
            A[i,j] = 0
        j=j+1
    i=i+1

where

k and p are an arbitrary fences

Assume

k<m
p<m

This ends up looking like a diagonal slice + a horizontal slice + a vertical slice. Can it be done without the for loops above?

Further question on new_data = data[mask] @tcaswell, using my "for" created mask "removes" some rows and columns and somehow leaves me with a vector, not an array of the original data.shape. Why? How can I prevent this? — Hokiexterra
– Hokiexterra, Commented Nov 19, 2012 at 21:21

unutbu · Accepted Answer · 2012-11-19 19:40:00Z

3

In [1]: import numpy as np

In [2]: k = 2

In [3]: i, j = np.ogrid[0:5,0:5]

In [4]: mask = (j-i-k < 0)

In [5]: mask
Out[5]: 
array([[ True,  True, False, False, False],
       [ True,  True,  True, False, False],
       [ True,  True,  True,  True, False],
       [ True,  True,  True,  True,  True],
       [ True,  True,  True,  True,  True]], dtype=bool)

In [6]: mask.shape
Out[6]: (5, 5)

In [7]: mask.dtype
Out[7]: dtype('bool')

answered Nov 19, 2012 at 19:40

unutbu

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Comments

tacaswell · Accepted Answer · 2012-11-19 21:37:18Z

1

xdim,ydim = data.shape
k = 2
a, b = np.meshgrid(range(ydim),range(xdim))
mask = (b - a -k) < 0

new_data = data[mask]

new_data2 = np.array(data) # to force a copy
new_data2[~mask] = 0

new_data is a vector because the masking processes (done this way) flattens the array. Your are selecting a shape with a ragged that cannot be represented as an array. If you just want to set the non-selcted values to 0, use new_data2.

edited Nov 19, 2012 at 21:37

answered Nov 19, 2012 at 19:42

tacaswell

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John Vinyard · Accepted Answer · 2012-11-19 19:45:43Z

1

Here's yet another way using np.indices:

>>> import numpy as np
>>> a = np.arange(90).reshape(10,9)
>>> b = np.indices(a.shape)
>>> k = 2
>>> i = b[1] - b[0] - k
>>> a[i < 0]
array([ 0,  1,  9, 10, 11, 18, 19, 20, 21, 27, 28, 29, 30, 31, 36, 37, 38,
   39, 40, 41, 45, 46, 47, 48, 49, 50, 51, 54, 55, 56, 57, 58, 59, 60,
   61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78,
   79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89])

answered Nov 19, 2012 at 19:45

John Vinyard

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