CODEWITHSUNDEEP
pythonarraysnumpy
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Reputation: 327

How to check is numpy 2d array "surrounded" by zeros

Is there any neat way to check is numpy array surrounded by zeros.

Example:

[[0,0,0,0],
 [0,1,2,0],
 [0,0,0,0]]

I know I can iterate it element wise to find out but I wonder is there any nice trick we can use here. The numpy array is of floats, n x m of arbitrary size.

Any ideas are welcome.

Upvotes: 6

Views: 2904

Answers (3)

Pranjal Jain
Pranjal Jain

Reputation: 86

You can use numpy.any() to test if there is any non-zero element in numpy array.

Now, to test if a 2D array is surrounded by zeroes, you can get first and last columns as well as first and last rows and test if any of those contains a non-zero number.

def zero_surrounded(array):
    return not (array[0,:].any() or array[-1,:].any() or array[:,0].any() or array[:,-1].any())

Upvotes: 4

jmd_dk
jmd_dk

Reputation: 13090

Not the fastest, but perhaps the shortest (and hence a "neat") way of doing it:

surrounded = np.sum(a[1:-1, 1:-1]**2) == np.sum(a**2)
print(surrounded)  # True

Here, a is the array.

This compares the sum of all squared elements to the sum of all squared elements except for those on the boundary. If we left out the squaring, cases where positive and negative boundary values add up to zero would produce the wrong answer.

Upvotes: 2

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476503

We can check this by constructing two submatrices:

  • A[[0,-1]] the first and the last row, including the first and last column; and
  • A[1:-1,[0,-1]] the first and last column, excluding the first and last row.

All the values of these matrices should be equal to zero, so we can use:

if np.all(A[[0,-1]] == 0) and np.all(A[1:-1,[0,-1]] == 0):
    # ...
    pass

This works for an arbitrary 2d-array, but not for arrays with arbitrary depth. We can however use a trick for that as well.

For an arbitrary matrix, we can use:

def surrounded_zero_dim(a):
    n = a.ndim
    sel = ([0,-1],)
    sli = (slice(1,-1),)
    return all(np.all(a[sli*i+sel] == 0) for i in range(n))

Using the slice is strictly speaking not necessary, but it prevents checking certain values twice.

Upvotes: 3

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