Python Numpy manipulate matrix

Question

I have a matrix that looks like this:

[[ 0 0 0 0 ]
 [ 0 0 0 0 ]
 [ 0 1 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 1 0 ]
 [ 0 0 0 0 ]]

and I want to change it in a way, that if there is 1 value in a row, then make two rows above and after to 1 as well, so the output should be:

[[ 0 1 0 0 ]
 [ 0 1 0 0 ]
 [ 0 1 0 0 ]
 [ 0 1 0 0 ]
 [ 0 1 0 0 ]
 [ 0 0 0 0 ] 
 [ 0 0 0 0 ]
 [ 0 0 1 0 ]
 [ 0 0 1 0 ]
 [ 0 0 1 0 ]
 [ 0 0 1 0 ]]

any idea or help please?

thank you

The binary_dilation function with an appropriate array for the 'structure' kwarg might be a good option in that case. — Alex Riley
– Alex Riley, Commented Jan 13, 2017 at 12:58

Dennis Sakva · Accepted Answer · 2017-01-12 10:48:46Z

1

for (x,y) in np.argwhere(n==1):
    xfrom=max(x-2,0)
    xto=min(n.shape[0]-1,x+2)
    for i in range(xfrom, xto+1):
        n[i,y]=1

edited Jan 12, 2017 at 10:48

answered Jan 12, 2017 at 10:06

Dennis Sakva

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2 Comments

TheDaJon Over a year ago

what is the n? is it the matrix? if so, I get an error of index out of bounds

Dennis Sakva Over a year ago

You're right. n - matrix, I corrected the upper bound to n.shape[0]-1. Sorry for that.

Ethan Kong · Accepted Answer · 2017-01-13 02:27:58Z

When dealing with an array of 0s and 1s, you may want to convert it to an boolean array and make use of the logical operators. And, slicing and indexing in numpy is very handy.

def expand_in_place(bool_matrix, expand_up=2, expand_down=2):
    assert expand_up >= 0 and expand_down >= 0

    # expands upward
    for _ in range(expand_up):
        bool_matrix[:-1] |= bool_matrix[1:].copy()

    # expands downward
    for _ in range(expand_down):
        bool_matrix[1:] |= bool_matrix[:-1].copy()

Example usage:

matrix = np.array([[0, 0, 0, 0],
                   [0, 0, 0, 0],
                   [0, 1, 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, 1, 0],
                   [0, 0, 0, 0]])
bool_matrix = matrix.astype(bool)
expand_in_place(bool_matrix)
print()  # newline
print(bool_matrix.astype(int))

Output:

In [64]: 
[[0 1 0 0]
 [0 1 0 0]
 [0 1 0 0]
 [0 1 0 0]
 [0 1 0 0]
 [0 0 0 0]
 [0 0 0 0]
 [0 0 1 0]
 [0 0 1 0]
 [0 0 1 0]
 [0 0 1 0]]

Explanation:

The key is what happens when you call bool_matrix[:-1] |= bool_matrix[1:] every time.

I add .copy() because bool_matrix[1:] |= bool_matrix[:-1] doesn't work as expected.

In [108]: small = np.array([False, False, True, False, False])

In [109]: small[1:] |= small[:-1]

In [110]: small
Out[110]: 
array([False, False,  True,  True,  True], dtype=bool)

You can easily extend it to expand in four directions.

Ricardo Cárdenes · Accepted Answer · 2017-01-13 02:40:33Z

0

Much easier:

for row, col in np.argwhere(n == 1):
    n[row-2:row+3,col] = 1

Let Python's and NumPy's slicing semantics work for you. No need to calculate boundaries.

answered Jan 13, 2017 at 2:40

Ricardo Cárdenes

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