Extract separate non-zero blocks from array

Question

having an array like this for example:

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

What\'s the fastest way in Pytho

Answer 1

A trivial change to my answer at Finding the consecutive zeros in a numpy array gives the function find_runs:

def find_runs(value, a):
    # Create an array that is 1 where a is `value`, and pad each end with an extra 0.
    isvalue = np.concatenate(([0], np.equal(a, value).view(np.int8), [0]))
    absdiff = np.abs(np.diff(isvalue))
    # Runs start and end where absdiff is 1.
    ranges = np.where(absdiff == 1)[0].reshape(-1, 2)
    return ranges

For example,

In [43]: x
Out[43]: array([1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1])

In [44]: find_runs(1, x)
Out[44]: 
array([[ 0,  4],
       [ 9, 12],
       [14, 16],
       [20, 22]])

In [45]: [range(*run) for run in find_runs(1, x)]
Out[45]: [[0, 1, 2, 3], [9, 10, 11], [14, 15], [20, 21]]

If the value 1 in your example was not representative, and you really want runs of any non-zero values (as suggested by the text of the question), you can change np.equal(a, value) to (a != 0) and change the arguments and comments appropriately. E.g.

def find_nonzero_runs(a):
    # Create an array that is 1 where a is nonzero, and pad each end with an extra 0.
    isnonzero = np.concatenate(([0], (np.asarray(a) != 0).view(np.int8), [0]))
    absdiff = np.abs(np.diff(isnonzero))
    # Runs start and end where absdiff is 1.
    ranges = np.where(absdiff == 1)[0].reshape(-1, 2)
    return ranges

For example,

In [63]: y
Out[63]: 
array([-1,  2, 99, 99,  0,  0,  0,  0,  0, 12, 13, 14,  0,  0,  1,  1,  0,
        0,  0,  0, 42, 42])

In [64]: find_nonzero_runs(y)
Out[64]: 
array([[ 0,  4],
       [ 9, 12],
       [14, 16],
       [20, 22]])