How to find where in numpy array a zero element is preceded by at least N-1 consecutive zeros?

Approach #1

We can use 1D convolution -

def nzeros(a, n):
    # Define kernel for 1D convolution
    k = np.ones(n,dtype=int)

    # Get sliding summations for zero matches with that kernel
    s = np.convolve(a==0,k)

    # Look for summations that are equal to n value, which will occur for
    # n consecutive 0s. Remember that we are using a "full" version of
    # convolution, so there's one-off offsetting because of the way kernel
    # slides across input data. Also, we need to create 1s at places where
    # n consective 0s end, so we would need to slice out ending elements.
    # Thus, we would end up with the following after int dtype conversion
    return (s==n).astype(int)[:-n+1]

Sample run -

In [46]: a
Out[46]: array([0, 0, 0, 0, 1, 0, 0, 0, 1, 1])

In [47]: nzeros(a,3)
Out[47]: array([0, 0, 1, 1, 0, 0, 0, 1, 0, 0])

In [48]: nzeros(a,2)
Out[48]: array([0, 1, 1, 1, 0, 0, 1, 1, 0, 0])

Approach #2

Another way to solve and this could be considered as a variant of the 1D convolution approach, would be to use erosion, because if you look at the outputs, we can simply erode the mask of 0s from the starts until n-1 places. So, we can use scipy.ndimage.morphology's binary_erosion that also allow us to specify the portion of kernel center with its origin arg, hence we will avoid any slicing. The implementation would look something like this -

from scipy.ndimage.morphology import binary_erosion

out = binary_erosion(a==0,np.ones(n),origin=(n-1)//2).astype(int)

Using for loop:

def nzeros(a, n):
  #Create a numpy array of zeros of length equal to n
  b = np.zeros(n)

  #Create a numpy array of zeros of same length as array a
  c = np.zeros(len(a), dtype=int)

  for i in range(0,len(a) - n):
    if (b == a[i : i+n]).all():  #Check if array b is equal to slice in a
      c[i+n-1] = 1

  return c

Sample Output:

print(nzeros(a, 3))
[0 0 1 1 0 0 0 1 0 0]