Here's a vectorized approach taking inspiration from NumPy based forward-filling
for the forward-filling part in this solution alongwith masking
and slicing
-
def forward_fill_ifsame(x):
# Get mask of non-zeros and then use it to forward-filled indices
mask = x!=0
idx = np.where(mask,np.arange(len(x)),0)
np.maximum.accumulate(idx,axis=0, out=idx)
# Now we need to work on the additional requirement of filling only
# if the previous and next ones being same
# Store a copy as we need to work and change input data
x1 = x.copy()
# Get non-zero elements
xm = x1[mask]
# Off the selected elements, we need to assign zeros to the previous places
# that don't have their correspnding next ones different
xm[:-1][xm[1:] != xm[:-1]] = 0
# Assign the valid ones to x1. Invalid ones become zero.
x1[mask] = xm
# Use idx for indexing to do the forward filling
out = x1[idx]
# For the invalid ones, keep the previous masked elements
out[mask] = x[mask]
return out
Sample runs -
In [289]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,3,1,0,1])
In [290]: np.vstack((x, forward_fill_ifsame(x)))
Out[290]:
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 3, 1, 0, 1],
[1, 1, 1, 1, 0, 0, 2, 0, 3, 3, 3, 3, 3, 1, 1, 1]])
In [291]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,1,1,0,1])
In [292]: np.vstack((x, forward_fill_ifsame(x)))
Out[292]:
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 1],
[1, 1, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 1, 1]])
In [293]: x = np.array([1,0,1,1,0,0,2,0,3,0,0,0,1,1,0,2])
In [294]: np.vstack((x, forward_fill_ifsame(x)))
Out[294]:
array([[1, 0, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2],
[1, 1, 1, 1, 0, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2]])