Fill zero values of 1d numpy array with last non-zero values

Here's a solution using np.maximum.accumulate:

def fill_zeros_with_last(arr):
    prev = np.arange(len(arr))
    prev[arr == 0] = 0
    prev = np.maximum.accumulate(prev)
    return arr[prev]

We construct an array prev which has the same length as arr, and such that prev[i] is the index of the last non-zero entry before the i-th entry of arr. For example, if:

>>> arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2])

Then prev looks like:

array([ 0,  0,  0,  3,  3,  5,  6,  7,  7,  7,  7,  7, 12])

Then we just index into arr with prev and we obtain our result. A test:

>>> arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2])
>>> fill_zeros_with_last(arr)
array([1, 1, 1, 2, 2, 4, 6, 8, 8, 8, 8, 8, 2])

Note: Be careful to understand what this does when the first entry of your array is zero:

>>> fill_zeros_with_last(np.array([0,0,1,0,0]))
array([0, 0, 1, 1, 1])

Inspired by jme's answer here and by Bas Swinckels' (in the linked question) I came up with a different combination of numpy functions:

def fill_zeros_with_last(arr, initial=0):
     ind = np.nonzero(arr)[0]
     cnt = np.cumsum(np.array(arr, dtype=bool))
     return np.where(cnt, arr[ind[cnt-1]], initial)

I think it's succinct and also works, so I'm posting it here for the record. Still, jme's is also succinct and easy to follow and seems to be faster, so I'm accepting it :-)

If the 0s only come in strings of 1, this use of nonzero might work:

In [266]: arr=np.array([1,0,2,3,0,4,0,5])
In [267]: I=np.nonzero(arr==0)[0]
In [268]: arr[I] = arr[I-1]
In [269]: arr
Out[269]: array([1, 1, 2, 3, 3, 4, 4, 5])

I can handle your arr by applying this repeatedly until I is empty.

In [286]: arr = np.array([1, 0, 0, 2, 0, 4, 6, 8, 0, 0, 0, 0, 2])

In [287]: while True:
   .....:     I=np.nonzero(arr==0)[0]
   .....:     if len(I)==0: break
   .....:     arr[I] = arr[I-1]
   .....:     

In [288]: arr
Out[288]: array([1, 1, 1, 2, 2, 4, 6, 8, 8, 8, 8, 8, 2])

If the strings of 0s are long it might be better to look for those strings and handle them as a block. But if most strings are short, this repeated application may be the fastest route.