How to fill elements between intervals of a list

Here's a numpy based approach using np.cumsum:

a = np.array([np.NaN, np.NaN, 1, np.NaN, np.NaN, np.NaN, 0, np.NaN, 
              1, np.NaN, 0, 1, np.NaN, 0, np.NaN,  1, np.NaN])

ix0 = (a == 0).cumsum()
ix1 = (a == 1).cumsum()
dec = (ix1 - ix0).astype(float)
# Only necessary if the seq can end with an unclosed interval
ix = len(a)-(a[::-1]==1).argmax()
last = ix1[-1]-ix0[-1]
if last > 0:
    dec[ix:] = a[ix:]
# -----
out = np.where(dec==1, dec, a)

---

print(out)
array([nan, nan,  1.,  1.,  1.,  1.,  0., nan,  1.,  1.,  0.,  1.,  1.,
        0., nan,  1., nan])

Pandas solution:

s = pd.Series(list_1)
s1 = s.eq(1)
s0 = s.eq(0)
m = (s1 | s0).where(s1.cumsum().ge(1),False).cumsum().mod(2).eq(1)
s.loc[m & s.isna()] = 1
print(s.tolist())
#[nan, nan, 1.0, 1.0, 1.0, 1.0, 0.0, nan, 1.0, 1.0, 0.0, 1.0, 1.0, 0.0, nan, 1.0, 1.0]

---

but if there is only 1, 0 or NaN you can do:

s = pd.Series(list_1)
s.fillna(s.ffill().where(lambda x: x.eq(1))).tolist()

output

[nan,
 nan,
 1.0,
 1.0,
 1.0,
 1.0,
 0.0,
 nan,
 1.0,
 1.0,
 0.0,
 1.0,
 1.0,
 0.0,
 nan,
 1.0,
 1.0]