More efficient fillna(numpy) - numpy

Question

I need an array version of a function similar to Pandas.fillna, in the forum I collected a lot of answers to create the following function, but it is still 3 times times slower than Pandas.fillna, I want to know if there is a better way to optimize, thank you.
def fillna(self,axis=None,mask=None,value=None,method='pad'):
""" array fillna
Parameters
----------
self : 1d/2d
axis : axis(0 or 1)
mask : Custom mask, or Built np.isfinite(x)
value : int
method : 'back', 'pad', 'mean'
--------
"""
x = np.asarray(self)
if mask is None: mask = np.isfinite(x)
if (not value is None)|(method=='mean'):
out = x.copy()
if x.ndim == 1:
if method=='mean':
out[~mask] = np.nanmean(x)
else: out[~mask] = value
else:
vask = ~mask * (np.nanmean(x,1)[:,None] if axis==1 else np.nanmean(x,0))
out[~mask] = vask[~mask]
else:
if axis is None: axis = 0
if x.ndim==1:
if method=='pad':
idx = np.where(mask,np.arange(mask.shape[0]),0)
np.maximum.accumulate(idx,axis=0,out=idx)
return x[idx]
elif method=='back':
idx = np.where(mask[::-1],np.arange(mask.shape[0]),0)
np.maximum.accumulate(idx,axis=0,out=idx)
return x[mask.shape[0]-idx[::-1]-1]
else: return x
if axis==1:
if method=='back': mask = mask[:, ::-1]
idx = np.where(mask,np.arange(mask.shape[1]),0)
else:
if method=='back': mask = mask[::-1,:]
idx = np.where(mask,np.arange(mask.shape[0])[:,None],0)
np.maximum.accumulate(idx,axis=axis,out=idx)
if axis==1:
if method=='back': idx = idx.shape[1]-idx[:, ::-1] - 1
out = x[np.arange(idx.shape[0])[:,None], idx]
else:
if method=='back': idx = idx.shape[0]-idx[::-1, :] - 1
out = x[idx,np.arange(idx.shape[1])]
return out