AttributeError with divmod(Series, Series) with 0 in the denominator

[TomAugspurger]

Strange one. This works,

In [39]: a = pd.Series([1, 1])

In [40]: divmod(a, pd.Series([1, 2]));

This raises

In [41]: divmod(a, pd.Series([0, 2]));
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-41-04277abe6f5a> in <module>
----> 1 divmod(a, pd.Series([0, 2]));

~/sandbox/pandas/pandas/core/ops.py in wrapper(left, right)
   1714             rvalues = rvalues.values
   1715
-> 1716         result = safe_na_op(lvalues, rvalues)
   1717         return construct_result(left, result,
   1718                                 index=left.index, name=res_name, dtype=None)

~/sandbox/pandas/pandas/core/ops.py in safe_na_op(lvalues, rvalues)
   1660         try:
   1661             with np.errstate(all='ignore'):
-> 1662                 return na_op(lvalues, rvalues)
   1663         except Exception:
   1664             if is_object_dtype(lvalues):

~/sandbox/pandas/pandas/core/ops.py in na_op(x, y)
   1640             result = masked_arith_op(x, y, op)
   1641
-> 1642         result = missing.fill_zeros(result, x, y, op_name, fill_zeros)
   1643         return result
   1644

~/sandbox/pandas/pandas/core/missing.py in fill_zeros(result, x, y, name, fill)
    544             mask = ((y == 0) & ~np.isnan(result)).ravel()
    545
--> 546             shape = result.shape
    547             result = result.astype('float64', copy=False).ravel()
    548

AttributeError: 'tuple' object has no attribute 'shape'

[TomAugspurger]

This sends us down the bad path:

pandas/pandas/core/missing.py

Line 540 in dfcd2b2

if (y == 0).any():

Haven't looked into it. Don't plan to look further right now.

[jbrockmendel]

I've got a patch for the immediate issue by just checking for divmod in fill_zeros and operating element-wise, but it surfaces another question about what we expect this to operation to return.

a = pd.Series([1, 1])
b = pd.Series([0, 2])

result = divmod(a, b)
expected = a // b, a % b

result[0] matches expected[0], but result[1] and expected[1] are:

>>> res1[1]
0    inf
1    1.0
dtype: float64
>>> expected[1]
0    NaN
1    1.0
dtype: float64

[jreback]

see what numpy does for float args

[jbrockmendel]

Looks like #23293 made it so this doesn't raise, but does use incorrect fill value. Will fix in a variant of #27130.