cudf.DataFrame.sort_values - `ValueError: Cannot convert value of type NotImplementedType to cudf scalar`

Mehdi 0
Viewed 17

I get an error when using sort_values on cudf DataFrame (Version: 22.2.0) :

>>> import cudf
>>> df = cudf.DataFrame()
>>> df['a'] = [0, 1, 2]
>>> df['b'] = [-3, 2, 0]
>>> df.sort_values('b')
ValueError: Cannot convert value of type NotImplementedType to cudf scalar

Do you know why I get this kind of error? (PS: the example above is from the documentation)

For information, here are the last calls from Traceback:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/core/indexed_frame.py:554, in IndexedFrame._gather(self, gather_map, keep_index, nullify, check_bounds)
    551 if not is_integer_dtype(gather_map.dtype):
    552     gather_map = gather_map.astype("int32")
--> 554 if not libcudf.copying._gather_map_is_valid(
    555     gather_map, len(self), check_bounds, nullify
    556 ):
    557     raise IndexError("Gather map index is out of bounds.")
    559 return self._from_columns_like_self(
    560     libcudf.copying.gather(
    561         list(self._index._columns + self._columns)
   (...)
    568     self._index.names if keep_index else None,
    569 )

File cudf/_lib/copying.pyx:65, in cudf._lib.copying._gather_map_is_valid()

File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/core/scalar.py:264, in Scalar.__ge__(self, other)
    263 def __ge__(self, other):
--> 264     return self._scalar_binop(other, "__ge__")

File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/core/scalar.py:346, in Scalar._scalar_binop(self, other, op)
    344 else:
    345     result = self._dispatch_scalar_binop(other, op)
--> 346     return Scalar(result, dtype=out_dtype)

File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/core/scalar.py:75, in Scalar.__init__(self, value, dtype)
     73     self._device_value = value
     74 else:
---> 75     self._host_value, self._host_dtype = self._preprocess_host_value(
     76         value, dtype
     77     )

File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/core/scalar.py:156, in Scalar._preprocess_host_value(self, value, dtype)
    153 if isinstance(value, decimal.Decimal) and dtype is None:
    154     dtype = cudf.Decimal128Dtype._from_decimal(value)
--> 156 value = to_cudf_compatible_scalar(value, dtype=dtype)
    158 if dtype is None:
    159     if not valid:

File ~/miniconda3/envs/devs/lib/python3.9/site-packages/cudf/utils/dtypes.py:246, in to_cudf_compatible_scalar(val, dtype)
    243     return val
    245 if not cudf.api.types._is_scalar_or_zero_d_array(val):
--> 246     raise ValueError(
    247         f"Cannot convert value of type {type(val).__name__} "
    248         "to cudf scalar"
    249     )
    251 if isinstance(val, Decimal):
    252     return val

ValueError: Cannot convert value of type NotImplementedType to cudf scalar

Thank you very much for your help

cudf dataframe python
1 Answers

This issue does not present in the current release of cuDF (22.08) and likely does not present in some of the more recent previous versions. You may be using an out-of-date version compared to the documentation you're viewing (or this was a brief bug).

import cudf
df = cudf.DataFrame()
df['a'] = [0, 1, 2]
df['b'] = [-3, 2, 0]
df.sort_values('b')
    a   b
0   0   -3
2   2   0
1   1   2
Nick Becker0
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