Cudf: [FEA] Series or DataFrame support for equivalent of numpy.isclose

[paul-tqh-nguyen]

Is your feature request related to a problem? Please describe.

It's useful for testing to be able to perform the equivalent of numpy.isclose on instances of cudf.DataFrame and cudf.Series.

Describe the solution you'd like

This would be a nice thing to have:

>>> import cudf
>>> s1 = cudf.Series([1.9876543,   2.9876654,   3.9876543])
>>> s2 = cudf.Series([1.987654321, 2.987654321, 3.987654321])
>>> rel_tol=1e-5
>>> abs_tol=0.0
>>> s2.isclose(s1, rel_tol, abs_tol)
0    True
1    True
2    True
dtype: bool
>>> 

Describe alternatives you've considered

Here's my current hand-rolled solution:

>>> import cudf
>>> s1 = cudf.Series([1.9876543,   2.9876654,   3.9876543])
>>> s2 = cudf.Series([1.987654321, 2.987654321, 3.987654321])
>>> rel_tol=1e-5
>>> abs_tol=0.0
>>> s2.abs().mul(rel_tol).add(abs_tol).sub(s1.sub(s2).abs()).gt(0)
0    True
1    True
2    True
dtype: bool
>>> 

There's nothing wrong with using this approach pervasively. I figured it'd be more convenient to have isclose as a built-in method.

I'm happy to help in whatever way I can.

[beckernick]

Could you use cupy.isclose for this? I believe with pandas you would still use np.isclose.

[kkraus14]

Could you use cupy.isclose for this? I believe with pandas you would still use np.isclose.

@paul-tqh-nguyen this would go from cudf --> cupy zero copy so it wouldn't cause any performance degradation. Could you give that a shot?

[paul-tqh-nguyen]

Using cupy.isclose solves all of my problems!

My apologies; I should've looked there earlier.

Thanks for the quick and helpful responses!

[kkraus14]

One note that this won't handle null values so I'm going to reopen this to provide some syntactic sugar in cuDF surrounding the cupy function in the future.