pandas cut a series with nan values

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I would like to apply the pandas cut function to a series that includes NaNs. The desired behavior is that it buckets the non-NaN elements and returns NaN for the NaN-elements.

import pandas as pd
numbers_with_nan = pd.Series([3,1,2,pd.NaT,3])
numbers_without_nan = numbers_with_nan.dropna()

The cutting works fine for the series without NaNs:

pd.cut(numbers_without_nan, bins=[1,2,3], include_lowest=True)
0      (2.0, 3.0]
1    (0.999, 2.0]
2    (0.999, 2.0]
4      (2.0, 3.0]

When I cut the series that contains NaNs, element 3 is correctly returned as NaN, but the last element gets the wrong bin assigned:

pd.cut(numbers_with_nan, bins=[1,2,3], include_lowest=True)
0      (2.0, 3.0]
1    (0.999, 2.0]
2    (0.999, 2.0]
3             NaN
4    (0.999, 2.0]

How can I get the following output?

0      (2.0, 3.0]
1    (0.999, 2.0]
2    (0.999, 2.0]
3             NaN
4      (2.0, 3.0]
pandas nan python
1 Answers

This is strange. The problem isn't pd.NaT, it's the fact your series has object dtype instead of a regular numeric series, e.g. float, int.

A quick fix is to replace pd.NaT with np.nan via fillna. This triggers series conversion from object to float64 dtype, and may also lead to better performance.

s = pd.Series([3, 1, 2, pd.NaT, 3])

res = pd.cut(s.fillna(np.nan), bins=[1, 2, 3], include_lowest=True)

print(res)

0    (2, 3]
1    [1, 2]
2    [1, 2]
3       NaN
4    (2, 3]
dtype: category
Categories (2, object): [[1, 2] < (2, 3]]

A more generalized solution is to convert to numeric explicitly beforehand:

s = pd.to_numeric(s, errors='coerce')
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