Numpy: Checking if a value is NaT

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余生分开走 2020-12-23 21:12
nat = np.datetime64(\'NaT\')
nat == nat
>> FutureWarning: In the future, \'NAT == x\' and \'x == NAT\' will always be False.

np.isnan(nat)
>> TypeError:         


        
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  • 2020-12-23 21:46

    Very simple and surprisingly fast: (without numpy or pandas)

        str( myDate ) == 'NaT'            # True if myDate is NaT
    

    Ok, it's a little nasty, but given the ambiguity surrounding 'NaT' it does the job nicely.

    It's also useful when comparing two dates either of which might be NaT as follows:

       str( date1 ) == str( date1 )       # True
       str( date1 ) == str( NaT )         # False
       str( NaT )   == str( date1 )       # False
    
    wait for it...
    
       str( NaT )   == str( Nat )         # True    (hooray!)
    
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  • 2020-12-23 21:49

    This approach avoids the warnings while preserving the array-oriented evaluation.

    import numpy as np
    def isnat(x):
        """ 
        datetime64 analog to isnan.
        doesn't yet exist in numpy - other ways give warnings
        and are likely to change.  
        """
        return x.astype('i8') == np.datetime64('NaT').astype('i8')
    
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  • 2020-12-23 21:54

    Since NumPy version 1.13 it contains an isnat function:

    >>> import numpy as np
    >>> np.isnat(np.datetime64('nat'))
    True
    

    It also works for arrays:

    >>> np.isnat(np.array(['nat', 1, 2, 3, 4, 'nat', 5], dtype='datetime64[D]'))
    array([ True, False, False, False, False,  True, False], dtype=bool)
    
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  • 2020-12-23 22:02

    pandas can check for NaT with pandas.isnull:

    >>> import numpy as np
    >>> import pandas as pd
    >>> pd.isnull(np.datetime64('NaT'))
    True
    

    If you don't want to use pandas you can also define your own function (parts are taken from the pandas source):

    nat_as_integer = np.datetime64('NAT').view('i8')
    
    def isnat(your_datetime):
        dtype_string = str(your_datetime.dtype)
        if 'datetime64' in dtype_string or 'timedelta64' in dtype_string:
            return your_datetime.view('i8') == nat_as_integer
        return False  # it can't be a NaT if it's not a dateime
    

    This correctly identifies NaT values:

    >>> isnat(np.datetime64('NAT'))
    True
    
    >>> isnat(np.timedelta64('NAT'))
    True
    

    And realizes if it's not a datetime or timedelta:

    >>> isnat(np.timedelta64('NAT').view('i8'))
    False
    

    In the future there might be an isnat-function in the numpy code, at least they have a (currently open) pull request about it: Link to the PR (NumPy github)

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  • 2020-12-23 22:07

    Another way would be to catch the exeption:

    def is_nat(npdatetime):
        try:
            npdatetime.strftime('%x')
            return False
        except:
            return True
    
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  • 2020-12-23 22:10

    INTRO: This answer was written in a time when Numpy was version 1.11 and behaviour of NAT comparison was supposed to change since version 1.12. Clearly that wasn't the case and the second part of answer became wrong. The first part of answer may be not applicable for new versions of numpy. Be sure you've checked MSeifert's answers below.


    When you make a comparison at the first time, you always have a warning. But meanwhile returned result of comparison is correct:

    import numpy as np    
    nat = np.datetime64('NaT')
    
    def nat_check(nat):
        return nat == np.datetime64('NaT')    
    
    nat_check(nat)
    Out[4]: FutureWarning: In the future, 'NAT == x' and 'x == NAT' will always be False.
    True
    
    nat_check(nat)
    Out[5]: True
    

    If you want to suppress the warning you can use the catch_warnings context manager:

    import numpy as np
    import warnings
    
    nat = np.datetime64('NaT')
    
    def nat_check(nat):
        with warnings.catch_warnings():
            warnings.simplefilter("ignore")
            return nat == np.datetime64('NaT')    
    
    nat_check(nat)
    Out[5]: True
    


    EDIT: For some reason behavior of NAT comparison in Numpy version 1.12 wasn't change, so the next code turned out to be inconsistent.

    And finally you might check numpy version to handle changed behavior since version 1.12.0:

    def nat_check(nat):
        if [int(x) for x in np.__version__.split('.')[:-1]] > [1, 11]:
            return nat != nat
        with warnings.catch_warnings():
            warnings.simplefilter("ignore")
            return nat == np.datetime64('NaT')
    


    EDIT: As MSeifert mentioned, Numpy contains isnat function since version 1.13.

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