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nat = np.datetime64('NaT') nat == nat >> FutureWarning: In the future, 'NAT == x' and 'x == NAT' will always be False. np.isnan(nat) >> TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''
How can I check if a datetime64 is NaT? I can’t seem to dig anything out of the docs. I know Pandas can do it, but I’d rather not add a dependency for something so basic.
>>> 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)
Since NumPy version 1.13 it contains an
>>> 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)
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: FutureWarning: In the future, 'NAT == x' and 'x == NAT' will always be False. True nat_check(nat) Out: 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: 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.
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!)
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')
Another way would be to catch the exeption:
def is_nat(npdatetime): try: npdatetime.strftime('%x') return False except: return True