# python – Numpy: Checking if a value is NaT

## python – Numpy: Checking if a value is NaT

pandas can check for `NaT`

with `pandas.isnull`

:

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

If you dont 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 cant be a NaT if its not a dateime
```

This correctly identifies NaT values:

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

And realizes if its 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 `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)
```

#### python – Numpy: Checking if a value is NaT

**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 wasnt 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 youve checked MSeiferts 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 wasnt 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.