using lower_cast (or finding some other way to divide datetimes)

[Anthony Morgan]

I'm sure I'm doing something daft here, but perhaps someone could enlighten me as to why this doesn't work:

import numpy as np

from numba import jit

from numba import types

from numba.extending import lower_cast

@lower_cast(types.NPDatetime('ns'), types.int64)

def dt_val(context, builder, fromty, toty, value):

    return value

@jit(nopython=True, locals={'x': types.int64})

def test(x):

    return x // 1000000000

res = test(np.datetime64(60*1000000000,'ns'))

fails with, 

numba.errors.TypingError: Failed at nopython (nopython frontend)

No conversion from datetime64(ns) to int64 for 'x', defined at None

or perhaps any alternative approach for taking the floor of a datetime64, that doesn't involve taking a view of the item outside of numba would be appreciated. 

thanks

[Chris Bartak]

It seems like you can directly convert inside your calling function

@jit(nopython=True)

def test(x):

    return np.int64(x) // 1000000000

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[Anthony Morgan]

That's really great, thank you. It's a bit annoying that the inverse doesn't work, because the datetime64 constructor ins't exposed to numba, but I can live with that easily enough if there isn't an easy solution to that:

import numpy as np

from numba import jit

from numba import types

from numba.extending import lower_cast

@lower_cast(types.NPDatetime('ns'), types.int64)

def dt_val(context, builder, fromty, toty, value):

    return value

@lower_cast(types.int64, types.NPDatetime('ns'))

def val_dt(context, builder, fromty, toty, value):

    return value

@jit(nopython=True)

def test(x):

    y = np.int64(x) // 1000000000

    return np.datetime64(y,'ns')

res = test(np.datetime64(60*1000000000,'ns'))

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