Stack Overflow Asked by Anton Philippoff on November 17, 2021
For example, I have a piece of code:
import numpy as np
from numba import njit
d = np.array(['2001-01-01T12:00', '2002-02-03T13:56:03.172'],
dtype='datetime64')
@njit
def datetime_operand(date):
x = date[1] - date[0]
return x
datetime_operand(d) // the result is numpy.timedelta64(34394163172,'ms')
Simple option to type np.int64(x)
doesn’t help.
Actually, the answer is:
import numpy as np
dt64 = np.datetime64('2009-01-01')
print((dt64 — np.datetime64('1970-01-01T00:00:00Z')) / np.timedelta64(1, 's'))
Answered by Anton Philippoff on November 17, 2021
I am not sure about the result you look for, but I suppose pandas can help here:
import pandas
@njit
def datetime_operand(date):
df = pandas.DataFrame(d.T)
df[0]=pd.to_datetime(df[0])
return (df[0].iloc[1]-df[0].iloc[0]).to_numpy()
The result is like this:
datetime_operand(d)
umpy.timedelta64(34394163172,'ms')
Answered by Nose on November 17, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP