convert numpy.datetime64 into epoch time - numpy

Question

I am trying to convert my numpy array new_feat_dt containing numpy.datetime64 into epoch time. I want to make sure when the conversion happens the date stays in utc format?
I am using numpy 1.16.4 and python3.6
I have tried two ways of conversion as shown in code below.
import numpy as np
new_feat_dt = [np.datetime64('2019-07-25T14:23:01'), np.datetime64('2019-07-25T14:25:01'), np.datetime64('2019-07-25T14:27:01')]
final= [(x - np.datetime64('1970-01-01T00:00:00Z')) / np.timedelta64(1, 's') for x in new_feat_dt]
print (final)
print(type(final[0]))
final2= [np.datetime64(x,'s').astype(int) for x in new_feat_dt]
print (final2)
print(type(final2[0]))
Output of the above code:
[1564064581.0, 1564064701.0, 1564064821.0]
<class 'numpy.float64'>
[1564064581, 1564064701, 1564064821]
<class 'numpy.int32'>
The above is happening because the times in new_feat_dt array is considered as GMT. I want it to be considered as my local time which is ('US/Eastern').
The correct conversion should be this:
[1564078981,1564079101,1564079221]

Answer 1

The numpy.datetime64 is a timezone naive datetime type. To add timezone information into the datetime, try use python's datetime with pytz module.
import numpy as np
import pytz
from datetime import datetime
new_feat_dt = [np.datetime64('2019-07-25T14:23:01'), np.datetime64('2019-07-25T14:25:01'), np.datetime64('2019-07-25T14:27:01')]
eastern = pytz.timezone('US/Eastern')
final = [int(eastern.localize(dt.astype(datetime)).timestamp()) for dt in new_feat_dt]
print(final)
The output:
[1564078981, 1564079101, 1564079221]
It's probably better to initialize all your new_feat_dt using datetime.datetime.