纪元时间和正常日期之间的差异(以秒为单位) - python

Question

我有两个变量：一个是纪元时间，另一个是正常日期时间，

df['epochtime']
0       1457586382
1       1457586382
2       1457586391
3       1457586692
4       1457586391
5       1457586391
6       1457586692
7       1457586692
8       1457572611
df['time']
0      2016-03-10 00:06:21.903
1      2016-03-10 00:06:21.908
2      2016-03-10 00:06:30.895
3      2016-03-10 00:06:30.895
4      2016-03-10 00:06:30.899
5      2016-03-10 00:06:30.899
6      2016-03-10 00:06:31.045
7      2016-03-10 00:06:31.094
8      2016-03-10 01:16:51.390 

我想要这两个时间之间的差异（以秒为单位）。以下是我的尝试，

pd.to_datetime(df['epochtime'], unit ='s' ) - df['time']

以下是我的输出，

0               00:00:00.097000
1               00:00:00.092000
2               00:00:00.105000
3               00:05:01.105000
4               00:00:00.101000
5               00:00:00.101000
6               00:05:00.955000
7               00:05:00.906000
8      -1 days +23:59:59.610000   

我想以秒为单位将其设置为整数。同样对于第 8 个值，由于某种奇怪的原因，差异为 -1 天，它应该只是秒。谁能帮我把秒差作为整数？

Answer 1

您必须将日期时间转换为纪元时间，然后进行算术减法。

正如您所说，您始终使用 UTC 时间，您需要使用 calendar.timegm()，因为它需要 UTC 时间作为参数。

import time
import datetime
import calendar
TIME_FORMAT = '%Y-%m-%d %H:%M:%S'
def local_datetime_2_timestamp(local_datetime):
    '''strptime takes local datetime as parameter'''
    timestamp = str(time.mktime(datetime.datetime.
        strptime(local_datetime, TIME_FORMAT).timetuple()))[:-2]
    return timestamp
def utc_2_timestamp(utc_datetime):
    '''timegm takes UTC time as parameter'''
    return calendar.timegm(utc_datetime.timetuple())
print local_datetime_2_timestamp("2016-03-10 00:06:21.903".split(".")[0])
print utc_2_timestamp(datetime.datetime(2016,3,10,0,6,21))

Answer 2

from time import mktime, strptime
date_a = '20-01-2015 11:05:02'
epoch_b = 1457586692
epoch_a = int(mktime(strptime(date_a, '%d-%m-%Y %H:%M:%S')))
print(epoch_a - epoch_b)