Python 3.x Python-计算两个datetime.time对象之间的差异

我有两个datetime.time对象，我想计算它们之间的小时差。比如说

a = datetime.time(22,00,00)
b = datetime.time(18,00,00)

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我希望能够减去这些，从而得到值4。

要计算差值，必须将

datetime.time

对象转换为

datetime.datetime

对象。然后当你减法时，你得到一个

timedelta

对象。为了找出

timedelta

对象的小时数，您必须找到总秒数并将其除以

3600

# Create datetime objects for each time (a and b)
dateTimeA = datetime.datetime.combine(datetime.date.today(), a)
dateTimeB = datetime.datetime.combine(datetime.date.today(), b)
# Get the difference between datetimes (as timedelta)
dateTimeDifference = dateTimeA - dateTimeB
# Divide difference in seconds by number of seconds in hour (3600)  
dateTimeDifferenceInHours = dateTimeDifference.total_seconds() / 3600

我就是这样做的

a = '2200'
b = '1800'
time1 = datetime.strptime(a,"%H%M") # convert string to time
time2 = datetime.strptime(b,"%H%M") 
diff = time1 -time2
diff.total_seconds()/3600    # seconds to hour 

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输出：4.0

我从这个问题中得到了结果：

a='2017-10-10 21:25:13'

b='2017-10-02 10:56:33'

a=pd.to_datetime(a)

b=pd.to_datetime(b)

c.total_seconds()/3600

但在一系列不起作用的情况下：

table1['new2']=table1['new'].total_seconds()/3600