Python计算两个datetime.time对象之间的差异

Question

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)

我希望能够减去这些，这样就得到了4的值

Answer 1

要计算差异，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

Answer 2

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 

产出：4.0

Answer 3

我从这个问题中得到了我的结果：

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