Question

我的df看起来像这样：

   Time_Start               CODE    time
0  2018-01-31 16:45:04.263  B76     2018-01-31 16:48:06
1  2018-01-31 16:10:26.000  974     2018-01-31 16:50:06

我从time中减去Time_Start。对于第二个，我得到39:49这是预期的输出但是对于第一个我得到的数字像2.737000（不是我想要的）。

Time_Start为datetime64[ns]，time为object。这就是我进行转换和减法的方式：

waiting['time'] = pd.to_datetime(waiting['time']) 
waiting['Duration'] = waiting['time'] - waiting['Time_Start']
waiting['Duration'] = waiting['Duration'].apply(lambda x: str(x)[-8:])

我对python很新，所以有人能告诉我我做错了什么以及为什么我得到了2个不同的输出？我正在使用python 2.x。

完整代码：

query that returns the dataframe
.......



end_time = np_server_date.ix[0]['data'] """current time"""
end_time = end_time.strftime('%Y-%m-%d %H:%M:%S')
waiting['time'] = end_time """it's now the current time"""

waiting['time'] = pd.to_datetime(waiting['time']) 
waiting['Duration'] = waiting['time'] - waiting['Time_Start']
waiting['Duration'] = waiting['Duration'].apply(lambda x: str(x)[-8:])

Answer 1

好的，所以我觉得我现在有足够的信息来回答你的问题。

蝙蝠的权利，你需要create a connection with pyodbc。

然后，您希望循环遍历结果集，然后通过构造函数或使用两次strptime（）创建datetime对象。

然后，您将减去日期时间对象以获得作为timedelta对象的时间之间的差异。

然后，根据您需要对该信息执行的操作，您可以通过数学运算将其转换为您的格式。

import datetime
import pyodbc

#create connection
with pyodbc.connect(db, password) as cnxn:

    #create cursor
    cursor = cnxn.cursor()
    cursor.execute('YOUR QUERY')

    #loop through results
    while true:

        #use fetchone generator to optimize memory usage
        row = cursor.fetchone()

        #create datetime objects
        start_time = datetime.strptime(row[0], '%Y-%m-%d %H:%M:%S.%f')
        end_time = datetime.strptime(row[2], '%Y-%m-%d %H:%M:%S')

        #the following creates a timedelta object
        time_diff = end_time - start_time
            #a couple ways to deal with the result
            #printing time_diff yields Days:Seconds:Microseconds
            #the following yields the difference in seconds
            #difference = time_diff.totalseconds()
            #alternatively, you can get days, seconds, microseconds.
            #either way you have to do some math to convert to hours and 
            #minutes if desired
            #difference = time_diff.days + ":" + time_diff.seconds + ":" 
            #                            + time_diff.microseconds

            #TODO your division to format the seconds how you'd like
            #using divmod, you can get hours, min, second

            m, s = divmod(time_diff.seconds, 60)
            h, m = divmod(m, 60)
            print "%d:%d:%02d:%02d" % (time_diff.days, h, m, s) 

            #TODO whatever you need to do with this information
        if not row:
           break

Python用时间格式减去

1 个答案: