将日期和时间(小时x分钟x秒)转换为仅时间 [英] Convert Days and Time (Hours x Minutes x Seconds) to Time only

问题描述

我有一个数据框，在其中我对两个不同的日期进行了区别，以获取小时和分钟的区别，例如:

I have a Dataframe in which I am making the difference between two different dates to get the difference in Hours and Minutes, for example:

 start_date = '2018-07-03 16:03:00'
 data_final = '2018-07-05 00:00:00'
 duration = data_final - start_date

我要查找的结果是 '31:57:00'，即两个日期之间的总时差.但是我得到的结果是:"1天，7:57:00" (每24小时写为1天).

The result I'm looking for is '31: 57: 00 ', or the total time difference between the two dates. But the result I have is: '1 day, 7:57:00' (Every 24 hours it writes as 1 day).

我尝试使用以下语句将其转换为XMinutesHours格式:

I tried converting it to an XMinutesHours format with the statement:

print (datetime.datetime.strptime (duration, "%H:%M:%S"))

但是我得到了错误:

ValueError:时间数据"1天，7:57:00"与格式％H:％"不匹配M:％S'

ValueError: time data '1 day, 7:57:00' does not match format '% H:% M:% S'

有什么主意吗?

推荐答案

您需要以小时，分钟和秒为单位计算等效项，您可以实现一个函数来获取该值，例如:

You need to calculate the equivalent in hours, minutes and seconds, you could implement a function to get this value, for example:

from datetime import datetime

def get_duration(duration):
    hours = int(duration / 3600)
    minutes = int(duration % 3600 / 60)
    seconds = int((duration % 3600) % 60)
    return '{:02d}:{:02d}:{:02d}'.format(hours, minutes, seconds)

format_str = '%Y-%m-%d %H:%M:%S'
start_date_str = '2018-07-03 16:03:00'
end_date_str = '2018-07-05 00:00:00'

start_date = datetime.strptime(start_date_str, format_str)
end_date = datetime.strptime(end_date_str, format_str)
duration = (end_date - start_date).total_seconds()

print(get_duration(duration))

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