添加24小时到python的负时间差 [英] Add 24 hours to a negative time difference in python
本文介绍了添加24小时到python的负时间差的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
下面是一些解释问题的表：
CallDate BeginningTime EndingTime Interval
75 1/8/2009 1900-01-01 07:49:00 1900-01-01 08:19:00 00： 30:00
76 1/11/2009 1900-01-01 14:37:00 1900-01-01 14:59:00 00:22:00
77 1/9/2009 1900- 01-01 09:29:00 1900-01-01 09:56:00 00:27:00
78 1/11/2009 1900-01-01 09:20:00 1900-01-01 10： 13:00 00:53:00
79 1/16/2009 1900-01-01 15:11:00 1900-01-01 15:50:00 00:39:00
80 1 / 17/2009 1900-01-01 22:52:00 1900-01-01 23:26:00 00:34:00
81 1/19/2009 1900-01-01 05:48:00 1900- 01-01 06:32:00 00:44:00
82 1/20/2009 1900-01-01 23:46:00 1900-01-01 00:21:00 -23：25：00
83 1/20/2009 1900-01-01 21:29:00 1900-01-01 22:08:00 00:39:00
84 1/23/2009 1900-01-01 07 ：33：00 1900-01-01 07： 55:00 00:22:00
85 1/30/2009 1900-01-01 19:33:00 1900-01-01 20:01:00 00:28:00
更新：这是导致我这一点的代码
df ['BeginningTime'] = pd.to_datetime（df ['BeginningTime']，format ='％H：％M'）
df ['EndingTime'] = pd。 to_datetime（df ['EndingTime']，format ='％H：％M'）
df ['Interval'] = df ['EndingTime'] - df ['BeginningTime']
df [['CallDate'，'BeginningTime'，'EndingTime'，'Interval']]
解决方案
如果你只是想添加一天到timedelta，如果它是否定的：
df ['Interval'] = df ['Interval']。apply（lambda x：x + Timedelta（days = 1）if x< 0 else x）
如果可以安全假设结束时间将在24小时内，您可以检查结束时间是否早于开始时间，并使用timedelta将一天添加到结束时间，而不是间隔时间。
从datetime导入datetime，timedelta
d1 = datetime.strptime（1900-01-01 23:46:00 ，％Y-％m-％d％H：％M：％S）
d2 = datetime.strptime（1900-01-01 00:21:00，％Y-％m- ％d％H：％M：％S）
如果d2 < d1：
d2 + = timedelta（days = 1）
打印d2 - d1
＃0:35:00
使用熊猫你可以这样做：
import pandas as pd
from pandas import Timedelta
d = {
CallDate：[
1/8/2009，
2009年11月11日，
1/9/2009，
1/11/2009，
1/16/2009，
1 / 17/2009，
1/19/2009，
1/20/2009，
1/20/2009，
1/23 / 2009，
1/30/2009
]，
BeginningTime：[
1900-01-01 07:49:00，
1900-01-01 14:37:00，
1900-01-01 09:29:00，
1900-01-01 09:20:00，
1900-01-01 15:11:00，
1900-01-01 22:52:00，
1900-01-01 05:48:00，
1900-01-01 23:46:00，
1900-01-01 21:29:00，
1900-01-01 07:33:00
1900-01-01 19:33:00
]，
EndingTime：[
1900-01-01 08:19:00，
1900-01-01 14:59:00，
1900-01-01 09:56:00，
1900-01-01 10:13:00，
1900-01-01 15:50:00
1900-01-01 23:26:00，
1900-01-01 06:32:00，
1900-01-01 00:21:00 ，
1900-01-01 22:08:00，
1900-01-01 07:55:00，
1900-01-01 20:01:00
}
df = pd.DataFrame（data = d）
df ['BeginningTime'] = pd.to_datetime（df [ 'beginning']，format =％Y-％m-％d％H：％M：％S）
df ['EndingTime'] = pd.to_datetime（df ['EndingTime']，format = ％Y-％m-％d％H：％M：％S）
def interval（x）：
如果x ['EndingTime']< x ['BeginningTime']：
x ['EndingTime'] + = Timedelta（days = 1）
return x ['EndingTime'] - x ['BeginningTime']
df ['Interval'] = df.apply（interval，axis = 1）
在[2]中：df
Out [2]：
BeginningTime CallDate EndingTime Interval
0 1900-01-01 07 ：49：00 1/8/2009 1900-01-01 08:19:00 00:30:00
1 1900-01-01 14:37:00 1/11/2009 1900-01-01 14 ：59：00 00:22:00
2 1900-01-01 09:29:00 1/9/2009 1900-01-01 09:56:00 00:27:00
3 1900 -01-01 09:20:00 1/11/2009 1900-01-01 10:13:00 00:53:00
4 1900-01-01 15:11:00 1/16/2009 1900 -01-01 15:50:00 00:39:00
5 1900-01-01 22:52:00 1/17/2009 1900-01-01 23:26:00 00:34:00
6 1900-01-01 05:48:00 1/19/2009 1900-01-01 06:32:00 00:44:00
7 1900-01-01 23:46:00 1 / 20/2009 1900-01-01 00:21:00 00:35:00
8 1900-01-01 21:29:00 1/20/2009 1900-01-01 22:08:00 00 ：39：00
9 1900-01-01 07:33:00 1/23/2009 1900-01-01 07:55:00 00:22:00
10 1900-01-01 19:33:00 1/30/2009 1900-01-01 20:01:00 00:28:00
I am using python to calculate time intervals between two events. Each event has a 'beginning time' and an 'ending time.' I have found the difference between the two in a new column, 'interval', but have negative values when the beginning and ending time are on different days (for instance begin 23:46:00 and end 00:21:00 gives -23:25:00). I would like to create an if-statement to run through the 'interval' column and add 24 hours to any negative values. However, I have had problems with adding 24 hours to the 'interval' values. Currently my 'interval' dtype=timedelta64[ns].
Here is a little bit of the table to clarify the problem:
CallDate BeginningTime EndingTime Interval
75 1/8/2009 1900-01-01 07:49:00 1900-01-01 08:19:00 00:30:00
76 1/11/2009 1900-01-01 14:37:00 1900-01-01 14:59:00 00:22:00
77 1/9/2009 1900-01-01 09:29:00 1900-01-01 09:56:00 00:27:00
78 1/11/2009 1900-01-01 09:20:00 1900-01-01 10:13:00 00:53:00
79 1/16/2009 1900-01-01 15:11:00 1900-01-01 15:50:00 00:39:00
80 1/17/2009 1900-01-01 22:52:00 1900-01-01 23:26:00 00:34:00
81 1/19/2009 1900-01-01 05:48:00 1900-01-01 06:32:00 00:44:00
82 1/20/2009 1900-01-01 23:46:00 1900-01-01 00:21:00 -23:25:00
83 1/20/2009 1900-01-01 21:29:00 1900-01-01 22:08:00 00:39:00
84 1/23/2009 1900-01-01 07:33:00 1900-01-01 07:55:00 00:22:00
85 1/30/2009 1900-01-01 19:33:00 1900-01-01 20:01:00 00:28:00
Update: Here is the code that had led me to this point
df['BeginningTime']=pd.to_datetime(df['BeginningTime'], format='%H:%M')
df['EndingTime']=pd.to_datetime(df['EndingTime'], format='%H:%M')
df['Interval']=df['EndingTime']-df['BeginningTime']
df[['CallDate','BeginningTime','EndingTime','Interval']]
解决方案
If you just want to add 1 day to the timedelta if it is negative:
df['Interval']=df['Interval'].apply(lambda x: x + Timedelta(days=1) if x < 0 else x)
If it is safe to make the assumption that the end time will be within 24 hours, you can check to see if the end time is earlier than the start time and use timedelta to add a day to the end time rather than the interval time.
from datetime import datetime, timedelta
d1 = datetime.strptime( "1900-01-01 23:46:00", "%Y-%m-%d %H:%M:%S" )
d2 = datetime.strptime( "1900-01-01 00:21:00", "%Y-%m-%d %H:%M:%S" )
if d2 < d1:
d2 += timedelta(days=1)
print d2 - d1
# 0:35:00
With pandas you can do something like this:
import pandas as pd
from pandas import Timedelta
d = {
"CallDate": [
"1/8/2009",
"1/11/2009",
"1/9/2009",
"1/11/2009",
"1/16/2009",
"1/17/2009",
"1/19/2009",
"1/20/2009",
"1/20/2009",
"1/23/2009",
"1/30/2009"
],
"BeginningTime": [
"1900-01-01 07:49:00",
"1900-01-01 14:37:00",
"1900-01-01 09:29:00",
"1900-01-01 09:20:00",
"1900-01-01 15:11:00",
"1900-01-01 22:52:00",
"1900-01-01 05:48:00",
"1900-01-01 23:46:00",
"1900-01-01 21:29:00",
"1900-01-01 07:33:00",
"1900-01-01 19:33:00"
],
"EndingTime": [
"1900-01-01 08:19:00",
"1900-01-01 14:59:00",
"1900-01-01 09:56:00",
"1900-01-01 10:13:00",
"1900-01-01 15:50:00",
"1900-01-01 23:26:00",
"1900-01-01 06:32:00",
"1900-01-01 00:21:00",
"1900-01-01 22:08:00",
"1900-01-01 07:55:00",
"1900-01-01 20:01:00"
]
}
df = pd.DataFrame(data=d)
df['BeginningTime']=pd.to_datetime(df['BeginningTime'], format="%Y-%m-%d %H:%M:%S")
df['EndingTime']=pd.to_datetime(df['EndingTime'], format="%Y-%m-%d %H:%M:%S")
def interval(x):
if x['EndingTime'] < x['BeginningTime']:
x['EndingTime'] += Timedelta(days=1)
return x['EndingTime'] - x['BeginningTime']
df['Interval'] = df.apply(interval, axis=1)
In [2]: df
Out[2]:
BeginningTime CallDate EndingTime Interval
0 1900-01-01 07:49:00 1/8/2009 1900-01-01 08:19:00 00:30:00
1 1900-01-01 14:37:00 1/11/2009 1900-01-01 14:59:00 00:22:00
2 1900-01-01 09:29:00 1/9/2009 1900-01-01 09:56:00 00:27:00
3 1900-01-01 09:20:00 1/11/2009 1900-01-01 10:13:00 00:53:00
4 1900-01-01 15:11:00 1/16/2009 1900-01-01 15:50:00 00:39:00
5 1900-01-01 22:52:00 1/17/2009 1900-01-01 23:26:00 00:34:00
6 1900-01-01 05:48:00 1/19/2009 1900-01-01 06:32:00 00:44:00
7 1900-01-01 23:46:00 1/20/2009 1900-01-01 00:21:00 00:35:00
8 1900-01-01 21:29:00 1/20/2009 1900-01-01 22:08:00 00:39:00
9 1900-01-01 07:33:00 1/23/2009 1900-01-01 07:55:00 00:22:00
10 1900-01-01 19:33:00 1/30/2009 1900-01-01 20:01:00 00:28:00
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