pandas 熊猫日期列减法

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时间:2020-09-14 00:44:02  来源:igfitidea点击:

pandas date column subtraction

pythondatetimepandas

提问by Neil

I have a pandas dataframe like this..

我有一个像这样的Pandas数据框..

       created_time  reached_time
2016-01-02 12:57:44      14:20:22
2016-01-02 12:57:44      13:01:38
2016-01-03 10:38:51      12:24:07
2016-01-03 10:38:51      12:32:11
2016-01-03 10:38:52      12:23:20
2016-01-03 10:38:52      12:51:34
2016-01-03 10:38:52      12:53:33
2016-01-03 10:38:52      13:04:08
2016-01-03 10:38:52      13:13:40

I want to subtract these two date columns and want to get time

我想减去这两个日期列并想得到 time

I am doing following in python

我在 python 中进行以下操作

speed['created_time'].dt.time - speed['reached_time']

But it gives me following error

但它给了我以下错误

TypeError: ufunc subtract cannot use operands with types dtype('O') and dtype('<m8[ns]')

TypeError: ufunc subtract cannot use operands with types dtype('O') and dtype('<m8[ns]')

datatype of created_timeis objectand datatype of reached_typeis timedelta64[ns]

的数据类型created_timeobject与数据类型reached_typeIStimedelta64[ns]

回答by unutbu

You could drop down to NumPy arrays and do the datetime/timedelta arithmeticthere. First, create an array of dates of dtype datetime64[D]:

您可以下拉到 NumPy 数组并在那里执行datetime/timedelta 算术。首先,创建一个 dtype 日期数组datetime64[D]

dates = speed['created_time'].values.astype('datetime64[D]')

Then you have two options: you could convert reached_timeto dates, and subtract dates from dates:

然后您有两个选择:您可以转换reached_time为日期,并从日期中减去日期:

speed['reached_date'] = dates + speed['reached_time'].values
speed['diff'] = speed['created_time'] - speed['reached_date']

or you could convert created_timeto timedeltas, and subtract timedeltas from timedeltas:

或者您可以转换created_time为 timedeltas,并从 timedeltas 中减去 timedeltas:

speed['created_delta'] = speed['created_time'].values - dates
speed['diff'] = speed['created_delta'] - speed['reached_time']


import pandas as pd

speed = pd.DataFrame(
    {'created_time': 
     ['2016-01-02 12:57:44', '2016-01-02 12:57:44', '2016-01-03 10:38:51',
      '2016-01-03 10:38:51', '2016-01-03 10:38:52', '2016-01-03 10:38:52',
      '2016-01-03 10:38:52', '2016-01-03 10:38:52', '2016-01-03 10:38:52'],
     'reached_time': 
     ['14:20:22', '13:01:38', '12:24:07', '12:32:11', '12:23:20', 
      '12:51:34', '12:53:33', '13:04:08', '13:13:40']})
speed['reached_time'] = pd.to_timedelta(speed['reached_time'])
speed['created_time'] = pd.to_datetime(speed['created_time'])

dates = speed['created_time'].values.astype('datetime64[D]')

speed['reached_date'] = dates + speed['reached_time'].values
speed['diff'] = speed['created_time'] - speed['reached_date']

# alternatively
# speed['created_delta'] = speed['created_time'].values - dates
# speed['diff'] = speed['created_delta'] - speed['reached_time']

print(speed)

yields

产量

         created_time  reached_time        reached_date              diff
0 2016-01-02 12:57:44      14:20:22 2016-01-02 14:20:22 -1 days +22:37:22
1 2016-01-02 12:57:44      13:01:38 2016-01-02 13:01:38 -1 days +23:56:06
2 2016-01-03 10:38:51      12:24:07 2016-01-03 12:24:07 -1 days +22:14:44
3 2016-01-03 10:38:51      12:32:11 2016-01-03 12:32:11 -1 days +22:06:40
4 2016-01-03 10:38:52      12:23:20 2016-01-03 12:23:20 -1 days +22:15:32
5 2016-01-03 10:38:52      12:51:34 2016-01-03 12:51:34 -1 days +21:47:18
6 2016-01-03 10:38:52      12:53:33 2016-01-03 12:53:33 -1 days +21:45:19
7 2016-01-03 10:38:52      13:04:08 2016-01-03 13:04:08 -1 days +21:34:44
8 2016-01-03 10:38:52      13:13:40 2016-01-03 13:13:40 -1 days +21:25:12


Using HRYR's improvement, you can do the computation without dropping down to NumPy arrays (i.e. no need to access .values):

使用HRYR 的改进,您可以在不下降到 NumPy 数组的情况下进行计算(即无需访问.values):

dates = speed['created_time'].dt.normalize()
speed['reached_date'] = dates + speed['reached_time']
speed['diff'] = speed['created_time'] - speed['reached_date']

回答by HYRY

Convert created_timecolumn to datetime first:

首先将created_time列转换为日期时间:

df["created_time"] = pd.to_datetime(df["created_time"])

Then use df["created_time"] - df["created_time"].dt.normalize()to get the time part as timedeltatype.

然后用于df["created_time"] - df["created_time"].dt.normalize()获取时间部分作为timedelta类型。