pandas 日期栏减法 [英] pandas date column subtraction

问题描述

我有一个像这样的熊猫数据框.

I have a pandas dataframe like this..

       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

我想减去这两个日期列，并想获得time

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

我正在用python进行跟踪

I am doing following in python

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

但这给了我以下错误

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

created_time的数据类型是object，reached_type的数据类型是timedelta64[ns]

datatype of created_time is object and datatype of reached_type is timedelta64[ns]

推荐答案

您可以下拉至NumPy数组，然后执行

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

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

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

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

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

或者您可以将created_time转换为timedeltas，并从timedeltas中减去timedeltas:

or you could convert created_time to timedeltas, and subtract timedeltas from 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)

收益

         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

---

使用 HRYR的改进，您可以进行计算而不会下降到NumPy数组(即无需访问.values):

---

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

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

这篇关于 pandas 日期栏减法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！